CHEM 103
MODULE 3 EXAM 2022
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
A reaction between HCl and NaOH is being studied in a styrofoam coffee cup
with NO lid and the heat given off is measured by means of a thermometer
immersed in the reaction mixtu...
chem 103 module 3 exam 2022 click this link to access the periodic table this may be helpful throughout the exam a reaction between hcl and naoh is being studied in a styrofoa
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CHEM 103
MODULE 3 EXAM 2022
Question 1
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
A reaction between HCl and NaOH is being studied in a styrofoam coffee cup
with NO lid and the heat given off is measured by means of a thermometer
immersed in the reaction mixture. Enter the correct thermochemistry term
to describe the item listed.
1. The type of thermochemical process
2. The amount of heat released in the reaction of HCl with NaOH
1. Heat given off = Exothermic process
2. The amount of heat released = Heat of reaction
Question 2
Click this link to access the Periodic Table. This may be helpful throughout the exam.
1. Show the calculation of the final temperature of the mixture when a
22.8 gram sample of water at 74.6oC is added to a 14.3 gram sample of
water at 24.3oC in a coffee cup calorimeter.
c (water) = 4.184 J/g oC
2. Show the calculation of the energy involved in freezing 54.3 grams of
water at 0oC if the Heat of Fusion for water is 0.334 kJ/g
1. - (mwarn H2O x cwarn H2O x ∆twarn H2O) = (mcool H2O x ccool H2O x ∆tcool H2O)
, - [22.8 g x 4.184 J/g oC x (Tmix - 74.6oC)] = [14.3 g x 4.184 J/g oC x (Tmix -
24.3oC)]
- [95.3952 J/oC x (Tmix - 74.6oC)] = [59.8312 J/oC x (Tmix - 24.3oC)]
Tmix = 55.2oC
2. ql↔s = m x ∆Hfusion = 54.3 g x 0.334 kJ/g = 18.14 kJ (since heat is
removed) = - 18.14 kJ
Question 3
Click this link to access the Periodic Table. This may be helpful throughout the exam.
Show the calculation of the amount of heat involved if 35.6 g of H2S is
reacted with excess O2 to yield sulfur trioxide and water by the following
reaction equation. Report your answer to 4 significant figures.
2 H2S (g) + 3 O2 (g) → 2 SO2 (g) + 2 H2O (g) ΔH = -
1124 kJ
1 mol H2s = 34.1 g of H2S = 603.2 kj
(35.6g/34.1 g) x -603.2 kJ = (1.0439) x (-603.2 kJ) = -629. 7 kj
2 H2S (g) + 3 O2 (g) → 2 SO2 (g) + 2 H2O (g) ΔH = - 1124 kJ
ΔHrx is for 2 mole of H2S
reaction uses 35.6 g of H2S = 35.6/34.086 = 1.044 mole of H2S
q = ΔHrx x new moles / original moles
q = -1124 kJ x 1.044 mole of H2S / 2 mole H2S = 586.7 given off
Question 4
Click this link to access the Periodic Table. This may be helpful throughout the exam.
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