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trasher [3.6K]
2 years ago
12

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

e of the gas from 43 atm to 17 atm. Assume that the volume of the tank and the temperature of the oxygen are constant during this operation. Answer in units of g.
Chemistry
1 answer:
dsp732 years ago
7 0

Answer:

12.8 g of O_{2} must be withdrawn from tank

Explanation:

Let's assume O_{2} gas inside tank behaves ideally.

According to ideal gas equation- PV=nRT

where P is pressure of O_{2}, V is volume of O_{2}, n is number of moles of O_{2}, R is gas constant and T is temperature in kelvin scale.

We can also write, \frac{V}{RT}=\frac{n}{P}

Here V, T and R are constants.

So, \frac{n}{P} ratio will also be constant before and after removal of O_{2} from tank

Hence, \frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}

Here, \frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm} and P_{after}=17atm

So, n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol

So, moles of O_{2} must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of O_{2} = 32 g/mol

So, mass of O_{2} must be withdrawn = (32\times 0.40)g=12.8g

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Answer:

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Explanation:

A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).

We use the formula for dilutions:

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