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CHEM 103 MODULE 3 EXAM

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CHEM 103 MODULE 3 EXAM

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M3: Exam- Requires Respondus
LockDown Browser
● Due No due date
● Points 100
● Questions 10
● Time Limit 120 Minutes
● Requires Respondus LockDown Browser

Attempt History
Attempt Time Score

LATEST Attempt 1 120 minutes 78 out of 100
Score for this quiz: 78 out of 100
Submitted Sep 6 at 10:59am
This attempt took 120 minutes.

Question 1
pts

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 a 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 calorimeter and mixture of HCl + NaOH



2. The air above the cup

Your Answer:

1 . Adiabatic system because there is transfer of heat between the mixture of HCl and NaOH to
the calorimeter

2. Vapor pressure

1. With lid = closed system

, 2. The air above the cup = Surroundings


Question 2
pts

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 42.7 gram sample of
water at 82.1oC is added to a 18.5 gram sample of water at 20.6oC in a coffee cup calorimeter.

c (water) = 4.184 J/g oC



2. Show the calculation of the energy involved in freezing 75.4 grams of water at 0oC if the
Heat of Fusion for water is 0.334 kJ/g

Your Answer:

1.

m1= 42.7 g T1 = 82.1

m2 = 18.5 g T2 = 20.6



Heat lost = Heat gained

Let Tf be the final temperature

m1c (T1 - Tf ) = m2c (Tf - T2)




42.7 g × 4.184 J / g ∘ C ( 82.1 ∘ C − T ) = 18.5 g × 4.184 J / g ∘ C ( T −
20.6 ∘ C )




( 42.7 g × 4.184 J g ∘ C ) ( 18.5 g × 4.184 J g ∘ C ) × ( 82.1 ∘ C − T ) = ( T
− 20.6 ∘ C )

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