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kumpel [21]
3 years ago
14

Plumbers, welders, or glass blowers often use the gas acetylene (C2H2) because it burns in oxygen with a very hot flame. The pro

ducts of the combustion of acetylene are carbon dioxide and water vapor. Write the unbalanced chemical equation for this process. (Include states-of-matter under the given conditions in your answer.)
Chemistry
1 answer:
solmaris [256]3 years ago
5 0

Answer:

Unbalanced equation: C_{2}H_{2}(g)+O_{2}(g)\rightarrow CO_{2}(g)+H_{2}O(g)

Explanation:

  • Acetylene reacts with oxygen to produce carbon dioxide and water vapor.
  • Here reactants are acetylene and oxygen. Here products are carbon dioxide and water vapor.
  • We know that physical state of water vapor is gas. Therefore each and  every reactants and products are gaseous.
  • States of the species in unbalanced equation is represented inside parentheses with an abbreviation "g" for gaseous state.
  • Unbalanced equation: C_{2}H_{2}(g)+O_{2}(g)\rightarrow CO_{2}(g)+H_{2}O(g)
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Answer:

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Explanation:

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its molecular formula can also be written as CrH3O12S3

molar mass of Cr (HSO4)3 can be calculated by following method;

atomic mass of Cr = 51.9961 u

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molar mass of Cr(HSO4)3 =  51.9961+ 1.00784×3 + 32.065×3 + 15.999×12

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3 years ago
Assume that silver and gold form ideal, random mixtures. Calculate the mass of pure Ag needed to cause an entropy increase of 20
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Answer:

m_{Ag}=2,265.9g

Explanation:

Hello!

In this case, since the definition of entropy in a random mixture is:

\Delta S=-n_TR\Sigma[x_i*ln(x_i)]

For this silver-gold mixture we write:

\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]

By knowing the moles of gold:

n_{Au}=100g*\frac{1mol}{197g} =0.508mol

It is possible to write the aforementioned formula in terms of the variable x representing the moles of silver:

20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]

Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:

x=n_{Ag}=21.0molAg

So the mass is:

m_{Ag}=21.0mol*\frac{107.9g}{1mol}\\ \\m_{Ag}=2,265.9g

Best regards!

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