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2 years ago

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An empty water bottle is full of air at 15°C and standard pressure. The volume of the bottle is 0. 500 liter. How many moles of

air are in the bottle? The water bottle contains mole of air.

1 answer:

drek231 [11]2 years ago

5 0

The moles of gas in the bottle has been 0.021 mol.

The ideal gas has been given as the gas where there has been negligible amount of interatomic collisions. The ideal gas equation has been given as:

$PV=nRT$

<h3>Computation for the moles of gas</h3>

The gi<em>ve</em>n gas has standard pressure, $P=1\rm atm$

The volume of the gas has been, $V= 0. 500 \;\rm L$

The temperature of the gas has been, $T=15^\circ \text C\\
T=288\;\rm K$

Substituting the values for the moles of gas, <em>n:</em>

<em />

<em /> $\rm 1\;\times\; 0. 500 =\textit n\;\times\;0.08214\;atm.L/mol.K\;\times\;288\;K\\\\
\textit n=\dfrac{0. 500}{0.08214\;\times\;288} \;mol\\\\
\textit n=0.021\;mol$

The moles of gas in the bottle has been 0.021 mol.

Learn more about ideal gas, here:
brainly.com/question/8711877

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1 year ago

Determine the empirical formulas for compounds with the following percent compositions:

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<u>Answer:</u>

<u>For a:</u> The empirical formula of the compound is $P_2O_5$

<u>For b:</u> The empirical formula of the compound is $KH_2PO_4$

<u>Explanation:</u>

<u>For a:</u>

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.

For Phosphorus = $\frac{1.406}{1.406}=1$

For Oxygen = $\frac{3.525}{1.406}=2.5$

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = $1\times 2=2$

For Oxygen = $2.5\times 2=5$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is $P_2O_5$

<u>For b:</u>

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Potassium = $\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles$

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.

For Potassium = $\frac{0.736}{0.735}=1$

For Hydrogen = $\frac{1.5}{0.735}=2.04\approx 2$

For Phosphorus = $\frac{0.735}{0.735}=1$

For Oxygen = $\frac{2.9375}{0.735}=3.99\approx 4$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is $KH_2PO_4$

3 0

3 years ago