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dedylja [7]
3 years ago
14

Water’s heat of vaporization is 2,257 joules/gram. How much energy is released when 11.2 grams of water vapor condenses into liq

uid?
A.
202 J
B.
2,246 J
C.
3,840 J
D.
9,950 J
E.
25,300 J
Chemistry
1 answer:
atroni [7]3 years ago
6 0

Heat of vaporization is the heat required to vaporize unit mass of a substance at it's boiling point. The heat of vaporization or condensation of water is 2257 J/g.

It means that 1 g of water involves 2257 J energy during condensation or vaporization at its boiling point (100 C)

Given mass of water vapor that is condensing = 11.2 g

Calculating the heat released when 11.2 g of water vapor condenses in to liquid water at 100^{0}C:

11.2 g *2257\frac{J}{g} = 25,278 J

Rounding the final answer 25,278 J to three significant figures we get, 25,300 J

Therefore the correct answer is E.  25,300 J


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A flask with a volume of 125.0mL contains air with a density of 1.269 g/L. What is the mass of the air contained in the flask?
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3 years ago
55 L of a gas at 25oC has its temperature increased to 35oC. What is its new volume?
ladessa [460]

Answer:

Approximately 56.8 liters.

Assumption: this gas is an ideal gas, and this change in temperature is an isobaric process.

Explanation:

Assume that the gas here acts like an ideal gas. Assume that this process is isobaric (in other words, pressure on the gas stays the same.) By Charles's Law, the volume of an ideal gas is proportional to its absolute temperature when its pressure is constant. In other words

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1},

where

  • V_2 is the final volume,
  • V_1 is the initial volume,
  • T_2 is the final temperature in degrees Kelvins.
  • T_1 is the initial temperature in degrees Kelvins.

Convert the temperatures to degrees Kelvins:

T_1 = \rm 25^{\circ}C = (25 + 273.15)\; K = 298.15\; K.

T_2 = \rm 35^{\circ}C = (35 + 273.15)\; K = 308.15\; K.

Apply Charles's Law to find the new volume of this gas:

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 55\;L \times \frac{308.15\; K}{298.15\; K} = 56.8\; L.

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