Fundamentals of General, Organic, and Biological Chemistry 8th edition by John E. McMurry test bank

Diff: 2
Section:  1.11
LO:  1.11
Global LO:  G4
111) Why is the number 1.8 (and not some other value) used in the formula for converting between Celsius and Fahrenheit temperatures?
Answer:  This value compensates for the different size of the two degrees. In the Celsius scale there are 100 degrees between the freezing point and the boiling point of water, but in the Fahrenheit scale there are 180 degrees to cover the same interval. The ratio of 180 to 100 is 1.8, so this correction factor is used in the formula.
Diff: 2
Section:  1.11
LO:  1.11
Global LO:  G4
 
112) Why is the number 273.15 (and not some other value) used in the formula for converting between Celsius and Kelvin temperatures?
Answer:  This value compensates for the fact that the freezing point of water on the Celsius scale is 273.15 degrees lower than on the Kelvin scale. Since kelvins are the same size as Celsius degrees, no other correction factors are needed.
Diff: 2
Section:  1.11
LO:  1.11
Global LO:  G4
 
113) How many calories are released when 500 g of water cools from 95.0°C to 25.0°C?
A) 35.0 cal
B) 70.0 cal
C) 1.25 ×  cal
D) 3.50 ×  cal
E) 4.75 ×  cal
Answer:  D
Diff: 1
Section:  1.11
LO:  1.12
Global LO:  G4
 

114) If 55.0 g of olive oil has 877 cal of heat added to it at a room temperature of 26.0°C, what will be the final temperature of the olive oil?  The specific heat of olive oil is 2.19 cal/g°C.
A) 33.3°C
B) 7.3°C
C) 19.0°C
D) 26.0°C
E) 3.3 °C
Answer:  A
Diff: 3
Section:  1.11
LO:  1.12
Global LO:  G4
115) If 75.0 g of water at 30.0°C absorbs 900 calories, the new temperature will be
A) 18.0°C.
B) 22.0°C.
C) 42.0°C.
D) 105°C.
E) 160°C.
Answer:  C
Diff: 3
Section:  1.11
LO:  1.12
Global LO:  G4
 
116) What is the specific heat of a metal if it takes 26.5 calories to raise the temperature of a piece weighing 50.0 g by 5.00 degrees Celsius?
A) 250 cal/g °C
B) 133 cal/g °C
C) 6.63 cal/g °C
D) 1.89 cal/g °C
E) 0.106 cal/g °C
Answer:  E
Diff: 3
Section:  1.11
LO:  1.12
Global LO:  G4
 

117) What is the specific heat of a metal if it takes 48.4 calories to raise the temperature of a 45.0 g sample by 5.0°C?
A) 0.186 cal/g °C
B) 0.215 cal/g °C
C) 5.34 cal/g °C
D) 225 cal/g °C
E) 242 cal/g °C
Answer:  B
Diff: 3
Section:  1.11
LO:  1.12
Global LO:  G4
 
118) What is the density of a 6.0 × 102 mL liquid sample that weighs 450 g?
A) 1050 g/mL
B) 270 g/mL
C) 1.33 g/mL
D) 0.75 g/mL
E) 0.37 g/mL
Answer:  D
Diff: 1
Section:  1.12
LO:  1.13
Global LO:  G4
119) Calculate the density of cyclohexane if a 50.0 g sample has a volume of 64.3 mL.
A) 114.3 g/mL
B) 14.3 g/mL
C) 1.29 g/mL
D) 0.778 g/mL
E) 0.322 g/mL
Answer:  D
Diff: 1
Section:  1.12
LO:  1.13
Global LO:  G4
 

120) A 35.0 mL sample of a liquid weighs 27.2 g. What is the density of the liquid?
A) 62.2 g/mL
B) 7.80 g/mL
C) 1.29 g/mL
D) 0.952 g/mL
E) 0.777 g/mL
Answer:  E
Diff: 1
Section:  1.12
LO:  1.13
Global LO:  G4
 
121) What is the specific gravity of a liquid sample with a mass of 35.0 g and a volume of 14.00 mL?
A) 14.0 g/mL
B) 2.50 g/mL
C) 21.0
D) 14.0
E) 2.50
Answer:  E
Diff: 1
Section:  1.12
LO:  1.13
Global LO:  G4
 
122) What is the volume of a gold nugget that weighs 2.20 kg? The density of gold is
A) 8.60 ×  
B) 116
C) 11.6
D) 8.60
E) 0.116
Answer:  B
Diff: 2
Section:  1.12
LO:  1.13
Global LO:  G4

123) What is the mass of 30.0 mL of a solution with a density of 1.60 g/mL?
A) 53.3 g
B) 48.0 g
C) 31.6 g
D) 18.8 g
E) none of the above
Answer:  B
Diff: 2
Section:  1.12
LO:  1.13
Global LO:  G4
 
124) A 2.36  sample of an unknown metal weighs 18.5 g.  What is the sample's density?
A) 7.84 g/
B) 0.127 /g
C) 15.75 g/
D) 0.784 g/
E) 1.27 /g
Answer:  A
Diff: 2
Section:  1.12
LO:  1.13