Dictionary Directory

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1. Cascading Dilutions - This is a method whereby one dilutes a liquid into another, then uses the resulting diluted liquid to create another, and then again from the resulting dilution etc.
2. Concentration, Matter, & Volumes - n (amount of mass in mols) can be linked to a concentation. Notice here the notation C is used for molar concentration for simplicity of notation. M (molar mass g/mol) could have and is often used instead.
   $$ n = {C}\cdot{V} \implies n = {\frac{mol}{L}}\cdot{L} \implies n = \frac{mol\cdot{\cancel{L}}}{\cancel{L}} = mol $$ $$ n = {C}\cdot{V} \implies n = \frac{mol\cdot{\cancel{L}}}{\cancel{L}} = mol $$ $$ C = \frac{n}{V} \implies C = \frac{mol}{L} $$ $$ V = \frac{n}{C} \implies V = \frac{\cancel{mol}\cdot{L}}{\cancel{mol}} = L $$

   Where:

   • n - Amount of mass in mols
   • C - Concentration, molarity, or molar concentration mol/L
   • V - Volume officially: m³ we use litres L

   Note that mol/L is often written as M for short-hand; not to be confused with M = g/mol.

3. Constants - Here's a quick list on common physical constants often used in science. These are used throughout the site, do not hesitate to reference this sheet when needed.

   Biological Constants:

   • Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144…
     $$ F_{n} = F_{n - 1} + F_{n - 2} $$
     • Where: \( F_{1} = 1, \space F_{2} = 1 \)
     • Or: \( F_{1} = 0, F_{1} = 1 \)
   • Golden Ratio: \( \approx 1.6180339887 \ldots \)
     $$ \frac {a+b}{a} = \frac{a}{b} = \frac {1 + \sqrt {5}}{2} = \varphi $$

   

   • Physiological Temperature: 36.5-37.5°C

   Chemical Constants:

   • Avogrado's Number: \( 6.022140857 \cdot 10^{23} mol^{−1} \)
   • Dissociation constant: \( K_w = 1.0 \cdot 10^{-14} \)

   Physical Constants:

   • Gravitational Constant:
     $$ G = 6.67408\cdot10^{-11}\cdot{m^3}\cdot{kg^{-1}}\cdot{s^{-2}} $$
   $$ F = G \cdot \frac{m_1 \cdot m_2}{r^2} $$
4. Conversions - For a simple method on converting unit prefixes from one to another refer to the below table.

   Prefix	Symbol	Modifier	Conversion
   Peta	P	1015	⬇︎ Multiply by positive exponent
   Tera	T	1012
   Giga	G	109
   Mega	M	106
   Kilo	k	103
   Hecto	h	102
   Deca	da	101
   N/A	N/A	1 = 100
   Deci	d	10-1	Multiply by negative exponent ⬆︎
   Centi	c	10-2
   Milli	m	10-3
   Micro	μ	10-6
   Nano	n	10-9
   Pico	p	10-12
   Femto	f	10-15

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5. Dilutions - In order to calculate the new concentrations of solutions:
   $$ C_1 \cdot V_1 = C_2 \cdot V_2 $$

   Variables:

   • C₁ - Initial concentration.
   • V₁ - Volume taken from initial solution at aforementioned concentration.t
   • C₂ - New concentration after V₁ has been placed in new volume V₂.
   • V₂ - New volume into which V₁ is placed.

   Use the formula by isolating the variable.

   One use of the formula is to find the new concentration of a solution when introduced to a new container containing another solution. Think of C₁ as the concentration of the chemical or solution one wants to dilute, which will be placed into another liquid. In order to transfer this C₁ solution to another liquid one must take a determined volume of it, this is what is marked as V₁. This V₁ sample of the C₁ solution can be taken and introduced into a new volume V₂. Thus, it is so we know to isolate C₂ and find our new concentration.

   Make sure to always verify that all units are in accord, and use dimensional analysis in order to verify.

   If:

   • C - μg/mL
   • V - μL
   $$ \frac{C_1 \cdot V_1}{V_2} = C_2 $$
   $$ \frac{\frac{g}{mL} \cdot \cancel{\mu L} }{\cancel{\mu L}} = \frac{g}{mL} $$

   If one were to want to find the amount of a solution containing a product to be placed into another volume necessary for the creation of a new solution of the new volume with a given concentration then one would need the following form of the formula.

   $$ V_1 = \frac{C_2 \cdot V_2}{C_1 \cdot 10^{3}} \implies \frac{g/mL \cdot mL}{g/mL \cdot 10^{3}} = mL $$

   If:

   • C - μg/mL
   • V - mL
6. Dimensional Analysis - The process of setting up a formula for use, and running a calculation only using the necessary units instead of values. If the units are in accord and equal each other after elimination then the formulation is correct.

   Example: Concentrations, Matter, & Volumes

   $$ n = {C}\cdot{V} \implies n = {\frac{mol}{L}}\cdot{L} \implies n = \frac{mol\cdot{\cancel{L}}}{\cancel{L}} = mol $$

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7. [Physics] Force - A force can be mesured and calculated in a number of different ways. As per Newton's second law of motion, a force is described by the following formula:
   $$ F = m \cdot a $$

   Where:

   • F - In Newtons (units); kg·m/s^2.
   • m - Mass in Kg.
   • a - Acceleration in m/s².

   Indeed a force is the result of an accelerating object encountering a mass.

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8. [Chemistry/Physics] Ideal Gas Law - This law is used in relation to gasses. It
   $$ PV = nRT $$

   Where:

   • P - The amount of pressure
   • V - Volume of the gas
   • n - The amount of particles in the system in mols
   • R - Ideal gas constant 8.314459848 kg·m²·mol^-1·K^-1·s^-2

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9. Matter & Mass - For dealing with quantities of matter, use the following formulas and know the following units. Pay attention and be careful not to confuse the symbol for molar mass M with the symbol for molarity M these are the same except that molar mass is to be italicised. For the sake of clarity will always use M to represent molar mass.
   $$ \require{cancel} n = \frac{m}{M}\implies n =\frac{g}{\frac{g}{mol}}\implies n = \frac{\cancel{g}\cdot{mol}}{\cancel{g}} \implies n = mol $$ $$ m = M\cdot{n} \implies m = \frac{g}{mol} \cdot mol \implies m = g $$ $$ M = \frac{m}{n} \implies M = \frac{g}{mol} $$

   Where:

   • n - Amount of mass in mols
   • m - Amount of mass in grams
   • M - Molar mass of the material in g/mol

   With this formula you can find the amount of mass in mols, the amount of mass in grams or even the molar mass of a material.

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10. Percentage of Powder In Solution - And for calculating the percentage of a powder in a solution use the following formula. Here we take the mass to roughly be equivalent to the volume, and present an example with a solution of agar.

    As an example we will use 400 mL of TAE x1, and 0.8% agar.

    $$ \% = \frac{m}{V} \cdot 100 \implies m = \frac{\%}{100} \cdot V $$

    Were:

    • m is mass of agar in g
    • v is volume of buffer in mL

    As per our formula the mass to use for 0.8% is:

    $$ m = \frac{0.8 \%}{100} \cdot 400 = 3.2 g $$

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11. Units - You will often encounter two different kinds of units. Fundamental units which cannot be expressed in any other way, such as grams, times, and distances. These fundamental units can be combined in different ways to express more complex ideas. These are known as derived units; Such as joules or even concentrations J = (kg.m²)/s² and M = mol/L respectively.

    For a run down on common and essential units used in science refer to the tables below:

    Fundamental Units

    Measurement	Measurement's Symbol	Dimensional Symbol	SI Unit	Associated Symbol
    Mass	m	M	kilogram	kg
    Amount of matter	n	N	mole	mol
    Time	t	T	second	s
    Distance	l, x, r...	L	metre	m
    Temperature	T	θ	kelvin	K
    Electrical current	I, i	I	ampere	A
    Luminous intensity	Iv	J	candela	cd

    Derived Units

    Name	Measurement	Measurement's Symbol	Fundamental SI units	Other Derived SI Units
    Hertz	Frequency	Hz	1/s
    Newton	Force, weight	N	kg·m·s−2
    Pascal	Pressure, stress	Pa	kg·m−1·s−2	N/m2
    Joule	Energy, work, heat	J	kg·m2·s−2	N·m
    Watt	Power, radiant flux	W	kg·m2·s−3	J/s
    Coulomb	Electrical charge/Quantity of Electricity	C	s·A
    Volt	Voltage (electrical potential difference), electromotive force	V	kg·m2·s−3·A−1	W/A
    Farad	Capacitance	F	kg−1·m−2·s4·A2	C/V
    Ohm	Electrical resistance, impedance, reactance	Ω	kg·m2·s−3·A−2	V/A
    Siemens	Electrical conductance	S	kg−1·m−2·s3·A2	A/V
    Weber	Magnetic flux	Wb	kg·m2·s−2·A−1	V·s
    Tesla	Magnetic flux density	T	kg·s−2·A−1	Wb/m2
    Henry	Inductance	H	kg·m2·s−2·A−2	Wb/A
    Degrees Celsius	Temperature relative to 273.15 K	°C	1°C = 273.15 K
    Lumen	Luminous flux	lm	cd	cd·sr
    Lux	Illuminance	lx	m−2·cd	lm/m2
    Becquerel	Radioactivity (decays/time)	Bq	s−1
    Gray	Absorbed dose (ionising radiation)	Gy	m2·s−2	J/kg
    Sievert	Equivalent dose (ionising radiation)	Sv	m2·s−2	J/kg
    Katal	Catalytic activity	kat	mol·s−1

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12. Variance (Statistics) -
13. Volume (Geometry) - A volume is the measure of the space inside a three dimensional object; it is measured using height, length and width. These measurements multiplied together equal a volume.

    In order to measure complex shapes, one can simplify the complex shape into smaller more simple shapes, this of course only gives an estimation of the volume of the sum of the shapes. A more precise way of measuring complex shapes is by liquid displacement; whereby one submerges the object into a known volume. The displaced volume is thus equal to the volume of the submerged object.

    Formulas for basic geometrical shapes:

    Volume of a Cube

    $$ vol(cube) = h \cdot w \cdot l $$

    Volume of a Pyramid

    $$ vol(pyramid) = \frac{A_b \cdot h}{3} $$

    For finding the area of a rectangular base and a rectangular base:

    Reminder: Area Pyramid Bases

    $$ area(rectangle) = l \cdot w $$ $$ area(triangle)* = \frac{l \cdot w}{2} $$

    Note* only works if l and w are perpendicular to one another, ie. a 90° angle is formed.

    Volume of a Cylinder

    $$ vol(cylinder) = π \cdot r^2 \cdot h $$

    The volume of a cylinder is found by first finding the area of the circle that composes the shape then multiplying by the height of the shape.

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