Chemistry Chapter 7-8 Flashcards
Combination
A+B = AB
Decomposition
AB = A+B
Single Replacement
A+BC = AC+B
Double Replacement
AB+CD = AD+CB
Combustion
Cx + Hy + O2 = CO2 + H2O
Steps of Combustion
Step 1: Put 2 in front of CxHy and distribute
Step 2: Go to H and H2O, balance and distribute
Step 3: Balance C in CO2 and distribute
Step 4: Add up Os on the right and the balance on the left
Step 5: Reduce if everything is reducable
Add Ratio and Balanced
OIL RIG
Oxidation is loss of electrons
Reduction is gain of electrons
Rules of Oxidation Reduction
1: Assign free elements a charge of 0
Ex: Cu or Fe2
Diatomic elements are free elements
2: Assign other charges based on oxidation numbers charges
ex: 2Cu^0 + O2^0 = 2Cu^+2O-2
3: Compare the charges and identify as oxidation or reduction
Oxidation reaction is when
0 goes to positive charge
Loses electrons
ex: 0 to 2+
Reduction reaction is when
0 goes to negative charge
Gains electrons
ex: 0 to 2-
If there is an e- on the left it is
reduction
If there is an e- on the right it is
oxidation
Oxidation always
loses electrons, may also be seen as an addition of oxygen or loss of hydrogen
Reduction always
gains electrons, may also be seen as loss of oxygen or gain of hydrogen
Half Redox reaction
Cu^0 + 2AgNO3 = Cu(NO3)2 + 2Ag
Cu goes from 0 to 2+
Ag goes from 1+ to 0
Half Redox:
Cu^0 = Cu^+2 + 2e-
Oxidation
Ag^+1 + e- = Ag^0
Reduction
What is a mole?
Unit of measurement, amount of a pure substance containing the same number of chemical units (atoms or molecules)
Avagadro’s Number
6.02 x 10^23 particles
1 mole / 6.02 x 10^23 or 6.02 x 10^23 / 1 mole
Moles of an element in a compound
Subscripts are used to write conversion factors for moles of an element
How many atoms of O are in 0.150 mole of C9H8O4?
3.61 x 10^23 atoms of O
How many CO2 molecules are in 0.500 mole of CO2?
3.01 x 10^23 molecules of CO2
What is the number of moles in 1.8 x 10^24 atoms of S?
3.0 moles of S
Molar Mass
quantity in grams that equals atomic mass
ex: Na molar mass = 23g
H2 molar mass = 2g
Molar mass of compounds
Add all element’s molar mass together
ex: C2H6O
C = 2 x 12.01
H = 6 x 1.01
O = 1 x 16.0
= 46.0 g/mol
Calculations using molar mass
A box of salt contains 737g of NaCl, how many moles of NaCl are in the box?
12.6 moles of NaCl
Law of conservation of mass
Matter cannot be created or destroyed
No change in total mass occurs
Mass of products is equal to mass of reactants
Mole-Mole Factors, how many moles of each reactant and product are in 2Fe + 3S = Fe^2S^3
2 moles Fe, 3 moles S, 1 mole Fe2S3
What is the mole mole factor for 3H2 + N2 = 2NH3
1 mole N2 / 3 mole H2 or 3 mole H2 / 1 mole N2
Calculations using mole factors
4Fe + 3O2 = 2Fe2O3
How many moles of Fe2O3 can form from 6.0 moles of O3?
4.0 mole of Fe2O3
How many mole of Fe are needed for the reaction of 12.0 mole of O2?
4Fe + 3O2 = 2Fe2O3
16.0 mole of Fe
Converting mass calculations (g to mole to g)
Multiply the top first, then the divide by the bottom
4Na + O2 = 2Na2O
How many grams of Na2O are produced when 57.5 grams of Na react?
77.5 g NaO2
Limiting Reactant (LR)
Substance that is used up first
Limits the amount of product that can form
Reactant that runs out first
Excess reactant
What doesn’t run out/what is left over
Calculating LR
Ex: You have 3 mole of CO and 5 mole of H2, how many mole of CH4O?
CO + 2H2 = CH4O
H2 is the LR
What is the LR in this equation
N2 + 3H2 = 2NH3
You have 2.50 g N2 and 2.00 g H2, what is the LR of 2NH3?
N2 is the LR
Exothermic
Heat is released, negative #
ex: H2 + Cl2 = 2HCl + 185 kj
Heat = -185
Endothermic
Heat is absorbed, positive #
ex: N2 + O2 + 180 kj = 2NO
Heat = 180 kj
Heat of reaction is
-the amount absorbed or released during a reaction that takes place at constant pressure
-Change occurs when reactants interact, bonds break apart or when products form
Heat of reaction formula
Heat = H (products) - H (reactants)
How much heat in Kj is released when nitrogen and hydrogen react to form 50.0 g of ammonia?
Heat = -92.2 Kj
N2 + 3H2 = 2NH3
- 135 Kj
Exothermic
Combined Gas Law formula
P1V1/T1 = P2V2/T2
Boyle’s Law
T and n are constant
P1V1 = P2V2
Charle’s Law
P and n are constant
V1/T1 = V2/T2
Gay - Lussac’s Law
V and n are constant
P1/T1 = P2/T2
In combined gas laws, what is always constant?
n = moles
Temp always needs to be in Kelvin
Dalton’s law of partial pressure
P total = P1 + P2
Rearrange to solve for unknown
Ideal Gas Law
Pv = nRT
P = pressure
v = volume
n = moles
R = gas law constant
T = temp
How many L in 1 mole?
1 mole = 22.4 L
Standard pressure
1 atm
How many mmHg is in 1 atm
760 mmHg
How many kpa is in 1 atm
101.3 kpa
Absolute zero =
0 Kelvin or -273 C
