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beks73 [17]
3 years ago
7

What is the volume of 5.8g of H2?

Chemistry
1 answer:
BartSMP [9]3 years ago
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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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:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

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

7 0
2 years ago
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Answer:

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Explanation:

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f=\dfrac{n}{t}\\\\f=\dfrac{3}{5}\\\\f=0.6\ Hz

So, the frequency of a wave is 0.6 Hz.

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