Module Worksheet Solutions
- https://docs.google.com/document/d/
1tDiR6HgYL241alWeukmN3dUDuDKZ3M1XeEFE78SJdkA/edit (2023 Y11 Chem
Syllabus Notes)
- CHEM Term 1 Document; CHEM Term 3 Document
PRE 1.11 Physical and Chemical Changes 😊
HSC 1.15 Experiment: Galvanic Cells 😐
HSC 1.16 Electrochemistry Problems 😐
PRE 2.5 Reactions of Metals 😊
PRE 2.8 Experiment: Reactions of Metals 😊
PRE 3.18 Dilution Calculations 😐
PRE 3.19 Experiment: Preparing and Diluting Solutions 😡
PRE 4.15 Reaction Rates 😐
.・゜-: ✧ :-・‥…━☆1☆━…‥・-: ✧ :-゜・．
,26 Apr | Conservation of Mass to Determine [HCl]
Aim
To determine the concentration of HCl using the conservation of mass.
Equipment
- 1x 50mL measuring cylinder - A scientific balance
- 1x 250mL beaker - ≈ 25mL HCl
- 1x 25mL beaker - ≈ 2g NaHCO3
Method
1. Measure ≈25mL of HCl in a 50mL measuring cylinder, and ≈2g of NaHCO 3 in a
250mL beaker.
2. Weigh the HCl and NaHCO3 on the scientific balance to find the mass of the beakers.
3. Conduct the reaction between HCl and NaHCO3 in the 250mL beaker (so the products
don’t overspill during the reaction).
4. Wait for the reaction to finish, and record the mass of the NaCl and H2O by subtracting
the mass of the beakers.
5. Calculate the concentration of HCl.
Results
Balanced equation: NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
Products Reactants
HCl NaHCO3 NaCl H2O CO2
Mass (g) 24.75 1.99 23.69 (sum) 1.06
n(CO2) = m/M = 1..01 ≈ 0.024 mol
n(HCl) = n(CO2) due to 1:1 ratio in the balanced equation
n(HCl) = 0.024 mol
n = cv
n(HCl) = [HCl] x V(HCl) → [HCl] = n(HCl) / V(HCl)
[HCl] = 0.024 / () because it has to be in litres
Therefore, [HCl] = 0.96 mol/L
Discussion
- It is assumed that the density of HCl was 1g/mL, which lowers the validity of the
experiment.
- We had 96% accuracy to the results because the expected result was 1 mol/L (reliability).
- Our measurements were conducted to the most accurate extent excluding human error.
✧༺ 2 ༻✧
, 27 Apr | Gas Laws
- https://docs.google.com/presentation/d/
1ghixSN1ILJpKtVr8Yb4pJLw7MOMawrKW7W1bbPr7MHY/
edit#slide=id.g3afdc3e624_0_666 (Presentation)
Pressure Units
Pressure = Force / Area
Name of unit Symbol for unit Conversion to Nm-2
newtons per square metre Nm-2
pascal Pa 1 Pa = 1 Nm-2
kilopascal kPa 1 kPa = 1x103 Pa = 1x103 Nm-2
atmosphere atm 1 atm = 101.3 kPa = 1.013 x 105 Nm-2
bar bar 1 bar = 100 kPa = 1x105 Nm-2
millimetres of mercury mmHg 760 mmHg = 1 atm = 1.013 x 105 Nm-2
torr torr 1 torr = 1 mmHg
- Volume is inversely proportional to pressure, with constant temperature and molar mass
- Volume is proportional to temperature (K = Kelvin), with constant pressure and molar
mass
- Volume is proportional to molar mass, with constant temperature and pressure (see
below “Calculating Gas Volumes”)
✧༺ 3 ༻✧
- https://docs.google.com/document/d/
1tDiR6HgYL241alWeukmN3dUDuDKZ3M1XeEFE78SJdkA/edit (2023 Y11 Chem
Syllabus Notes)
- CHEM Term 1 Document; CHEM Term 3 Document
PRE 1.11 Physical and Chemical Changes 😊
HSC 1.15 Experiment: Galvanic Cells 😐
HSC 1.16 Electrochemistry Problems 😐
PRE 2.5 Reactions of Metals 😊
PRE 2.8 Experiment: Reactions of Metals 😊
PRE 3.18 Dilution Calculations 😐
PRE 3.19 Experiment: Preparing and Diluting Solutions 😡
PRE 4.15 Reaction Rates 😐
.・゜-: ✧ :-・‥…━☆1☆━…‥・-: ✧ :-゜・．
,26 Apr | Conservation of Mass to Determine [HCl]
Aim
To determine the concentration of HCl using the conservation of mass.
Equipment
- 1x 50mL measuring cylinder - A scientific balance
- 1x 250mL beaker - ≈ 25mL HCl
- 1x 25mL beaker - ≈ 2g NaHCO3
Method
1. Measure ≈25mL of HCl in a 50mL measuring cylinder, and ≈2g of NaHCO 3 in a
250mL beaker.
2. Weigh the HCl and NaHCO3 on the scientific balance to find the mass of the beakers.
3. Conduct the reaction between HCl and NaHCO3 in the 250mL beaker (so the products
don’t overspill during the reaction).
4. Wait for the reaction to finish, and record the mass of the NaCl and H2O by subtracting
the mass of the beakers.
5. Calculate the concentration of HCl.
Results
Balanced equation: NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
Products Reactants
HCl NaHCO3 NaCl H2O CO2
Mass (g) 24.75 1.99 23.69 (sum) 1.06
n(CO2) = m/M = 1..01 ≈ 0.024 mol
n(HCl) = n(CO2) due to 1:1 ratio in the balanced equation
n(HCl) = 0.024 mol
n = cv
n(HCl) = [HCl] x V(HCl) → [HCl] = n(HCl) / V(HCl)
[HCl] = 0.024 / () because it has to be in litres
Therefore, [HCl] = 0.96 mol/L
Discussion
- It is assumed that the density of HCl was 1g/mL, which lowers the validity of the
experiment.
- We had 96% accuracy to the results because the expected result was 1 mol/L (reliability).
- Our measurements were conducted to the most accurate extent excluding human error.
✧༺ 2 ༻✧
, 27 Apr | Gas Laws
- https://docs.google.com/presentation/d/
1ghixSN1ILJpKtVr8Yb4pJLw7MOMawrKW7W1bbPr7MHY/
edit#slide=id.g3afdc3e624_0_666 (Presentation)
Pressure Units
Pressure = Force / Area
Name of unit Symbol for unit Conversion to Nm-2
newtons per square metre Nm-2
pascal Pa 1 Pa = 1 Nm-2
kilopascal kPa 1 kPa = 1x103 Pa = 1x103 Nm-2
atmosphere atm 1 atm = 101.3 kPa = 1.013 x 105 Nm-2
bar bar 1 bar = 100 kPa = 1x105 Nm-2
millimetres of mercury mmHg 760 mmHg = 1 atm = 1.013 x 105 Nm-2
torr torr 1 torr = 1 mmHg
- Volume is inversely proportional to pressure, with constant temperature and molar mass
- Volume is proportional to temperature (K = Kelvin), with constant pressure and molar
mass
- Volume is proportional to molar mass, with constant temperature and pressure (see
below “Calculating Gas Volumes”)
✧༺ 3 ༻✧
