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Review of Chemistry 11

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Prior Knowledge Needed for Chemistry 12

- Naming molecules and writing formulas
- Balancing Chemical Equations
- Counting Significant figures
- Calculating with significant figures
- Mole calculations
- Basic Stoichiometry

What do you remember?

Categorize the following according to what elements are involved and the types of intramolecular bonds:

HCl, NaCl, CO2, Cu(ClO) 2, CaO, HF, Ca(OH) 2, HIO3, Sr(NO3) 2, Pb(NO3) 2, AgNO3, KBr, NaOH, N2O3, FeO, CH3COOH,

Analyze each pair (name = formula) and come up with some rules…

Hyroiodic Acid = HI

Nitrous Acid = HNO2

Nitric Acid = HNO3

Sulphur rules…diflouride = SF2

Dinitrogen trioxide = N2O3

Why? rules…

General Guidelines rules…

- Balance any polyatomic ions first
- Balance metals
- Balance nonmetals
- Balance “O” and “H” last
- Recount both sides to make sure!!

Al + O rules…2 Al2O3

C rules…2H6 + O2 CO2 + H2O

Answers… rules…

100 = 1

0.001 = 1

0.010 = 2

10.04 = 4

1.0 x 102 = 2

197 = 3

0.0000200910 = 6

100.20 = 5

10.0 = 3

1090 = 3

What are the general rules for what zero’s are significant?

Calculating with Significant Figures significant?

Answers… significant?

Adding and Subtracting…

10.00 + 2.0 = 12.0

36.98 – 0.0013= 36.98

General Rule?

Answers… significant?

Multiplying and Dividing…

190 / 3 = 60

1403 / 7.0 = 2.0 x 102

0.09700 x 1.2 = 0.12

General Rule?

Mole significant?

Calculations

Answer… significant?

How many grams of NaCl are in 4.50 moles of NaCl?

# g NaCl = 4.50 mol x 58.5 g = 263 g NaCl

1 mol

What steps were used to answer this question?

Answer… significant?

How many moles of NH4Br are in 15.87 g of NH4Br?

# mol NH4Br= 15.87g x 1 mol =0.162 mol NH4Br

97.9 g

What steps were used to answer this question?

General Rules… significant?

Basic significant?

Stoichiometry

Answers… significant?

Given the following balanced equation, answer the questions following it:

9Na + 4ZnI2 8NaI + NaZn4

a.) If 2.50 moles of Na are reacted, how many moles of ZnI2 will be consumed?

# mol ZnI2 = 2.50 mol Na x 4 mol ZnI2 = 1.11 mol ZnI2

9 mol Na

b.) If you completely react 526.68 g of ZnI2, what mass of NaZn4 will be produced?

# g NaZn4 = 526.68g of ZnI2x 1 mol ZnI2 x 1 mol NaZn4 x 284.6g NaZn4 = 117.4g NaZn4

319.2 g ZnI2 4 mol ZnI2 1 mol NaZn4

What steps were taken to answer these questions? significant?