Chapter 4 – Problem Solving in Chemistry – Answer Key, Schemes and Mind Maps of Chemistry

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Chemistry Name: _____________________ Date: _________

Chapter 4 – Problem Solving in Chemistry – Answer Key

Chapter 4: 1 – 5, 7, 9 – 17, 19, 22 – 24, 26 – 29, 33, 34, 38, 40, 51, 52, 56, 70 (30 total)

Practice Problems

1. The density of silicon is 2.33 g/ cm^3. What is the volume of a piece of silicon that has a mass of 62.9 g? Rearrange d = m/ V to V = m/ d; V = 62.9 g/ 2.33 g/ cm^3 ; V = 26.9 cm^3
2. Helium has a boiling point of 4 K. This is the lowest boiling point of any liquid. Express this temperature in degrees Celsius.

Use ̊C = K – 273. ̊C = 4 K – 273; Answer is -269 ̊C

Section Review 4.

1. List three useful problem-solving skills.

a. Write out any problem, list things that are known and things unknown. b. Draw diagrams, pictures, and maps, if necessary. c. Collect as much data as possible, from the problem. d. Allow yourself plenty of time, don’t wait until the last minute to complete homework.

1. State in your own words the three suggested step for solving word problems.

a. Analyze the problem first – list what is known and what is unknown. b. Calculate – set up the problem mathematically – solving for unknowns. c. Evaluate – ask yourself if the answer makes any sense.

1. Identify the statements that correctly complete the sentence: Good problem solvers

a. read a problem only once.

b. check their work.

c. break complex problems down into one or more simpler problems.

d. look for relationships among pieces of information.

1. Calculate normal body temperature (37 ̊ C) on the Kelvin scale.

Wanted: temperature in kelvins. Given: body temperature in degrees Celsius.

Equation: K = ̊C + 273; Solve: 37 + 273 = 310 K

Practice Problems

1. An experiment requires that each student use an 8.5-cm length of magnesium ribbon. How many students can perform the experiment if there is a 570-cm length of magnesium ribbon available?

𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 = (570 𝑐𝑚 )((1 student)/(8.5 cm))= 67 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠

1. A 1.00-degree increase on the Celsius scale is equivalent to a 1.80-degree increase on the Fahrenheit scale. If a temperature increases by 48.0 ̊ C, what is the corresponding temperature increase on the Fahrenheit scale?

Consider the ratio or conversion factor provided, of 1.00 ̊C/ 1.80 ̊F

Practice Problems

1. Using tables from Chapter 3, convert the following.

a. 0.044 km to meters (0.044 𝑘𝑚) (1,000 𝑚1 𝑘𝑚 ) = 44 𝑚

b. 4.6 mg to grams (4.6 𝑚𝑔) ( (^) 1000 𝑚𝑔1 𝑔𝑟𝑎𝑚) = 0.0046 𝑔

c. 8.9 m to decimeters (8.9 𝑚) (10 𝑑𝑒𝑐𝑖𝑚𝑒𝑡𝑒𝑟𝑠1 𝑚 ) = 89 𝑑𝑚

d. 0.107 g to centigrams (0.107 𝑔) (100 𝑐𝑔1 𝑔 ) = 10.7 𝑐𝑔

1. Convert the following.

a. 15 cm^3 to liters (15 𝑐𝑚^3 ) ( (^) 1 𝑐𝑚1 𝑚𝐿 3 ) ( (^) 1000 𝑚𝐿1 𝐿 ) = 0.015 𝐿

b. 7.38 g to kilograms (7.38 𝑔) ( (^) 1000 𝑔1 𝑘𝑔) = 0.00738 𝑘𝑔

c. 0.67 s to milliseconds (0.67 𝑠) (1000 𝑚𝑠1 𝑠 ) = 670 𝑠

d. 94.5 g to micrograms (94.5 𝑔) (1,000,000 𝜇𝑔1 𝑔 ) = 9.45 𝑥 10^7 𝜇𝑔

1. (continued)

d. 68.9 m to decimeters (68.9 𝑚) ( 10 𝑑𝑚 1 𝑚 ) = 689 𝑑𝑚, 𝑜𝑟 6.89 𝑥 10

e. 3.72 x 10-3^ kg to grams (3.72 𝑥 10−3^ 𝑘𝑔) (1000 𝑔1 𝑘𝑔 ) = 3.72 𝑔 𝑜𝑟 3.72 𝑥 10^0 𝑔

f. 66.3 L to cubic centimeters (66.3 𝐿) (1000 𝑚𝐿1 𝐿 ) (1 𝑐𝑚

3 1 𝑚𝐿 ) = 66,300 𝑐𝑚

(^3) 𝑜𝑟 6.63 𝑥 10 (^4) 𝑐𝑚 3

g. 0.0371 m to kilometers (0.0371 𝑚) ( (^) 1000 𝑚1 𝑘𝑚) = 0.0000371 𝑘𝑚 𝑜𝑟 3.71 𝑥 10−5^ 𝑘𝑚

1. A 2.00-kg sample of bituminous coal is composed of 1.30 kg of carbon, 0.20 kg of ash, 0. kg of water, and 0.35 kg of volatile (gas-forming) material. Using this information, determine how many kilograms of carbon are in 125 kg of this coal.

(125 𝑘𝑔 𝑐𝑜𝑎𝑙) (1.30 𝑘𝑔 𝑐𝑎𝑟𝑏𝑜𝑛2.00 𝑘𝑔 𝑐𝑜𝑎𝑙 ) = 81.3 𝑘𝑔 𝑐𝑎𝑟𝑏𝑜𝑛, 𝑜𝑟 8.13 𝑥 10^1 𝑘𝑔 𝑐𝑎𝑟𝑏𝑜𝑛

1. An atom of gold (Au) has a mass of 3.271 x 10-22^ g. How many atoms of gold are in 5.00 g of gold?

) = 1.53 𝑥 10^22 𝑎𝑡𝑜𝑚𝑠 𝐴𝑢

Practice Problems

1. How many minutes are there in exactly one week?

(1 𝑤𝑒𝑒𝑘) (7 𝑑𝑎𝑦𝑠1 𝑤𝑒𝑒𝑘) (24 ℎ𝑜𝑢𝑟𝑠1 𝑑𝑎𝑦 ) (60 𝑚𝑖𝑛𝑢𝑡𝑒𝑠1 ℎ𝑜𝑢𝑟 ) = 10, 080 𝑚𝑖𝑛𝑢𝑡𝑒𝑠, 𝑜𝑟 1.008 𝑥 10^4 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

1. How many seconds are there in exactly a 40 hour work week?

(40 ℎ𝑜𝑢𝑟𝑠) (60 𝑚𝑖𝑛𝑢𝑡𝑒𝑠1 ℎ𝑜𝑢𝑟 ) ( (^) 1 𝑚𝑖𝑛𝑢𝑡𝑒60 𝑠 ) = 144,000 𝑠𝑒𝑐𝑜𝑛𝑑𝑠, 𝑜𝑟 1.44 𝑥 10^5 𝑠𝑒𝑐𝑜𝑛𝑑𝑠

Practice Problems

1. Gold has a density of 19.3 g/ cm^3. What is the density in kilograms per cubic meter?

(19.3 𝑔 𝐴𝑢𝑐𝑚 3 ) ( (^) 1000 𝑔1 𝑘𝑔) (1,000,000 𝑐𝑚

3 1 𝑚^3 ) = 19,^

𝑘𝑔 𝑚^3 𝑜𝑟 1.93 𝑥 10

4 𝑘𝑔 𝑚^3

Section Review 4.

1. How can you solve a complicated problem more easily?

First analyze the problem by listing the knowns and unknowns. Then consider conversion

factors, ratios, and unit relationships and set up the problem (always including units in

intermediate steps), and calculate. Finally, evaluate your answer, does it make sense?

1. How are complex units dealt with in calculations?

You may want to use many conversion factors. Again, include units always in those

intermediate steps, treat units with as much importance as the numbers themselves.

1. Convert the following. Express your answers in scientific notation.

a. 7.5 x 10^4 nm to kilometers (7.5 𝑥 10^4 𝑛𝑚) ( (^) 1 𝑥 101 𝑚 (^9) 𝑛𝑚) ( (^) 1,000 𝑚1 𝑘𝑚 ) = 7.5 𝑥 10−8^ 𝑘𝑚

b. 3.9 x 10^5 mg to decigrams (3.9 𝑥 10^5 𝑚𝑔) ( 1 𝑔 1000 𝑚𝑔) (

10 𝑑𝑔 1 𝑔 ) = 3.9 𝑥 10

c. 0.764 km to centimeters (0.764 𝑘𝑚) (1000 𝑚1 𝑘𝑚 ) (100 𝑐𝑚1 𝑚 ) = 7.64 𝑥 10^4 𝑐𝑚

d. 2.21 x 10-4^ dL to microliters (2.21 𝑥 10−4𝑑𝐿) ( (^) 10 𝑑𝐿1 𝐿) (1 𝑥 10

(^6) 𝜇𝐿 1 𝐿 ) = 2.21 𝑥 10

(^1) 𝜇𝐿

1. Light travels at a speed of 3.00 x 10^10 cm/ s. What is the speed of light in kilometers per hour?

(3.00 𝑥 10

(^10) 𝑐𝑚 𝑠 ) (^

60 𝑠 1 𝑚𝑖𝑛) (

60 𝑚𝑖𝑛 1 ℎ𝑟 ) (^

1 𝑚 100 𝑐𝑚) (^

1 𝑘𝑚 1000 𝑚) = 1.08 𝑥 10

9 𝑘𝑚 ℎ𝑟

Chapter 4 Review

1. A volume of 5.00 mL of mercury (Hg) is added to a beaker that has a mass of 87.3 g. What is the mass of the beaker with the added mercury? 4.

First calculate the mass of Hg added: (5.00 𝑚𝐿) (1 𝑐𝑚

3 1 𝑚𝐿 ) (

13.56 𝑔 𝐻𝑔 𝑐𝑚^3 ) = 67.8 𝑔 𝐻𝑔

Then, add the mass of 67.8 g Hg to that of the beaker: 87.3 g + 67.8 g = 155 g.

1. What is the name given to a ratio of two equivalent measurements? 4.

Conversion Factor

1. Earth is approximately 1.5 x 10^8 km from the sun. How many minutes does it take light to travel from the sun to Earth? The speed of light is 3.00 x 10^8 m/s.

(1.5 𝑥 10^8 𝑘𝑚) (1000 𝑚1 𝑘𝑚 ) ( (^) 3.00 𝑥 10𝑠 (^8) 𝑚) (1 𝑚𝑖𝑛60 𝑠 ) = 8.2 𝑚𝑖𝑛

1. Choose the term that best completes the second relationship.

a. journey: route problem: ____________

(1) unknown (3) known

(2) plan (4) calculate

b. meter: 100 cm gram: _____________

(1) 0.001 kL (3) 1000 mg

(2) 100 cm (4) 100 kg

1. The density of dry air at 20 ̊ C is 1.20 g/ L. What is the mass of air, in kilograms, of a room that measures 25.0 m by 15.0 m by 4.0 m?

First calculate the volume of the room: (25.0 m) (15.0 m) (4.0 m) = 1,500 m^3

(1,500 𝑚^3 ) (1 𝑥 10

(^6) 𝑐𝑚 3 1 𝑚^3 ) (^

1 𝑚𝐿 1 𝑐𝑚^3 ) (^

1 𝐿 1000 𝑚𝐿) (

1.20 𝑔 𝑑𝑟𝑦 𝑎𝑖𝑟 1 𝐿 ) (^

1 𝑘𝑔 1000 𝑔) = 1.8 𝑥 10