Grade 11 Review Unit Flashcards by Happy Gill

Nomenclature & Reactions

Ionic & Covalent Compounds

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Nomenclature & Reactions

Naming Binary Ionic Compounds

Binary Ionic Compounds:

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

Naming From Formula:

Write name of metal
Write name of non-metal
Change non-metal suffix to IDE

Formula From Name:

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

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Nomenclature & Reactions

Naming Binary Molecular Compounds

Binary Molecular Compounds:

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

Naming From Formula:

Name the first element
Name the second element
Add the suffix –ide to the second element

Formula From Name:

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

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Nomenclature & Reactions

Polyatomic Ions And Radicals

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

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Nomenclature & Reactions

Other Common Radicals

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

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Nomenclature & Reactions

Adding and Removing Oxygen - Polyatomic Ions

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Nomenclature & Reactions

Multivalent Cations Special Names

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Nomenclature & Reactions

Naming Binary Acids

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

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Nomenclature & Reactions

Synthesis Reactions

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)}

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Nomenclature & Reactions

Decomposition Reactions

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)+ CO₂ _(g)

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Nomenclature & Reactions

Single Displacement Reactions

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.

General Formula: A + BC -> AC + B

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Nomenclature & Reactions

Double Displacement Reactions

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

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)

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

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Nomenclature & Reactions

Combustion Reactions

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

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Stoichiometry

Scientific Notation

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

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Stoichiometry

Significant Digits

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

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Stoichiometry

The Mole Variables

Stoichiometry

The Mole Formula Questions

Stoichiometry

Mole 2 Step Process

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

Stoichiometry

Mole-Mole, Mass-Mass

Stoichiometry

Percent Yield

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.

Stoichiometry

Limiting & Excess Reagents

Solutions

Molar Concentration

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

Solutions

Preparing Solutions - From a Solid

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.

Solutions

Dilutions of a Stock Solution to Form a Weaker Solution

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
*

# Solutions Molar Concentration/Titration Calculations

* 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) *

# Solutions Calculating Ion Concentrations by Mixing 2 Solutions

* 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

# Solutions Ideal Gas Law

* 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