SAC 1

Question 1

renewable fuels

Answer 1

• are produced faster than used
  adv - reduced CO2 emissions, can be made from organic waste
  dis - lower energy content, requires land for production

Question 2

non-renewable fuels

Answer 2

• are used faster than produced
  adv - cheap, high energy content and efficiency
  dis - high CO2 emissions, heavy pollutants, leaks can cause explosions

Question 3

fermentation equation

Answer 3

C6H12O6(aq)————->2C2H5OH(aq)+2CO2(g)

Question 4

complete combustion

Answer 4

produces CO2 and H2O

Question 5

incomplete combustion

Answer 5

• limited oxygen
  produces CO/C and H2O

Question 6

exothermic

Answer 6

• negative △H value
• released
• high to low

Question 7

endothermic

Answer 7

• positive △H value
• absorbed
• low to high

Question 8

glucose equation

Answer 8

6CO2(g)+6H2O(l)———>C6H12O6(aq)+6O2(g)

Question 9

cellular respiration equation

Answer 9

C6H12O6(aq)+6O2(g)——->6CO2(g)+6H2O(l)

Question 10

% efficiency

Answer 10

= desired ➗ before x 100

Question 11

stoichiometry ratios

Answer 11

N2(g)+3H2(g)——->2NH3(g)
mol1 3 2
vol 1 3 2
mL 10 30 20

Question 12

density and mass equation

Answer 12

density = mass ➗ volume
mass = density x volume

Question 13

graph (example)

Answer 13

x axis = various fuels
y axis = energy (kJg-1) (enthalpy)

Question 14

energy in MJ, released by complete combustion of 1.00L of fuel

Answer 14

1. m=dxv x 1000 = no° in g
2. heat of combustion no° x no° of mols ➗ 1000
3. answer

(to get mols no° use mass(g) ➗ molar mass

Question 15

mass of (CO2) in grams produced by complete combustion of 1.00L of fuel

Answer 15

1. ratio CO2 and ratio of 1st reactant
2. multiply ratio no° of CO2 by mols of 1st reactant
3. gives us CO2 no°of mols
4. use m=nxM
5. answer

Question 16

convert celcius to kelvin

Answer 16

Celcius no° + 273

Question 17

gas equation not at SLC

Answer 17

pV=nRT
R=8.31
V=L
T=K
p=kPa
n=mols

V=nRT ➗ p
T = pxV➗Rxn

Question 18

at SLC (25° and 100kPa)

Answer 18

Vm=24.8L
n=V➗Vm

Question 19

q=mc△T

Answer 19

q=energy (J)
m=mass of water
c=specific heat capacity (4.18)
T=temperature change

Question 20

energy released from food (in kJ)
1.08g of almonds, heating 150.0g of water, temp change = 13 degrees, answer is kJg

Answer 20

1. no° from q=mc△T➗1000
   (8151➗1000=8.15)
2. that no°➗mass of food in g
   (8.15 ➗ 1.08)
3. answer
   (77.55kJg)

Question 21

calculate energy per gram

Answer 21

1. 1.0➗change is mass
   (1.0➗1.9)
2. that no° x q=mc△T no° ➗ 1000
   (0.526 x 20482 ➗ 1000)
3. answer
   (10.78kJg-1)

Question 22

calculate energy per mole

Answer 22

1. 1.0➗change is mass
   (1.0➗1.9)
2. that no° x q=mc△T no° ➗ 1000
   (0.526 x 20482 ➗ 1000)=(10.78kJg-1)
3. multiply by molar mass
   (10.78 x 74.12)
4. answer
   (799.01 kJmol)

Question 23

should produce … kJ of energy, knowing this calculate energy loss

Answer 23

1. energy per mol number ➗ … kJ number
   (799.01 ➗ 2676)
2. times by 100
   (799.01 ➗ 2676 x 100)
3. answer
   (29.86%)

Question 24

explain structure difference

Answer 24

• where the OH group is attached/located, and which carbon it is attached to
• which will cause △H values to vary

Question 25

gauze mat v clay triangle

Answer 25

- gauze mat has more metal and absorbes heat meaning less heat get to the water, therefore resulting in a lower temperature - the clay triangle allows more heat to reach the 100ml of water - using a clay triangle you will receive higher △H values

Question 26

conclusion

Answer 26

- refer to aim - refer to results - what does it show