Choose the correct statement regarding the behavior of water.

Chemistry

1 answer:

AfilCa [17]3 years ago

5 0

Answer:

d. The energy required to evaporate 1 kg of liquid water equals the energy released when 1 kg of water vapor condenses into liquid.

Explanation:

Hello,

Since we're considering the same amount of water, the vapor phase has a higher energy content than the liquid phase, thus, for the specified amount of water particles (those contained in the given 1 kg) the energy MUST be same when taking them either to a gaseous phase or to a liquid phase, the only difference is the sign which is negative from gaseous to liquid (heat withdrawal) and positive from liquid to gaseous (heat adding).

Best regards.

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A sample of neon occupies a volume of 752 ml at 25 0

lyudmila [28]

$V_{initial} = 752\:mL$
$T_{initial} = 25.0^0C$
converting to Kelvin
TK = TC + 273
TK = 25.0 + 273 → TK = 298.0 → $T_{initial} = 298.0\:K$
$V_{final} = ? (in\:milliliters)$
$T_{final} = 50.0^0C$
TK = TC + 273
TK = 50.0 + 273 → TK = 323.0 → $T_{final} = 323.0\:K$

By the first Law of Charles and Gay-Lussac, we have:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$

Solving:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$
$\frac{ 752 }{ 298.0 } = \frac{ V_{f} }{ 323.0 }$
Product of extremes equals product of means:
$298.0* V_{f} = 752*323.0$
$298.0 V_{f} = 242896$
$V_{f} = \frac{242896}{298.0}$
$\boxed{\boxed{V_{f} \approx 815.08\:mL}}\end{array}}\qquad\quad\checkmark$