CHEM 103 MUDULE 1,2,3,4,5& 6 EXAM QUESTIONS AND ANSWERS VERIFIED 100% CORRECT| PORTAGE LEARNING|

CHEM 103 MUDULE 1,2,3,4,5& 6 EXAM
QUESTIONS AND ANSWERS VERIFIED 100%
CORRECT| PORTAGE LEARNING|

MODULE 2 EXAM

Question 1
Click this link to access the Periodic Table. This may be helpful throughout the exam.


Show the calculation of the molecular weight for the following compounds,
reporting your answer to 2 places after the decimal.


1. Al2(CO3)3


2. C8H6NO4Cl


1. 2Al + 3C + 9O = 233.99
2. 8C + 6H + N + 4O + Cl = 215.59

Question 2
Click this link to access the Periodic Table. This may be helpful throughout the exam.


Show the calculation of the number of moles in the given amount of the following
substances. Report your answerto 3 significant figures.


1. 13.0 grams of (NH4)2CO3

2. 16.0 grams of C8H6NO4Br

1. Moles = grams / molecular weight = 13..09 = 0.135 mole

,Portage Learning / Chem 103 / module 1-6 Exam with Answers


3.

2. Moles = grams / molecular weight = 16..04 = 0.0615 mole

Question 3
Click this link to access the Periodic Table. This may be helpful throughout the exam.


Show the calculation of the number of grams in the given amount of the following
substances. Report your answer to 1 place after the decimal.


1. 1.20 moles of (NH4)2CO3


2. 1.04 moles of C8H6NO4Br

1. Grams = Moles x molecular weight = 1.20 x 96.09 = 115.3 grams


2. Grams = Moles x molecular weight = 1.04 x 260.04 = 270.4 grams

Question 4
Click this link to access the Periodic Table. This may be helpful throughout the exam.

,Portage Learning / Chem 103 / module 1-6 Exam with Answers


Show the calculation of the percent of each element present in the following
compounds. Report your answer to 2 places after the decimal.


1. (NH4)2CrO4


2. C8H8NOI


1. %N = 2 x 14.01/152.08 x 100 = 18.43% x 100 = %H = 8 x 1.008/152.08
5.30%
%Cr = 1 x 52.00/152.08 x 100 = 34.20% %O = 4 x
16.00/152.08 x 100 = 42.08%



2. %C = 8 x 12.01/261.05 x 100 = 36.80% x 100 = %H = 8 x 1.008/261.05
3.09%
%N = 1 x 14.01/261.05 x 100 = 5.37% 16.00/261.05 %O = 1 x
x 100 = 6.13%
%I = 1 x 126.9/261.05 x 100 = 48.61%

Question 5
Click this link to access the Periodic Table. This may the be helpful throughout
exam.


Show the calculation of the empirical formula for each compound whose
elemental composition is shown below.


38.76% Ca, 19.87% P, 41.27% O



38.76% Ca / 40.08 = 0..6416 = 1.5 x 2 = 3
19.87% P / 30.97 = 0..6416 = 1 x 2 = 2

41.27% O / 16.00 = 2..6416 = 4 x 2 = 8 → Ca3P2O8

Question 6

, Portage Learning / Chem 103 / module 1-6 Exam with Answers


Click this link to access the Periodic Table. This may be helpful throughout the exam.


Balance each of the following equations by placing coefficients in front of each
substance.


1. C6H6 + O2 → CO2 + H2O

2. As + O2 → As2O5

3. Al2(SO4)3 + Ca(OH)2 → Al(OH)3 + CaSO4

1. 2 C6H6 + 15 O2 → 12 CO2 + 6 H2O

2. 4 As + 5 O2 → 2 As2O5

3. Al2(SO4)3 + 3 Ca(OH)2 → 2 Al(OH)3 + 3 CaSO4

Question 7
Click this link to access the Periodic Table. This may be helpful throughout the exam.


Classify each of the following reactions as either:
Combination
Decomposition
Combustion
Double Replacement
Single Replacement


1. H2SO4 → SO3 + H2O

2. S + 3 F2 → SF6

3. H2 + NiO → Ni + H2O


1. H2SO4 → SO3 + H2O = Decomposition, One reactant → Two Products

2. S + 3 F2 → SF6 = Combination. Two reactants→ One product
3. H2 + NiO → Ni + H2O = Single Replacement, Hydrogen displaces metal ion