UNIT 3 STUDY NOTES CHEMISTRY - A1 Tuition - Unit 3 (Sample).pdf · GRAVIMETRIC ANALYSIS...

CH

EM

ISTR

Y

UN

IT 3 ST

UD

Y N

OT

ES

“C

HE

MIC

AL

PAT

HW

AY

S”

©2007 - 2008 A1 TUITION All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise without written permission from the publisher with the exception of fair dealing for the purpose of research or study, criticism or review as permitted under the Copyright Act 1968.

PRINTED IN FULL COLOUR. $29.95

2nd edition

FORMULAS & EQUATIONS. DIAGRAMS & TABLES. GUIDELINES FOR SCIENTIFIC WRITING. UPDATED FOR 2008 SYLLABUS CHANGES.

100% AUSTRALIAN PRODUCT.

WARNING: Keep away from water, moisture and dampness, otherwise ink may run or smudge.

I

1 2 5 5

14 17 19 34 35 36 42 43 49 54 58 60 62

63

65 69 70 71 72 76

77

78

80

82

87

www.a1tuition.com.au I

Hi and welcome to Year 12 chemistry. I hope that you found last years work helpful and it is still somewhere in your minds cause guess what its back. Over the next year we will be exploring concepts that you learned in Units 1 & 2 and continue to build upon them. Now I know there are a few of you thinking ‘help! I did not do Units 1 & 2’ or ‘help! I am still having trouble with the old concepts’. Do not worry, if you have not yet check them out there are study guides available fro Units 1 and 2 in this series. I suggest you have a look at them. If you still are having trouble I suggest you enlist the help of a tutor. This agency have really good highly qualified tutors that are more than happy to help you throughout the year so not hesitate to ask for help.

Towards the end of this guide is a set of 8 appendices which are designed for a reference point

for different topics. I have placed in these appendices a lot of tables, formula you have already learnt and conversion factors. This should help you throughout the year. If you have previous guides for chemistry from us you will notice these appendices are standard every guide and increase each unit.

The first area of study is all about analysis. Here we will be revising all the calculations we

explored with the gas laws and stoichiometry, volumetric analysis and a few new techniques. All these techniques assume you have a good knowledge of different types of reactions and a good firm grasps on calculations so pay attention to the hints in this guide.

The second area study is devoted to organic chemistry. In unit 1 we learnt a lot about the

different compounds and how they react together with other molecules, this year we will explore more reactions and also how organic chemistry works in medicine.

I hope you enjoy this guide as much as I enjoyed writing it for you. I wish you the best of luck

for your VCE year and hope you do well.

Amy Paul

www.a1tuition.com.au 1

CHEMICAL ANALYSISCHEMICAL ANALYSIS VOLUMETRIC ANALYSIS

GRAVIMETRIC ANALYSIS

STOICHIOMETRY AND THE GAS LAWS

REDOX EQUATIONS

CHROMATOGRAPHY

SPECTROSCOPIC TECHNIQUES

MATCHING ANALYTICAL TECHNIQUES TO A PARTICULAR TASK


VOLUMETRIC ANALYSIS

Involves a titration where a measured volume of solution of known concentration (standard solution) is reacted with a measured volume of solution whose concentration is to be determined;

Indicators are used to signal the point at which the reaction is complete according to the equation (the equivalence point);

Acid-base indicators are weak acids whose conjugate base is a different colour from the acid form. A standard solution:

Has a known concentration Can be prepared by:

Reacting it with a solution whose concentration is known or Preparing directly from primary standard.

A primary standard:

1. is available in pure form 2. has a known formula 3. is stable enough not to react with the atmosphere 4. has a reasonably high molar mass to minimise weighing errors. 5. is preferably inexpensive.

To prepare a standard solution from a primary standard:

1. weigh the primary standard accurately 2. transfer to a volumetric flask 3. half fill the flask with water 4. stopper and shake the flask to dissolve the solid. 5. fill the flask so that the bottom meniscus is on the calibration mark. 6. stopper and shake several times.

Direct and Back Titrations

A direct titration is when the test solution is titrated directly with the standard solution. This is commonly used for strong acid/strong base reactions.

In back titrations, excess standard solution is added to the test solution. The amount of left-over standard is then determined by titration.

Back titrations are used for weak acid or base titrations when there is no sharp colour change at the endpoint. Equipment used in titration is:

Volumetric Flask: is the flask we mix the standard solution in. Pipette: is used to measure a specific volume of our unknown solution. This measured volume is known as an aliquot and is poured into a conical flask. Burette: is a measuring device that allows us to control accurately the amount of standard solution we place in our reaction. The volume measured in the reaction is known as a titre.

Solution rinsed with: Glassware Used for Test solution Standard solution Water

Burette Accurately transferring and measuring the standard solution to the test solution. Low Correct High

Pipette Used for transferring diluted test solution into the conical flask. Correct Low High

Conical Flask Contains our diluted test solution. High Low Correct Volumetric Flask Used to prepare our primary standard solution. Low High Correct


Calculations involved in titrations: Calculations involved in titrations: Calculate number moles of primary standard used. Calculate number moles of primary standard used. Calculate number of moles in 1 aliquot of test solution. Calculate number of moles in 1 aliquot of test solution. Calculate total number of moles in undiluted solutions. (Times by the ratios in volumes of the aliquot and diluted

solution.) Calculate total number of moles in undiluted solutions. (Times by the ratios in volumes of the aliquot and diluted

solution.) Find the concentration of undiluted test solution. Find the concentration of undiluted test solution.

Indicators are plant extracts that change colour when they come into contact with an acid or a base. They are extremely useful because they can tell us information during an experiment. Indicators are usually acids or bases themselves which is why they work so well. To the left is a chart of common indicators and their colourings. You will be given a similar table in your exams; though not so colourful.

Indicators are plant extracts that change colour when they come into contact with an acid or a base. They are extremely useful because they can tell us information during an experiment. Indicators are usually acids or bases themselves which is why they work so well. To the left is a chart of common indicators and their colourings. You will be given a similar table in your exams; though not so colourful. You know an equations represents an acid/base reaction is: You know an equations represents an acid/base reaction is:

An H+ ion is transferred from one species to another during the reaction. An H The reaction is one of the following types: The reaction is one of the following types:

+ ion is transferred from one species to another during the reaction.

Metal + acid → a salt + hydrogen gas; Metal + acid → a salt + hydrogen gas; Eg: Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) Eg: Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

Metal hydroxide + acid → a salt + water; Metal hydroxide + acid → a salt + water; Eg: 2KOH(aq) + H2CO3(aq) → K2CO3(aq) + 2H2O(l) Eg: 2KOH(aq) + H2CO3(aq) → K2CO3(aq) + 2H2O(l)

Metal oxide + acid → a salt + water; Metal oxide + acid → a salt + water; Eg: Al2O3 + 6HNO3(aq) → 2Al(NO3)3(aq) + 3H2O(l) Eg: Al2O3 + 6HNO3(aq) → 2Al(NO3)3(aq) + 3H2O(l)

Metal carbonate + acid → a salt + water + carbon dioxide; Metal carbonate + acid → a salt + water + carbon dioxide; Eg: 3MgCO3(aq) + 2H3PO4(aq) → Mg3(PO4)2(s) + 3H2O(l) + 3CO2(g) Eg: 3MgCO3(aq) + 2H3PO4(aq) → Mg3(PO4)2(s) + 3H2O(l) + 3CO2(g)

Metal hydrogen carbonate + acid → a salt + water + carbon dioxide; Metal hydrogen carbonate + acid → a salt + water + carbon dioxide; Eg: NaHCO3(aq) + CH3COOH(aq) → NaCH3COO(aq) + H2O(1) + CO2(g) Eg: NaHCO3(aq) + CH3COOH(aq) → NaCH3COO(aq) + H2O(1) + CO2(g)

Metal sulphide + acid → a salt + hydrogen sulphide; Metal sulphide + acid → a salt + hydrogen sulphide; Eg: CaS(aq) + HCl(aq) → CaCl2(aq) + H2S(g) Eg: CaS(aq) + HCl(aq) → CaCl2(aq) + H2S(g)

Acidic oxide (non metal oxide) + base → a salt + water. Acidic oxide (non metal oxide) + base → a salt + water. Eg: SO3(aq) + 2NaOH(aq) → Na2SO4(aq) + H2O(l) Eg: SO3(aq) + 2NaOH(aq) → Na2SO4(aq) + H2O(l)

At 25oC, [H3O+] × [OH-] = 10-14 M2 At 25

oC, [H3O+] × [OH-] = 10-14 M2

pH = -log10[H3O+] pH = -log10[H3O+]

[H3O+] = 10-pH [H3O+] = 10-pH

(Heinemann, 2006)


Example: 60 mL of 0.10 M of HCl is diluted to 600 mL. Find concentrations of H3O+ and OH- ions in an acid, and hence, find pH of an acid.

C1 = 0.10M V1 = 60mL

V2 = 600mL C1V1 = C2V2

[HCl] = 010.0600

6010.0=

× M

HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq) 1 : 1 : 1 : 1

0.010M : 0.010M : 0.010M : 0.010M

[H3O+] = 0.010 M = 10-2 M Now [H3O+] × [OH-] = 10-14

10-2 × [OH-] = 10-14

[OH-] = 10-12 M = 1.0 × 10-12 M

pH = -log10[H3O+] = -log1010-2 = 2.0


GRAVIMETRIC ANALYSIS

Component to be analysed is a precipitate and its mass is determined. We can predict a precipitation reaction by consulting a solubility table. Ionic salts that are not

soluble will form a precipitate in solution. I have provided a solubility table for you in appendix E. Method:

1. Weigh sample accurately 2. Add water to dissolve the component under analysis. 3. Add excess of a suitable reactant to precipitate the component under analysis. 4. Filter precipitates and wash with a small amount of water. 5. Repeatedly dry and weigh precipitate until a constant mass is obtained.

Calculations involved in gravimetric problems From mass of precipitate containing X, calculate:

1. moles of precipitate 2. moles of X (from the formula of precipitate) 3. mass of X analysed 4. % substance X in sample.

Sources of error in gravimetric analysis:

ERROR EFFECTS ON % X X incompletely precipitated Lower than actual value.

Precipitate weighed wet Higher than actual value. Other ions in sample also precipitating Higher than actual value.

STOICHIOMETRY AND THE GAS LAWS

Stoichiometry is the term we use that is similar to accounting in book keeping. Though instead of working in money; we work with that little unit I introduced in Unit 1 called the mole. In units 1 & 2 I taught you all the formulas you need for stoichiometry. The quick ways of deriving them are outlined on the next page, if you are a little confused I suggest you have a flip through appendix H, I have a pretty comprehensive section there telling you how to derive the formulas. After the formulas I have given you a pretty comprehensive outline for what the variables stand for and also the constants you need for a few of the equations

Basically with the triangle method you just cover the variable you are looking for and it gives

you the exact variation of the formula we need. With the square, cross out the variables that are constant, then write the variables in the same order like a dilution.

Below are some rules for doing most stoichiometric problems. Follow these steps and you

cannot go wrong!

1. Always define you variables at the start and if not already given write a fully balanced equation. 2. Look at the information given, can you find a way to find the number of moles for one of the substances in

your equation. If you can do it. 3. Using the moles you have found in the last step, do a moles ratio, what we are looking for is the number of

moles of the substance we are looking for. 4. Now we have the number of moles of the substance we are looking at, find the quantity we need being it

mass, volume, concentration, density, number of particles, etc… 5. Reread the question and make sure you have answered it!


Sometimes when doing calculations the chemical reaction does not go to completion if the exact theoretical values are placed in the reacting vessel. So to overcome this problem, scientists will often place more of the lower costing reactant in the reaction to make sure the reaction completely reacts. The reactant that we use more of is known as the excess reactant. The other reactant in the equation that we use the exact amount and therefore the lesser number of moles in ratio to the excess is known as the limiting reactant.

Sometimes when doing calculations the chemical reaction does not go to completion if the exact theoretical values are placed in the reacting vessel. So to overcome this problem, scientists will often place more of the lower costing reactant in the reaction to make sure the reaction completely reacts. The reactant that we use more of is known as the excess reactant. The other reactant in the equation that we use the exact amount and therefore the lesser number of moles in ratio to the excess is known as the limiting reactant.

When doing stoichiometric questions we need to be wary of limiting reactants and excesses. To

know when to look for limiting reactants and excesses and how to find them, follow these steps: When doing stoichiometric questions we need to be wary of limiting reactants and excesses. To

know when to look for limiting reactants and excesses and how to find them, follow these steps: You will know that there is possibly an excess and a limiting reactants when it is possible in the question to

calculate the number of moles for both; You will know that there is possibly an excess and a limiting reactants when it is possible in the question to

calculate the number of moles for both; Find the number of moles in both samples; Find the number of moles in both samples; Compare these samples to the equation for the reaction. Using the moles ratios we can easily figure out which is in

excess and which is the limiting; Compare these samples to the equation for the reaction. Using the moles ratios we can easily figure out which is in

excess and which is the limiting; Using the reactant you have determined is the limiting reactant and carry out the calculations the answer the

question; Using the reactant you have determined is the limiting reactant and carry out the calculations the answer the

question; You can find out the remainder of the excess by finding the theoretical number of moles needed to fully carry out

the reaction and subtracting this value from the total number of moles from the excess. You can find out the remainder of the excess by finding the theoretical number of moles needed to fully carry out

the reaction and subtracting this value from the total number of moles from the excess.

This is about as much advice on this topic I can give you, the best thing now is to familiarise you with plenty of examples. I have given you quite a few examples, study each of them and have a go at them yourself. You will notice in these examples that I have set out very formally, try to set out similarly to the way I have done it.

This is about as much advice on this topic I can give you, the best thing now is to familiarise you with plenty of examples. I have given you quite a few examples, study each of them and have a go at them yourself. You will notice in these examples that I have set out very formally, try to set out similarly to the way I have done it.

PV = nRT PV = nRT

EXPLANATION OF TERMS EXPLANATION OF TERMS SI UNITS FOR THESE TERMS SI UNITS FOR THESE TERMS


n is the number of moles mol N is the number of particles This depends on what you are looking for.

M or Mw is relative molecular weight g⋅mol-1

m is the mass of a substance g C is the molar concentration of a substance mol⋅L-1

D is the density of a compound g⋅L-1

V is the volume of a substance L P is the pressure of a gas. atm or kpa (depending on gas constants used)

T is the temperature K

CONSTANTS DEFINITIONS NA is Avogadro’s Number NA = 6.022 × 1023 mol-1

R = 8.314 J⋅K-1⋅mol-1R is the general gas constant R = 0.0820578 L⋅atm⋅K-1⋅mol-1

22.4 L⋅mol-1 @ STP Vm = molar volume

24.5 L⋅mol-1 @ SLC Example 1: What mass of gas would be released if you mixed 10.0g of baking soda in vinegar until it all reacted?

If you are a little confused baking soda is actually sodium hydrogen carbon and vinegar is another name for ethanoic acid or acidic acid.

NaHCO3(s) + CH3COOH(aq) → NaCH3COO(aq) + H2O(l) + CO2(g) m(NaHCO3) = 10.0g

M(NaHCO3) = 22.99 + 1.008 + 12.01 + 16 × 3 = 84.008 g⋅mol-1

n(NaHCO3) = mol119.0008.84

10)NaHCO()NaHCO(

3

3 ==Mm

n(NaHCO3) : n(CO2) = 1:1 n(NaHCO3) = n(CO2) = 0.119 mol

M(CO2) = 12.01 + 16 × 2 = 44.01 g⋅mol-1

m(CO2) = n(CO2) × M(CO2) = 0.119 × 44.01 = 5.24g Example 2: The solid fuel in the booster stage of a space shuttle is a mixture of ammonium perchlorate and aluminum powder. Upon ignition, one reaction that occurs is:

6NH4ClO4(s) + 10Al(s) → 5 Al2O2(s) + 3N2(g) + 6HCl(g) + 9H2O(g)

a) What mass in kilograms, of aluminum should be mixed with 2.0 × 104 kg of NH4ClO4 for this reaction? b) What mass in kilograms, of alumina (aluminum oxide) is released in this reaction?

6NH4ClO4(s) + 10Al(s) → 5 Al2O2(s) + 3N2(g) + 6HCl(g) + 9H2O(g)

m(NH4ClO4) = 2.0 × 104 kg = 2.0 × 107 g M(NH4ClO4) = 14.01 + 1.008 × 4 + 35.45 + 16 × 4 = 117.492 g⋅mol-1

mol101.7492.117100.2

)ClO(NH)ClO(NH)ClO(NH 5

7

44

4444 ×=

×==

Mmn

a) n(Al) : n(NH4ClO4) = 10 : 6 = 5 : 3

n(Al) = 35

n(NH4ClO4) = 35

× 1.7 × 105 = 2.8 × 105 mol

M(Al) = 26.98 g⋅mol-1

M(Al) = n(Al) × M(Al) = 2.8 × 105 × 26.98 = 7.6 × 106 g = 7600 kg b) n(Al2O3) : n(NH4ClO4) = 5 : 6


n(Al2O3) = 65

n(NH4ClO4) = 65

× 1.7 × 105 = 1.4 × 105 mol

M(Al2O3) = 26.98 × 2 + 16 × 3 = 101.96 g⋅mol-1

M(Al2O3) = n(Al2O3) × M(Al2O3) = 1.4 × 105 × 101.96 = 1.4 × 107 g = 1.4 × 104 kg Example 3: Slaked lime, Ca(OH)2, is formed from quick lime, CaO, by the addition of water:

CaO(s) + H2O(l) → Ca(OH)2(s) What mass of slaked lime can be produced from a mixture of 15.0g of CaO and 5.0g of water?

You should notice that this question gives you possibilities on finding the number of moles in both samples. It does not mention which is in excess or is the limiting reactant, so we need to calculate this value first.

CaO H2O m(CaO) = 15.0 g

M(CaO) = 40.08 + 16.00 = 56.08 g⋅mol-1

n(CaO) = 267.008.560.15

CaO)(CaO)(

==Mm

mol

m(H2O) = 5.0 g M(H2O) = 2.016 + 16.00 = 18.016 g⋅mol-1

n(H2O) = 278.0016.180.5

O)H(O)H(

2

2 ==Mm

mol

From the equation the moles ration of CaO : H2O = 1 : 1, so therefore, the substance with the

highest number of moles is the excess an the lowest is the limiting reactant.

Excess reactant: H2O Limiting reactant: CaO

n(Ca(OH)2) : n(CaO) = 1 : 1 n(Ca(OH)2) = n(CaO) = 0.267 mol

M(Ca(OH)2) = 40.08 + 32 + 2.016 = 74.096 g⋅mol-1

m(Ca(OH)2) = n(Ca(OH)2) × M(Ca(OH)2) = 0.267 × 74.096 = 19.8 g

Example 4: A reaction vessel contains 10.0 g of white phosphorus and 10.0g of oxygen. The first reaction to occur is the formation of phosphorus (III) oxide, P4O6:

P4(s) + 3O2(g) → P4O6(s) If enough oxygen is present, it can further react with the oxide to produce phosphorus (V) oxide P4O10:

P4O6(s) + 2O2(g) → P4O10(s) a) What is the limiting reagent for the formation of P4O10? b) How much in grams of P4O10 is produced? c) How many grams of excess reactant (use the second equation) remain in the vessel? a) Okay I can probably guess that a lot of you are already confused by the question. It is very sneaky, as we need to compare two sets of limiting reagents. The question focuses on the second equation; however we need to figure out what we have left from the first before we can even attempt the second equation. So with the first equation we have:

P4 O2m(P4) = 10.0 g

M(P4) = 30.97 × 4 = 123.88 g⋅mol-1

n(P4) = 0807.088.1230.10

)P()P(

4

4 ==Mm

mol

m(O2) = 10.0 g M(O2) = 16.00 × 2= 32.00 g⋅mol-1

ne(O2) = 313.000.320.10

)O()O(

2

2 ==Mm

mol

n(P4) : n(O2) : n(P4O6) = 1 : 3 : 1 nu(O2) = 3n(P4) = 3 × 0.0807 = 0.242 mol


In this reaction O2 is clearly in excess. Now what we need to is figure out is the remainder. So

subtracting the excess from what was used, I get:

nr(O2) = ne(O2) - nu(O2) = 0.313 – 0.242 = 0.0710 mol ne(P4O6) = n(P4) = 0.0807 mol

n(P4O6) : n(O2) : n(P4O10) = 1 : 2 : 1

n(O2) = 2ne(P4O6) = 2 × 0.807 = 1.614 mol

Clearly from this calculation O2 is now the limiting reactant. b)

nu(P4O6) = n(P4O10) = 21

n(O2) = 21

× 0.0710 = 0.0355 mol

M(P4O10) = 123.88 + 160 = 283.88 g⋅mol-1

m(P4O10) = n(P4O10) × M(P4O10) = 0.0355 × 283.88 = 10.1g c)

nr(P4O6) = ne(P4O6) - nu(P4O6) = 0.807 – 0.0355 = 0.772 mol M(P4O6) = 123.88 + 96 = 219.88 g⋅mol-1

m(P4O6) = n(P4O6) × M(P4O6) = 0.772 × 219.88 = 170 g Example 5: A helium balloon has a volume of 10L at 1.0 atm. As the balloon rises the pressure decreases until the balloon bursts at 0.9 atm. What is the final volume of the balloon just before it bursts?

V1 = 10 L P1 = 1.0 atm V2 = ? P2 = 0.9 atm

P1V1 = P2V2

V2 = L119.0100.1

2

11 =×

=PVP

Example 6: A balloon is measured to have 10 L of gas at 10oC on a particular morning. If the temperature by midday has risen to 18oC and the pressure remains the same what is the new volume of the balloon?

V1 = 10 L T1 = 10 oC = 283 K V2 = ? T2 = 18oC = 291 K

2

2

1

1

TV

TV

=

L10283

29110

1

212 =

×==

TTVV

Example 7: A chemist prepares 0.345 mol of krypton gas at a certain pressure and temperature in an expandable container. The chemist then adds another 0.100 mol of krypton gas into the container. What is the change in volume of the container to keep constant pressure and temperature?

I have done this example for a reason. This is a difficult question but if you think about it logically it should be okay. Sometimes looking at questions that lead to straight theory can seem daunting and difficult to answer. Using equations can help; just leave your answer in variables.

Let the initial volume of container be V V1 = V L n1 = 0.345 mol

V2 = ? n2 = 0.345 mol + 0.100 mol = 0.445 mol


2

2

1

1

nV

nV

=

VVVnnVV 30.1

6989

345.0445.0

1

212 ====

Now if you think of 1.30 it could represent 130%, meaning the volume rises by 30% its original volume.

Example 8: A relief valve on an industrial storage tank operates if the pressure of the carbon dioxide gas exceeds 200 atm. On a hot day in December the tanks pressure is 190 atm at a temperature of 20oC at 9:00am in the morning. By 3:00pm the temperature of the tank has risen to 50oC. Should the relief valve have operated?

P1 = 190 atm T1 = 20oC = 293 K P2 = ? T2 = 50oC = 323 K

2

2

1

1

TP

TP

=

210293

323190

2

212 =

×==

PTPP atm

The pressure would have been high and the valve relief should have operated. Example 9: Simon’s lungs have a volume of 360cm3 at a pressure of 1.08 atm at a temperature of 37oC. In a clinic Simon blew into a lung capacity machine, however the machine measures his lung capacity at standard lung conditions putting the measurement out. Find the value that the machine measures.

V1 = 360 cm3 P1 = 1.08 atm T1 = 37oC = 310 K V2 = ? P2 = 1.00 atm T2 = 25oC = 298 K

2

22

1

11

TVP

TVP

=

37431000.1

29836008.1

21

2112 =

×××

==PTTVPV cm3

Example 10: A dentist has a certain sample of laughing gas in his dental clinic. On a particular patient he uses half the sample. What should the dentist do to the temperature of the container to maintain the same volume and pressure of the laughing gas?

This is another example that you need to think clearly about. You will notice that there are no numerical values at all in this question; we need to assign pro-numerals, and think what is happening initially and in the end.

Let the initial number of moles be n and the initial temperature be T in Kelvin. n1 = n T1 = T

n2 = n21

T2 = ?

n1T1 = n2T2

Tn

nTnTnT 2

212

112 ===

Hence, the dentist needs to double the temperature to keep the pressure and volume constant.


Example 11: A bottle of oxygen gas is in a cylindrical container that measures 20 cm across and 1.00m in height. If the pressure is at 5 atm and at a temperature of 25oC, calculate the mass of oxygen in the tank.

r = 10 cm = 0.10 m h = 1.00 m V = πr2h = π × 0.102 × 1.00 = 0.01π m3 = 10π L

P = 5 atm = 506.5 kpa T = 25 oC = 298 K

R = 8.314 J K-1 mol-1

n = ? PV = nRT

mol16.0π105.506

298314.8=

××

==PVRTn

M = 16 × 2 = 32 g⋅mol m = nM = 0.16 × 32 = 5.0g

Example 12: Calculate the number of molecules and the mass of oxygen in a balloon, which is spherical in shape and has a diameter of 20 cm at: a) STP b) SLC

V = 4.2LmL102.4cm102.410π34π

34 33333 =×=×=×=r

M = 16 × 2 = 32 g⋅mol

a) Vm = 22.4 n = VVm = 4.2 × 22.4 = 94 mol

N = nNA = 94 × 6.02 ×1023 = 5.7 × 1025 molecules m = nM = 94 × 32 = 3000g = 3.0 kg

b) Vm = 25.4

n = VVm = 4.2 × 25.4 = 110 mol N = nNA = 110 × 6.02 ×1023 = 6.4 × 1025 molecules

m = nM = 110 × 32 = 3400g = 3.4 kg

Before I leave this section I thought it might be prudent to remind you how to work with percentage composition. We learnt about this calculation back in unit 1, and it was handy to work out chemical compositions of unknown compounds. While it is quite easy to do, a lot of students forget how to go about it so I have outlined this method again below. Molecular Formula: The exact ratio of elements of a compound. Eg C2H6 Empirical Formula: the molecular formula in it simplest form. Eg CH3 is the empirical formula for C2H6. Percentage Composition: the amount each element is in compound in a percentage.

% composition = 100compound of mass total

element of mass×

Example 13: Determine the percentage composition of borax, (Na2BB4O7).

Sodium 22.99 × 2 = 45.98Boron 10.81 × 4 = 43.24Oxygen 16.00 × 7 = 112.00

Total 201.22


% sodium = %85.2210022.20198.45

=×

% boron = %49.2110022.201

24.43=×

% oxygen = %66.5510022.20100.112

=×

Now we have explored finding the percentage composition, let’s look going in reverse. Taking

the percentage composition we can find the empirical formula. If we know the molar mass of the compound we can also find the molecular formula. How we accomplish this goal is by following these steps:

1. Assume that in 100g of sample you will have your total % in grams of the sample. 2. Divide each elemental sample by its relative atomic mass. 3. Place each element into a ratio; divide the ratio by the lowest number combination found. 4. You should now have whole numbers or almost whole numbers. Be careful with rounding, if it is pretty

much exact, you do not have to worry. However, sometimes you may need to times your ratio by a number to get every part an exact number. Common numbers for these are outlined below. a. If the decimal is halfway, that is a number such as 3.5, doubling the ration will give you whole numbers b. If the decimal is repeating 3 or 6, then times by 3 to obtain a whole number.

5. If the molar mass of the compound is known, find the molar mass of you empirical formula. Divide the molar mass by the empirical formula to find the common ration between the two. Then times your empirical ratio by this number.

A quick way to remember what I just wrote is via the quick poem below:

Percentage to mass. Mass to mole.

Divide by small. Times till whole.

Example 14: A compound is determined to have the percentage composition of 55.68% of zinc, 20.46% carbon and the rest nitrogen. a. Find the empirical formula for the compound. b. Given that the molar mass is 117.42 g⋅mol-1 find the molecular formula for this compound.

a. Assume that in 100g of compound there is 55.68g of Zinc, 20.46g of carbon and 23.86g of nitrogen (100 – 55.68 –

20.46) Zinc : Carbon : Nitrogen

38.6568.55 :

01.1246.20 :

01.1486.23

0.8516 : 1.704 : 1.703 1 : 2 : 2

∴ZnC2N2

b. Molar mass of empirical = 65.38 + 12.01 × 2 + 14.01 × 2 = 117.42 g⋅mol-1 This is the same as the molar mass so the molecular formula is ZnC2N2.


Example 15: A certain hydrocarbon is composed of 83.62% carbon with the rest being hydrogen. a. Find the empirical formula of the hydrocarbon b. Given that the molar mass of the formula is 86.172 g⋅mol-1. Find the molecular formula of the hydrocarbon. a. Assume that in 100g of the hydrocarbon there is 83.62g of carbon and 16.38% of hydrogen.

Carbon : Hydrogen

01.1262.83

: 008.1

38.16

6.963 : 16.25 1 : 2.334 3 : 7

∴C3H7 b. Molar mass of empirical = 12.01 × 3 + 1.008 × 7 = 43.086 g⋅mol-1

Molar mass: empirical = 86.172: 43.086 = 2 : 1 ∴molecular formula: C6H14

This last example is based on the fact that we are not starting with the percentage composition. It is more to do with actually testing an unknown compound by combustion and then calculating the empirical formula from that.

Example 16: 10.00g of a hydrocarbon that contains carbon, hydrogen and oxygen is burnt in an excess of oxygen. After all the hydrocarbon was combusted, the levels of carbon dioxide and water were measured. If there was 4.091g of water and 14.99g of carbon dioxide left calculate a. The mass of carbon atoms b. The mass of hydrogen atoms c. The mass of oxygen atoms d. The empirical formula of the hydrocarbon. e. The molecular formula of the hydrocarbon given that the molar mass is 88.062 g⋅mol-1. PLEASE NOTE: the combustion of a hydrocarbon gives water and carbon dioxide.

a. m(CO2) = 14.99g

M(CO2) = 12.01 + 16 × 2 = 44.01 g⋅mol-1

n(CO2) = ( ) 3406.001.4499.14

CO)CO(

2

2 ==Mm

mol

CO2 → C + 2O (means for 1 molecule of CO2 I have 1 atom of carbon and 2 atoms of oxygen. These ratios can be used in moles) n(C) = n(CO2) = 0.3406 mol m(C) = n(C) × M(C) = 0.3406 × 12.01 = 4.091g

b. m(H2O) = 4.091g

M(H2O) = 1.008 × 2 + 16 = 18.016 g⋅mol-1

n(H2O) = ( ) 2271.0016.18091.4

OH)OH(

2

2 ==Mm

mol

H2O → 2H + 1O (means for 1 molecule of H2O I have 2 atoms of hydrogen and 1 atom of oxygen. This can be represented by moles) n(H) = 2n(H2O) = 0.4542 mol m(H) = n(H) × M(H) = 0.4542 × 1.008 = 0.4578g

c. m(O) = m(unknown) – m(C) – m(H) = 10.00 – 4.091 – 0.4578 = 5.451g


d. m(O) = 5.4512, m(C) = 4.091, m(H) = 0.4578 d. m(O) = 5.4512, m(C) = 4.091, m(H) = 0.4578 C C : : H H : : O O

01.124091.0 :

008.14578.0 :

00.164512.5

0.3406 : 0.4542 : 0.3407 1 : 1.334 : 1 3 : 4 : 3

∴C3H4O3

e. M(empirical unknown) = 12.01 × 3 + 1.008 × 4 + 16.00 × 3 = 88.062 g⋅mol-1 This is the same as the molar mass of the hydrocarbon. So the molecular formula must be the same as the empirical formula. ∴C3H4O3

REDOX EQUATIONS

Involves the exchange of electrons between substances; Oxidation involves the loss of electrons; reduction involves gaining electrons (oilrig); The oxidised substance is the reducing agent or reductant because it causes reduction in another substance; The reduced substance is the oxidising agent or oxidant because it causes reduction in another substance.

Oxidant

Oxidises

Reduces

Electrons Reductant

Example 1: Identify the oxidant and reductant in the following equations and hence state which reactant is being oxidised and which is being reduced.

Cu2+(aq) + Ni(s) → Ni2+(aq) + Cu(s)

Oxidant: Cu2+

Reductant: Ni(s) Oxidisation: Ni(s) → Ni2+(aq) Reduction: Cu2+(aq) → Cu(s)

An oxidisation number is the number of electrons transferred in the reaction; Rules for Assigning Oxidation numbers:

In uncombined elements each atom has an oxidation number of zero. The oxidation number of a monatomic ion is equal to the charge of the ion. Oxygen has an oxidation number of -2 in compounds (excepts in peroxides, such as H2O2 and Ba2O, where it is -1 and in F2O where it is +2).

Hydrogen has an oxidation number of +1 in compounds (except hydrides were it is -1). The sum of oxidation numbers of all atoms in a neutral molecule is zero. The sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge on the ion.

An element is oxidised if its oxidation number increases;


An element is reduced if its oxidation number decreases; Combustion is an oxidation/reduction reaction:

Combustion is where oxygen gas is present in its elemental state as one of the reactants and as part of a compound as one of the products.

Example 2: Find the oxidisation numbers of the italicised element in the following: a) Fe(s) b) SO4

2-(aq) c) P4O10(s) d) NO3

-

The trick here is to turn them into mini equations and solve. For example let x be the oxidisation number of the italicised element: a) Fe(s) x = 0 (Element, has not charge, so its oxidisation number equals to charge. b) SO4

-2(aq) x + -2 × 4 = -2 (Ion has a charge, so the equation equals the charge) x = 6 (Oxygen has an oxidisation number of -2) c) P4O10(s) 4x + -2 × 10 = 0 (A neutral molecule has a oxidation number of zero.)

4x = 20 x = 5

d) NO3- x + -2 × 3 = -1

x = 5

You know an equation represents and oxidation/reduction reaction if: 1. An atom exists in its elemental form on one side of the equation and as part of a compound on the other side. 2. An atom exists as an ion with particular charge on one side of the equation and a different charge on the other side. 3. Oxygen is chemically added to the substance. 4. Any of the following changes occur during the reaction:

The permanganate ion ( )−4MnO is involved in the reaction and is reduced to the manganese ion (M2+) or manganese(IV) oxide (MnO2), or vice versa.

The dichromate ion ( )272OCr − is involved in the reaction and is reduced to chromium ion (Cr+3), or vice versa.

5. Concentrated sulfuric or nitric acids are involved in the reaction as oxidising agents. (These are sometimes known as oxidising acids.) If this is the case:

The concentrated sulfuric acid usually reduces to sulphur dioxide and water, although it can go further to elemental sulfur or hydrogen sulphide.

The nitric acid may reduce to an oxide of nitrogen, hydroxylamine or ammonia.

WRITING CHEMICAL EQUATIONS

When writing balanced equations if you follow these steps you should not go wrong:

1. Isolate the reactants and the products from the information in the question. (Sometimes you might only be given the reactants, so remember the different reaction types listed in this guide to help with finding the products)

2. Write down your equation, reactants on the left and products on the right, noting the states of each of the substances. (BE CAREFUL not the miss a precipitation reaction!).

3. Balance both sides by counting the number of elements on each side. Sometimes it is helpful to balance the polyatomic ions than the elements separately. You should also leave elements such as hydrogen and oxygen til last.

4. If equation will not balance with the hydrogen, oxygen and charge: Balance out the oxygen by adding water, H2O to the side with fewer oxygen atoms. Balance out the hydrogen by adding H+ to the side with the fewer hydrogen atoms. Balance the charge by adding electrons to the side with the more positive charge.

If you have two half equations you can form an ionic equation by multiply each equation by a

factor so that, which they are added together the electrons cancel out.


Example 3: Balance the following skeletal equations. Identify the oxidizing agent and the reducing agent in each reaction. a) H2S(aq) + Cl2(g) → S(s) + Cl-(aq) b) Cl2(g) → HOCl(aq) + Cl-(aq) c) Fe2+(aq) + Cr2O7

2-(aq) → Fe3+(aq) + Cr3+(aq) d) C2H5OH(aq) + MnO4

-(aq) → Mn2+(aq) + CH3COOH(aq)

Okay when doing a problem like this pair up like elements to form two half equations. a) H2S(aq) + Cl2(g) → S(s) + Cl-(aq)

If you look at this equation you can see I have highlighted two half equations in two different reactions.

H2S(aq) → S(s)

We need to balance this equation by adding 2H+(aq) to the products, then add 2e-(aq) to the products to balance the charge.

H2S(aq) → S(s) + 2H+(aq) + 2e-

Cl2(g) → Cl-(aq)

This equation is pretty easy to balance; we just need to add two electrons to the reactants to balance the charge.

Cl2(g) + 2e- → 2Cl-(aq)

Now we need to join these two half equations together. These two half equations have the same number of electrons, so they will cancel which is what we are looking for.

H2S(aq) + Cl2(g) → S(s) + 2Cl-(aq) + 2H+(aq)

b) Cl2(g) → HOCl(aq) + Cl-(aq)

For this equation think of two chlorine gas’ and split accordingly as show below:

Cl2(g) + Cl2(g) → HOCl(aq) + Cl-(aq)

Cl2(g) → HOCl(aq)

To balance this equation we need to add water molecules to the reactants and to compensate H+(aq) to the products. To balance the charge also we add electrons to the products.

Cl2(g) + 2H2O(l) → 2HOCl(aq) + 2H+(aq) + 2e-

Cl2(g) → Cl-(aq)

This equation is pretty easy to balance; we just need to add two electrons to the reactants to balance the charge.

Cl2(g) + 2e- → 2Cl-(aq)

Now we need to join these two half equations together. These two half equations have the same number of electrons, so they will cancel which is what we are looking for.

2Cl2(g) + 2H2O(l) → 2HOCl(aq) + 2Cl-(aq) + 2H+(aq)

c) Fe2+(aq) + Cr2O7

2-(aq) → Fe3+(aq) + Cr3+(aq) If you look at this equation you can see I have highlighted two half equations in two different reactions.

Fe2+(aq) → Fe3+(aq)

This equation is easy to balance out, just add an electron to the products

Fe2+(aq) →Fe3+(aq) + e-

Cr2O72-(g) → Cr3+(aq)

To balance this equation we need to add water to the products, and then balance the hydrogens by adding H+(aq) to the reactants. Then balance the electrons.

Cr2O72-(g) + 14H+(aq) + 9e- → 2Cr3+(aq) + 7H2O(l)

Now we need to join these two half equations together. These two half equations do not have the

same number of electrons, so we need to multiply the Fe by 9 to counteract this.

9Fe2+(aq) + Cr2O72-(aq) + 14H+(aq) → 9Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)


d) C2H5OH(aq) + MnO4-(aq) → Mn2+(aq) + CH3COOH(aq) d) C

If you look at this equation you can see I have highlighted two half equations in two different reactions. If you look at this equation you can see I have highlighted two half equations in two different reactions.

C2H5OH(aq) → CH3COOH(aq) C

To balance this reaction we need to add water to the reactants, hydrogen to the products and balance out the electrons to give:

To balance this reaction we need to add water to the reactants, hydrogen to the products and balance out the electrons to give: C2H5OH(aq) + H2O(l) → CH3COOH(aq) + 4H+(aq) + 4e-

MnO4-(aq) → Mn2+(aq) MnO

To balance this equation we need to add water to the products, and then balance the hydrogens by adding H+(aq) to the reactants. Then balance the electrons.

To balance this equation we need to add water to the products, and then balance the hydrogens by adding H

MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l) MnO

2H5OH(aq) + MnO4-(aq) → Mn2+(aq) + CH3COOH(aq)

2H5OH(aq) → CH3COOH(aq)

C2H5OH(aq) + H2O(l) → CH3COOH(aq) + 4H+(aq) + 4e-

4-(aq) → Mn2+(aq)

+(aq) to the reactants. Then balance the electrons.

4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)

Now we need to join these two half equations together. These two half equations do not have the

same number of electrons, so we need to multiply one by 4 and the other by 5. Beware of the fact that we have a lot of cancelling out to do, so I am going to rewrite the equations before adding them together.

Now we need to join these two half equations together. These two half equations do not have the same number of electrons, so we need to multiply one by 4 and the other by 5. Beware of the fact that we have a lot of cancelling out to do, so I am going to rewrite the equations before adding them together.

5C2H5OH(aq) + 5H2O(l) → 5CH3COOH(aq) + 20H+(aq) + 20e- 5C H OH(aq) + 5H O(l) → 5CH COOH(aq) + 20H

2 5 2 3

+(aq) + 20e-

4MnO4-(aq) + 32H+(aq) + 20e- → 4Mn2+(aq) + 16H2O(l) 4MnO4-(aq) + 32H+(aq) + 20e- → 4Mn2+(aq) + 16H2O(l)

5C2H5OH(aq) + 4MnO4

-(aq) + 12H+(aq) → 4Mn2+(aq) + 5CH3COOH(aq) + 11H2O(l) 5C2H5OH(aq) + 4MnO4-(aq) + 12H+(aq) → 4Mn2+(aq) + 5CH3COOH(aq) + 11H2O(l)

CHROMATOGRAPHY

Chromatography is a technic that is used to separate the substances in a mixture. It is also used to identify different substances. Some uses of chromatography today are:

Drugs present in blood; Sugars in fruit juice; Hydrocarbons in oil; Pollutants gases in exhaust fumes; Pesticides in water and soil.

Chromatography has many different methods, all of which I will be exploring a little later. What

these methods all have in common is that they have a: Stationary phase; A moving or mobile phase.

As the components are swept over the stationary phase by the solvent, they undergo a process of

adsorption onto the stationary phase, followed by desorption back into the liquid mobile phase. The rate of movement of each component depends mainly upon:

How strongly it absorbs onto the stationary phase; How readily it dissolves in the mobile phase.

PAPER CHROMATOGRAPHY & THIN-LAYER CHROMATOGRAPHY

In paper chromatography: Absorbent paper is the stationary phase A small dot of the solution is place on the bottom as shown in the diagram. The position of this dot is called the origin. The dot is submerged in solvent Solvent rises up the paper and carries the components of the substance with it.

In thin layer chromatography: (Lukins, 2001 pg 15)


Very similar to paper chromatography Very similar to paper chromatography Stationary phase is a thin layer of fine powder such as alumnia (aluminum oxide) spread on a glass or plate. Stationary phase is a thin layer of fine powder such as alumnia (aluminum oxide) spread on a glass or plate.

In both thin layer and paper chromatography, the mixture can be identified by: In both thin layer and paper chromatography, the mixture can be identified by:

solventby origin from moved Distancecompoundby origin from moved distance

=fR

Rf values are always less than one; The component most strongly adsorbed onto the stationary phase moves the shortest distance and has the lowest Rf

value; Each component has a distinct and different Rf values under the same conditions making it easy to identify.

Comparison of paper and thin-layer chromatography:

PAPER CHROMATOGRAPHY THIN-LAYER CHROMATOGRAPHY

Cheap Little preparation

Most efficient for polar and water-soluble compounds Easy to handle and store

Faster Detects smaller amounts

Better separation of less polar compounds Corrosive materials can be used.

A wide range of stationary phases are available.

HIGH PERFORMANCE LIQUID CHROMATOGRAPHY (HPLC)

In column chromatography the stationary phase is a solid or a solid that has been coated in a viscous liquid, which is packed into a glass column. The sample is applied carefully to the top of the packing and the solvent is slowly dripped onto the column from a reservoir above. A tap down the bottom allows the solvent which is known as eluent to leave the column at the same rate as it enters from the other end.

HPLC is a modern version of column chromatography of which allows a more sensitive analysis of a wide variety of compounds.

To the right is a picture of HPLC which differs from normal chromatograph

n is often 10-20

ows for the more frequent

The small particle size creates

y processes. For instance in HPLC: The size of the particles in the solid used in the colum

times smaller than the traditional column. The very small size solid particles alladsorption and desorption of the components, giving much better separation of similar compounds;

a considerable resistance to the flow of the mobile phase and the solvent is pumped through under high pressure – up to about 14000kPa;

A range of solids is available fro use in HPLC columns, some chemicals specially bonded to their surfaces to improve the separation of particular classes.

(Commons Et al, 1991 pg 61)


as;

as;

GAS-LIQUID CHROMATOGRAPHY (GLC) GAS-LIQUID CHROMATOGRAPHY (GLC)

Capable of detecting as little of 10-23 g of a compound; Capable of detecting as little of 10 Limited to compounds that can be vaporised without

decomposing; Limited to compounds that can be vaporised without

decomposing;

-23 g of a compound;

The mobile phase is a gas, generally nitrogen, called a carrier gas;

The mobile phase is a gas, generally nitrogen, called a carrier gas;

A small amount of the sample is injected into the top of the column through an injection port;

A small amount of the sample is injected into the top of the column through an injection port;

The injection port is heated to a temperature sufficient to instantly vaporise the sample which is then swept into the column by the carrier g

The injection port is heated to a temperature sufficient to instantly vaporise the sample which is then swept into the column by the carrier g

The column is a loop or series of loops, of glass that has an internal diameter of about 4 mm and is, in total 2-3 m long. The column is packed with porus solid that has been coated with a high-boiling liquid hydrocarbon or ester. This liquid acts as a liquid stationary phase. The column is mounted in an oven and heated;

The column is a loop or series of loops, of glass that has an internal diameter of about 4 mm and is, in total 2-3 m long. The column is packed with porus solid that has been coated with a high-boiling liquid hydrocarbon or ester. This liquid acts as a liquid stationary phase. The column is mounted in an oven and heated;

(Commons Et al, 1991 pg 63)

The components of the sample repeatedly pass into and out of solution with the stationary phase. The least soluble are swept out first by the gas into the detector.

The components of the sample repeatedly pass into and out of solution with the stationary phase. The least soluble are swept out first by the gas into the detector.

INTERPRETING THE PRINTOUTS FROM CHROMATOGRAMS INTERPRETING THE PRINTOUTS FROM CHROMATOGRAMS

Things to remember: Things to remember: The chromatogram is a graph that records the response of detector against time; The chromatogram is a graph that records the response of detector against time; From this graph we can then measure Rt which is similar to Rf. Under the same conditions for each experiment the

recorder should give the same Rt value for the same compound making identification easy. However, sometimes we can have peaks collide so we cannot get good results;

From this graph we can then measure R

The area under each peak is the proportion of how much substance is in the sample. The area under each peak is the proportion of how much substance is in the sample.

t which is similar to Rf. Under the same conditions for each experiment the recorder should give the same Rt value for the same compound making identification easy. However, sometimes we can have peaks collide so we cannot get good results;


Duration of 4 hours (including a 30 minute break). Focused on consolidating the semester’s coursework and exam preparation. Exam hints and tips provided. Complementary full colour study notes provided – email version only. Hardcopy costs $15 extra. Small groups – up to 30 per class. Location: Collingwood College in Collingwood – ample free parking and right next to bus stop and train station. Members: $45. Public: $50. Check website for details.

Call on 9527 – 7527 or

Visit the “VCE Revision Classes” section of our website for an enrolment form. www.a1tuition.com.au

UNIT 3 ENGLISH

UNIT 3 LEGAL STUDIES UNIT 3 MATHS METHODS

UNIT 4 ACCOUNTING

MEMBERS: $9.95 PUBLIC: $14.95


UNIT 1 & 2 MATHS METHODS

UNIT 3 FURTHER MATHS UNIT 3 & 4 SPECIALIST MATHS

UNIT 1 CHEMISTRY & UNIT 3 CHEMISTRY UNIT 3 ACCOUNTING UNIT 3 ECONOMICS

UNIT 3 & 4 MATHS METHODS



Printed in full colour. Easy to read and understand, using simple and non academic language. Unique in style, design and presentation. Limited print runs during the year. 100% Australian product – printed and bound by A1 TUITION. Compiled using black comb binding with a see through plastic cover and coloured hard card back.

Due to limited quantities printed, we no longer distribute notes through academic bookstores.

WHERE TO BUY

VCE and Careers EXPO Event held annually in April at the Caulfield Racecourse

www.vceandcareers.com.au

A1 Tuition VCE Revision Classes Events held annually in May & July at Collingwood College

www.a1tuition.com.au

A1 Tuition online shop www.a1tuition.com.au/shop


First edition published 2007. Second edition published 2008.

A1 TUITION

P.O. BOX 1188 CAULFIELD NTH 3161

VICTORIA, AUSTRALIA

Designed in Australia. Bound in Australia. Printed in Australia.

CRAM BOX®

Study Notes. Group Tuition. Sample Essays. Private Tuition.

Business Consulting. VCE Advantage Classes. VCE Revision Classes.

9527 – 7527 0413 696 979


AUTHOR MISS AMY PAUL Bachelor of Science

Bachelor of Teaching (Primary and Secondary)

Amy has been current with the VCE system throughout its changes since 2000. She is experienced, up to date with the current curriculum and this experience has enabled her to identify key problem areas that students face when dealing with the subject at the VCE level. Amy attained her first paid tutoring job in 2000. In 2003 Amy also gained a position as an author for Heinemann Education, writing 4 mathematics CD-ROMs, to help students and teachers with their schoolwork. In 2005 Amy began her formal training as a teacher. One of Amy’s ethos is not only to help the students with the subject she is paid to teach, but also to help them to formulate good study routines, for every subject they study. She also has a strong ethos about having up to date information and strives to find more resources to help her students in their studies. Amy offers full service private and group tuition with comprehensive study notes in the following VCE subjects:

Chemistry Maths Methods Specialist Maths

TUITION FEES

1 STUDENT 2 STUDENTS 3 STUDENTS 4 STUDENTS STANDARD FEE $75/1½ hrs $112.50/1½ hrs $131.25/1½ hrs $140.00/1½ hrs

FEE PER STUDENT $75/1½ hrs $56.25/1½ hrs $43.75/1½ hrs $35.00/1½ hrs

CONTACT AMY 0433 308 208

UNIT 3 STUDY NOTES CHEMISTRY - A1 Tuition - Unit 3 (Sample).pdf · GRAVIMETRIC ANALYSIS...

Documents

Transcript of UNIT 3 STUDY NOTES CHEMISTRY - A1 Tuition - Unit 3 (Sample).pdf · GRAVIMETRIC ANALYSIS...