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Anestetic [448]
3 years ago
9

1.42 mol sample of neon gas at a temperature of 13.0 °C is found to occupy a volume of 25.5 liters. The pressure of this gas sam

ple is mm Hg. ?
Chemistry
1 answer:
Dima020 [189]3 years ago
3 0

Answer: 996 mmHg

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

According to the ideal gas equation:

PV=nRT

P = Pressure of the gas = ?

V= Volume of the gas = 25.5 L

T= Temperature of the gas = 13°C = (273+13) K  = 286K

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 1.42

P=\frac{nRT}{V}=\frac{1.42\times 0.0821\times 286}{25.5}=1.31atm=996mmHg      (760mmHg=1atm)

Thus pressure of this gas sample is 996 mm Hg.

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2 years ago
Determine the empirical formula of a compound containing 48.38 grams of carbon, 6.74 grams of hydrogen, and 53.5 grams of oxygen
Darya [45]

Answer: The empirical formula of the compound becomes CH_2O

<u>Explanation:</u>

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

We are given:

Mass of C = 48.38 g

Mass of H = 6.74 g

Mass of O = 53.5 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

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

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

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of C}=\frac{48.38g}{12g/mol}=3.023 mol

\text{Moles of H}=\frac{6.74g}{1g/mol}=6.74 mol

\text{Moles of O}=\frac{53.5g}{1g/mol}=3.34 mol

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

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 3.023 moles

\text{Mole fraction of C}=\frac{3.023}{3.023}=1

\text{Mole fraction of H}=\frac{6.74}{3.023}=2.23\approx 2

\text{Mole fraction of O}=\frac{3.34}{3.023}=1.105\approx 1

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

The ratio of C : H : O = 1 : 2 : 1

Hence, the empirical formula of the compound becomes CH_2O

5 0
3 years ago
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