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xz_007 [3.2K]
2 years ago
12

The volume of a gas at 99.6 kPa and 24°C is 4.23 L. What volume will it occupy at 93.3 kPa ?

Chemistry
1 answer:
yuradex [85]2 years ago
6 0

The volume is increased to 4.52 L on decreasing the pressure to 93.3 kPa.

Explanation:

As per Boyle's law, the volume occupied by gas particles will be inversely proportional to the pressure experienced by those particles at constant temperature.

V=\frac{1}{P}

So in the present problem, the volume of gas at pressure P₁ = 99.6 kPa is given as V₁ = 4.23 L. The temperature is kept constant at 24°C. Then, if the pressure is decreased to 93.3 kPa, then the volume is tend to increase due to Boyle's law.

So let us consider the new pressure be P₂ = 93.3 kPa and the new volume has to be found.

Then using Boyle's law, P_{1} V_{1} = P_{2}  V_{2}

Then, V_{2}=\frac{P_{1} V_{1} }{P_{2} }

So, V_{2}=\frac{99.6*1000*4.23}{93.3*1000}=4.52 L

Thus, the volume is increased to 4.52 L on decreasing the pressure to 93.3 kPa.

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Explanation:

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Consider the balanced equation below. 4NH3 + 3O2 --> 2N2 + 6H2O What is the mole ratio of NH3 to N2?
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The balanced equation given is:
4NH3 + 3O2 .....> 2N2 + 6H2O

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How many chloride ions are in 0.486 moles of chloride ions?​
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Answer:

Since in a chloride ion, we have an additional electron

you might think that it will affect the mass but the mass of an electron is almost negligible so we will ignore that

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Amount of ions in 0.486 moles = 0.486 * (6.022*10^23)

Amunt of ions in 0.486 moles = 2.9 * 10^23 ions

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6 0
2 years ago
A gas cylinder filled with nitrogen at standard temperature and pressure has a mass of 37.289 g. The same container filled with
andrew-mc [135]

Answer:

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

Explanation:

Let the moles of nitrogen gas = x moles

Moles of carbon dioxide = x moles ( As both are filled at same temperature and pressure conditions )

Given:

Mass_{Container}+Mass_{Nitrogen\ gas}=37.289\ g

Molar mass of nitrogen gas, N_2 = 28.014 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

x\ moles= \frac{Mass}{28.014\ g/mol}

Mass of nitrogen gas = 28.014x g

So,

Let, Mass_{Container}=y

y+28.014x=37.289

Similarly,

Mass_{Container}+Mass_{Carbon\ dioxide\ gas}=37.440\ g

Molar mass of nitrogen gas, CO_2 = 44.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

x\ moles= \frac{Mass{44.01\ g/mol}

Mass of nitrogen gas = 44.01x g

So,

y+44.01x=37.440

Solving the two equations, we get :

Mass_{Container}=y=37.025\ g

x = 0.00943 moles

Thus, Given:

Mass_{Container}+Mass_{Unknown\ gas}=37.062\ g

37.025\ g+Mass_{Unknown\ gas}=37.062\ g

Mass of the gas = 0.037 moles

Moles = 0.00943 moles

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.00943\ moles= \frac{0.037\ g}Molar mass}

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

7 0
2 years ago
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