CHEM 1001/CHM 1020/CHM 1025C/CHM 1032C

Module Two Sample Pretest: Part M Multiple Choice 10 points

 

1 . The density of a silver coin can be calculated from the following data:

Mass of a silver coin

6.5081 g

Volume of the coin and water

23.7 mL

Volume of the water alone

23.1 mL

The density of the coin should be reported as using strict rules of significant figures


 

a.

 

10.847 g/mL


 

b.

 

10.85 g/mL


 

c.

 

10.8 g/mL


 

d.

 

11 g/mL


 

e.

 

1 x 101 g/mL

2. What mass of sodium chloride is dissolved in 85.0 g of water to make a 15.0% solution?


 

a.

 

13.2 g


 

b.

 

15.0 g


 

c.

 

17.6 g


 

d.

 

18.0 g


 

e.

 

19.3 g

3. The number of significant figures in 0.0060501000 is

___

a

5

___

b

6

___

c

8

___

d

9

___

E

11

 

 

 

 

4. Multiply 1.05 cm times 2.2 cm and round off the product to the proper number of significant digits.

a) 2 cm2
b) 2.0 cm2
c) 2.3 cm2
d) 2.31 cm2
e) 3 cm2

 

5. One cubic centimeter of helium gas contains 26,900,000,000,000,000,000 atoms at standard conditions. Express this number of helium atoms in scientific notation.

a) 2.69 * 1017
b) 2.69 * 1018
c) 2.69 * 1019
d) 26.9 * 1018
e) 26.9 * 1019

 

6. According to the metric system, 1000 g = 1 ______.

a) cg
b) kg
c) mg
d) Mg
e) g ( stands for Greek letter Mu)

 

 

7. A tall glass cylinder contains four separate liquid layers: mercury (d = 13.6 g/mL), chloroform (d = 1.49 g/mL), acetic acid (d = 1.05 g/mL), ether (d = 0.708 g/mL). If ice

(d = 0.99 g/mL) is dropped into the cylinder, where does it come to rest?

a) on top of the mercury layer
b) on top of the chloroform layer
c) on top of the acetic acid layer
d) on top of the ether layer
e) on the bottom of the cylinder

 

 

 

 

8. The term that refers to the close clustering of similar laboratory measurements is


a.

precision.


b.

repeatability.


c.

accuracy.


d.

exactness.


e.

the number of significant figures

 

 

 

9. How many scruples are there in 100. lbs? Some equivalents which may be helpful are given below:

1.00 scruple = 20.0 grains
1.00 g = 15.4 grains
1.00 grain = 0.0648 g
1.00 lb = 453.6 g
1.00 kg = 2.205 lb


a.

1.00 scruple

x

1.00 grain

x

1.00 kg

x

100 lb

x

1000 g

x 100. lbs.

20.0 grains

0.0648 g

2.205 lb

453.6 g

1 Kg


b.

20.0 grains

x

1.00 g

x

1.00 grain

x

453.6 g

x 100. lbs.

1.00 scruple

15.4 grains

0.0648 g

1.00 lb


c.

1.00 scruple

x

15.4 grains

x

0.0648 g

x

453.6 g

x 100. lbs.

20.0 grains

1.00 g

1.00 grain

1.00 lb


d.

1.00 scruple

x

15.4 grains

x

453.6 g

x 100. lbs

20.0 grains

1.00 g

1.00 lb


e.

1.00 scruple

x

1.00 grain

x

1.00 Kg

x 100. lbs

20.0 grains

0.0648 g

2.205 lb



 

10. The average velocity of carbon dioxide molecules at 0oC is 3.62 x 104 cm/sec. Which of the indicated calculations below would convert this value to the velocity in miles per hour?


a.

1 sec

x

2.54 cm

x

12 in

x

5280 ft

x

1 hr

3.62 x 104 cm

1 in

1 ft

1 mi

3600 sec


b.

1 sec

x

2.54 cm

x

1 ft

x

1 mi

x

3600 sec

3.62 x 104 cm

1 in

12 in

5280 ft

1 hr


c.

3.62 x 104

cm

x

2.54 cm

x

12 in

x

5280 ft

x

3600 sec

sec

1 in

1 ft

1 mi

1 hr


d.

3.62 x 104

cm

x

1 in

x

12 in

x

5280 ft

x

1 hr

sec

2.54 cm

1 ft

1 mi

3600 sec


e.

3.62 x 104

cm

x

1 in

x

1 ft

x

1 mi

x

3600 sec

sec

2.54 cm

12 in

5280 ft

1 hr

 

 

 

 

ANSWERS: 1.this is tricky: 6.5081g/0.6mL = E

(five significant figures divided by one significant figure number equals one significant #answer technically)

2.b ; 3.c ; 4 c; 5.c; 6.b; 7.c; 8. a; 9. d; 10. e