Review Unit 11

Calculate the amount of energy required to melt 35.0 grams of ice.
(not in sig figs)  H_f = 80.0 cal/g,  H_v = 560.0 cal/g, S_H = 1.00 cal/g ºC

• 19,600 cal
• 35.0 cal
• 2,800 cal 
• 1,568,000 cal 

If a 8.50g ice cube at -10^o C sits out on the counter, completely melts, and then warms to room temperature (25^o C) how much energy did the ice cube absorb? Use C = 2.108 J/g^oC for ice, C = 4.184 J/g^oC for water, and H_f = 334 J/g.

• 179.18 J
• 2839 J
• 889.1 J
• 3907.3 J

Describe the substance between letters C and D. 

• Gas
• Liquid
• Melting
• Evaporating

Between which points is the temperature of the substance remaining constant?

• A-B only. 
• A-B, C-D, E-F
• B-C only. 
• B-C, D-E

What is the melting point for this substance?

• 60 °C
• 40 °C
• 80 °C
• 100 °C

What letter on the diagram represents the triple point?

• A

• B

• C

• D

What is the melting point of this substance at 1 ATM of pressure?

• 40

• 60

• 100

• 110

What would the state of this substance be at 0.5 ATM of pressure at a 100 degrees Celsius?

• Solid

• Liquid

• Gas

• Michigan

Based on what you have learned... Is this substance likely water?

• Yes

• No

Which substance would have the highest boiling point at standard pressure?

• H₂O

• CCl₄

• CO₂

• NBr₃