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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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Explanation:

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Answer:

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2 years ago
a sample of an oxide of antimony (sb) contain 19.75 g of antimony combine with 6.5 g of oxygen . what is the simplest formula fo
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Explanation:

The given data is as follows.

      Mass of antimony = 19.75 g

      Molar mass of Sb = 121.76 g/mol

Therefore, calculate number of moles of Sb as follows.

                    Moles of Sb = \frac{mass}{\text{molar mass}}

                                         = \frac{19.75 g}{121.76 g/mol}

                                         = 0.162 mol

Mass of oxygen given is 6.5 g and molar mass of oxygen is 16 g/mol. Hence, moles of oxygen will be calculated as follows.

           Moles of oxygen = \frac{mass}{\text{molar mass}}

                                         = \frac{6.5 g}{16 g/mol}

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Hence, ratio of moles of Sb and O will be as follows

                          Sb : O

                      \frac{0.162}{0.162} : \frac{0.406}{0.162}

                           1 : 2.5

We multiply both the ratio by 2 in order to get a whole number. Therefore, the ratio will be 2 : 5.

Thus, we can conclude that the empirical formula of the given oxide is Sb_{2}O_{5}.

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3 years ago
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