What is the volume of 5.8g of H2?

1 answer:

4 0

At standard temperature and pressure, one mole of gas occupies 22.4 L. Knowing this, we must convert the grams to moles and moles to liters to get your answer. Here's how you do it!
$\frac{5.8gH2}{1} | \frac{1moleH2}{2.0gH2}| \frac{22.4LH2}{1moleH2} \\ \frac{5.8*1*22.4LH2}{2.0} \\ \frac{129.92LH2}{2.0} \\ 64.96LH2 \\ 65LH2$
With significant figures accounted for, 5.8 grams of hydrogen gas occupies 65 L.

Hope this helps!

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Natasha2012 [34]

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is: