OpenLLM-France/wikimedia huggingface dataset:
# AP Chemistry/Solutions
## Molarity versus Molality

* Molarity (M) - moles of solute per liter of solution.
* Molality (m) - moles of solute per kilogram of solvent.

## Solubility Rules

These will NOT be given to you on the test. You must memorize them.
* Compounds with Group 1 elements are soluble or the ammonium ion are soluble.
* Nitrates, acetates, chlorates, and perchlorates are soluble.
* Chlorides, bromides, and iodides are generally soluble, unless they contain copper, silver, lead, or mercury.
* Sulfates are soluble, unless they contain barium, lead, silver, strontium, or calcium.
* Most silver salts are insoluble unless listed in rule #2.
* Most hydroxide salts are insoluble unless stated in rule #1 or contain calcium, strontium, or barium.
* Most sulfides are insoluble unless stated in rule #1, or contain Group 2 elements.
* Carbonates and phosphates are insoluble unless listed in rule #1.

## Forgot Your Polyatomic Ions?

* Ammonium - ${\ce {NH4+}}$
* Nitrate - ${\ce {NO3-}}$
* Acetate - ${\ce {CH3COO-}}$
* Chlorate - ${\ce {ClO3-}}$
* Perchlorate - ${\ce {ClO4-}}$
* Sulfate - ${\ce {SO4^2-}}$
* Carbonate - ${\ce {CO3^2-}}$
* Phosphate - ${\ce {PO4^3-}}$

## Solubility

* Solids are more soluble when they are warm, but gases are the opposite.
* Supersaturated solutions will settle if disturbed.

## Colligative Properties

Colligative properties are properties of solutions that depend on the amount of solute particles present rather than the identity of the particles.
* Vapor pressure lowering
* Freezing point depression and boiling point elevation (molality)
* Osmotic pressure

## Not Solutions

* Colloids look "milky" when light is passed through them (Tyndall Effect)
* Suspensions settle if left long enough, and also look milky.
The Basics · Periodicity · Solids and Liquids · Gasses · Reactions · Stoichometry · Thermodynamics · Solutions · Acids and Bases · Equilibrium · Electrochemistry · Nuclear Chemistry · Organic Chemistry
Constants and Equations · About the AP Exam
# AP Chemistry/Solids and Liquids
## Phase Diagrams

All three phases exist at the triple point. Critical point is where the substance is neither gas nor liquid.

## Intermolecular Forces That Cause Ideal Gas Theory to Fail

* Polar substances - are more positive at one end and more negative at the other. Note that a compound may have polar bonds without being polar as a whole (if the poles cancel out).
* Hydrogen bonding - occurs when hydrogen bonds to N, O, or F. This is what gives water its unusual characteristics (high boiling point, heat capacity, and heat of vaporization, the fact that ice floats)
* London Dispersion - this occurs in massive, nonpolar substances, where a temporary electron asymmetry creates a dipole.

## Liquids

* Surface tension is stronger in polar compounds.
* Viscosity is how "sticky" the substance is.
* Capillary action - the attraction of a polar substance to a nonpolar causing it to rise up a narrow tube for instance

## Solids

* Amorphous solids are just viscous liquids. Very viscous liquids. E.g. Petroleum-derived pitch.
* Crystal lattices are made up of repeated unit cells.
* Metallic solids have a "sea" of decentralized electrons, which is why metals conduct electricity.
* Covalent network solids are basically one huge molecule made from repeating smaller molecules. Examples include (SiO$_2$)$_x$, SiC, graphite
The Basics · Periodicity · Solids and Liquids · Gasses · Reactions · Stoichometry · Thermodynamics · Solutions · Acids and Bases · Equilibrium · Electrochemistry · Nuclear Chemistry · Organic Chemistry
Constants and Equations · About the AP Exam
# AP Chemistry/Stoichometry
## Examples

A certain compound has the empirical formula ${\ce {CH}}$. When 4 moles of it is burned with excess oxygen (combustion), it produces 72.08g H$_2$O. What is the molecular formula of the compound?
Start by writing a balanced equation of the combustion of ${\ce {CH}}$.
${\ce {4CH + 5O2 -> 4CO2 + 2H2O}}$
Now use stoichometry to find the mass of the ${\ce {CH}}$.
72.08gH$_2$O
$72.08gH_{2}O\times {\frac {1molH_{2}O}{18.02gH_{2}O}}\times {\frac {4molCH}{2molH_{2}O}}\times {\frac {13.02gCH}{1molCH}}={104.16g}CH$
Remember that four moles were burned, so the mass you found is found in four moles. You can now use this to find the molecular formula.
${\frac {104.16g}{4molCH}}\times {\frac {1molCH}{13.02gCH}}=2$
This is number is the number you multiply the subscripts by, and so you find the molecular formula to be ${\ce {C2H2}}$
The Basics · Periodicity · Solids and Liquids · Gasses · Reactions · Stoichometry · Thermodynamics · Solutions · Acids and Bases · Equilibrium · Electrochemistry · Nuclear Chemistry · Organic Chemistry
Constants and Equations · About the AP Exam
# AP Chemistry/Nuclear Chemistry

* Nuclear chemistry is just arithmetic with two variables - charge and mass.
* proton + electron = neutron
* 2 protons + 2 neutrons = alpha particle (Helium-4 Nucleus)
* electron = beta particle
* energy = gamma ray emission
* proton = neutron + positron

## Nuclear Stability

* Anything with 83 protons (Bismuth) or more is radioactive. Small stable molecules have a neutron:proton ratio of approximately 1:1. More massive ones have a ratio of 1.5:1.
* Isotopes can also be radioactive, an example of this is carbon-14 which is used in carbon dating only has 6 protons but is radioactive .

## Half Life

The half life of an isotope is time it take for half of a sample to decay. It can vary from very small times(fraction of second) to millions of years. Knowing the half life of a substance allows radioactive dating to be done. The relative abundance of the isotopes in a sample compared to the original abundance can be used to determine the age of the specimen.
* ln(percent remaining) = -k(time)
* half-life = ln(2/k)
The Basics · Periodicity · Solids and Liquids · Gasses · Reactions · Stoichometry · Thermodynamics · Solutions · Acids and Bases · Equilibrium · Electrochemistry · Nuclear Chemistry · Organic Chemistry
Constants and Equations · About the AP Exam
# Adventist Youth Honors Answer Book/Arts and Crafts/Paper Quilling - Advanced

Paper Quilling - Advanced :
| Arts and Crafts North American Division | Skill Level 2 |
| Arts and Crafts North American Division | Year of Introduction: 2006 |

## References

* The Book of Paper Quilling: Techniques & Projects for Paper, 1995 Melinda Johnston,
Sterling Publishing Company, Inc
# Adventist Youth Honors Answer Book/Arts and Crafts/Plastic Canvas - Advanced

Plastic Canvas - Advanced :
| Arts and Crafts North American Division | Skill Level 2 |
| Arts and Crafts North American Division | Year of Introduction: 2006 |
* The requirements for this honor have not yet been published. The NAD has accepted the honor, but has not yet released the official requirements. The requirements listed below should be considered an advance, working copy, and they are therefore subject to change — perhaps significantly.
# A-level Mathematics/OCR/M2

This is the Mechanics 2 module of the OCR A-level Mathematics book. This module assumes that you are familiar with the C1, C2, C3, C4 and M1 modules. Please post all questions to A-level Mathematics/Discussion.

## Contents

* Centre of Mass
* Equilibrium of a Rigid Body
* Projectile Motion
* Uniform Circular Motion
* Impulse
* Work, Energy and Power

## Formulae

By the end of this module you will be expected to have learnt the following formulae:

### Projectile Motion

* Equation of Trajectory: $y=x\tan \theta -{\frac {gx^{2}}{2V^{2}\cos ^{2}\theta }}$

### Work, Energy and Power

* Gravitational Potential Energy (GPE): $E_{p}=mgh\,$
* Kinetic Energy (KE): $E_{k}={\frac {1}{2}}mv^{2}$
* Work done by a force: $W=Fd\cos \alpha \,$
* Power of a moving force: $P=Fv\,$

### Collisions

* Newton's Experimental Law: Separation speed $=e\times$ Approach speed
* It follows that the coefficient of restitution, $e={\frac {v_{2}-v_{1}}{u_{1}-u_{2}}}$

### Centripetal motion

* Using $F=ma\,$, and $a=r\omega ^{2}\,$ $,F=m\omega ^{2}r\,$
* $a={\frac {v^{2}}{r}}$
* $v=r\omega \,$
# A-level Mathematics/OCR/M3/Elastic Strings and Springs
## Hooke's Law and the Modulus of Elasticity

The natural length is the length of an elastic string or spring when it is not stretched or compressed. An elastic string or spring experiences a tension when its length is greater than the natural length. In addition, a spring experiences a compression when its length is less than its natural length. To simplify our analysis, we use the term tension to refer to both types of forces (i.e. tension and compression). Also, we use the term extension to refer to the change in the length of the string or spring. Thus, the extension for a compressed spring is negative.
According to Hooke's Law, the extension $x$ is proportional to the tension $T$ applied to the elastic string or spring. Although Hooke's Law holds only up to the limit of elasticity, we may safely assume its applicability unless otherwise told. We may write this relationship in terms of the natural length $l$ and the modulus of elasticity $\lambda$ (which is a property of the elastic string or spring independent of its length) as follows:
Note that the tension and the extension are in the same direction (i.e. the variables are either both positive or both negative). This should be intuitive since we are considering the force exerted on (i.e. NOT exerted by) the elastic string or spring.
If a mass attached to the end of the elastic string or spring is producing the extension, then by Newton's Third Law, the elastic string or spring exerts a force on the mass equal in magnitude and opposite in direction to its tension. To illustrate this, let us consider the system on the right.
Consider a particle P of mass $m$ suspended vertically from one end of a light (i.e. massless) elastic string of natural length $l$ and modulus of elasticity $\lambda$. The other end of the string is attached to a fixed point O. When the particle is at rest, the resultant force acting on it is zero according to Newton's Second Law. Therefore, the downward weight of the particle should balance the upward force exerted by the string on the particle (which is equal in magnitude to the tension of the string):
| | $mg$ | $=T$ |
| $\Rightarrow$ | $mg$ | $={\frac {\lambda }{l}}x$ |
| $\Rightarrow$ | $x$ | $={\frac {mgl}{\lambda }}$, which is the corresponding extension in the string. |

## Elastic Potential Energy

An elastic string or spring is able to store energy when it is extended (and compressed, in the case of a spring). This stored energy is termed the elastic potential energy (EPE). The EPE in an elastic string or spring is converted from the work done (by an external agent) in producing the required extension. This is just the work done against the force exerted by the elastic string or spring by virtue of its tension. Therefore, the EPE can be determined by integrating the tension $T$ wrt the extension $x$:
* $=$Work done to produce the extension $x$
* $=\int _{0}^{x}T\,dy$
* $=\int _{0}^{x}{\frac {\lambda }{l}}y\,dy$
* $={\frac {\lambda }{l}}\left[{\frac {y^{2}}{2}}\right]_{0}^{x}$
| $\Rightarrow$ | $\mathrm {EPE} ={\frac {\lambda }{2l}}x^{2}$ |
# A-level Mathematics/OCR/M2/Projectile Motion

A projectile is a body which moves under the action of gravity, and may also move horizontally. For M2 we only consider projectiles moving in one plane.
A projectile at launch can be thought as a vector, as either a velocity and angle or initial velocities in x and y planes, shown here.
As the object travels in only one plane, if we assume no air resistance (as is standard in M2), then the projectiles horizontal motion will remain constant while the vertical will only have the objects mass (gravity) acting upon it. This leads to the two equations below.
Below are the equations for the projectiles displacement. One can derive these from our equations of motion from M1 or integrate with respect to t the velocity equations as $\ x_{v}={\frac {d(x_{s})}{dt)}}\equiv \int x_{v}dt=x_{s}+c\!$ since we define our start point as the origin, c = 0
With these four equations, and applying some facets of symmetry we can solve many novel problem.
$\ y_{s}=usin(\theta )t-{\tfrac {1}{2}}gt^{2}\!$
$\ 0=t(usin(\theta )-{\tfrac {1}{2}}gt)\!$
$\ t=0\ and\ t={\frac {2usin(\theta )}{g}}\!$
The x_s at this point in time is the range,
$\ x_{s}=ucos(\theta )\ *\ {\frac {2usin(\theta )}{g}}\!$
C3 identity
$\sin(2x)=2\sin(x)\cos(x)\,$
$\ x_{\triangle }={\frac {u^{2}sin(2\theta )}{g}}\!$
$\ v^{2}=u^{2}+2as\!$
$\ 0^{2}=(usin(\theta ))^{2}+2(-g)h\!$
$\ 0=u^{2}sin^{2}(\theta )-2gh\!$
$\ 2gh=u^{2}sin^{2}(\theta )\!$
$\ h_{\triangle }={\frac {u^{2}sin^{2}(\theta )}{2g}}\!$

## Notes

* ↑ OCR C3 Trignometric Identities
# A-level Mathematics/OCR/M2/Uniform Circular Motion

Uniform circular motion describes a body which is following a circular path at a constant speed. For M2 you will only need to consider this where the circular path is horizontal.
Angular velocity is the rate at which a body turns through an angle. It is therefore defined as change in angle/time taken. If a body travels on a circular path, and turns through one revolution per second, its angular velocity is (2π radians) divided by (1 second), so its angular velocity is 2π rad/s.
The usual symbol for angular velocity is ω (a lower-case omega). The body's "tangential speed" and angular velocity are related by the equation v = rω, where v is in m/s and r is the radius of the circle (in m).
# A-level Mathematics/OCR/M2/Centre of Mass

The centre of mass of an object is the point through which its weight can be considered to act.
An object made of multiple parts, when a mean is taken of these from a fixed point, we can solve to find the centre of the objects mass.
# A-level Mathematics/OCR/M3/Impulse and Momentum
## Momentum in 2 dimensions

M2 builds on work done in M1, it doesn't introduce any new equations, however it introduces the use of vectors to describe Impulse, Momentum and the velocities involved.

### defining coordinate systems

Now we are working in 2 dimensions it will simplify problems if we break the vectors involved down into their component parts, but first we must define the coordinates we are going to use.

#### A uniform sphere impacting a smooth surface(such as a snooker ball hitting the side of a table)

In these kind of questions where a sphere is hitting a solid unmovable surface it will be useful to use the direction parallel to the surface as the x direction, and the direction perpendicular to the surface to be the y direction. With this convention we can see that an impulse that occurs on impact will only act in the y direction, therefore the x component of the velocity of the impacting sphere will be the same before and after the impact, and the impulse I will equal M(Uy+Vy) (I = M(V-U) I have taken Uy to have a negative velocity, having the positive y axes going away from the surface.)
I will not derive it here but if we are given the angle of approach α and the angle of separation β of the sphere we can work out the coefficient of restitution e using:
e = tan(β)/tan(α)

#### Oblique impacts

This is similar to the above instance, however it is now two uniform spheres colliding. So instead of using the surface of the spheres as the basis of our coordinate system, we should use axes that are the same for both spheres at the time of impact, this is called the line of centres. Where the two spheres collide, the line of centres is the line connecting the centres of the two spheres, we will use this now as our x direction, which makes a change to our above assertion that the velocity in the x direction is unchanging in the impact. we have moved our axes round so that now it is the velocities of the spheres in the y direction that are unchanged in the impact.
therefore we have that Vy = Uy for both spheres.
Now in the x direction we make a return to M1 and 2, in the x direction we use the ideas of conservation of momentum, because as with the example of the ball and surface, the impulse must act along the line of centres, hence only changing the x components of the velocities of the spheres.
# A-level Mathematics/OCR/M3/Kinematics
## Formulae

You are expected to know the following formulae, as they will not be provided in the formula book:
$a={\frac {dv}{dt}}={\frac {dv}{dx}}{\frac {dx}{dt}}=v{\frac {dv}{dx}}$
# A-level Mathematics/OCR/M2/Impulse

Impulse can be defined as Force x Time (i.e. the magnitude of a force multiplied by the time it acts for), and is measured in Ns (newton seconds).
If an impulse is applied to a body, the impulse is equal to the change in momentum of the object. This produces the impulse-momentum equation: Ft = mv - mu.
# A-level Mathematics/OCR/M3

This is the Mechanics 3 module of the A-level Mathematics book. This module assumes that you are familiar with the C1, C2, C3, C4, M1 and M2 modules. Please post all questions to A-level Mathematics/Discussion.

## Contents

* Elastic Strings and Springs
* Impulse and Momentum
* Circular Motion
* Kinematics
* Simple Harmonic Motion

## Formulae

By the end of this module you will be expected to have learnt the following formulae:

### Kinematics

* $a=v{\frac {dv}{dx}}$

### Simple Harmonic Motion

* ${\ddot {x}}=-\omega ^{2}x$
* $v^{2}=\omega ^{2}(a^{2}-x^{2})\,$
* $T={\frac {2\pi }{\omega }}$
* $x=a\sin(\omega t)\,$ or $x=a\cos(\omega t)\,$ or, generally, $x=a\cos(\omega t+\epsilon )\,$

### Elastic Springs and Strings

* Hooke's Law: $T={\frac {\lambda x}{l}}$
* Elastic potential energy: $E_{p}={\frac {\lambda x^{2}}{2l}}$

### Small-angle Approximations

* $\sin \theta \approx \theta$
* $\cos \theta \approx 1-{\frac {1}{2}}\theta ^{2}$
* $\tan \theta \approx \theta$
only applies when $\theta$ is in radians
# Ada Programming/Pragmas/Inline
## Summary

The pragma Inline specifies that a subprogram should be expanded inline when called in code.

## See also
### Wikibook

* Ada Programming
* Ada Programming/Pragmas
# Ada Programming/Pragmas/Admission Policy
## Description

An admission policy governs the order in which competing tasks are evaluated for acquiring the execution resource associated with a protected object.
# Ada Programming/Pragmas/Import
## Summary

The pragma Import directs the compiler to use code or data objects written in a foreign computer language.
Which foreign languages are supported depends on the compiler implementation. Typically C, C++, Cobol, and Fortran are supported.

## See also
### Wikibook

* Ada Programming
* Ada Programming/Pragmas
* Ada Programming/Pragmas/Export
* Ada Programming/Pragmas/Convention
* Ada Programming/Pragmas/Linker_Options
* Ada Programming/Pragmas/Interface (obsolescent)
# Ada Programming/Pragmas/Interface
## Summary

Interface is an Ada 83 interfacing pragma equivalent to pragma Import. However, it can be used just for subprograms while pragma Import allows also interfacing with global variables.
Some compilers for recent versions of the language still allow this pragma, but treat it just like pragma Import. In older code, pragma Import was usually followed by the implementation-defined pragma Interface_Name to specify the external name of the subprogram.
It is worth noting that "interface" is a new Ada 2005 keyword. However, a special exception is made in the standard to allow it as a pragma identifier for backwards compatibility.

## See also
### Wikibook

* Ada Programming
* Ada Programming/Pragmas
* Ada Programming/Pragmas/Import
* Ada Programming/Pragmas/Export
* Ada Programming/Pragmas/Convention
* Ada Programming/Pragmas/Linker_Options
* Ada Programming/Pragmas/Interface_Name (implementation defined)

### Ada Reference Manual
#### Ada 83

* 13.9: Interface to Other Languages
* Annex B: Predefined Language Pragmas

#### Ada 2005

* J.12 Pragma Interface (Annotated)

## References

* ↑ "J.12 Pragma Interface (Annotated)". Ada 2005 Reference Manual. {{cite book}}: External link in |chapter= (help)
# Ada Programming/Pragmas/Export
## Summary

The pragma Export directs the compiler to make available subprograms or data objects written in Ada to foreign computer languages. If a subprogram is exported, it is compiled with conventions expected by the foreign language. For example, if a subprogram is being exported to C, record types declared in the subprogram are compiled the same as C-style structs.
The set of supported foreign languages depends on the compiler implementation. Typically C, C++, Cobol, and Fortran are supported.

## See also
### Wikibook

* Ada Programming
* Ada Programming/Pragmas
* Ada Programming/Pragmas/Import
* Ada Programming/Pragmas/Convention
* Ada Programming/Pragmas/Linker_Options
* Ada Programming/Pragmas/External (implementation defined)
# Ada Programming/Pragmas/Controlled

Pragma Controlled is an old Ada 83 pragma that was never implemented by any compiler. Consequently, it was finally removed with Ada 2012.
The pragma was meant to prevent garbage collectors (also never implemented by any compiler) from tampering with the storage pool of the corresponding access type. The syntax was
Also see Deleting objects from a storage pool in the page about access types.

### Ada Reference Manual
#### Ada 83

* 4.8: Allocators

#### Ada 95

* 13.11.3: Pragma Controlled

#### Ada 2005

* 13.11.3: Pragma Controlled
# Ada Programming/Pragmas/Convention
## Summary

The pragma Convention directs the compiler to represent a type or subprogram using a foreign language convention.
The set of supported foreign languages depends on the compiler implementation. Typically C, C++, COBOL, and Fortran are supported.

## See also
### Wikibook

* Ada Programming
* Ada Programming/Pragmas
* Ada Programming/Pragmas/Export
* Ada Programming/Pragmas/Import
* Ada Programming/Pragmas/Linker_Options
# Ada Programming/Pragmas/Inspection Point
## Summary

The pragma Inspection_Point directs the compiler to ensure that the specified variable is available where the pragma appears.
This pragma aids in debugging.

## See also
### Wikibook

* Ada Programming
* Ada Programming/Pragmas

### Ada Reference Manual

* H.3.2 Pragma Inspection_Point (Annotated)

### Ada 95 Rationale

* H.2.2 The Pragma Inspection_Point
# Bartending/Cocktails/Tequila Sunrise
### Ingredients

* 45 ml (1½ oz) Tequila
* 90 ml (3 oz) Orange juice
* 15 ml (½ oz) Grenadine syrup

### Garnish

orange slice cherry straw swizzle stick

### Glass

highball glass

### Procedure

* Pour tequila and orange juice directly into highball with ice cubes.
* Add a splash of Grenadine syrup to create chromatic effect (sunrise), do not stir.
* Garnish with orange slice and cherry, straw and stirrer.
# Bartending/Cocktails/Three Dollar Shoes

"Three Dollar Shoes" is a gin cocktail made with Seagram's and Bombay Sapphire brand gin. It also includes Margarita mixer for sweetness, and grenadine to give the drink a pink tint.

## History

The Three Dollar Shoes cocktail was invented in Albuquerque, New Mexico on 21 July 2006 as a strong but sweet cocktail to drink before going to a local club for Salsa dancing. Originally, the drink included only Seagram's and Margarita mixer. Bombay Sapphire was added to give a higher alcohol content, and drawing inspiration from one of the inventors recently purchased pink high-heel shoes (for three dollars), the grenadine was added to give the cocktail a pink tint.
For serving the drink, a plastic water-bottle was used for ease and portability. In keeping with the original spirit of the cocktail, if at all possible the drink should be served from a large, cheap plastic bottle, though a drink mixer can be used instead.

## Ingredients

* 3 parts Seagram's brand Gin
* 2 parts Bombay Sapphire brand Gin
* 2 parts (or to taste) chilled margarita mixer
* 1 part Grenadine

## Glass

* As served originally, the drink should be poured from a plastic bottle, though a drink mixer may be substituted. The cocktail may be drunk strait from the water bottle, or poured into large glasses.[1]

## Procedure

* Add all ingredients, gin first, into the plastic bottle or mixer.
* Shake for 10 seconds, or until well blended.
* Drink directly from water-bottle, or pour into a large glass. It may also be served as is, chilled, or on the rocks.
* ↑ UR. (1964). Ur. (Revue.) Nouvelle série. no. 4. automne 1964. OCLC 504367676.