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Sveta_85 [38]
2 years ago
12

50.0 g of nitrogen gas (N2 ) are kept under pressure in a 3.00 L container. If the pressure is kept constant, how many moles of

nitrogen gas are present in the container if gas is added until the volume has increased to 5.00 L?
Chemistry
1 answer:
kotykmax [81]2 years ago
7 0

The resultant moles of nitrogen gas that are present in the container is 2.96 moles.

<h3>How do we convert mass into moles?</h3>

Mass of any substance will be converted into moles by using the below equation as:

n = W/M, where

  • W = given mass of nitrogen gas = 50g
  • M = molar mass of nitrogen gas = 28 g/mol
  • n = 50 / 28 = 1.78 moles

From the ideal gas equation of gas we know that moles and volume is directly proportional to each other and for this question equation becomes,

V₁/n₁ = V₂/n₂, where

  • V₁ & n₁ are the initial volume and moles of gas
  • V₂ & n₂ are the final volume and moles of gas.

On putting all values, we get

n₂ = (5L)(1.78mol) / (3L) = 2.96 moles

Hence required moles of nitrogen gas are 2.96.

To know more about ideal gas equation, visit the below link:

brainly.com/question/1056445

#SPJ1

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Formula to find the percent composition of an element in a compound is \frac {MolarMassOfElement}{MolarMassOfCompound}*100.

Now enter the variables into the formula.

\frac {12.4}{78.0}*100 = 15.897%
Round to 3 significant figures. 15.897% = 15.9%

The answer is 15.9% H
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Sulfuric acid dissolves aluminum metal according to the following reaction:
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Answer:

m_{H_2SO_4}=81.7gH_2SO_4

m_{H_2}=1.67gH_2

Explanation:

Hello,

Based on the given undergoing chemical reaction is is rewritten below:

2Al (s) + 3H_2SO_4 (aq)\rightarrow  Al _2(SO4)_3 (aq) + 3H_2 (g)

By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:

m_{H_2SO_4}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2SO_4}{2molAl}*\frac{98gH_2SO_4}{1molH_2SO_4} \\m_{H_2SO_4}=81.7gH_2SO_4

Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:

m_{H_2}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2}{2molAl}*\frac{2gH_2}{1molH_2} \\m_{H_2}=1.67gH_2

Best regards.

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