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3 years ago

Convert 16.8 L of nitrogen monoxide gas at STP to grams .

Chemistry

1 answer:

german3 years ago

8 0

Answer:

See explanation

Explanation:

1 mole of a gas occupies 22.4 L

x moles occupies 16.8 L

x = 1 mole * 16.8 L/22.4 L

x = 0.75 moles

number of moles = mass/molar mass

mass = number of moles * molar mass

mass = 0.75 moles * 30.01 g/mol = 22.5075 g = 2.25 * 10^1 g

the coefficient of the scientific notation answer = 2.25

the exponent of the scientific notation answer = 1

significant figures are there in the answer = 6

the right most significant figure in the answer = 3

number of moles = 12.5g/38g/mol = 0.3289 moles

1 mole occupies 22.4 L

0.3289 moles occupies 0.3289 moles * 22.4 L/1 mole

= 7.36736 L = 7.36736 * 10^0 L= 7.37 * 10^0 L

the coefficient of the scientific notation answer =7.37

the exponent of the scientific notation answer = 0

significant figures are there in the answer = 6

the right most significant figure in the answer= 3

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Explanation:

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3 years ago

A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the

Viefleur [7K]

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V$

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V$

The mole fraction of nitrogen in the mixture is:

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Learn more: brainly.com/question/2060778

5 0

3 years ago

How many grams of H2SO4 are needed to prepare 500. mL of a .250M solution?

zavuch27 [327]

Answer:

We need 12.26 grams H2SO4

Explanation:

Step 1: Data given

Volume of a H2SO4 solution = 500 mL = 0.500 L

Concentration of the H2SO4 solution = 0.250 M

Molar mass of H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = concentration * volume

Moles H2SO4 = 0.250 M * 0.500 L

Moles H2SO4 = 0.125 moles

Step 3: Calculate mass of H2SO4

Mass of H2SO4 = moles * molar mass

Mass of H2SO4 = 0.125 moles * 98.08 g/mol

Mass of H2SO4 = 12.26 grams

We need 12.26 grams H2SO4

7 0

3 years ago