Chemistry Unit 8 Test A

Consider two chemical reactions that can be combined to form a new reaction. If the Δ H values for these reactions are –500 kJ and +300 kJ, what is the Δ H for the resulting reaction?

• –800 kJ

• –200 kJ

• +200 kJ

• +800 kJ

Find the specific heat of a material if a 6-g sample absorbs 50 J when it is heated from 30°C to 50°C.

• 6000 J

• 6000 J/g°C

• 0.4 J

• 0.4 J/g°C

What is the amount of heat absorbed by 25. g of silver when it is heated from 20ºC to 30ºC? The specific heat of silver is 0.24 J/g•ºC.

• 60 J

• 60 J/g•ºC

• 0.0012 J

• 0.0012 J/g•ºC

What is the heat of formation of 1 mol of sulfur and 1 mol of oxygen to form SO₂?

• –593.6 kJ/mol

• –296.8 kJ/mol

• 0 kJ/mol

• +296.8 kJ/mol

Which of the following is a valid unit for specific heat (or specific heat capacity)?

• cal/g°C

• cal

• cal/g

• °C

Assume that 491.81 J of heat is added to 5.00 g of water originally at 23.0°C. What would be the final temperature of the water? (Specific heat capacity of water = 4.184 J/g°C.)

• 23.5 °C

• 61.5 °C

• 74.5 °C

• 46.5 °C

If 350.00 J of heat is added to 10.00 g of water initially at 15.0°C, what will be the final temperature of the water? (Specific heat capacity of water = 4.184 J/g°C.)

• 18.4 °C

• 23.4 °C

• 20.4 °C

• 25.4 °C

C(s) + O2(g) → CO2(g) ΔH = -393 kJ

2CO(g) + O2(g) → 2CO2(g) ΔH= -566 kJ

These two reactions will be combined to give a final reaction of 2C(s) + O2(g) → CO(g).

How could this be done?

• Reverse equation 1 and change ΔH to +393 kJ

• Double equation 1 and reverse equation 2

• Use equation 1 as is and keep ΔH as -393 kJ

• Reverse equation 1 and double it, changing ΔH to +786 kJ

Given the thermochemical equations

Pb+Cl2→PbCl2 ΔH=−223kJ

PbCl2+Cl2→PbCl4 ΔH=−87kJ

determine ΔH for

2PbCl2→Pb+PbCl4

• -310 kJ

• +310 kJ

• -136 kJ

• +136 kJ

What kind of equation is this: 2KI + MgCl₂ → 2KCl + MgI₂ + 200kJ

• Endothermic

• Exothermic