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CHEM 103 FINAL EXAM 2024: PORTAGE LEARNING (ALREADY GRADED A+) $10.49   Add to cart

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CHEM 103 FINAL EXAM 2024: PORTAGE LEARNING (ALREADY GRADED A+)

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CHEM 103 FINAL EXAM 2024: PORTAGE LEARNING (ALREADY GRADED A+) Question 1 Complete the two problems below: 1. Convert 0.0000726 to exponential form and explain your answer. 2. Convert 5.82 x 103 to ordinary form and explain your answer. Answer: 1. Convert 0.0000726 = smaller than 1 = negat...

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  • October 23, 2024
  • 5
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers
  • CHEM 103
  • CHEM 103
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Lectpearl
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CHEM 103 FINAL EXAM 2024: PORTAGE LEARNING
(ALREADY GRADED A+)

Question 1
Complete the two problems below:
1. Convert 0.0000726 to exponential form and explain your answer.


2. Convert 5.82 x 103 to ordinary form and explain your answer.

Answer:

1. Convert 0.0000726 = smaller than 1 = negative exponent, move decimal 5
places = 7.26 x 10-5

2. Convert 5.82 x 103 = positive exponent = larger than 1, move decimal 3
places = 5820

Question 2
Do the conversions shown below, showing all work:

1. 358oK = ? oC

2. 53oC = ? oF

3. 158oF = ? oK


Answer:



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

2. 53oC x 1.8 + 32 = 127.4 oF oC → oF (make larger) x
1.8 + 32

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

, Question 3
Show the calculation of the number of moles in the given amount of the following substances.
Report your answerto 3 significant figures.
1. 12.0 grams of Ca3(PO4)2

2. 15.0 grams of C9H8NO4Cl


Answer:

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


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

Question 4
Show the calculation of the percent of each element present in the following compounds. Report
your answer to 2 places after the decimal.


1. Al2(SO4)3


2. C7H5NOBr
Answer:
1. %Al = 2 x 26.98/342.17 x 100 = 15.77% %S = 3 x 32.07/342.17
x 100 = 28.12%
%O = 12 x 16/342.17 = 56.11%


2
%C = 7 x 12.01/ 199.02 x 100 = 42.24% %H = 5 x 1.008/ 199.02 x 100
= 2.53%
%N = 1 x 14.01/199.02 = 7.04% %O = 1 x 16.00/199.02 x 100
= 8.04%
%Br = 79.90/199.02 x 100 = 40.15%

Question 5
1. List and explain which of the following atoms holds its valence electrons less tightly.

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