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Flura [38]
3 years ago
7

What happens to the entropy of a sample of matter when it changes state from a solid to a liquid?

Chemistry
1 answer:
Sidana [21]3 years ago
5 0

Answer:

  • <em>When a sample of matter changes state from a solid to a liquid, its entropy</em> <u>increases</u>.

Explanation:

The <em>entropy</em> is a thermodinamic variable that measures the disorder of radomness of a substance or system.

In the <em>solid </em>stated the particles (molecules, atoms or ions) are tightly packed, forming an organized structure. So, the particles barely can move, just vibrating around their fixed positions.

In the <em>liquid </em>state the particles are more free to move (that is why liquids flow and take the shape of the vessel in which they are contained).

Then, you can predict that <em>when a matter changes state from a solid to a liquid</em>, the parciles will gain motion leading to a higher disorder, which means that the entropy increases.

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Part C<br> Number of molecules in 8.437x10-2 mol C6H6
N76 [4]

Answer:

There are 5.08\times 10^{22}\ \text{molecules of}\ C_6H_6  

Explanation:

In this problem, we need to find the number of molecules in 8.437\times 10^{-2} mol of C_6H_6.

The molar mass of C_6H_6 is 6\times 12+1\times 6=78\ g/mol

No of moles = mass/molar mass

We can find mass from above formula.

m=n\times M\\\\m=8.437\times 10^{-2}\ mol\times 78\ g/mol\\\\m=6.58\ g

Also,

No of moles = no of molecules/Avogadro number

N=n\times N_A\\\\N=8.437\times 10^{-2}\times 6.023\times 10^{23}\\\\N=5.08\times 10^{22}\ \text{molecules}

Hence, there are 5.08\times 10^{22}\ \text{molecules of}\ C_6H_6  

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3 years ago
What is the percent yield if 107.50 g NH3 reacts with excess O2 according to the
Maksim231197 [3]

Answer:

81.59%

Explanation:

  • 4NH₃ + 5O₂ → 4NO + 6H₂O

First we <u>convert 107.50 g of NH₃ into moles</u>, using its <em>molar mass</em>:

  • 107.50 g NH₃ ÷ 17 g/mol = 6.32 mol NH₃

Now we <u>calculate how many moles of NO would have been formed by the complete reaction of 6.32 moles of NH₃</u>:

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Then we <u>convert 6.32 moles of NO to grams</u>, using its <em>molar mass</em>:

  • 6.32 mol NO * 30 g/mol = 189.60 g NO

Finally we <u>calculate the percent yield</u>:

  • 154.70 g / 189.60 g * 100% = 81.59%

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Vinvika [58]

Answer:

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