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exis [7]
3 years ago
8

Calculate the second volumes. 955 L at 58 C and 108 KPa to 76 C and 123 KPa

Chemistry
1 answer:
Dmitriy789 [7]3 years ago
3 0

The second volume :   V₂= 884.14 L

<h3>Further explanation </h3>

Given

955 L at 58 C and 108 KPa

76 C and 123 KPa

Required

The second volume

Solution

T₁ = 58 + 273 = 331 K

P₁ = 108 kPa

V₁ = 955 L

P₂ = 123 kPa

T₂ = 76 + 273 = 349

Use combine gas law :

P₁V₁/T₁ = P₂V₂/T₂

Input the value :

108 x 955/331 = 123 x V₂/349

V₂= 884.14 L

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Vitek1552 [10]

<u>Answer:</u> The mass of decane produced is 1.743\times 10^2g

<u>Explanation:</u>

To calculate the number of moles, we use the equation:  

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Mass of hydrogen gas = 2.45 g

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Putting values in equation 1:, we get:

\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol

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C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)

As, decene is present in excess. So, it is considered as an excess reagent.

Thus, hydrogen gas is a limiting reagent because it limits the formation of products.

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1 mole of hydrogen gas produces 1 mole of decane.

So, 1.225 moles of hydrogen gas will produce = \frac{1}{1}\times 1.225=1.225mol of decane

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Putting values in equation 1, we get:

1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g

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<u>Explanation:</u>

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