CHEM 103 Final Exam (100 out of 100) Questions and Answers (New Update) Verified Elaborations.

CHEM 103 Final Exam (100 out of 100) Questions and t t t t t t t t t




Answers (New Update) Verified Elaborations.
t t t t t




Question 1 t




Complete the two problems below: t t t t




1. Convert 0.0000726 to exponential form and explain your answer. t t t t t t t t




2. Convert 5.82 x103 toordinary form and explain your answer. t t t
t
t t t t t t




Answer:
t



1. Convert 0.0000726 = smaller than 1 = negative exponent, move decimal 5
t t t t t t t t t t t



places = 7.26 x 10-5 t t t t




2. Convert 5.82 x 103 = positive exponent = larger than 1, move decimal 3
t t t
t
t t t t t t t t t



places = 5820 t t




Question 2 t




Do the conversions shown below, showing all work:
t t t t t t t




1. 358oK = ? oC t t t




2. 53oC = ? oF t t t




3. 158oF = ? oK t t t




Answer:


1. 358oK - 273 = 85 oC t t t t t K → oC (make smaller)
o
t t t t



-273
2. 53oC x 1.8 + 32 = 127.4 oF t t t t t t t
o
C → oF (make larger) x
t t t t t



1.8 + 32 t t




3. 158oF - 32 ÷ 1.8 = 70 + 273 = 343 oK t t t t t t t t t t t t F → oC → oK
o
t t t t




Question 3 t




Show the calculation of the number of moles in the given amount of the following
t t t t t t t t t t t t t t



substances. Report your answerto 3 significant figures.
t t t t t t t




1. 12.0 grams of Ca3(PO4)2 t t t




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t t t t t t t t t t t t t




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, 2. 15.0 grams of C9H8NO4Cl t t t




Answer:
1. Moles = grams / molecular weight = 12..18 = 0.0387 mole t t t t t t t t t t t t




2. Moles = grams / molecular weight = 15..61 = 0.0653 mole
t t t t t t t t t t t t




Question 4 t




Show the calculation of the percent of each element present in the following
t t t t t t t t t t t t



compounds. Report your answer to 2 places after the decimal.
t t t t t t t t t t




1. Al2(SO4)3


2. C7H5NOBr
Answer:
1. %Al = 2 x 26.98/342.17 x 100 = 15.77% t t t t t t t t %S = 3 x 32.07/342.17
t t t t



x 100 = 28.12%
t t t




%O = 12 x 16/342.17 = 56.11% t t t t t t




2.
%C=7x12.01/199.02 x 100= 42.24%
t t t t t t t t t %H = 5 x 1.008/ 199.02 x 100
t t t t t t t



= 2.53% t




%N = 1 x 14.01/199.02 = 7.04%
t t t t t t %O = 1 x 16.00/199.02 x 100 t t t t t t



= 8.04% t




%Br = 79.90/199.02 x 100 = 40.15%
t t t t t t




Question 5 t




Show the calculation of the heat of reaction (ΔHrxn) for the reaction: CH4
t t t t t t t t t t t t




t (g) + 2 O2 (g) → t t CO2 (g) + t 2 H2O (l)
t t




by using the following thermochemical data:
t t t t t




ΔHf0 CH4 (g) = -74.6 kJ/mole, ΔHf0 CO2 (g) = -393.5 kJ/mole, ΔHf0 H2O (l) = -285.8
t
t t t t t
t
t t t t t
t
t t t



kJ/mole
Answer:
t




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t t t t t t t t t t t t t




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