Grade 11 Review Unit

Question 1

Nomenclature & Reactions

Ionic & Covalent Compounds

Answer 1

Question 2

Nomenclature & Reactions

Naming Binary Ionic Compounds

Answer 2

Binary Ionic Compounds:

• Binary -> 2 elements
• Ionic -> metal ion + non-metal ion

Naming From Formula:

1. Write name of metal
2. Write name of non-metal
3. Change non-metal suffix to IDE

Formula From Name:

1. Write out name with space.
2. Write symbols & charge of elements.
3. Criss-cross charges as subscripts.
4. Combine as formula unit.

Question 3

Nomenclature & Reactions

Naming Binary Molecular Compounds

Answer 3

Binary Molecular Compounds:

• Compounds consisting of molecules that are made from two elements are called binary molecular compounds.

Naming From Formula:

1. Name the first element
2. Name the second element
3. Add the suffix –ide to the second element

Formula From Name:

1. Add the correct prefix in front of each element but if mono- is the prefix we do not add it to the name

Question 4

Nomenclature & Reactions

Polyatomic Ions And Radicals

Answer 4

• Polyatomic Ions -> a group of covalently bonded atoms that has a net charge (positive or negative)
  
  • NO₃^1- -> Nitrate
  • ClO₃^1- ->Chlorate
  • CO₃^2- ->Carbonate
  • SO₄^2- -> Sulfate
  • PO₄^3- -> Phosphate
• Radicals -> group of atoms that tend to stay together during most chemical reactions
• Compounds containing radicals must be electrically neutral. A metal combined with a radical forms a metal salt.
  
  • Sodium sulfate = Na¹⁺ SO₄^{2- ->} Na₂SO₄
  • Calcium phosphate = Ca²⁺ PO₄^2- -> CaPO₄

Question 5

Nomenclature & Reactions

Other Common Radicals

Answer 5

• MnO₄1 -> Permanganate
• C₂H₃O₂1- -> Acetate
• OH1- -> Hydroxide
• NH₄1+ -> Ammonium

Question 6

Nomenclature & Reactions

Adding and Removing Oxygen - Polyatomic Ions

Answer 6

Question 7

Nomenclature & Reactions

Multivalent Cations Special Names

Answer 7

Question 8

Nomenclature & Reactions

Naming Binary Acids

Answer 8

• Only H and another element are present
• The anion does not contain oxygen
• Prefix: hydro- suffix: -ic
  
  • H₁NO₃ ->Nitric acid
  • H₁ClO₃ -> Chloric acid
  • H₂CO₃ -> Carbonic acid
  • H₂SO₄ ->Sulphuric acid
  • H₃PO₄ -> Phosphoric acid
• The anion contains oxygen
• If anion ends in ate – suffix is ic
• If the anion ends in ITE – the suffix is ous
• If the anion is a per…ate – the acid is a per…ic acid
• If the anion is a hypo…ite – the acid is a hypo…ous acid

Question 9

Nomenclature & Reactions

Synthesis Reactions

Answer 9

• Direct combination reaction
• Simple substances combine to form more complex compounds
• Product can either be ionic or molecular
• Element + element à compound
• Compound + element à complex compound
• Compound + compound à very complex compound
• General Formula: A + B à AB
• E.g. 2Na + Cl₂ -> 2NaCl
• E.g. solid sodium nitride by burning sodium metal in nitrogen gas: 6Na_(s) + N_2(g) -> 2Na₃N_(s)

Specific Examples of Synthesis Reactions:

• Combustion – means burning with oxygen
  
  • Incomplete combustion
• Metal + oxygen -> metal oxide
  
  • 2Mg + O_{2 (g) ->} 2MgO
• Note: metal oxide in water à base
  
  • MgO + H₂O -> Mg (OH)₂
• Non-metal + oxygen à non- metal oxide
  
  • 2S + O₂ -> 2SO₃
• Note: non-metal oxide + water à acid
  
  • SO₃ + 2H₂O -> 2H₂SO_{4 (aq)}

Question 10

Nomenclature & Reactions

Decomposition Reactions

Answer 10

• This type of reaction often requires energy (ex. heat)
• A and B can either be elements of compounds
• Occurs a compound is broken down into simpler substances
  
  • There can be 2 or more simpler substances (elements or compounds)
• Special Case -> metal carbonates decompose to produce the metal oxide and carbon dioxide gas
• General Formula: AB à A + B
• E.g. 2HCl -> H_2(g) + Cl_{2 (g)}
• E.g. Metal CARBONATES decompose to produce the metal oxide and carbon dioxide gas: MgCO_{3 (aq) ->} MgO_(s) + CO_{2 (g)}

Question 11

Nomenclature & Reactions

Single Displacement Reactions

Answer 11

• Also known as: oxidation-reduction reactions
• An element reacts with a compound and displaces a second element from the compound to form an element and a new compound
• Always follow the activity series (will always be given) because some reactions will not occur.

1. General Formula: A + BC -> AC + B

Question 12

Nomenclature & Reactions

Double Displacement Reactions

Answer 12

• General Formula: AB + CD à AD + CB
• E.g. 2K(S) + 2H2O(l) à 2KOH(aq) + H2(g)

1. Precipitation Reactions

• Two compounds which are water soluble react to form two new compounds, one of which is a precipitate (ie. insoluble in water)
• In order to determine which one of the products will be the precipitate requires knowledge of the solubilities of salts in water (see solubility table)
  
  • AgNO3(aq) + NaCl(aq) -> NaNO3(aq) + AgCl(s)
  • BaCl2(aq) + K2SO4(aq) -> BaSO4(s) + 2KCl(aq)

2. Neutralization Reactions

• Sometimes called acid-base reactions
• Occurs between an acidic compound and a basic compound to form a chemical salt and water
• The pH is equal to 7 if all acid/base completely reacts
• Use bromothymol blue as indicator – turns green
  
  • HCl(aq) + NaOH(aq) -> H2O + NaCl(aq)
  • H2SO4(aq) + KOH(aq) -> 2H2O + K2SO4(aq)
• Note: The valence of multivalent elements stays the same on both sides of the equation

Question 13

Nomenclature & Reactions

Combustion Reactions

Answer 13

• Burning a hydrocarbon
• A hydrocarbon is a molecular compound containing only carbon, hydrogen and sometimes oxygen
• Complete combustion of a hydrocarbon always produces only 2 products: CO₂ and H₂O
• E.g. combustion of propane (C₃H₈): C₃H₈ + 5O₂ -> 3CO₂ + 4H₂O

Question 14

Stoichiometry

Scientific Notation

Answer 14

• Multiples of 10 are used to write very large or small numbers in scientific notation
  
  • 100 = 1 x 10 x 10 = 10²
  • 20 000 = 2 x 10 x 10 x 10 x 10 = 2 x10⁴
• The exponent on the 10 tells you how many places to move the decimal to obtain the ORIGINAL number
  
  • 4.5 x 10⁵ à move decimal 5 places to the right à 450 000
• When the number is less than 1, (e.g. 0.0025), the exponent on the 10 is negative
• Note: the base number is always written as a single digit followed by decimals if necessary
  
  • 65 000 = 6.5 x 10⁴ -> NOT: 65 x 10³

Question 15

Stoichiometry

Significant Digits

Answer 15

• Any digit of a number that is known with certainty
• Any digit that is obtained by actual measurement & thus is not simply a placeholder used to position the decimal point

General Rules:

• All of the digits from 1 to 9 are significant
  
  • 1.234 -> 4 sig figs, 1.2 -> 2 sig figs
• Zeros in between non-zero digits are significant
  
  • 1002 -> 4 sig figs, 3.07 -> 3 sig figs
• Any zero to the right of a nonzero digit (trailing zeroes) is significant
  
  • 0.230 -> 3 sig figs, 0.40 -> 2 sig figs
• All zeros to the left of the first non-zero digit (leading zeros) are not significant
  
  • 0.001 -> 1 sig fig
• Numbers obtained by counting are not considered when determining significant digits
  
  • 1341 students -> N/A
• When whole numbers end in 0, they are not necessarily significant
  
  • 50 600 -> can be 3, 4 or 5 sig figs
• The ambiguity in the last rule can be avoided by the use of standard or scientific notation
  
  • 1.90 x 10^{2 ->} 2 sig figs

Rules For Rounding off Numbers:

• If the digit to be dropped is greater than 5, the last retained digit is increased by one
  
  • 12.6 -> rounded up -> 13
• If the digit to be dropped is less than 5, the last remaining digit is left as it is
  
  • 12.4 -> rounded down -> 12
• If the digit to be dropped is 5 and if any digit following it is not zero, the last remaining digit is increased by one
  
  • 12.51 -> rounded up -> 13
• If the digit to be dropped is exactly 5 and is followed only by zeroes, round even
  
  • 11.5 -> round even -> 12 … 12.5 -> round even -> 12

​Significant Digits – Multiplying & Dividing:

• The number of sig figs in your final answer should be equal to the LOWEST number of significant digits in any of the given quantities
  
  • 0.5 x 3.08 = 1.52 -> 2 -> 1 sig fig
  • 2.00 x 3.5 = 7.0 -> 7.0 -> 2 sig figs

Significant Digits – Adding & Subtracting:

• The number of digits after the decimal in your final answer should be equal to the LOWEST number of digits after the decimal in any of the given quantities
  
  • 1.26 + 2.3 = 3.56 -> 3.6 -> 1 decimal place
  • 1.26 + 103.3 = 103.56 -> 103.6 -> 1 decimal place

Question 16

Stoichiometry

The Mole Variables

Answer 16

Question 17

Stoichiometry

The Mole Formula Questions

Answer 17

Question 18

Stoichiometry

Mole 2 Step Process

Answer 18

• Use 1 of the 2 formulas to find the value of a given formula. Then sub in that value into the second formula.
• 1. Calculate the number of Potassium atoms in 2.5 moles of K₃PO₄
     * 1) N= n(N_A)

N= 2.5mol (6,02 x10²³molecules/mol)

N= 1.505 x10²³molecules

• 2) N_K= 3 x N_K3PO4

N_K= 3 atoms/mol (1.505 x10²³molecules)

N_K= 4.151 x10²⁵

Question 19

Stoichiometry

Mole-Mole, Mass-Mass

Answer 19

Question 20

Stoichiometry

Percent Yield

Answer 20

• Percent yield -> tells us how “efficient” a reaction is. It cannot be bigger than 100%.
• Theoretical yield -> The expected or calculated amount of product formed is called the
• Actual yield ->The amount of product actually formed in the experiment is called the
  
  • P.C = (actual yield/theoretical yield) x100
• Step 1 -> Find the actual yield of product (usually given in question, or can be calculated).
• Step 2 -> Find the theoretical yield of product.
• Step 3 -> Use the formula to calculate percent yield.

Question 21

Stoichiometry

Limiting & Excess Reagents

Answer 21

Question 22

Solutions

Molar Concentration

Answer 22

• c = concentration (mol/L or M)
• n = # of moles (mol)
• V = volume (L)
• [] square bracket also a symbol of concentration

Question 23

Solutions

Preparing Solutions - From a Solid

Answer 23

• How would you create 500 mL of a 2.5 mol/L solution of sodium hydroxide?
• Calculate the number of moles in the solution

n = C x V

= 2.5 mol/L x 0.500 L

= 1.25 mol

• The number of moles cannot be measured?
  
  1. What can we measure? -> mass
  2. Find the mass of solid required

m = n x M

= 1.25 mol x 40.0 g/mol

= 50 g of NaOH (2 sig digits)

• Use a balance to mass out this amount of solid NaOH, add it to 500 mL of water (distilled).
• Obtain a 500 mL volumetric flask. Add the measured amount of solid to the flask. (50g)
• Add approx 250 mL of distilled water to the flask. Cap and invert to mix (several times)
• Add remaining distilled water to the fill line. Cap and invert again.

Question 24

Solutions

Dilutions of a Stock Solution to Form a Weaker Solution

Answer 24

• n₁ = n₂
• c₁v₁ = c₂v₂

Method:

• Obtain a 1000 mL volumetric flask. Add about 500 mL of distilled water.
• Measure 250 mL of concentrated HCl(aq) with a graduated solution. Add carefully to the flask.
• Stopper, swirl and invert to mix solute and solvent
• Add more distilled water to fill line. Stopper, swirl and mix.

Concentration

• Concentration -> the quantity of a given solute in a solution
• Dilute -> a relatively small amount of solute per unit volume of solution
• Concentrated -> a large amount of solute per unit volume
• Concentrations can be calculated and expressed in numerous ways
  *

Question 25

Solutions

Molar Concentration/Titration Calculations

Answer 25

• C = n/V, n = m/M, m = n(M)
• in an equation if the concentration (M) is given of the reactant and the volume of the product, find concentation of the other reactant by multiplying the concentration (M) of reactants given by the coeffient of the unknown reactant over the coefficent of the given reactant
• The concentration of the product is the sum of the 2 reactants concentration multiplied by the products coefficient over the reactants coefficient
  
  • applies only if the volume is of the product
• If volume and concentration of a reactant is given and only volume of another reactant, find moles then use other formulas to find concentration
• How to find moles:
  
  • moles of given/coefficient multiplied by, unknown moles/coefficient (cross multiply)
    *

Question 26

Solutions

Calculating Ion Concentrations by Mixing 2 Solutions

Answer 26

• volume & concentration of one substance is given and volume & concentration of another substance is given
• write chemical equation of both substances and state givens underneath (product)
• multiply concentation of first solution by the volume of the product over the total volume (a+b)
  
  • gives concentration of 1 molecule (multiple by coefficent)
• repeat for second substance

Question 27

Solutions

Ideal Gas Law

Answer 27

• An ideal has (perfect gas) is one which obeys all gas laws exactly
• There are no attractive forces between molecules and condense to liquid when cooled
• They can be compressed to a volume of zero
• Constants
  
  • R = 8.314 kPa L/ K mol
  • R = 0.0821 l atm/ K mol
• PV = nRT
  
  • P -> kPa or atm
  • V -> L
  • n -> mol
  • R -> constant
  • T -> K
• PM = DRT