20) In an experiment, 50 cm3 of 1 mol dm-3 NaCl is added to 50 cm3 of 1 mol dm-3 AgNO3 solution . The following data is obtained

∆H = mcO/mol = 2940/1000x 0.05 = - 58.8 kJ mol-1

Initial temp of NaCl 28.0 o CInitial temp of AgNO3 solution 29.0 o C

Highest temperature of mixture 35.5 o C

b) Calculate heat change c) Calculate heat of reaction

Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1 o C -1

mcO = 100 x 4.2 x 7 = 2940 J

Precipitation reaction a) What is the name of this reaction?

d) If NaCl solution is replaced with HCl solution, predict the heat of reaction . Justify your answer

e) If NaCl solution is replaced with Na2CO3 solution, predict the heat of reaction . Justify your answer

Heat of rex is the SAME ,because the precipitate is still the same, that is AgCl

Heat of rex is the DIFF , because the precipitate is NOT the same, that is Ag2CO3

21) In an experiment, 100 cm3 of 1.0 mol dm-3 CaCl2 solution is added to 100 cm3 of 1.0 mol dm-3 Na2CO3 solution . The following data is

obtained

Ca 2+ + CO3 2- CaCO3 ∆H = + 12.6 kJ mol-1

Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1 o C -1

= 1.5 oC

Calculate the temperature change of the mixture

∆H = mcO / mol therefore O = ∆H x mol / mc = 12.6 x 1000 x 0.1 / 200 x 4.2

22) In an experiment, 1 g of zinc powder is added to 50 cm3 of 0.2 mol dm-3 CuSO4 solution . The following data is obtained

∆H = mcO / mol = 1050 / 1000x 0.01 = - 105 kJ mol-1

Initial temp of CuSO4 solution 28.0 o C

Highest temperature of mixture 33.0 o C

b) Calculate heat change

c) Calculate heat of reaction

Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1 o C -1

mcO = 50 x 4.2 x 5 = 1050 J

Displacement reaction a) What is the name of this reaction?

23) In an experiment, 100 cm3 of 2.0 mol dm-3 HCl is added to 100 cm3 of 2.0 mol dm-3 NaOH solution . The following data is obtained

∆H = mcO / mol = 10080 / 1000x 0.2 = - 50.4 kJ mol-1

Initial temp of HCl solution 29.5 o C

Initial temp of NaOH solution 29.5 o C

Highest temperature of mixture 41.5 o C

b) Calculate heat change

c) Calculate heat of reaction

Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1 o C -1

mcO = 200 x 4.2 x 12 = 10080 J

Neutralisation reaction a) What is the name of this reaction?

21) The thermochemical equation for the reaction between ethanoic acid and sodium hydroxide solution is given below

CH3COOH + NaOH CH3COONa + H2O ∆H = -55 kJ mol-1

Given Density of solution : 1 g cm-3 , specific heat of solution is 4.2 J g-1 o C -1

= 2750 J

Calculate the heat given out when 200 cm 3 of ethanoic acid is added to 100 cm3 of sodium hydroxide solution . Both solutions have a concentration of 0.5 mol dm-3

Mol of ethanoic acid = MV/1000 = 0.5 x 200/1000 = 0.1 mol

Mol of sodium hydroxide = MV/1000 = 0.5 x 100/1000 = 0.05 mol

∆H = mcO/mol Therefore mcO =∆H x mol = 55x1000 x 0.05

25) In an experiment, the following data is obtained

∆H = mcO/mol = 9249 /1000x 0.0156 = - 592.3 kJ mol-1

Volume of water used 100 cm3 Initial temp of of water 29.0 o CHighest temperature of water reached 51.0 o CMass of spirit lamp and methanol before combustion 156.55 gMass of spirit lamp and methanol after combustion 156.05 g

b) Calculate heat change

c) Calculate heat of reaction

[Given Density of water : 1 g cm-3 , specific heat capacity of water is 4.2 J g-1 o C -1

RAM : H,1; C,12; O,16 ]

mcO = 100x 4.2 x 22 = 9240 J

Combustion a) What is the name of this reaction?

25) The heat of combustion of ethanol is -1376 kJ mol-1 What is its fuel value?

Fuel value = heat of combustion/ Molar mass

Fuel value = 1376/ 46 = - 29.9 kJ g -1

C3H7OH + O2 CO2 + H2O3 49/2

Heat released = mcO =200 x 4.2 x 31 =26040 Joule

Mole = mass/molar mass = 0.84/60 = 0.014

0.014 mole of propanol released heat 26040 jouleSo 1 mol propanol 1x 26040/0.014 =1860 kJ

Heat of combustion = -1860 kJ mol -1

= -1860 kJ mol -1

……………………………………….when 1 mol of…………………………………. ….. is formed from ……………………………………………..in solution

Magnesium carbonate

Heat change

magnesium ions and carbonate ions

Mg 2+ + CO3 2- MgCO3

mco = 50 x 4.2 x 5.5 = 1155J

Mole = MV/1000 = 25 x2 /1000 = 0.05 mol

No of mole of MgCO3 = no of mole of magnesium ion = 0.05 mol

0.05 mol of MgCO3 release heat 1155 JSo 1 mole of MgCO3 1 x 1155/0.05 Joule = 23.1 kJ

Heat of precipitation = + 23.1 kJ mol -1

In thermochemistry, do not use from equation

H= positive

Same precipitate is formed, that is magnesium carbonate/ sodium ion or potassium ion does not take part in the reaction

Oxidised: iron (II) sulphate / Fe 2+

Reduced : Acidified potassium manganate(VII)

Oxidation:

Reduction:

K Mn O4 = 0

A)AIM OF EXPERIMENTAble to give the aim of the experiment correctlySample answersTo investigate the effect of Metal Mg and Cu when it in contact with iron in the rusting of iron// To study the effect of Metal Cu can faster the rusting of iron while metal Mg prevent the iron from rusting

a. variable

Manipulated variable : Metal X and metal Y//two different metals(one metals is less electropositive and one is more electropositive than iron)//pairs of metal X / iron and Y/iron

Responding variable : Rusting of iron//iron rust//the formation of brown solid//formation of blue spotConstant variables:Iron nail//jelly solution//temperature

c. hypothesisAble to state the relationship between the manipulated variable and the responding variable with direction correctly. Sample answerWhen a more electropositive metal is in contact with iron, the metal inhibits rusting. // When a less electropositive metal is in contact with iron, the metal speeds up rusting.

d. Material & apparatus  Materials/substances Two Iron nails,Magnesium/zinc/aluminium strip, tin/copper/lead/silver stripPotassium hexacyanoferrate(III) solution + phenolphthaleinApparatus Test tube/boiling tube, Sand paper

e. procedureSample answer:•Clean the iron nails and metals strip with sand paper.•Coil iron nails with magnesium ribbon and copper strips.•Put/place the coiled iron nail into different test tube. •Pour/add/fill the hot jelly solution containing potassium hexacyanoferrate(III)solution and phenolphthalein into the test tube.

e. procedureSample answer:•Leave the test tube in a test tube rack for few days.•Record the observation.•Steps 1 to 6 are repeated using different metal/Y with iron(if steps 2 does not mention two different test tube).

f. Tabulated data

Pair of metals Observation

Mg/Fe

Cu/Fe

Revision redox

Redox reaction or non redox?• Which one is REDOX? State your reason

a) AgNO3 + NaCl AgCl + Na NO3

b) Cl2 +2 KI 2KCl + I2

Reaction a) is not a redox reaction because oxidation number

of Ag does not change, that is from +1 to +1

Reaction b) is a redox reaction because oxidation number

of Cl change, from 0 to -1

State oxidation number for the underlined • KMnO4

• K I• Fe Cl2

• I 2

• Cu• CuSO4

• MnO4 –

• H3O +

(+1 + Mn + 4O = 0 ) , (+1 + Mn -8 = 0 ), Mn = +7

(+1 + I = 0 ) , I = -1

+2

0

0

+2

(Mn + 4O = -1) , Mn = +7

(3H -2 = + 1) , H = +1

0+2

+3 +2reduction

oxidation

Mg + 2FeCl3 2FeCl2 + MgCl2

Which is oxidized ? reason

Which is reduced? Reason

Which is oxidizing agent, why

Which is reducing agent?why

Mg because ON of Mg increase from 0 to +2

FeCl3 because ON of Fe decrease from +3 to +2

FeCl3 because it oxidise Mg to MgCl2

Mg because it reduce FeCl3 to FeCl2

0 -1

-1 0oxidation

reduction

Br2 + 2NaI 2NaBr + I2

Which is oxidized ? reason

Which is reduced? Reason

Which is oxidizing agent, why

Which is reducing agent? why

Sodium iodide because ON of iodine increase from -1 to 0

Bromine because ON of Br decrease from 0 to -1

Bromine because it oxidise NaI to I2

NaI because it reduce Br2 to NaBr

0 +2

+2 0reduction

oxidation

Zn + CuSO4 ZnSO4 + Cu

Which is oxidized ? reason

Which is reduced? Reason

Which is oxidizing agent, why

Which is reducing agent? why

Zinc because ON of zinc increase from 0 to +2

CuSO4 because ON of copper decrease from +2 to 0

CuSO4 because it oxidise Zn to ZnSO4

Zinc because it reduce CuSO4 to Cu

Reaction A is NOT a redox reaction because Oxidation number of Na does not change, that is from +1 to +1

Reaction B is a REDOX reaction because Oxidation number of Mg change from 0 to +2

In P, +2

In Q, +1

P is copper(II) oxideBecause ON of copper is +2

Q is copper(I) oxide because ON of copper is +1

+2 0

0+1

Reducing agent

oxidation

Oxidising agent

reduction

H 2

CuO

CuO

H 2

Because ON of hydrogen increased from 0 to +1

Because ON of copper decreased from +2 to 0

Because it oxidise hydrogen to hydrogen oxide

Because it reduce CuO to Cu

+2 0reduction

0 +1oxidation

State the fuction ofi) Glass wool

ii) Potassium manganate (VII) powder

Prevent metal from mixing with KMnO4 ( can cause explosion)

release oxygen when heated to react with metal powder