8/9/2019 Study Sheet for Biochemistry

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Avogadros Number

o 1 mole of an element = 6.02 *

1023

Oxidation-Reduction

o LEO GER

Loss of Electrons is

Oxidation

Gain of Electrons is

Reduction

o OIL RIG

Oxidation Is Loss of

electrons

Reduction Is Gain of

electrons

Three Conditions Required for a

Reaction to Occur

o

Collision The reactants mustcollide

o Orientation The reactants

must align properly to break and

form bonds

o Energy The collision must

provide the energy of activation

Exothermic Reactions

o The energy of the products is

lower than that of the reactants

and heat is given off

Endothermic Reactions

o The energy of the products is

higher than the energy of the

reactants and heat must be

absorbed for products to form

Gas Pressure

o Pressure (P) = Force / Area

Units for Measuring Pressure

o Atmosphere = atm = 1 atm

o Millimeters of Hg = mmHg = 760

mmHg

o Torr = torr = 760 torr

Boyles Law

o P1V1 = P2V2

Charless Law

o V1 / T1 = V2 / T2

Gay-Lussacs Law

o P1 / T1 = P2 / T2

Combined Gas Law

o

P1V1 / T1 = P2V2 / T2 Avogadros Law

o V1 / n1 = V2 / n2

STP Conditions

o Standard temperature is 0C

(273 K)

o Standard pressure is 1 atm (760

mmHg)

Daltons Law

o P total = P1 + P2 + P3 +

Percent Concentration

o Concentration of a solution =

Amount of Solute / Amount of

Solution

Mass Percent (% m/m)

o Mass Percent (% m/m) = Mass

of Solute (g) / Mass of Solute (g)

+ Mass of Solvent (g) * 100%

Volume Percent (% v/v)

o Volume Percent (% v/v) =

Volume of Solute / Volume of

Solution * 100&

Mass/Volume Percent (% m/v)

o Mass/Volume Percent (% m/v) =

Grams of Solute / Milliliters of

Solution * 100%

Percent Concentrations as

Conversion Factors

o The value of 100 in the

denominator of a percent

expression is an exact number

o

(Ex.) 10 g KCI / 100 g Solutionor 100 g Solution / 10 g KCI

Molarity (M)

o Molarity (M) = Moles of Solute /

Liters of Solution

Dilution

o C1V1 = C2V2

Molarity ofDiluted

o M1V1 = M2V2

Comparison ofSolutions, Colloids,

and Suspensions

o Solution:

Small particles such as

atoms, ions, or small

molecules

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Particles do not settle

Particles cannot be

separated by filters or

semipermeable

membranes

o Colloid

Larger molecules or

groups of molecules or

ions

Particles do not settle

Particles can be

separated by

semipermeable

membranes but not by

filters

o Suspension

Very large particles thatmay be visible

Particles settle rapidly

Particles can be

separated by filters

Calculating pH

o pH = - log [H3O+]

Calculating the [H3O+] Concentration

from pH

o [H3O+] = 10-

pHor [H3O

+] =

antilog (- pH)

Calculating pOH

o pOH = - log [OH-]

Calculating [OH-] = 10-pOH or [OH-] =

antilog (- pOH)

Relationship Between pH and pOH

o pH + pOH = 14

Ion-Product Constant of Water (Kw)

o Kw = [H3O+] * [OH

-]

o [OH-] = Kw / [H3O+]

o [H3O+] = Kw / [OH

-]