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Tanzania [10]
3 years ago
15

A sample of gas in a balloon has an initial temperature of 21 ∘C and a volume of 1310 L . If the temperature changes to 70 ∘C ,

and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas? Express your answer with the appropriate units.
Chemistry
1 answer:
bixtya [17]3 years ago
7 0

Answer:

1528.3L

Explanation:

To solve this problem we should know this formula:

V₁ / T₁ = V₂ / T₂

We must convert the values of T° to Absolute T° (T° in K)

21°C + 273 = 294K

70°C + 273 = 343K

Now we can replace the data

1310L / 294K = V₂ / 343K

V₂ = (1310L / 294K) . 343K → 1528.3L

If the pressure keeps on constant, volume is modified directly proportional to absolute temperature. As T° has increased, the volume increased too

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3 years ago
What is the molarity of solution that is 5.50 percentage by mass oxalic acid and has a density of 1.024 g/ml
Y_Kistochka [10]

Answer:

0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.

Explanation:

Mass percentage of oxalic acid = 5.50%

This means that in 100 grams of solution there are 5.50 grams of oxalic acid.

Mass of solution , m = 100

Volume of the solution = V

Density of the solution = d = 1.024 g/mL

V=\frac{m}{d}=\frac{100 g}{1.024 g/mL}=97.66mL

V = 97.66 mL = 0.09766 L

(1 mL = 0.001 L)

Moles of oxalic acid = \frac{5.50 g}{90 g/mol}=0.06111 mol

Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}

The molarity of the solution :

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0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.

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