[Music]

an integer partition of an integer n is

a list of integers l1 l2 up to lk

satisfying the properties that n is the

sum of the integers in the list and the

integers in the list are weakly

decreasing and greater than or equal to

1. the elements in the list are called

parts

for example the list 43211 is an integer

partition of eleven because eleven is

equal to four plus three plus two plus

one plus one this is a partition of

eleven into five parts

the young diagram of a partition is an

array of boxes with li boxes in row i

so for example the young diagram of the

partition 43211 is pictured here because

there are four boxes in the first row

then three then two then one and then

one

let's let p of n be the number of

integer partitions of n and p sub k of n

be the number of integer partitions of n

with k parts

if we're trying to compute p of n we can

list out all of the young diagrams for n

in this case we have n is nine and here

are all the partitions

natural way to organize the partitions

is to group them together with other

partitions with the same number of parts

as we've done here

[Music]

the smallest number of parts a partition

can have is one and the largest number

of parts it can have is n where each

part has size one this visual shows us

that the number of partitions of n is

equal to the sum of the number of

partitions of n into k parts where k

ranges from 1 to n

now we can present our primary result

about the number of partitions of n into

k parts

if we have all the partitions of n into

k parts we can divide these into two

disjoint sets the first set are the

partitions that have a part of size one

and the second set consists of the

partitions with no part of size one the

partitions that have a part of size one

are in one-to-one correspondence with

those partitions where we remove one of

the parts of size one

but these partitions are just the

partitions of n minus one into k minus

one parts so we can count them using p

sub k minus one of n minus one for each

of the partitions that have no part of

size one we can remove one from each

part resulting in a partition of n minus

k into k parts these can be counted by p

sub k of n minus k both of these

processes are reversible and because the

two sets have no elements in common we

see that p sub k of n is equal to p sub

k minus one of n minus one plus p sub k

of n minus k

this is a recursive formula for p sub k

of n the number of partitions of n into

k parts we now see that we do not need

to enumerate all of the partitions to

know how many there are

instead we can use the two formulas that

we just encountered to find the number

of partitions of size n

namely we can realize that the number of

partitions of n is equal to the sum of

the number of partitions of n into k

parts where k ranges from one to n and

that the number of partitions of n into

k parts satisfies the recursion given

here

of course this is only one such

technique to count the number of

partitions of n can you think of a

different way to count them

you

Recurrence for partitions into k parts (visual proof)

In this video, we define partitions and show how to think visually about integer partitions. We then visually prove a recurrence formula for the number of integer partitions of n with exactly k parts.

If you like integer partitions, check out these videos:
https://youtu.be/p0FKjVA_sRE (synthwave enumeration)
https://youtu.be/M4TmnYxS4gk from @behackl

If you like this video,  please check out my others and consider subscribing. Thanks!

Here is a nice resource if you want to know more:

https://www.whitman.edu/mathematics/cgt_online/book/section03.03.html

#integerpartitions #integers #partition #youngdiagram #counting #count #mathvideo #math #mtbos #manim #animation #theorem #visualproof #proof #iteachmath #mathematics #discretemath #combinatorics #enumeration #synthwave #algorithm

To learn more about animating with manim, check out:
https://manim.community
_________________________________________
Music in this video:
Elysium by Alexander Nakarada | https://www.serpentsoundstudios.com
Music promoted by https://www.free-stock-music.com
Attribution 4.0 International (CC BY 4.0)
https://creativecommons.org/licenses/by/4.0/


ok guys for this problem we're going to
want to generate this compound starting

from benzene using any any other
compounds you want. so as a hint we're probably

gonna be using some electrophilic
aromatic substitution so go ahead and

see if you can build this molecule

ok so let's take a look at the two substituents that are on the target
molecule so first we have an alkyl group

we know that we can attach an alkyl
group onto a benzene ring by Friedel

Crafts alkylation and then we also have
nitration. but what order are we

gonna do them in? in order to answer this
question we have to understand something

about the directing abilities of these
groups. now we know that an alkyl group

is an ortho/para director and we know
that a nitro group is a meta director

and we also understand that this spatial
relationship between them on the

molecule is ortho, so this is gonna tell us
what order we have to do the reactions

in because given that they have an ortho
relationship to one another it must be

the ortho/para director that directed
the other group to its position and so

what we're saying is that we must
starting with benzene have done the

Friedel Crafts alkylation here with this alkyl chloride to get this

intermediate because if we have this
intermediate then upon nitration we will

get a mixture of ortho and para

nitrated products whereas if we were to
do the nitration first if we did the

nitration first and let's say this was
our nitrated product then if we went

and did the alkylation we would get meta
direction from the nitro group and so

the nitro group cannot have been the
compound that was added first because that

will not give us a structure that
matches our target molecule, instead

we're recognizing that alkyl is ortho/para, nitro group is meta, so it must

have been the alkyl group that directed
the nitro group to the ortho position, we

can get this mixture, separate them out
and there is going to be our target molecule

thanks for watching guys, subscribe to my channel for more tutorials, and as always feel free to email me

Practice Problem: Electrophilic Aromatic Substitution Retrosynthesis

You gotta know your EAS reactions for this one! Make sure you do them in the right order, too.

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number five asks draw the correct form

of the resonance hybrid given the

following resonance structures so let's

take a look at what we've got here we

have an oxyanion right there so we have

an O minus right there and then how are

we getting to each of these resonance

structures we can take this negative

charge and form a CO Pi Bond right there

if we then take this Pi Bond and push

that up into the carbon atom to get that

next annion right there so as we can see

we've got the pie Bond there and then we

have the new location for the anion so

we've shifted that o minus over to a c

minus now let's take that and let's put

that right there we can make another

piie Bond there if we take this piie

Bond and we push it up onto that carbon

so now we have a pi Bond here and we

have a new location for the carbanion

then to get to that fourth resonance

structure we are going to put that right

here

Structure and Bonding OCHEM Exam Question 5

This is an abbreviated version of this question for YouTube Shorts. Check out the full version in the linked video.


[Music] button

[Music] Professor Dave and Chegg here we've
learned a lot about proton NMR Spectra and this

is indeed the single most important technique in
organic chemistry for the characterization of any

molecule however there are other related tools
that can also be quite useful and one of them

is carbon 13 NMR spectroscopy let's learn how
this works as well as another form of nuclear

magnetic resonance spectroscopy this operates by
precisely the same principles as proton NMR it is

simply focused on a different nuclide instead
of protium an isotope of hydrogen data will be

collected regarding carbon 13 nuclides now it
is true that only about one percent of carbon

atoms will be carbon 13 Isotopes so most of
the individual molecules in a sample will

not contain any carbon 13. but this is okay
because given a macroscopic sample trillions

upon trillions of molecules within that sample
will contain a carbon 13 atom at a particular

location in the molecule thus generating data
for us to examine regarding that position on the

molecule and the same goes for every other carbon
in the molecule so we can therefore gather data

regarding the chemical environment experienced
by every carbon atom in any structure this

approach can be described as signal averaging
as performed by Fourier transform NMR or ft NMR

the tricky thing about signal averaging is that
there is a lot of noise to deal with the signals

from the 13c nuclides are weak in comparison
to background electronic noise but we can run

hundreds or even thousands of Trials and average
them together hence the term signal averaging and

digitally lock on to any data that is non-random
and therefore due to 13c nuclides the random noise

can then be minimized so that the important data
can stand out the problem is that taking so many

measurements is extremely time consuming that's
where the Ft NMR comes into play a short pulse

of RF energy irradiates the sample all 1H and
13c nuclides resonate at once and the resulting

complex signal is manipulated using Fourier
transforms allowing for data to be collected

in mere seconds we will examine carbon 13 Spectra
later but for now there is one interesting fact

to be aware of we will not see splitting patterns
the way we do with proton NMR the reason for this

has to do with the low abundance of this isotope
it is unlikely that carbon 13 nuclides will be

adjacent in any particular molecule so there is
no opportunity for them to split each other's

signals the way protons do coupling with protons
themselves also does not occur because these

Spectra are recorded with Broadband decoupling a
second irradiation makes hydrogen's spin flip so

rapidly that their local magnetic fields average
to zero and no coupling with carbon will occur

so with that we understand roughly
how carbon 13 NMR Works through

signal averaging and Fourier transforms
Professor day for Chegg see you next time

What is 13C-NMR Spectroscopy? Ft. Professor Dave

1-H NMR spectroscopy is the most important technique in organic chemistry for the characterization of any molecule. But there are other useful related tools, and one of them is 13C-NMR spectroscopy. @ProfessorDaveExplains explores the topic of 13C-NMR spectroscopy in this video, delving into how it works, and explaining the process of signal averaging as performed through Fourier transform NMR.

Get more homework help from Chegg at https://che.gg/3HbtG8Y
Watch the whole Organic Chemistry playlist: https://www.youtube.com/playlist?list=PLllCkartRRdMPnATf1X8qVdo5RzM8gA_M

About Chegg:
Chegg is a leading education technology company that provides students with affordable and accessible study resources. Chegg's mission is to help students save time, save money and get smarter.

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00:00 Intro
00:30 1-H NMR spectroscopy
00:42 13C-NMR spectroscopy
01:27 Signal averaging / Fourier transform NMR
02:26 13C-NMR spectra features


مرحبا معكم البروفيسور ديف, لنتحدث عن المُعَايَرَة

إن معايرة الحمض والقاعدة هي طريقة لتحديد كمية الحمض أو القاعدة الموجودة في المحلول

من خلال تحديد الحجم الدقيق للحمض أو القاعدة التي ستتفاعل معه بشكل تام

إنه مثل نوع من الحسابات الكيميائية

على منحنى معايرة لمحلول حمضي

يتم تخطيط الرقم الهيدروجيني (pH) مقابل حجم القاعدة التي تتم إضافتها تدريجياً

عند اضافة قاعدة يرتفعPH ببطء

ثم يرتفع بشكل حاد نحو نقطة التكافؤ

هذه هي النقطة التي تمت فيها إضافة قاعدة بكمية كافية

لتحييد كمية الحمض الفعلية في المحلول

عند إجراء المعايرة، نتعرف على نقطة التكافؤ باستخدام شيء يسمى المؤشر.

المؤشر هو مادة ستتحول إلى لون واضح بمجرد بلوغ نقطة التكافؤ

عندما نحيد حمضًا قويًا بقاعدة قوية، تكون نقطة التكافؤ 7

إذا كان أي من النوعين أو كليهما ضعيفًا ، فسيختلف هذا العدد

على سبيل المثال، لنفترض أن هناك محلول هيدروكسيد الصوديوم ونريد أن نعرف تركيزه

افترض إننا نفاعل مع حمض الكبريتيك 3.0 م بحجم 25 مل  من القاعدة

ونجد أنه يستغرق 11.6 مل من الحمض للوصول إلى نقطة التكافؤ.

إذا حولنا إلى لترات ثم لدينا عوامل تحويل لتركيز الحمض

ثم نسبة القياس الكيميائي ، يمكننا الوصول إلى تركيز المواد المتفاعلة

لنتحقق من مدى الإستيعاب

شكرًا على المشاهدة اشترك في قناتي لمزيد من الدروس

وكما هو الحال دائمًا ، لا تتردد في مراسلتي عبر البريد الإلكتروني

Acid-Base Titration

Any introductory chemistry class will include titrations, and to do these, you have to do math. But you get to see pretty colors, too! Here's a quick run through of the stoichiometry of it all.

Watch the whole General Chemistry playlist: http://bit.ly/ProfDaveGenChem

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Book Depository: http://bit.ly/3aOVDlT


five asks how many stereoisomers are

possible for the following compound we

need to understand that this depends on

how many stereo centers there are on the

compound for example if you have one

chyro Center that chyro Center can be r

or it can be S and then if you have two

chyro centers now we have more

possibilities so what the formula is is

2 to the N the number of chyro centers n

you take two and raise it to that power

and that's how many stereo isomers are

possible how many KY Centers do we have

well we have one right here there's that

bromine the implied hydrogen and then

the two directions of the Ring we have

one right here as well because we have

this substituent and then the two

directions of the ring as well as the

implied hydrogen and then this also is a

chyro center because that chlorine can

be on the dash or the wedge and that

will be different because we've got

chlorine hydrogen methyl and then the

rest of the molecule so if we have three

chyal centers then we're saying that's

two

Determining All Possible Stereoisomers


Professor Dave here, let’s prove the Pythagorean
theorem.

As I mentioned before when we initially learned
the Pythagorean theorem, there are very elegant

ways to prove that the theorem is true, and
now that we know how to calculate area, we

are ready to perform one such proof ourselves.

Let’s say we draw a right triangle like
this one, and label the sides A, B, and C,

like we are used to.

Now let’s make three more copies of this
triangle and arrange them like this, so that

we get two squares.

The larger square has a side length of A plus
B, while the smaller inner square has a side

length of C.

We can calculate the areas of all of these
shapes.

We know that the area of one of the triangles
will be one half base times height, or one

half A times B. The area of the large square
will be the side length squared, or A plus

B quantity squared.

And the small square will have an area of
C squared.

Now let’s set up a relationship between
these areas.

The area of the large square is equal to A
plus B quantity squared as we just said, but

it is also equal to the area of the small
square plus the area of all the triangles,

since these shapes comprise the larger square.

So we can make an equation that states the
following.

A plus B quantity squared, the area of the
big square, is equal to four times one half

AB, which is four times the area of one triangle,
since there are four triangles, plus C squared,

the area of the small square.

We know how to square a binomial by using
the FOIL method, so let’s take away the

exponent and just write the binomial out twice.

First gives us A squared, outer gives us AB,
inner gives us AB, and last gives us B squared.

Combining like terms, that’s A squared plus
two AB plus B squared.

On the right, we can multiply four and one
half, and we get two AB plus C squared.

As it happens, we have two AB on both sides,
so those will cancel out, and wouldn’t you

know it, take a look at what we are left with.

The Pythagorean theorem.

I hope this result makes you feel warm and
fuzzy inside, because proofs like this are

what math is really all about.

There are many other proofs of this theorem
that involve similar calculations with triangles

and squares and other shapes, so if you’re
just dying to know them all, it’s time to

hit the internet.

Otherwise, let’s move forward and wrap things
up with geometry.

Proving the Pythagorean Theorem

We learned about the Pythagorean Theorem, but where did it come from? How do we know it's definitely true? What if old Pythag just made it up off the top of his mystical skull? Lucky for us, in math we can proof that things are definitely true, and there are tons of ways to prove that the Pythagorean theorem is true. Here's one of the simpler ones for you to memorize, so you can bust it out at a party and be the coolest kid in the room!

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okay let's do some practice regarding
enthalpy of combustion so the question

asks how much heat is released upon the
combustion of 1.00 litres of isooctane

and that has the formula C8H18. now in
order to answer this question we will

need to know that the density of
isooctane is zero point six nine two

grams per milliliter and we also need to
know that the Delta H of combustion for

iso-octane is negative 5461 kilojoules
per mole so this is all the data you

need go ahead and give this a try.

so let's start out with our one point
zero zero liters of isooctane what are

we gonna do first well we know we're
gonna do some kind of stoichiometry so

we're gonna want to try to get two moles
so how do we do this well we've got

liters and our density was in grams per
milliliter so we're gonna need in

milliliters so let's just convert this
into milliliters first we've got one

point zero zero liters times a thousand
and since we can only have three sig

figs that is going to be one point zero
zero times 10 to the 3 milliliters. now

as we said we have that density so we
can convert to grams one point zero zero

times 10 to the 3 milliliters times zero
point six nine two grams per milliliter

milliliters will cancel and we'll get
692 grams so now that we have a mass of

isooctane we want to use that mass to
get moles because that way we can use

the Delta H of combustion that we were
given so 692 grams times 1 mol / 114

grams that's what we get just by adding
up all the carbon and hydrogen atoms

that is going to give us six
point zero seven moles so we have six

point zero seven moles of isooctane in
that one liter of isooctane and then

lastly we can use our thermochemical
data we can do six point zero seven

moles times negative five thousand 461
kilojoules that are released for every

mole of iso-octane and we get negative
three point three one times ten to the

four kilojoules that is how much energy is associated with the combustion of one liter of isooctane.

Practice Problem: Enthalpy of Combustion

Stoichiometry for combustion? You bet! If we know about moles, we can calculate energy changes. Give it a try!

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Hey it’s Professor Dave; let’s work with
percentages.

We are all aware of percentages, as they are
ubiquitous in society.

There is a sale today, thirty percent off.

Leave a twenty percent tip after the meal.

Today there is a fifty percent chance of rain.

So what are these things?

Well in Latin, centum means one hundred, so
percent literally means out of a hundred.

It’s just a way of conveying magnitude,
by modifying some total into a hundred, and

then reporting a value of interest as some
fraction of that hundred.

If you get a ninety percent on your math quiz,
you could have gotten nine points out of ten,

eighteen points out of twenty, four hundred
fifty out of five hundred, or any other number

that obeys this ratio of ninety out of a hundred.

Whatever we have to do to the larger number
to make it one hundred, we do that same thing

to the other number.

Ten must be multiplied by ten to get a hundred,
so the nine is also multiplied by ten.

In this way, percentages are ratios, they
are just very specific ratios that compare

numbers to a hundred, because although this
number is totally arbitrary, it is an ideal

size for comparison.

One hundred is small enough that our brains
can easily comprehend its magnitude, but also

large enough that there are sufficient gradations
between zero and a hundred so as to make meaningful

comparisons.

In this way, we can look at percentages as
simple fractions.

Eighty percent is eighty one-hundredths.

We learned how to simplify fractions, so let’s
divide both of these numbers by twenty.

We should get four fifths.

This means that eighty percent and four fifths
are equivalent; they mean the same thing.

Moving to another example, fifty percent is
fifty one hundredths, and dividing both numbers

by fifty, we get one half.

This makes sense, as fifty is half of a hundred.

If you have something like three percent,
that will be three one hundredths, and that

can’t be simplified any further.

We should be able to convert between percentages,
fractions, and even decimals in this way with

relative ease, so let’s check comprehension.

Working With Percentages

Oh my! There's a big sale at your favorite store! 30% off, in fact! But wait, what does that mean exactly? Percentages are convenient ways of conveying magnitude by comparing some value to one hundred. Why a hundred? Because it's not too big and not too small. Let's learn how to work with these babies.

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[Music]

Professor Dave and CH here we know about

markovnikov's rule which dictates that

in a reaction like electrophilic

addition when a carbocation intermediate

is formed it is specifically the more

substituted carbocation of the two that

are possible which will form but why is

this let's look at carbocation stability

to understand how this works first a few

details about carbocations these are

carbons with a formal positive charge

because they have lost an electron

domain because of this they are SP2

hybridized and trigonal planer so they

are totally flat about the carbocation

with 120° Bond angles this means that

there is one remaining unhybridized p

orbital and this will be vacant

extending on loes perpendicular to the

plane of the molecule now let's let's

revisit the mechanism for an addition

reaction to contextualize carbocation

stability recall that in the first step

of a hydrohalogenation a new CH Bond was

produced leaving us with a carbocation

but it was the case that the new

hydrogen ended up exclusively on the

less substituted carbon leaving us with

this specific carbocation which we can

call a tertiary carbocation since the

carbon that bears the formal positive

charge is bonded to three other carbons

in this case three methyl groups we did

not see the reverse situation where the

hydrogen ended up on the more

substituted carbon which would leave us

with this carbocation which we would

call a primary carbocation since the

carbon would be bonded to only one other

carbon the rest of the molecule only the

first pathway occurs and never the

second the reasoning behind this is that

carbocations become more stable as they

are more substituted meaning when they

have more adjacent alkal groups this has

to do with a phenomenon called hyper

conjugation where electrons in adjacent

CH bonds which are roughly parallel to

the lobes of the vacant unhybridized p

orbital lend some electron density to

the carbocation thereby stabilizing it

the more alkal groups there are adjacent

the greater the effect so the methyl

cation would be the least stable

carbocation possible followed by a

primary carbocation with one alkal group

then a secondary carbocation with two

alkal groups and the most stable would

be a tertiary carbocation with three

alkal groups this same phenomenon is

also behind the stabilization of alkenes

with greater substitution another reason

for this stabilization has to do with

inductive effects this refers to

electronic effects that are occurring

through Sigma bonds the electron density

surrounding alkal groups is more

polarizable through the sigma Bond

system than that around a hydrogen atom

so more alal results in a greater

inductive effect which is

stabilizing with that we understand how

hyper conjugation and inductive effects

both contribute to the carbocation

stability Trend which says that more

highly substituted carbocations meaning

those with more alkal groups will be

more stable Professor Dave for che see

you next

time

go

Understanding Carbocation Structure and Stability Ft. Professor Dave

Now that we've covered Markovnikov's rule, let's delve further into the concept of carbocation. We'll take a look at carbocation properties, we'll revisit the mechanism for a specific addition reaction to contextualize carbocation stability, and we'll finally talk about hyperconjugation and inductive effects in the context of carbocation stability.

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00:00 Intro
00:31 Carbocation properties
00:59 Addition mechanism
01:50 Hyperconjugation
02:42 Inductive effects


when you study chemistry there's a lot

of things that come up when you're

making measurements and will refer to it

being an intensive or an extensive

property or variable and what we mean by

these is we really mean it's a property

that changes with amount or it's a

property that's more intensive which

doesn't change with them out and right

there is really what your definitions

are any property that you can measure

that as you change the amount of

substance but that property doesn't

change with the amount is called

intensive a really great example of this

is temperature it's one of our premier

intensive variables if you will a great

example if you had a glass of water

that's 25 degrees C and you pour half

the water into another glass and ask

yourself now what's the temperature

temperature still 25 degrees because

it's an intensive property it absolutely

does not matter about the amount that

you have it's intensive now let's

compare that to the mass of the water if

I have a glass of water

I know it's mass is say 500 grams and I

pour half of it out now the mass is not

500 grams it's 250 grams that's because

that's an extensive variable extensive

properties vary with the amount and we

love to quantify amounts we quantify in

mass and grams we quantify ammount

itself in moles of substance we quantify

length how long is something how tall is

something volume is a great example of

an extensive property that actually

brings in length three times height

depth and width you multiply all three

of them you get yourself a volume which

is very extensive you get more amount

you get a bigger volume an interesting

thing though about extensive properties

if you ratio one extensive property

against another extensive property you

get an intensive property a great

example of this is density density is

mass per unit volume the mass is

extensive the more you

have the more mass you have volume is

extensive the more volume you have the

more the number goes up if you Rea show

them mass divided by volume you get

density and that's intensive the water

example I gave earlier when I pour it

doesn't matter when I split it the

density is still approximately one gram

per milliliter even in the part I pour

it off it's one gram per milliliter so

this is an important concept you will

see over and over when you're answering

problems numerically in a chemistry

class is the answer an intensive or is

it an extensive property one last thing

extensive properties by themselves tend

not to reveal much about what the

substance actually is for example let's

say I have a substance and I tell you

it's 500 grams that really tells you

nothing about the substance other than

the amount I have but if I tell you I

have a substance that has a density of

one gram per milliliter it really limits

you on your choices of what that could

be and so what we tend to do in the

sciences and what we do in chemistry is

we list intensive properties and if you

get a good listing of them it truly

identify that substance as being a

unique version of those intensive

properties meaning we like to

characterize matter by listing intensive

properties and that's really what the

importance of both of these things are

and you need to keep track of them

Intensive Extensive Properites

Intensive and Extensive properties.


number four what is the iup pack name

for the following compound so now we

have a cyclohexene derivative so we have

a double bond inside the cyclohexane

ring which makes it cyclohexene and now

it's not arbitrary the way we number we

need for the two carbons participating

in the double bond to be carbons one and

two but there are two options possible

we could either have this be number one

and this be number two or we could have

this be number one and this be number

two and we need to choose the option

that will give any substituents on the

molecule occurring soonest what we end

up realizing is that we have to go 1 2 3

4 5 6 like that there's a few common

errors here some will try to label the

carbon with the chlorine as number one

that doesn't work again it's cyclohexene

which means the carbons participating in

the double bond needs to be carbons one

and two some

Naming a Compound With Multiple Functional Groups


four asks the following pair of

structures are identical enantiomers

diamer and constitutional isomers so we

have some straight chain alkanes with

bromo groups on them and why don't we go

ahead and number these parent chains so

on the left which direction would we

number this we would start from the

right and we would get 1 2 3 4 5 6 7 8

so that is an octane and then on the

right this one we would number from left

to right because that will give the

bromine occurring soonest 1 2 3 4 5 6 7

8 so where are the bromo groups over

here they're on carbons three and five

and then on the other structure they're

on carbons 2 and four so these are

actually completely different molecules

these are not stereoisomers these are

constitutional isomers they have

different connectivity remember that

stereoisomer have

Determining the Type of Isomer


Hey it's professor Dave, let's define internal energy.

We now understand

temperature as a measure of the

molecular kinetic energy present in the

particles of a substance. As we said, this

kinetic energy is distributed amongst

translational motion, rotational motion

and vibrational motion. There is also

molecular potential energy by virtue of

the electromagnetic force acting amongst

the atoms of an individual molecule and

between each separate molecule. The sum

of all of these types of energy that are

exhibited by the particles of a

substance is called the internal energy

of that substance, represented by an

upper case U. This is the energy

associated with atomic motion and it is

a quantity that is directly proportional

to the temperature of a sample. Higher

temperatures mean more internal energy,

lower temperatures mean less internal

energy. In this way, we must understand

that a substance does not contain heat,

it contains internal energy, like the

kinetic energy of all the particles, and

it is the transfer of this energy from

areas of high temperature to areas of

low temperature that we can label as

heat. Internal energy is an important

concept in the study of thermodynamics

which is the study of heat and

temperature and their relation to energy

and work. The internal energy of a system

can be increased due to heat transfer

but also due to things like friction or

structural deformation when bending a

piece of metal or stretching a rubber

band. But as we learned earlier, for a

particular system there will always be

conservation of energy. We already know

about the kinetic energy and potential

energy of an object moving through a

gravitational field so let's add the

internal energy of the particles in the

object to this list, and restate

conservation of energy with the

following equation: change in potential

energy plus

change in kinetic energy plus change in

internal energy will always be zero. That

means that whenever there is a change in

any one of these quantities the

difference in energy must transform into

or be provided by one of the other forms.

So when two objects participate in an

inelastic collision, some of the kinetic

energy is transferred into internal

energy, which will be absorbed by the

object. Internal energy can in turn be

used to do work through heat transfer,

pressure-volume work by an expanding gas

or some other process. We should also

note that internal energy is a state

function, which means that the internal

energy of a system depends only on the

state of the system and not how it got

to that state. We will learn more about

state functions later. These kinds of

concepts are precisely what we will be

examining when we discuss the laws of

thermodynamics, so let's move on to this

incredibly important subject.

Thanks for watching, guys. Subscribe to my channel

for more tutorials, support me on patreon

so I can keep making content, and as

always, feel free to email me:

Internal Energy

We know about kinetic energy and potential energy, which can interchange when an object moves through a gravitational field, so let's add to that list the internal energy of the object. This gives us a new and comprehensive way to describe conservation of energy. Check it out!

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number five what is the iup pack name

for the following compound here we have

an Aline and we need to number the

parent chain so as to give the triple

bond occurring soonest so that is going

to go like this 1 2 3 4 5 6 7 8 9 now

there are a couple common errors here it

does look funny sometimes with the

triple bond there there are students

that don't realize this is a carbon and

this is a carbon some students will have

a tendency to see this just as two

carbons with a very long triple bond but

remember that carbons participating in

triple bonds are SP hybridized and

therefore exhibit linear molecular

geometry so we do have carbons 2 and

four totally 180° apart as well as three

and five so this geometry the way that

we draw this is very deliberate that is

accurately depicting the geometry of the

molecule so make sure we understand that

and

IUPAC Nomenclature of Alkynes


okay let's practice using solubility
product constant fluorite or CaF2 is

slightly water soluble and dissolves
according to the following equation so

we have fluoride solid in equilibrium
with the calcium 2 plus ion and 2

fluoride ions now a saturated fluoride
solution is measured as having a calcium

ion concentration of 2.1 times 10 to the
negative 4 molar so what is the

solubility product for fluorite so
that's all the information we need go

ahead and give this a try.

so let's put
our equilibrium up top and once again we

know that the calcium ion concentration
is 2.1 times 10 to the negative 4 molar

so what can we do with this information
well we know that fluorite is going to

dissolve to give a calcium ion and to
fluoride ions so if enough fluoride has

dissolved so as to give a calcium ion
concentration of that magnitude then we

know that the fluoride ion concentration
must be double that because there are

two fluorides that form for every
calcium ion so that means the fluoride

ion concentration must be 4.2 times 10
to the negative 4 molar right that has

exactly double the calcium ion
concentration so we figured that out

just by stoichiometry now let's write
our Ksp expression so this is just like

any other equilibrium expression it
involves only aqueous or gaseous species

and so that means we're just going to
have the calcium ion concentration times

the fluoride ion concentration squared
and then we don't have to put that over

any reactants because the reactant is a solid

now we have all the information we need
we have the calcium ion concentration

and we have the fluoride ion
concentration so we just plug those

numbers in and then we do the arithmetic
and we get a Ksp value of three point

seven times ten to the negative eleven.

Practice Problem: Solubility Product Constant Calculations

We've learned that there are ionic solids that are insoluble in water, and while for the most part that's accurate, we are bending the truth a little bit. Even compounds that are totally water insoluble will still dissolve a teeny tiny bit. We can communicate just how much using solubility product constants, which are another type of equilibrium constant. Let's learn how to calculate these now!

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question number one says in terms of

intermolecular forces which of the

following compounds is only capable of

London dispersion so we have different

kinds of intermolecular forces possible

there could be molecules that do dipole

dipole interactions for example but one

of these is only capable of London

dispersion in order to check out what's

going on we want to draw these molecules

out so we can identify what kinds of

bonds are present ch3

ch2 we've got that o group there

ch3f we got that Central carbon atom and

then we've got the F and then we've got

some hydrogens so that's got that

tetrahedral geometry ch2 ch2 that means

that we have a double bond because every

carbon is going to have four bonds there

and then we have hbr so what do we

notice about the types of bonds that are

present in order for something to do

dipole dipole interactions we have to

have some Bond dipoles we need to have

some polar bonds there can we find some

polar bonds

most of these do have polar bonds so for

example on ethanol there we've got an O

bond this is partial minus this is

partial plus so that is a polar bond and

that Bond dipole is going to be able to

interact with other bond dipoles to make

dipole dipole interactions in this case

they're specifically hydrogen bonds and

then over here we have a polar CF Bond

right this is also polarized towards the

Florine the carbon is going to be

partially positive the Florine is

partially negative the same thing with

hbr we've got a partially positive H and

we've got a partially minus bromine and

so those are examples of polar bonds

that are going to be able to make

electrostatic interactions with other

polar bonds that will be dipole dipole

interactions or hydrogen bonds that are

stronger interactions than London

dispersion however over here we don't

have any of those with ch2 ch2 these are

all non-polar bonds so carbon carbon

bonds are obviously non-polar completely

non-polar and CH bonds there's not

enough an electro negativity difference

there to consider that a polar bond so

we would say that these are all

non-polar bonds and therefore it is not

able to do any dipo dipole interactions

or anything of that magnitude we're

going to say C right here that is ch2

ch2 that is the only molecule that is

only capable of London dispersion

Professor Dave Explains Intermolecular Forces

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welcome to MooMoo Math and Science in

this video let's talk about Solutions

suspensions and colloids a solution is a

mixture of two or more substances the

particles in a solution cannot be seen

light is not visible when it passes

through a solution

and it cannot be separated using a

filter

an example would be sugar and water

sugar is the solute and water is the

solvent the solute is the item being

dissolved and the solvent is what the

item is being dissolved into

this may help you remember

the difference between a solute and a

solvent think of placing the loot into a

vent to hide it so the solute goes into

the solvent

a suspension is a mixture of a fluid

with undissolved particles in it the

solid particles may be visible and will

settle over time in addition the

particles can be filtered out

light is visible when it passes through

a suspension

the solid particles are called the

internal phase and the fluid is called

the external phase

an example would be muddy water

and sand and water

the sand does not dissolve into the

water

but it's mixed into the water and over

time it will settle

a colloid is a mixture consisting of

very small insoluble particles dispersed

into a fluid

light is visible when it passes through

the colloid

you cannot use a filter to filter out

what is dispersed and the particles do

not settle over time

examples include glue

milk and fog

a colloid has a dispersed phase which

are the suspended particles and a

continuous phase which is what the

particles are suspended in

The Continuous phase can be a solid

liquid or gas

I hope that was helpful thanks for

watching and remember kindness

multiplies kindness be kind to someone

today

Solution Suspension Colloid

Learn the difference between a solution,suspension, and a colloid.
This video will help with the following Science  standard
S8P1. Obtain, evaluate, and communicate information about the structure and properties
of matter.
a. Develop and use a model to compare and contrast pure substances (elements and
compounds) and mixtures.
(Clarification statement: Include heterogeneous and homogeneous mixtures

In this video I compare these three types of mixtures. A solution and a colloid are homogenous mixtures and the suspension is a heterogenous mixture.


hi my name is Paul from physics hype

today I want to give you a very quick

overview of the concept of lenses law

please like share and subscribe and

maybe buy me a coffee the link is in the

description below if you want a more

fulsome discussion of lenses law please

check out the link there or the link in

the description below where I give a

more detailed analysis today is just a

short version of it what is lenses law

well lenses law is related to the

concept of Faraday's law and let's write

the Faraday's law down which basically

says that if you generate an EMF because

so we have a change of flux with respect

to time that a metal is experiencing

then there's a negative sign that refers

to the concept of lenses law and lenses

law says that whatever the EMF that is

generated will always produce a polarity

that opposes the one that generates it

and so one way we can describe this is

using a coil so let's say I have a coil

like this

and I'm going to connect this to a

voltmeter and I've insert a North Pole

and I push that into the coil we have a

magnetic field around my bar magnet and

it's going to enter the coil which

results in an EMF Genie being generated

and as a result in this case a current

is produced now the card that is

produced will always produce a magnetic

field that opposes it so as I push this

in I am going to produce a North Pole in

this end it's going to not want to go in

it's going to experience a force of

repulsion now because of the fact that

there's a North Pole over here that

means I will generate a South bolt over

here and So based on my diagram my code

is going to go from there to there so if

I pull that magnet out the north is now

moving away that will produce a polarity

here that is South to oppose it it's

moving away it will create a force that

wants to pull it back in which will

therefore oppose the change that causes

it and so therefore the kind therefore

that is in used is determined by what is

happening over here and it'll always

oppose and as a result that is going to

be consistent with the conservation of

energy because as you pull it in you

require energy to push it down and so

that requires energy to be produced over

here so that is lenses law please like

share and subscribe put a comment down

below and consider supporting me by

buying me a coffee take care bye for now

A quick "cheat sheet " Lenz's Law

This video review the nature of Lenz's as a subset of Faraday's Law

For a further elaboration, see my longer videos on the topic...

Eddy currents and Lenz's Law - https://youtu.be/yU_wwWC-K9c
Lenz's Law in Action - https://youtu.be/0QS3blZFzUg
electromagnetic Braking Explained  -  https://youtu.be/BIsKthqeKSo

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hey everyone Professor Davis here again

to talk to you a little bit about hyper

conjugation specifically how that

applies to carbocation

stability so carbocations are most

stable when they're more substituted we

know this from our lecture

course and we also know that alkal

groups are weekly electron donating and

that's the mechanism by which they

stabilize

carbocations because reducing or

delocalizing charges in small Organics

gives those ions greater stability so we

expect methyl carbocations to be very

rare because they're so

unstable primary is not much better but

when we get on to secondary and tertiary

carbocations we see these constantly in

organic chemistry these are Staples of

first order substitution and elimination

reactions and the reason is that those

alkal groups are increasing the

stability of the

carbocation so let's take a look at what

it is about these R groups that allows

them to donate electrons or to put

electrons close to that carbocation

Center so here I've drawn for you the

methyl cation and the ethyl cation and

these two carbocations you're not really

going to see in a genuine Bonafide

reaction because they're just not stable

enough but they're the simplest examples

that will allow me to demonstrate how

hyper conjugation works so let's set

aside the fact that these are

unreasonably stable and just take a look

at them right now

anyway so I'm going to turn them into

three-dimensional representations and

then I'm going to turn these

representations on their

side and when I do this I can place the

unoccupied p orbital of each carbocation

Center vertically in the plane of our

page now what's different between the

ethyl and methyl cations is that the

ethyl cation has a neighboring CH Sigma

bonding molecular orbital and this

bonding molecular orbital is actually

filled with electrons that's how it's

holding on to the

hydrogen but we also know that these

electrons though I've drawn them here in

between the two bonded atoms we know

that they actually have free run of that

entire volume that is defined by the

molecular

orbital so they can run all around up

and down and as you notice occasionally

they find their way over to the empty p

orbital on the carbocation

and when this happens they slightly

attenuate the positive charge thereby

decreasing it and remember what we said

previously anything that decreases or

delocalizes charge in small organic

molecules or ions is going to stabilize

them so this is why having additional

alkal groups around a carbocation center

makes them more stable and therefore

more likely to

form so next time we'll talk a little

bit about how this hyper conjugate ation

and the stabilizing effect of alkal

groups on carbocations leads to things

like hydride and methyl shifts and I'll

see you then

Hyperconjugation Explained

More substituted carbocations tend to be more stable.  Here, Professor Davis explains the role of hyperconjugation in the process of stabilizing this critical class of organic ions.  Find more videos like this one at www.youtube.com/chemsurvival


It’s Professor Dave, let’s check out Legionnaires’ Disease.

Legionnaires’ disease, also called Legionellosis,
is a severe, often lethal form of pneumonia,

which just means lung inflammation.

Symptoms may include cough, muscle aches,
diarrhea, fever, and chills.

In the U.S., somewhere between 10,000 and
18,000 people get infected each year, some

of which end up in intensive care units.

We first learned about this disease in 1976,
when a mysterious outbreak of pneumonia occurred

at an American Legion convention in Philadelphia,
hence the name.

Because of the fatal nature of some of the
cases, health officials launched an intensive

investigation that resulted in the definition
of a new family of pathogenic bacteria, the

Legionellaceae.

The family Legionellaceae consists of a single
genus, Legionella, with 48 species and 70

serogroups.

These bacteria are slender gram-negative rods
that are pleomorphic, which means they can

change shape in response to their environment.

Infection with Legionella pneumophila has
three possible outcomes: One is no symptoms at all.

Two, an influenza-like illness called Pontiac
fever.

And three, legionnaires’ disease, a severe
form of pneumonia.

Legionella is naturally found in freshwater
lakes and streams, but can be particularly

dangerous to humans when it spreads in man-made
building water systems like cooling towers,

plumbing systems, hot water heaters, or decorative
fountains.

In particular, these bacteria thrive in the
mist created by air-conditioning ducts, and

can infect an entire building of people, like
that fateful summer of 1976 in Philadelphia.

A person becomes infected by breathing in
small droplets of water that contain Legionella,

and typically can’t spread the disease to
other people.

Legionnaires’ disease most often develops
in people with an already increased likelihood

of developing respiratory sickness, such as
those who smoke cigarettes or those with a

compromised immune system.

The Legionella, if able to establish an infection,
replicate within cells of the immune system.

After an incubation period of two to ten days,
an infection of L. pneumophila can cause acute

illness, which can escalate to multi-organ
disease of the gastrointestinal tract, liver,

central nervous system, or the kidneys.

To diagnose a case of Legionnaires’ disease,
clinicians typically take a chest x-ray, and

might conduct a urine test or saliva test
to confirm the presence of the bacteria.

Typically, antibiotics can clear an infection
of legionnaires’ disease, but approximately

10% of people who get sick with the disease
will die due to complications.

The best way to prevent Legionnaires’ disease
is to properly maintain building water systems,

hot tubs, and fountains.

So if you have many hot tubs and fountains
in your mansion, be sure to take precautions now.

Legionnaires’ Disease: Legionella pneumophila

One day in 1976, there was a terrible outbreak of an unknown disease at an American Legion convention in Philadelphia. What was the pathogen responsible for this so-called Legionnaires' disease? Let's find out!

Script by Kellie Vinal

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[Music]

button

[Music]

Professor Dave and Chegg here an

interesting aspect of organic molecules

is that they can join together to form

polymers these are long chain-like

molecules consisting of many repeats of

a small subunit called a monomer let's

learn about these now the best way to

learn about polymers is by examining the

different types of polymers so that we

can see the monomers they are comprised

of and how they link together ethene can

polymerize to form polyethylene this is

just thousands of fully saturated ch2

units perpetuating like a never-ending

straight chain alkane this substance is

a flexible plastic with many

applications if we replace the hydrogen

atoms on ethene with fluorine atoms and

then allow this to polymerize we get a

material called Teflon this is what

gives certain containers their non-stick

coating because the CF bonds are quite

inert

they are almost entirely unreactive

making the material tough and

non-flammable here is a table depicting

many other synthetic polymers according

to the monomers they are comprised of

some of them may sound familiar and

their applications will certainly be

familiar as these materials are

ubiquitous

polymerization can occur by a variety of

mechanisms sometimes they go by addition

polymerization which is a radical

mechanism meaning that it employs

radical intermediates which have

unpaired electrons another mechanism is

condensation polymerization where each

time two monomers are joint a water

molecule is lost a homopolymer is one in

which there is only one repeating

monomeric Unit A copolymer is one in

which there are two different monomers

that must combine to form the chain

nylon is an example of a copolymer

polyethylenes are some of the most

important polymers and there are two

forms low density polyethylene or ldpe

and high density polyethylene or HDPE

the low density variety is as such

because of the extensive branching which

causes the molecule to not be able to

pack together as tightly as for the high

density variety which is comprised

predominantly of straight chain

molecules with these compounds if a

substituent is present like this methyl

group this can result in an isotactic

chain where the methyl groups are all on

the same side of the molecule or an

atactic chain where they alternate from

one side to the other as is depicted

with these Dash and wedge bonds if the

substituent is a chlorine atom we get

polyvinyl chloride or PVC if the

substituent is a phenyl group meaning a

Benzene ring we get polystyrene if

styrene monomers are combined with some

other monomer to make a copolymer we can

get a variety of other useful polymers

with various applications

with that we have taken a glimpse into

the Incredible World of polymers which

have transformed the way we make

industrial materials Professor Dave for

Chegg see you next time

all right

foreign

Introduction to Polymers | Professor Dave & Chegg Explain

In this video, @ProfessorDaveExplains provides a comprehensive introduction to polymers. Polymers are made up of monomers linked together in chains. He explains the different types of polymers, the mechanisms behind polymerization, and provides examples of polyethylenes such as low-density and high-density polymers.

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Watch the whole General Chemistry playlist: https://www.youtube.com/playlist?list=PLllCkartRRdPQdEE-oyt5Di31wIwJc5S6

About Chegg:
Chegg is a leading education technology company that provides students with affordable and accessible study resources. Chegg's mission is to help students save time, save money and get smarter.

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00:00 Intro
00:27 Types of polymers
01:28 Polymerization mechanisms
02:00 Polyethylenes


ok so for today's problem we've got this
compound here with a carbonyl and a

methyl group and our target molecule has
a methyl and a bromo group and what we

want to be able to do is figure out how
to get from here to here in two steps, give it a shot.

as we're analyzing this, let's take a look at the molecule

we're realizing that we have to lose oxygen, we
have to figure out a way to get that

oxygen atom off of there, and we also have to
figure out a way to get a bromine atom

on there, and so we have two steps to
play with, either one is gonna get rid of

the oxygen one is gonna add the bromine
or somehow both can happen at once but

what we definitely want to do is get rid of
that oxygen, so we don't really know of a

way to remove a carbonyl oxygen so why
don't we reduce it. let's take sodium

borohydride, just a nice soft reducing
agent, and let's as step 1 reduce the

carbonyl. we're gonna get an alcohol. now
that we have an alcohol we have some

different possibilities to think about.
now one possibility that we might wanna

use is some strong acid to
protonate that hydroxyl group, get it

to leave as a water molecule, but then what we want to

do is choose a strong acid selectively that
might perform the other transformation

that we also want to occur, so why don't we
use hydrobromic acid and see what happens.

so once again the first step was that
we had some hydride and that went ahead

and reduced it so here is our alcohol,
now if we use hydrobromic acid

we're gonna get that right there, so the
hydroxyl has been protonated, now water

is a good leaving group, we are in acidic
conditions, so carbocations are going to be

fine, so that can go ahead and leave. now
the interesting thing here is if we have

a secondary carbocation, and remember
that a secondary carbocation that is

adjacent to a tertiary or quaternary
carbon, you're probably gonna get some

kind of hydride or or methanide shift.
and so what's happening here is we've

got this implied hydrogen and what we
can do is it'll spontaneously just do a

hydride shift and so if this hydride
moves over there that's going to

neutralize that carbocation, it's gonna leave a
carbocation at the tertiary position. once

again this happens because the tertiary carbocation is
more stable than a secondary carbocation, so that

is going to happen spontaneously and
pretty much instantly. and now since we

used hydrobromic acid, when this
deprotonated, its gonna leave a lot of

bromide swimming around in solution and
so that's going to kill two birds with

one stone because if there's a lot of
bromide and one finds this carbocation

it is absolutely going to react and
generate the intended product. so because

we were clever in the selection of acid,
we've been able to induce this in two steps.

the first thing we did was reduce that
ketone to the alcohol and then because

we used HBr we are able to get that
water to leave, allow this carbocation

rearrangement to occur spontaneously and
then the free bromide ion swimming

around in solution finds the carbocation and there we go, we get our product

thanks for watching guys, subscribe to my
channel for more tutorials and as always

feel free to email me

Practice Problem: Synthetic Strategy

We've got starting material and we've got a target molecule, and we've gotta figure out how to make the transformation in just two steps. Sift through that bag of synthetic tricks!

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the sodium potassium pump is an active

transport mechanism that is driven by

the breakdown of ATP and works through a

series of conformational changes in a

transmembrane protein 3 sodium ions bind

to the cytoplasmic side of the protein

causing the protein to change its

conformation in its new conformation the

molecule becomes phosphorylated at the

expense of a molecule of ATP the

phosphorylation induces a second

conformational change the translocates

the three sodium ions across the

membrane in this new conformation the

protein has a low affinity for sodium

ions and the three bound sodium ions

dissociate from the protein and diffuse

into the extracellular fluid the new

conformation has a high affinity for

potassium ions two of which bind to the

extracellular side of the protein the

bound phosphate now dissociates and the

protein reverts to its original

conformation exposing the two potassium

ions to the cytoplasm on the inside of

the cell this conformation has a low

affinity for potassium ions so the two

bound potassium ions dissociate from the

protein and diffuse into the interior of

the cell

Sodium potassium pump animation

📌 𝐅𝐨𝐥𝐥𝐨𝐰 𝐨𝐧 𝐈𝐧𝐬𝐭𝐚𝐠𝐫𝐚𝐦:- https://www.instagram.com/drgbhanuprakash
📌𝗝𝗼𝗶𝗻 𝗢𝘂𝗿 𝗧𝗲𝗹𝗲𝗴𝗿𝗮𝗺 𝗖𝗵𝗮𝗻𝗻𝗲𝗹 𝗛𝗲𝗿𝗲:- https://t.me/bhanuprakashdr
📌𝗦𝘂𝗯𝘀𝗰𝗿𝗶𝗯𝗲 𝗧𝗼 𝗠𝘆 𝗠𝗮𝗶𝗹𝗶𝗻𝗴 𝗟𝗶𝘀𝘁:- https://linktr.ee/DrGBhanuprakash

SODIUM-POTASSIUM PUMP ANIMATED LECTURE

The #Sodium-PotassiumPump. The process of moving sodium and potassium ions across the cell membrance is an active transport process involving the hydrolysis of ATP to provide the necessary energy. It involves an enzyme referred to as Na+/K+-ATPase.

Na+-K+-ATPase (Sodium potassium ATPase)
------------------------------------------------------------------------
A transmembrane protein pump that actively transports three Na+ ions out of the cell for every two K+ ions that it moves into the cell. Uses ATP for energy. The ATP binding site is on the cytosolic side. Na+-K+-ATPase maintains the concentration gradients of Na+ and K+ and is thus responsible for the resting membrane potential of all cells. The pump is phosporylated when it transports sodium and dephosphorylated when it transports potassium.

#sodiumpotassiumpump #sodiumpotassiumpumpanimation #sodiumpotassiumpumpphysiology #activetransport #usmlevideos #membranephysiology #usmlestep1videos #neetpg #mbbs #fmge #drgbhanuprakash #physiologyvideos #animatedmedicalvideos


[Music]

Professor Dave and CH here we know that

meaning they possess the maximum

hydrogen content possible so every

carbon has as many bonds to hydrogen as

possible but some hydrocarbons are

unsaturated so let's learn about the

various degrees of unsaturation that are

possible if sat saturation means maximum

possible hydrogen content then

unsaturation means less than the maximum

possible hydrogen content in other words

carbon atoms are participating in

additional carboncarbon bonds or bonds

to other elements instead of bonding to

hydrogen so ethane c2h6 is fully

saturated because the two carbon atoms

can't have any more bonds to hydrogen

atoms ethine or ethylene c2h4 has one

degree of unsaturation due to the carbon

carbon Pi bond this Pi bond is

preventing two additional hydrogen atoms

from coordinating we can say the same

thing about straight chain butane or

c4h10 versus cyob butane or

c4h8 because the molecule closes in on

itself to form a ring the two end

carbons must have a bond between them

and this additional Bond results in a

degree of unsaturation in general every

Pi Bond and ring in a molecule adds

another degree of unsaturation which

will result in two hydrogens lost from

the maximum possible given by cnh h2n +2

which describes alkanes knowing the

degree of unsaturation can help us

predict molecular structure let's say

that we know the molecular mass of a

hydrocarbon as being 82 atomic mass

units which corresponds with a formula

of C6 h10 if a six carbon molecule were

fully Satur saturated it would have the

formula C6 h14 since there are only 10

hydrogens this molecule must have 2° of

unsaturation this does not tell us

precisely what the molecule is but we

know that the molecule must either have

two piie bonds two rings or one Pi Bond

and one ring so it could be

cyclohexene it could be hexine or many

other possibilities degrees of

unsaturation can extend to compounds

with hetero atoms as well for more

involved examples we will use this

equation instead where the number of

degrees of unsaturation will be equal to

twice the number of carbon atoms in the

molecule plus two plus the number of

nitrogen atoms minus the number of

halogens minus the number of hydrogens

all over two so for example take the

formula C6 h10 NCL how many degrees of

unsaturation must this molecule have

well twice the number of carbons is 12

plus 2 plus the one nitrogen minus the

one hogen minus 10 hydrogens all over

two so that's 4 over 2 which is equal to

two and in fact this could be a molecule

like this and with the one ring and the

one Pi Bond we do indeed see that there

are two degrees of

unsaturation this concept of degrees of

unsaturation will be useful for certain

applications especially spect R roscopy

which we will examine later for now we

must simply know what this term means

and be able to determine molecules which

exhibit a specific number of degrees of

unsaturation Professor Dave for check

see you next

time

How to Calculate Degrees of Unsaturation Ft. Professor Dave

We know that alkanes are fully saturated hydrocarbons. But not all hydrocarbons are fully saturated. With @ProfessorDaveExplains, let's take a look at the concept of degrees of unsaturation, including how to calculate the degree of unsaturation of a hydrocarbon, using a molecular formula.

Get more homework help from Chegg at https://che.gg/3HbtG8Y
Watch the whole Organic Chemistry playlist: https://www.youtube.com/playlist?list=PLllCkartRRdMPnATf1X8qVdo5RzM8gA_M

About Chegg:
Chegg is a leading education technology company that provides students with affordable and accessible study resources. Chegg's mission is to help students save time, save money and get smarter.

Subscribe: https://www.youtube.com/@chegg
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00:00 Intro
00:29 Degrees of unsaturation


in this video we'll look at how the Lac

operon functions in

eoli this is the Lac operon here you can

see the three structural genes Lac Z Lac

Y and Lac a Lac Z encodes the protein

beta

galactosidase Lac y encodes the enzyme

permease and Lac a encodes the protein

transacetylase all three of these

proteins work together to help the cell

bring in and break down lactose when

it's available in the environment

lactose is a sugar that the cell can use

for energy and

growth so these genes all share a

promoter and an operator further

Upstream of the Lac operon you see the

regulatory Gene Lac ey which encodes an

active

repressor the cell wants to be able to

break lactose down when it's present so

expression of these genes is necessary

when lactose is present in the

environment when lactose is not present

it would be a waste of the cell's energy

and resources to continue to make these

proteins so transcription of these

structural genes will be

limited so how does the cell alter

structural gene expression when lactose

is absent from the environment the Lac

eye repressor is active it binds to the

operator and prevents RNA polymerase

from binding to the promoter this limits

transcription of the lack genes in this

state the operon is said to be

repressed when lactose is present in the

cell the low concentration of

metabolizing proteins that are always

available start to convert lactose into

Alo lactose this Alo lactose acts as an

inducer binding to the Lac ey repressor

making it

inactive the repressor becomes inactive

because The Binding of alol lactose

causes its DNA binding domain to change

shape so it's unable to attach to the

DNA operator

in this state the repressor cannot block

The Binding of RNA polymerase to the

promoter and therefore transcription of

the structural genes can occur this

means that the lack proteins are

produced and the cell is able to break

down lactose at a higher rate in this

state the operon is said to be induced

by the presence of

lactose the Lac operon is said to be

inducible because lactose must be

present for the structural genes of the

operon to be

expressed that's the Lac operon when

lactose is present the repressor is

inactive and the genes that encode the

proteins that help break down lactose

are

expressed when lactose is absent the

repressor is active and the cell limits

production of those lactose metabolizing

proteins as a way to save energy and

resources if you'd like to delve deeper

into the Lac operon check out out my

Advanced La operon video if you want to

learn about another operon check out my

video on the trip operon

The lac Operon Explained

Basics of the lac operon, including how lactose presences alters repressor shape to modify lac gene expression.


[Music]

button

[Music]

Professor Dave and Chegg here the main

factor that influences reaction rates is

the concentration of each reactant we

can describe the relationship between a

reaction rate and the concentration of

its reactants by using something called

a rate law it will be important for us

to be able to derive these for any given

system but before we get to that let's

first make sure we understand precisely

what these are a rate law will take this

general form where the rate will be

equal to a rate constant K times the

product of the molar concentrations of

the reactants each raised to a

particular exponent these exponents are

almost always positive integers though

it is technically possible for them to

be fractions or negative numbers and

they are not evident from looking at the

Stoichiometry of the reaction let us

state that one more time for emphasis as

this is an extremely common error among

students the X exponents that we see

here in the rate law are not related to

and not necessarily equivalent to the

stoichiometric coefficients from a

balanced equation instead these

exponents must be determined

experimentally the rate constant is

independent of these concentrations but

it does depend on other parameters like

temperature and surface area the

exponents in the rate law tell us

something called the reaction order for

a reaction with two reactants the rate

law would look like this if m is 1 we

would say the reaction is first order

with respect to a if n is one it is

first order with respect to B if n is 2

it would be second order in B if it was

0 we would say it was Zero order in B

the overall reaction order is simply the

sum of the individual reaction orders so

if M and N are both one the overall

reaction order is two since one plus one

equals two

here are some examples of rate loss with

the reaction orders listed let's make

sure we can write these as they will be

important in our study of kinetics say

we are examining the reaction of

nitrogen dioxide and carbon monoxide to

form nitrogen monoxide and carbon

dioxide if we acquire kinetic data and

find that the reaction is second order

in NO2 and zero order in Co what will be

the rate law for this reaction well in

the rate law each concentration will be

raised to a certain power equal to the

reaction order for that reactant so

given the data we collected the

concentration of NO2 will be raised to

the second power and the concentration

of Co will be raised to the zero power

which makes it equal to one it can

therefore drop out of the rate law and

we will see that the rate law for this

reaction will simply involve the NO2

concentration squared let's try one more

for this reaction the rate law is as

follows what is the reaction order with

respect to each reactant and what is the

overall reaction order we can see from

the exponents that this reaction is

second order in no because of the two

and first order in H2 because if no

exponent is listed it's the same as an

exponent of one and the reaction is

therefore third order overall since two

plus one equals three

this may all seem pretty confusing but

for now we simply need to focus on these

definitions as we will soon put them to

good use when we derive reaction orders

experimentally which will make this

whole concept much more clear Professor

Dave for check see you next time

foreign

Introduction to Rate Laws | Professor Dave & Chegg Explain

In this video, @ProfessorDaveExplains explains rate laws and their importance in determining reaction rates. The concentration of each reactant is a key factor that affects the reaction rate. By understanding the definition of rate laws and examining some examples, we can progress to experimental derivation of reaction orders.

Get more homework help from Chegg at https://che.gg/3HbtG8Y
Watch the whole General Chemistry playlist: https://www.youtube.com/playlist?list=PLllCkartRRdPQdEE-oyt5Di31wIwJc5S6

About Chegg:
Chegg is a leading education technology company that provides students with affordable and accessible study resources. Chegg's mission is to help students save time, save money and get smarter.

Subscribe: https://www.youtube.com/@chegg
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00:00 Intro
00:22 What are rate laws?
01:27 Reaction order
02:16 Example


okay let's do some calculations
regarding the molar mass of a gas. if we

have a one point three one gram sample
of a hydrocarbon we vaporize it and it

is found to occupy two point zero liters
at 25 degrees Celsius and one point zero

atmospheres what is the gas we want to
identify this gas now we're gonna

probably need to use some information
regarding ideal gases to answer this

question so if you need to review ideal
gases go ahead and check out that

tutorial and when you're ready give this a try.

so we have our given information one

point three one grams 2.0 liters 25
degrees Celsius and 1 atmosphere so

we're definitely going to need to use
the ideal gas law in order to answer

this question
so that's PV equals nRT and if we want

to identify that gas we're probably
going to need the molar mass and molar

mass is grams per mole we have grams but
we don't have moles so let's solve for

it let's solve for n in the ideal gas
law so n is going to be equal to PV over

RT so now let's just plug in what we
know n or the number of moles in this

sample of gas are going to be equal to
one point zero zero atmospheres times

two point zero liters that's the volume
of the sample divided by the gas

constant R which in this case we're
going to use the version that is 0.0821

liters atmospheres per mole Kelvin
because we're going to want to cancel

out those other units and then the
temperature is 25 degrees Celsius but

remember we always want to express that
in Kelvin when we're doing calculations

so that is going to be 298 Kelvin and
then if we plug all that into the

calculator we get 0.08 to
so that is how many moles are in this

sample so now if we know the mass of
this sample of gas and we know how many

moles it represents all we have to do is
take the value in grams and divide by

the value in moles and we'll get grams
per mole so 1.31 grams over 0.08 two

moles is approximately 16 grams per mole
and now while this could be a few

different things we know it is a
hydrocarbon and so the only hydrocarbon

that is gonna have this molar mass is
gonna be methane because we need at

least one carbon atom that's 12 and then
we have no more room for a second one

and just add four hydrogen atoms on
there and we've gotten methane so this

gas must have been methane.

Practice Problem: Molar Mass of a Gas

We know about the ideal gas law, but what are some clever things we can do with it? Well, how about identifying an unknown gas by solving for its molar mass? Sure, why not!

Try all of the general chemistry practice problems: http://bit.ly/ProfDaveGCPP

General Chemistry Tutorials: http://bit.ly/ProfDaveGenChem

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PATREON► http://patreon.com/ProfessorDaveExplains

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Professor Dave again, let’s talk about midsegments.

We just looked at all kinds of bisectors and
other lines that pertain to triangles, so

let’s learn just one more type.

A midsegment of a triangle is a line segment
that joins the midpoints of two sides of the

triangle.

Every triangle will therefore have precisely
three midsegments, and these three line segments

form another triangle, the midsegment triangle.

Since the vertices of the midsegment triangle
are the midpoints of these sides, we can know

that the segments on either side of a midpoint
are equal to one another.

This is signified by these tick marks, one
line here and one line here means these are

the same.

Two lines here and two lines here means these
are the same as each other, but not necessarily

the same as the segments with one tick mark.

Three tick marks, same idea.

Furthermore, the triangle midsegment theorem
states that any midsegment of a triangle is

parallel to one of the sides of the triangle
and is also precisely half the length of that

side.

So here we can see one midsegment and the
side of the triangle that it is parallel to.

The midsegment is half the length of the side,
and since the side is split up into two equal

sections, the midsegment is equal to both
of these two sections.

We can say the same thing for the other two
midsegments and their corresponding parallel

sides.

This allows us to draw all kinds of figures
like this one, where some lengths are given

and others can be determined.

If we know the length of this midsegment,
we know this side length.

If we know this other side length, then we
know the length of this other midsegment.

There’s plenty of algebra we can do with
this theorem, so let’s check comprehension.

The Triangle Midsegment Theorem

We just went over a bunch of different theorems, but we need to do one more! It's kind of neat. I don't have anything else to say, just watch this, it's really short.

Watch the whole Mathematics playlist: http://bit.ly/ProfDaveMath

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in a document if we suspect we've

misspelled a word we can use the fine

function to highlight the error and

correct it or delete it within our DNA

that function is taken on by a system

called CRISPR casts nine CRISPR is short

for clustered regularly interspaced

short palindromic repeats CRISPR

consists of two components the caste

nine protein that can cut DNA and a

guide RNA that can recognize the

sequence of DNA to be edited to use

CRISPR casts nine scientists first

identify the sequence of the human

genome that's causing a health problem

then they create a specific guide RNA to

recognize that particular stretch of a's

t's g's and c's in the DNA the guide RNA

is attached to the DNA cutting enzyme

cast 9 and then this complex is

introduced to the target cells it

locates the target letter sequence and

cuts the DNA at that point scientists

can then edit the existing genome by

either modifying deleting or inserting

new sequences it effectively makes

CRISPR cast 9 a cut-and-paste tool for

DNA editing in the future scientists

hope to use CRISPR cast 9 to develop

critical advances in patient care or

even cure lifelong inherited diseases

CRISPR Explained

This video is an explanation of CRISPR-Cas 9.
____________________________________________
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hormones are chemical messengers

synthesized at one location that deliver

messages to a different location five

main types of hormones influence plant

growth oxen gibberellin cytokine in

ethylene and abscisic acid auxin is the

hormone that allows plants to bend

towards the light this is called

phototropism tropism is the turning of

an organism towards an external stimulus

auxin is synthesized at stem tips and

travels downward that's why if you

remove the stem tip the plant will lose

the ability to grow towards the light

aux-in makes the plant bend towards the

light by moving to the shaded part of

the stem gibberellins are important at

several stages of plant development for

instance they are required in order for

a plant to germinate and our hands up

regulated at that time they also promote

elongation of internodes

aka the stem between nodes which are

despot from which the leaves grow

cytokinin counter senescence in plants

it is also important in making new plant

organs aka roots versus shoots this

hormone is produced in the root apical

meristem in other words the tips of

roots and is passively carried up with

water through the plants xylem the plant

uses the ratio of aux-in versus

cytokinin to determine if it should form

roots or shoots more auxin promotes the

formation of roots while more cytokinin

promotes the formation of shoots

cytokinin also works with the next

hormone will discuss ethylene to cost

abscission or shedding of leaves flowers

and fruits ethylene the magical ripening

hormone if you put a ripe banana in a

bag with unripe bananas the unripe

bananas will ripen much more quickly

ethylene is gaseous with low solubility

in water it diffuses out of cells and

more of it gets produced by rapidly

growing and dividing cells this is

important for germinating plants since

the high production of this hormone

inhibits the expansion of leaves and

allows the chute to dig upwards through

the soil

until it reaches light abscisic acid is

a hormone that signals dehydration it is

synthesized in the chloroplasts of

leaves when a plant is water stressed

this hormone travels through its xylem

and phloem and causes the leaves to

close their stomata to reduce water loss

through photo transpiration these are

just five types of hormones that

influence plant growth however there are

many more if you liked this video please

like and subscribe it would help me make

more videos and make sure to comment

with any topics you'd like me to cover

in future videos also it would be really

nice if you could support me on patreon

thank you

PLANT HORMONES - Auxin Gibberellin Cytokinin Ethylene Abscisic Acid


يا شباب أنا أستاذ ديف وأريد
لأخبرك عن الكيمياء

ما هي الكيمياء لماذا يجب علينا
اعرف عنها

الكيمياء هي دراسة ما يهم
مصنوعة من وكيف يتغير

من المهم أن تعرف لأنه يصف
كل شيء من حولنا

كل المواد ، تقريبا كل عملية. إذا
نريد حقا أن نفهم كيف لدينا

الهيئات العمل

كيف تعمل الأمراض ، وكيفية علاجها
الأطعمة التي هي مغذية والتي ليست كذلك

يجب أن نتعلم الكيمياء. إلا انت
فهم أساسيات الجزيئات

كل ما يمكنك فعله هو تكرار الأشياء التي قمت بها
سمعت. ولكن كيف عرفت أنك سمعت

صحيح؟

إذا كنت تعرف الكيمياء يمكنك مقارنة
ما يقوله شخص ما كيف تعرف ذلك

الجزيئات تتصرف

إذا كان هناك تناقض صارخ لك
يمكن بسهولة رفض المعلومات الخاطئة

ولن يتم استغلالك من أجل
مكسب شخص آخر. ينبغي لنا

تكون قادرة على التفكير النقدي مثل هذا في
كل سبيل الحياة

سواء كان ذلك الصحة أو السياسة أو غير ذلك. مع
هذه السلسلة الكيمياء العامة

لدي هدفين. واحد هو مساعدة عالية
طلاب المدارس والجامعات أخذ

دورة الكيمياء الأولى.

يرجى استخدام هذه الدروس لتكملة
محاضراتك بأي طريقة مفيدة

لك

ولكن لأن هذا هو التمهيدية
المواد وبالنظر إلى أن الكيمياء هي

العلوم المركزية

هذه السلسلة الكيمياء العامة يعني أيضا
لمن يريد أن يتعلم العلم

مع عدم وجود معرفة مسبقة ولأي
السبب. أنا سوف كسر مفاهيم كما

ببساطة ممكن

بحيث يمكن لأي شخص تعلمها. سيكون هنالك
لحظات حيث نذهب أكثر من الممارسة

مشاكل

التي هي موجهة أكثر نحو الطلاب في
فئة ، ولكن لا تتردد في تعلم هذه أيضا.

لأن القوة التنبؤية لل
الرياضيات هي ما يثبت

شرعية علمنا

إذا كنت تشاهد هذا الكليب وتعتزم
لعرض السلسلة بأكملها

الرجاء الضغط على الكيمياء العامة
قائمة التشغيل تحت صفحة البروفيسور ديف الرئيسية

لعرض بالترتيب المقترح. عندما كنا
تحقق من فهمنا بعد الدرس

لا تتردد في إيقاف المقطع حتى يكون لديك
متسع من الوقت للتفكير

قبل ان اقول لك الجواب. اذا احتجت
لمشاهدة مقطع أكثر من مرة بالكامل

فهمته

حسنا. وإذا كان لديك أي
أسئلة حول ما تعلمته

لا تتردد في الكتابة لي

استمتع بمعرفتك المكتشفة حديثًا

Introduction to Chemistry

Hey, you! Yes, you there. Normal Jack or Jill. Do you want to learn science? What's that? Oh, you don't know anything about science? Yes, salt is salty. That's one thing, I guess. It doesn't matter. If you want to start from scratch, you've come to the right place. Learn chemistry first! There's no easier way than with Professor Dave.

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If 3A minus B equals 23, and 4B plus 3C equals 22.
What is the average of A, B and C? The average of

three items is the sum of the items divided by
three. To find the sum, add the given equations.

So you have 3A plus 3B plus 3C equals
forty-five. Now dividing both sides of

the equation by 3, will give us the sum
equals 15. Therefore, the average is 5.

Solve an Average of 3 Items from two give equations | SAT Math

This videos explains how to solve an average of three items from two give equations in an easy way.

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number one asks a bronzee lowy acid

blank while a Lewis acid blank we have

these two different paradigms the Bron

stead Lowry definition and the Lewis

definition and we need to know that for

Bron stead Lowry we're talking about

transferring protons while with Lewis

we're talking about electrons a bronzee

Lowry acid is one that donates protons

and a Lewis acid is one that accepts

electrons so this one is going to be

option b remember a bronze said y acid

donates protons alouis acid accepts

electrons

Important Acids and Bases Definitions #organicchemistry

📽This is a clip from Professor Dave's Acids and Bases Organic Chemistry Practice Exam which is only available on Chemmunity.com

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hey it's professor Dave, let's talk about
carbon

you may have heard that carbon is the

element of life, but what does that
mean? what determines whether an atom is

of carbon in the first place?

an atom consists of positively charged protons

neutrons which have no charge, and
negatively charged electrons

an element is defined by the number of
protons in the nucleus

one proton is hydrogen, two is helium

3 lithium, 4  beryllium, 5

boron and six is carbon. the number of
neutrons and electrons can change but

by definition any atom with six protons
is a carbon atom

how do they form? hydrogen the simplest and
most abundant of all

elements accumulates into clumps due to
gravity which become stars

the tremendous inward pressure triggers
nuclear fusion which is when protons and

neutrons smash together and fuse

many elements including carbon and
larger ones like iron

are made in stars. much larger ones need
the immense energy a supernova

explosion

or particle accelerators here on earth.
so what's so special about carbon?

let's look at the electrons. carbon atoms
have six. the first two are in the inner

shell

and the other four are the so-called
valence electrons in the outer shell

these are the ones available for bonding
with other atoms which happens when two

atoms share two electrons

because elements in this part of the
periodic table want to have eight

electrons in the outermost shell

carbon has a tendency to make four bonds.
these can be single

double or triple bonds and with many
different elements. in addition while

an atom that's bound to two or three other
atoms will adopt a linear or flat shape

a carbon atom that's bound to four other
atoms will take on a three-dimensional

shape

because that's how the electron clouds
on the four atoms can be as far away

from each other as possible

so it's the ability of carbon to bond in
a wide variety of shapes with so many

elements

including itself that makes it so
special. also

carbon-carbon bonds are strong enough to
be stable but not so strong that they

can't break and rearrange

which makes them excellent building
blocks. when by itself

carbon has many allotropes, or ways an
element can be arranged

there's the graphite in your pencil
which consists of slippery sheets of

carbon that rub off easily

there's diamond, the hardest substance known

and we can even make interesting
structures like nanotubes and fullerenes

that's just carbon by itself. when
combined with hydrogen

oxygen, and nitrogen and a few other
elements it makes all the diverse

structures in your body

including carbohydrates, proteins, and DNA.
we can take nature a step further and

synthesize novel materials with an array
of purposes

putting carbons in the empty spots in an iron
lattice makes it stronger

we call this steel which is used in most
large structures we know today

teflon, a carbon-fluorine polymer is a
non-stick substance that is highly

resistant to chemical degradation

and carbon based synthetic drugs like
inhibitors that can silence a faulty

enzyme

hold the prospect of curing so many
diseases. that's a lotta reasons to love

carbon

it really is the element of life. see you
next time

Carbon: The Element of Life

You may have heard that carbon is the element of life. What does that mean? Let's find out!

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EMAIL► ProfessorDaveExplains@gmail.com
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Check out "Is This Wi-Fi Organic?", my book on disarming pseudoscience!
Amazon: https://amzn.to/2HtNpVH
Bookshop: https://bit.ly/39cKADM

Barnes and Noble: https://bit.ly/3pUjmrn
Book Depository: http://bit.ly/3aOVDlT