Chem 2 Test 1 Equations

Temperature change

q = MC∆T. (M=Mass, C=specific heat)

Phase change

q = n∆H (n=moles)

Density

ρ = (nA)/(Na*V) = mass/volume. (n=number of atoms in unit cell, A=atomic weight, Na=Avo number, V=volume of unit cell)

Avogadro's number

6.022 × 10²³ atoms per mol (Na)

Volume of a sphere

V= 4/3πr³

Henry’s Law

c = kP (c = concentration (M), k = constant that depends on temperature (mol/L*atm), p = pressure (atm))

Molarity (M)

moles of solute/liters of solution

Molality (m)

moles of solute/kilograms of solvent

Volume percent (%v/v)

volume of solute/volume of solution *100 ▪ Volumes must be in the same units

Weight percent (%w/w)

Weight percent = mass of solute/mass of solution *100 ▪ Masses must be in same units

Weight by mass percent (%w/v)

mass of solute/volume of solution *100 ▪ Must be in g/mL or kg/L

Mole fraction

moles of solute/moles of total solution ▪ All mole fractions of individual solutes should total to one

Parts per million (ppm)

milligrams of solute/liters of solution x 10 ^6

Parts per billion (ppb)

micrograms of solute/liters of solution x 10^9

Vapor pressure lowering point

P1=X1P0 (P1=pressure change, P0=initial pressure, X1=mole fraction)

Boiling point elevation

∆Tb = kb(solute) * molality(solute) (Kb would be given to you)

Freezing point depression

∆Tf = kf(solute) * molality(solute)

Osmotic Pressure

Π = iMRT = (n/V)RT (Π=osmotic pressure in atm; M=molarity of the solution; R=gas law constant (0.08026 L*atm/mol*K); T=absolute temperature in kelvin

Gas law constant

0.08026 L*atm/mol*K

kilo

10³

centi

10^-2

milli

10^-3

micro (u)

10^-6

nano (n)

10^-9

F to C

Deduct 32, then multiply by 5, then divide by 9

C to F

Divide by 5, then multiply by 9, then add 32

C to K

Add 273.15

SC number of atoms in unit cell

1

BCC number of atoms in unit cell

2

FCC number of atoms in unit cell

4

R value for CC

8.314 J/(mol K)