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A sample containing 4.30 g of O2 gas has an initial volume of 13.0 L. What is the final volume, in liters, when each of the following changes occurs in the quantity of the gas at constant pressure and temperature.

A sample of 200 g of O, is removed from the 4.30 g of O, in the container.​

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Respuesta :

Answer:

Total amount of Oâ‚‚ occupy volume = 76.50 L

2 .30 gm Oâ‚‚ occupy volume = 7.48 L

Explanation:

We konw that,

Molar mass of Oâ‚‚ = 31.9988 gm / mol

1 mole of Oâ‚‚ = 31.9988 gm

Sample of oxygen = 0.600 mole

Computation:

0.600 mole = 31.9988 × 0.600

0.600 mole = 19.199 gm

Total amount of Oâ‚‚ = 4.30 + 19.199

Total amount of Oâ‚‚ = 23.499 gm

Total amount of O₂ occupy volume = 14 × 23.499/4.30

Total amount of Oâ‚‚ occupy volume = 76.50 L

Total gm of Oâ‚‚ = 4.30 - 2.00= 4.30 gm

2 .30 gm O₂ occupy volume = 14 × 2.30/4.30

2 .30 gm Oâ‚‚ occupy volume = 7.48 L

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