OpenLLM-France/wikimedia huggingface dataset:
# Open Education Handbook

* About this book
* Who is this book for?
* How this book was written
* Acknowledgements
* Licence

## Open Education

* What is open education?
* History of open education
* Education: traditional & open
* Benefits of open education
* Access, participation, collaboration

## Open Education Resources

* Open Educational Resources (OER)
* Types of OER
* Why use OER?
* History of the OER movement
* Finding and using OER
* Assessing OER quality
* Creating & developing OER
* Useful software for OER creation
* Publishing OER online
* Editor tools for building and remixing OERs
* Impact of OER
* OER & Accessibility
* MOOCs (Massively Open Online Courses)
* OER for the developing world
* OER communities and interest groups
* Open Textbooks
* OER Resources and Handbooks

## Open Licences and Copyright

* Intellectual Property, Rights & Licensing
* Open Licences
* Misperceptions about OER & copyright

## Open Education in Policy and Practice

* Practising open learning
* Open Educational Practices (OEP)
* Open Assessment
* Open Badges
* Open Policy
* Open Advocacy

## Open Data

* Open Education Data
* Open Data & Institutions
* Types of Open Data
* Creating Open Data
* Using Open Data
* Open Data and Learning Analytics
* Open Source Education Tools
* Open data for education: LinkedUp Challenge
* Open Data - Useful Links

## Open Communities

* Open Education - Conferences and Events
* Open Education Data - Conferences and Competitions
* Open Education Directory
* Openness and the developing world
* Wikipedia

## Other

* Glossary
* Frequently Asked Questions (FAQ)
This book was previously published by Open Knowledge and is reproduced here under the work's attribution-only licence.
# Open Education Handbook/About this book

"Open Education" is a topic which has become increasingly popular in a variety of contexts. This handbook has been written to provide a useful point of reference for readers with a range of different roles and interests who are interested in learning more about the concept of Open Education and to help them deal with a variety of practical situations.
As a "living" online document, we hope that it will continue to evolve, reflecting cutting edge research and innovation in this area and helping educational communities to come to an improved understanding of the value of open.
When the process of writing this book first started, the original intention was only to cover open data use in education. As the project progressed it was felt that a broader scope would enable readers to have a better understanding of the synergy and overlap between different aspects or facets of open education (such as open resources, open data, open practices and open culture).
We have been guided by the idea that the handbook will continue to grow and evolve through involvement with the learning communities it is intended to support. The latest version of this handbook includes:
* An overview of different elements of the open education ecosystem
* Information about useful tools and software
* Useful references
* A glossary of terms commonly used in open education
* Case studies and real-life examples
* Answers to frequently asked questions
* Discussion of key issues in open education
During the course of writing the handbook many organisations and individuals related to The Open Education Working Group have contributed. These are listed in the acknowledgements section.
This handbook is a deliverable of the LinkedUp Project (WP4: Dissemination and Community-building) with the work being led by Open Knowledge.
# Pinyin/Middle part

See also
# Pinyin/Midlife

See also
# Pinyin/Military

Search Pinyin military
Militaries are a general name for affairs related to wars, armies and soldiers, etc. Militaries are related to many areas, mainly related to wars. Also, militaries itself included various knowledges. Militaries are a part of politics, are violent methods that a country or organization use to pursue one's goals, and also the goals are always related to interests.
Jūnshì shì yǔ zhànzhēng, jūnduì, jūnrén děng yǒuguān shìwù de zǒngchēng. Jūnshì yǔ hěnduō fànchóu yǒuguān, zhǔyào yǔ zhànzhēng yǒuguān. Cǐwài, jūnshì běnshēn bāohánle gèzhǒng xuéwèn. Jūnshì shì zhèngzhì de yībùfèn, shì yīgè guójiā huòzhě jítuán yòng bàolì shǒuduàn dádào zìjǐ mùdì de fāngfǎ, érqiě mùdì wǎngwǎng yǔ lìyì yǒuguān.
# Pinyin/Milk

Search Pinyin milk
Milk is a white liquid made by mammals, for example cows, dogs, and humans. It is made in the mammary glands (breasts, udders, or teats) of female mammals. Because newborn babies have no teeth, they must be given milk before they can eat solid food. Milk has many nutrients which help babies to grow healthily.
naai shì bǔrǔdòngwù (lìrú niú, gǒu hé rén) shēngchǎn d báisè yètǐ, yóu cíxìng bǔrǔdòngwù d rǔxiàn (rǔfáng huò rǔtóu) zhìzào. Xīnshēng yīng'ér méiyǒu yáchǐ, néng chī gùtǐ shíwù zhīqián, bìxū chī nǎi. Nǎi yǒu xǔduō yíngyǎngsù, bāngzhù yīng'ér jiànkāng chéngzhǎng.
Milk is an important part of a balanced diet. It contains a lot of calcium that helps bones and teeth to grow. Also, milk provides the body with high quality proteins, and helps in meeting the body's requirement for vitamins. One glass of milk gives about 44% to the daily-recommended vitamins intake.
Nǎi shì jūnhéng yǐnshí d zhòngyào bùfèn, hányǒu dàliàng gàizhì, bāngzhù gǔgé hé yáchǐ shēngzhǎng. Cǐwài, nǎi tígōng shēntǐ yōuzhì dànbáizhì, yǒuzhù mǎnzú réntǐ suǒxū d vitamin. 1-bēi nǎi hányǒu yuē 44% měirì tuījiàn vitamin shèrùliàng.
# Pinyin/Middle East

The Middle East is a region that includes the southwest of Asia and the northeast of Africa.
Zhōngdōng shì yīgè bāokuò Yàzhōu xīnánbù hé Fēizhōu dōngběibù de dìqū.
# Pinyin/Militant

See also
# Pinyin/Micronutrient

See also
Micronutrients are essential elements needed by life in small quantities. They include microminerals and Vitamins. They are dietary minerals needed by the human body in very small quantities (generally less than 100mg/day) while macrominerals are required in larger quantities.
Wēiliàngyíngyǎngsù shì shēngwù shǎoliàng xūyào de zhòngyào yuánsù, bāokuò wēiliàngkuàngwùzhì hé wéishēngsù. Wēiliàngyíngyǎngsù shì réntǐ suǒxū hěn shǎoliàng (yībān shǎoyú 100mg/rì) de yǐnshí kuàngwùzhì, ér hóngliàngkuàngwùzhì zé dàliàng xūyào.
# Pinyin/Migrant

Search migrant
# Pinyin/Military exercise

Search Pinyin military
Search Pinyin military exercise
Search Pinyin exercise
# Solar System/Comets
## The Kuiper Belt and the Oort Cloud

The Kuiper Belt is a belt of objects starting just beyond the orbit of Neptune (with some objects' orbits overlapping that of Neptune). In addition to being the home of some dwarf planets and many smaller objects, it is the source of the short-period comets. The Oort Cloud is a group of objects located much farther from the Sun than the Kuiper Belt; it is the source of the long-period comets.

## Gallery
### Distance
### Surface

* 67P/Churyumov–Gerasimenko
* Surface of 67P/Churyumov–Gerasimenko
# Solar System/Dwarf planets

A dwarf planet is an object that is massive enough to be made spherical by its own gravity but has not cleared its neighboring region of planetesimals and is not a satellite. There are currently five recognized dwarf planets:
* Solar System/Ceres
* Solar System/Pluto
* Solar System/Eris
* Solar System/Makemake
* Solar System/Haumea
# Solar System/Earth

Earth is the third planet from the Sun and has a diameter of 12,756 kilometers, making it the fifth largest planet. It is the only place in the universe where at this time life is known to exist. Earth is one of the four terrestrial planets (planets having a solid surface).

## Physical characteristics
### Shape

The diameter of the Earth at the poles is 44 kilometers less than its diameter at the equator.

### Orbit

The orbit is nearly circular, with an eccentricity of only 0.0167. Earth orbits the Sun once every 365.2422 sidereal days (rotations relative to the distant stars) at an average distance of 149,597,870 kilometers, and this distance is defined as one AU (astronomical unit). This distance is often expressed approximately in terms of miles as 93 million miles. Earth is at its closest to the Sun each year (that is, is at perihelion) at a time in the range January 2 to January 5, and is farthest from the Sun (is at aphelion) at a time in the range July 4 to July 7. Light from the Sun takes an average of 499 seconds (8 minutes 19 seconds) to get to Earth.

### Rotation

The Earth rotates prograde (in the same direction as it travels around the Sun) relative to the distant stars once every 23 hours 56 minutes 4 seconds. Since the Earth is traveling around the Sun during this time, it takes an extra 3 minutes 56 seconds to complete a rotation relative to the Sun; so the solar day is 24 hours long. (Indeed, the "hour" is defined as 1/24 of a solar day.)
Earth's axis is tilted by 23.45° relative to the perpendicular of its orbital plane. This tilt is what causes seasons to occur on Earth. (Since Earth's orbit is virtually circular, the minor variations in the distance from the Sun are not the cause of the seasons.) The longest day of the year in the northern or southern hemisphere occurs at the time in the orbit when points on the Tropic of Cancer or Tropic of Capricorn (23.45° north or south latitude) are pointed directly at the Sun at noon (so the Sun appears directly overhead). This time is known as the Summer Solstice in that hemisphere and the Winter Solstice in the opposite hemisphere. The times when the points on the equator are pointed directly at the Sun at noon are called the equinoxes, because at those times all points on Earth experience 12 hours between sunrise and sunset and 12 hours between sunset and sunrise.
# Quantum Chemistry/Example 2
## Question

Starting with the wave equation of a 1D box:
$\Psi \left(x\right)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi }{L}}x\right)$
The classical probability of a particle existing in the region of the box [0, 1/L] is,
$P\left(1/L\right)_{classical}={\frac {1}{L}}$
Where $L$ is the length of the 1D box, $n$ is the principle quantum number, and $x$ is the position of the particle in a 1D box.
Derive an equation that calculates the probability of the particle being in the [0, 1/L] for the quantum mechanical particle in a 1D box. Is this result compatible with the correspondence principle?
# Quantum Chemistry/Example 20

An electronic transition of H atom from an energy level to the ground state is observed with a corresponding wavelength of 102.5 nm. Determine the initial state of the electron from which transition has occurred.
The Rydberg's phenomenological equation will be used to solve the energy transition problem
    Rydberg's phenomenological equation for hydrogen $\lambda =\mathbb {R} _{H}\left({\frac {1}{n_{1}^{2}}}\right)-\left({\frac {1}{n_{2}^{2}}}\right)$
where the $R_{H}$ is Rydberg constant for hydrogen and is equal to 109737 $cm^{-1}$. $n_{1}$ is the final energy level, $n_{2}$ is the initial energy level for hydrogen transition. Both $n_{1}$ and $n_{2}$ are integers and $n_{2}>n_{1}$.
The wavelength is given as 103nm
Concert to SI units:
$\lambda =103nm*1^{-7}{\frac {cm}{nm}}=1.03^{-5}cm$
The H atom's transition is from unknown energy level to the ground state.
So we can know the final state is the ground state which means $n_{1}=1$
The $n_{2}$ now can be calculated:
${\frac {1}{n_{2}^{2}}}={\frac {1}{n_{1}^{2}}}-{\frac {1}{\lambda *R_{H}}}$
${\frac {1}{n_{2}^{2}}}={\frac {1}{1}}-{\frac {1}{0.00103cm*109737cm^{-1}}}$
$n_{2}^{2}=8.67\approx {9}$
Because the energy level always $>0$
$n_{2}=3$
Therefore, the initial state of the electron from which transition has occurred is energy level 3.
# Quantum Chemistry/Example 21

Write a question and its solution that shows the calculation of the vibrational force constant of H19F. Show what the force constant would have to be for HF to absorb visible red light instead of IR light.
If H19F has a vibrational fundamental wavenumber of 4141.3 cm, What is the bond force constant? What would the force constant need to be to absorb visible red light instead of IR light? H has a n exact mass of 1.00782 u, and 19F has an exact mass of 18.998403 u.
Solution:
First, calculate the reduced mass for H19F.
$\mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}$
$\mu ={\frac {(1.00782)(18.998403)}{1.00782+18.998403}}$
$\mu =0.95705\ {\text{u}}\times 1.66\times 10^{-27}{\text{kg}}\ {\text{u}}^{-1}$
$\displaystyle \mu =1.589\times {10^{-27}}\ {\text{kg}}$
Now, solve for the force constant:
$\nu ={\frac {1}{2\pi }}\left({\frac {k}{\mu }}\right)^{\frac {1}{2}}$
$2\pi \cdot c\cdot {\tilde {\nu }}={\left({\frac {k}{\mu }}\right)^{\frac {1}{2}}}$
${\frac {k}{\mu }}=4\pi ^{2}c^{2}{\tilde {\nu }}^{2}$
$k=4\pi ^{2}c^{2}{\tilde {\nu }}^{2}\mu$
$k=4\pi ^{2}(2.998\times 10^{10}{\text{cm}}\cdot {\text{s}}^{-1})^{2}(4141.3\ {\text{cm}})^{2}(1.589\times 10^{-27}\ {\text{kg}})$
$k=966.99{\text{N}}\cdot {\text{m}}^{-1}$
Therefore, the bond force constant is 967 N m$^{-1}$.
Red light is around 700 nm, or 700×10$^{-9}$ m. Converted to wavenumber, that is 14285.71 cm$^{-1}$. Substituting this in,
$k=4\pi ^{2}(2.998\times 10^{10}{\text{cm}}\cdot {\text{s}}^{-1})^{2}(14285.71\ {\text{cm}})^{2}(1.589\times 10^{-27}\ {\text{kg}})$
$k=11506.7\ {\text{N}}\cdot {\text{m}}^{-1}$
Therefore, the bond force constant would have to be 11506 N m$^{-1}$ to absorb visible light.
# Quantum Chemistry/Example 28
## Question

For an electron in a 3D cube with a length of 6.00 Å, what is:

## Solution

(a) The mass of an electron must be known to determine the energy of the particle. To determine the energy state of an electron in a 3D box, the equation that relates the energy levels of a particle in a cube must be used with every variable in its SI units. The mass of an electron is known to be ${9.1093837{\text{×}}10^{-31}{\text{kg}}}$. Planck's constant is known to be ${6.62607015{\text{×}}10^{-34}{\text{m²·kg·s⁻¹}}}$. The length of the box is ${6.00{\text{x}}10^{-10}{\text{m}}}$ since one angstrom is known to be $10^{-10}{\text{m}}$. The energy of the electron in the cube can be calculated as follows,
    ${E}=\left({\frac {h^{2}}{8mL^{2}}}\right)\left({n_{\text{x}}^{2}+n_{\text{y}}^{2}+n_{\text{z}}^{2}}\right)$
    ${E}=\left({\frac {\left(6.62607015{\text{×}}10^{-34}{\text{m²·kg·s⁻¹}}\right)^{2}}{8{\text{×}}\left(9.1093837{\text{×}}10^{-31}{\text{kg}}\right){\text{×}}\left(6.00{\text{x}}10^{-10}{\text{m}}\right)^{2}}}\right)\left({{4^{2}}+{5^{2}}+{3^{2}}}\right)$
    ${E}=\left(1.68{\text{×}}10^{-19}{\text{J}}\right)\left(50\right)$
    ${E}=8.38{\text{×}}10^{-18}{\text{J}}$
(b) Degeneracy is defined as an energy state with an identical energy to that of a particle with a different set of quantum numbers. In this question, the energy of degenerate states must equal the following,
    ${E}=\left({\frac {h^{2}}{8mL^{2}}}\right)\left({{4^{2}}+{5^{2}}+{3^{2}}}\right)=\left({\frac {50h^{2}}{8mL^{2}}}\right)$
To determine the degeneracy of a system, the simplest method is to use a matrix to ensure a systematic process is followed. In this case, the sum of the square of all quantum numbers must equal 50. Therefore, the matrix is as follows,
    ${n_{\text{x}}}$ ${n_{\text{y}}}$ ${n_{\text{z}}}$ ${n_{\text{x}}^{2}+n_{\text{y}}^{2}+n_{\text{z}}^{2}}$ 4534² + 5² + 3² = 50 4354² + 3² + 5² = 50 5345² + 3² + 4² = 50 5435² + 4² + 3² = 50 3453² + 4² + 5² = 50 3543² + 5² + 4² = 50
This results in the identification of 6 different sets of quantum numbers, and thus, this energy state is said to be 6-fold degenerate.
(c) To determine the number of nodal planes, each axis must be considered individually. Mathematically, to calculate the number of nodal planes associated with the x-axis, the following equation is used,
    ${N_{\text{nodal planes x-axis}}=n_{\text{x}}-1}$
Therefore, in this example, the number of nodal planes in x-axis is,
    ${N_{\text{nodal planes x-axis}}=4-1=3}$
Similarly, for the y-axis and the z-axis, the number of nodal planes is,
    ${N_{\text{nodal planes y-axis}}=n_{\text{y}}-1=5-1=4}$
    ${N_{\text{nodal planes z-axis}}=n_{\text{z}}-1=3-1=2}$
Therefore, the total number of nodal planes in this system is 9. This was determined by summing the number of nodal planes associated with each individual axis.
# Structural Biochemistry/Liquid Dosage Forms
### Solutions:

are homogenous mixtures that include one or more soluble ingredients (can be a liquid, solid, or gas) dissolved in the solvent (water or water-miscible liquid).
The advantages of this dosage form is that it can be absorbed relative quickly in the gastrointestinal tract, easier to swallow so that it is usually prescribed for pediatric or geriatric doses. If drug is an external solutions form it must be carefully label.

### Syrups:

this type of drug is mostly used for antibiotics, antihistamines, antitussives, and vitamins. It is defined as sweet, viscous, concentration aqueous solutions of sugar.

### Elixirs:

This is probably the most widely used type of liquid dosage form because of its pleasant taste, relative stability, and ease of preparation characteristics. It is a sweetened hydroalcoholic solution in with the concentration of alcohol may vary (no more than 20%)

### Tinctures:

Tinctures is another form of alcoholic or hydroalcoholic solutions like elixers but with a higher concentration of alcohol. Vegetable, animal, or chemical materials are main ingredient for this form.

### Suspensions:

contains finely divided insoluble medicinal products (called internal phase) dispersed in aqueous external phase (usually contains additional flavoring agent).
Patients can take this drugs by mouth or being applied externally like lotions or be injected into the body depend on which type of suspensions was dispensed to them.

### Emulsions:

Available in both internal and external preparations. Must carefully read the instruction to see if the emulsion is to be used internally or externally. This medicinal products contains water dispersed in oil (stabilized by emulsifying agent) or oil dispersed in water ( both stabilized by emulsifying agent) and microemulsion or transparent emulsion.

### References

Reifman, Noah. Certification Review For Pharmacy Technicians. 9th ed. the United States of America: AuthorHouse, 2011. 61-88. Print.
# Structural Biochemistry/Lipids/Techniques to Study Membranes/Preparation of Bilayers and Vesicles

Calcium cation is required for the process of lipid vesicle fusion to form a phosphate lipid bilayer.
# Structural Biochemistry/Other Non-Common Amino Acids

There are a vast majority of non-common amino acids. There are typically presented as twenty to twenty two common amino acids. The most common amino acids are Glycine, Alanine, Valine, Leucine, Isoleucine, Methionine, Proline, Phenylalanine, Tyrosine, Tryptophan, Serine, Threonine, Asparagine, Glutamine, Cysteine, Lysine, Arginine, Histidine, Asparatate, and Glutamate. There are also two amino acids that aren't typically included in the standard and they are selenocysteine and pyrrolysine.

## Selenocysteine - Sel/U

Structure
This amino acid is considered one of the primary amino acids. There is typically considered twenty-two primary amino acids. This one is similar to Cysteine although it does not contain sulfur. Instead the R group on this amino acid is -CH2SeH. The sulfur that is typically present in the amino acid cysteine is now replaced by a selenium atom.

## Pyrrolysine - Pyl/O

Structure
This amino acid is also considered one of the primary amino acids but is relatively not included. It obviously has similarities to lysine in that its R group is lysine but it has an added extra bit to the end of the lysine. The added end ring is considered a pyrroline ring. This is where it gets its name pyrrolysine from by combining the two names together.

## Other Amino Acids

There are a number of other amino acids but are not considered primary amino acids. These amino acids are usually the result of changes are the amino acid is joined into a protein. These are called post translational modifications and when this happens it can change the structure of the amino acid.
# Structural Biochemistry/Organic Chemistry/Techniques in Organic Chemistry
## Overview

Techniques in organic chemistry applied to biology can produce many accommodations such as medicines that are useful to everyone. Methods of separating, isolating, purifying, and identifying unknown compounds as well as known compounds in unknown mixtures are helpful to the pharmaceutical industry. By isolating and purifying an unknown compound and then performing other analyses such as NMR or electron microscopy, the structure of the compound can be identified and with further analyses of the structure, its biological implications and effects can be understood using chemistry. Ever since the extraction of morphine from opium, many natural products have been isolated and their structures determined. Once the structure was determined, chemists were able to devise synthetic methods to synthesis these compounds, creating modern pharmaceutical chemistry. One example of this is Caffeine. By isolating, purifying, and identifying its structure, scientists have been able to understand how the consumption of caffeine results in wakefulness and alertness. These biological effects are the results of its structural similarity to adenosine and cyclic adenosine monophosphate (cAMP). This similarity allows caffeine to bind to the active site of receptors or enzymes that normally react with adenosine and its derivatives. Adenosine is important in the regulation of brain activity as build up of adenosine levels in the brain during the day causes the concentration to increase and if it is high enough, adenosine binds to brain receptors that activates mechanisms causing drowsiness and sleep. When caffeine binds to the adenosine receptors, it prevents the binding of adenosine and therefore delays sleepiness and increases alertness. Also, Caffeine inhibits the enzyme cyclic nucleotide phosphodiesterase (cAMP-PDE) that converts cAMP to noncyclic monophospate. cAMP is a secondary messenger responsible for increasing blood pressure and delivering more oxygen to the brain. When caffeine binds to cAMP-PDE, it stops the breakdown of cAMP and consequently prolongs the excitation effects. In conclusion, the use of organic chemistry techniques to isolate, purify, identify, and understand compound structures can lead to a chemical understanding of the biological phenomenon that result for chemical interactions which forms the basis for biochemistry.

## References

Mohrig, Jerry R. Techniques in Organic Chemistry. 2006, W.H. Freeman and Company
Organic Chemistry Lab, Chemistry 143A
# Structural Biochemistry/Organic Chemistry/Types of Marcomolecules
## Types of Macromolecules

Macromolecules are large polymer chains. They are made up of many monomers linked together. There are four main categories of macromolecules: Carbohydrates, Nucleic Acids (DNA), Lipids, and Proteins. The function of each macromolecule depends on their structure. Monomers are bonded together by a process called dehydration synthesis, while polymers are broken down by hydrolysis.
# Structural Biochemistry/Organic Chemistry/Types of Macromolecules
## Macromolecules

They are polymers made of preexisting monomers. They can form functional units by connecting together. Forming these functional unites requires energy. Examples of macromolecules include proteins, nucleic acid, lipids and polysaccharides.

## References

1 ^ "http://en.wikipedia.org/wiki/Macromolecule," 2009.
2 ^ Nelson, D.L., "Lehninger Principles of Biochemistry," 2008.
1) Proteins are made of smaller building blocks called amino acids. There are 20 different amino acids. The chains of amino acids fold up in complex ways, giving each protein a unique 3D shape. 2) Nucleic acids allow organisms to transfer genetic information from one generation to the next. 3) Carbohydrates have several roles in living organisms. Carbohydrate derivates are actively involved in fertilization, immune systems, the development of disease, blood clotting and development. 4) Lipids include fats, phospholipids and steroid. Lipids also have function in the body.