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3 years ago

CALCULATOR

Chemistry

1 answer:

suter [353]3 years ago

5 0

Answer: The possible molecular formula will be $CO_2$

Explanation:

Mass of C= 27.3 g

Mass of O = 72.7 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{27.3g}{12g/mole}=2.275moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{72.7g}{16g/mole}=4.544moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{2.275}{2.275}=1$

For O = $\frac{4.544}{2.275}=2$

The ratio of C : O = 1: 2

Hence the empirical formula is $CO_2$

The possible molecular formula will be= $n\times CO_2$

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a 5L container contains 3 moles of helium and 4 moles of hydrogen at a pressure of 9 atms maintaining a constant T and additiona

Stells [14]

Answer:

7.71 atm

Explanation:

Given the following data:

$V = 5 L$

$n_{He} = 3 mol$

$n_{H_2} = 4 mol$

$p_1 = 9 atm$

$T = const$

According to the ideal gas law, we know that the product between pressure and volume of a gas is equal to the product between moles, the ideal gas law constant and the absolute temperature:

$pV = nRT$

Since the temperature and the ideal gas constant are constants, as well as the fixed container volume of 5 L, we may rearrange the equation as:

$\frac{p}{n}=\frac{RT}{V}=const$

This means for two conditions, we'd obtain:

$\frac{p_1}{n_1}=\frac{p_2}{n_2}$

Given:

$p_1 = 9 atm$

$n_1 = n_{initial total} = n_{He} + n_{H_2} = 3 mol + 4 mol = 7 mol$

$n_2 = n_{final total} = n_{He} + n_{H_2} = 3 mol + 4 mol + 2 mol = 9 mol$

Solve for the final pressure:

$p_2 = p_1\cdot \frac{n_2}{n_1}$

Now, according to the Dalton's law of partial pressures, the partial pressure is equal to the total pressure multiplied by the mole fraction of a component:

$p_{H_2}=\chi_{H_2}p_2$