When 5.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at

Question

When 5.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling 178) point of 80.1°C, 169.5 kJ are absorbed and PV for the vaporization process is equal to 14.5 kJ then

A) E = 155.0 kJ and H = 169.5 kJ.

B) E = 169.5 kJ and H = 155.0 kJ.

C) E = 184.0 kJ and H = 169.5 kJ.

D) E = 169.5 kJ and H = 184.0 kJ.

179) When 10.00 moles of H2(g) reacts with 5.000 mol of O2(g) to form 10.00 mol of H2O(l) at 25°C and 179) a constant pressure of 1.00 atm. If 683.0 kJ of heat are released during this reaction, and PV is
equal to -37.00 kJ, then

A) H° = +683.0 kJ and E° = +646.0 kJ.

B) H° = -683.0 kJ and E° = -646.0 kJ.

C) H° = +683.0 kJ and E° = +720.0 kJ.

D) H° = -683.0 kJ and E° = -720.0 kJ.

Zinc reacts with aqueous sulfuric acid to form hydrogen gas: 156) Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)

In an experiment, 201 mL of wet H2 is collected over water at 27°C and a barometric pressure of

733 torr. The vapor pressure of water at 27°C is 26.74 torr. The partial pressure of hydrogen in this experiment is ________ atm.

A) 706 B) 0.929 C) 0.964 D) 1.00 E) 760

In the laboratory, hydrogen gas is usually made by the following reaction:

Zn(s) + 2 HCl(aq) H2(g) + ZnCl2(aq)

How many liters of H2 gas, collected over water at an atmospheric pressure of 752 mm Hg and a temperature of 21.0°C, can be made from 3.566 g of Zn and excess HCl? The partial pressure of water vapor is 18.65 mm Hg at 21.0 °C.

A) 1.36 L B) 0.0975 L

C) 1.30 L D) 1.33 L

Thanks ( Show the calculations please)

Explanation / Answer

178) option A is correct.

179) option B is correct

option B is correct

178) heat absorbed hence delta H is positive

then substitute delta H = delta u + P delta V ------ (1)

we get answer

179) heat released hence delta H is negative

then substitute above equation 1 we get answer

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