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julia-pushkina [17]
3 years ago
8

An experiment shows that 113 mL gas sample at pressure of 721mmHg changed to 901 mmHg.

Chemistry
1 answer:
maria [59]3 years ago
5 0

Answer:

90.4 mL

Explanation:

Step 1: Given data

  • Initial volume of the gas (V₁): 113 mL
  • Initial pressure of the gas (P₁): 721 mmHg
  • Final volume of the gas (V₂): ?
  • Final pressure of the gas (P₂): 901 mmHg

Step 2: Calculate the final volume of the gas

According to Boyle's law, the volume of a gas is inversely proportional to the pressure. We can calculate the final volume of the gas using the following expression.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 721 mmHg × 113 mL / 901 mmHg

V₂ = 90.4 mL

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T2  is  therefore  =  P2V2T1/P1V1
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How many grams of H2S is needed to produce 18.00g of PbS if the H2S is reacted with an
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Answer:

2.56 grams of H₂S is needed to produce 18.00g of PbS if the H2S is reacted with an  excess (unlimited) supply of Pb(CH₃COO)₂

Explanation:

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Pb(CH₃COO)₂ + H₂S → 2 CH₃COOH + PbS

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) they react and produce:

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In this case,  to know how many grams of H₂S are needed to produce 18.00 g of PbS, it is first necessary to know the molar mass of the compounds H₂S and PbS and then to know how much it reacts by stoichiometry. Being:

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The molar mass of the compounds are:

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So, by stoichiometry they react and are produced:

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Then the following rule of three can be applied: if 239 grams of PbS are produced by stoichiometry from 34 grams of H₂S, 18 grams of PbS from how much mass of H₂S is produced?

mass of H_{2} S=\frac{18 grams of PbS*34 grams of H_{2}S }{239 grams of PbS}

mass of H₂S= 2.56 grams

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