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2 years ago

if oxygen gas is collected over water at 23.5°c and 750.0 mm Hg, what is the pressure of the gas collected?

Chemistry

1 answer:

Alex73 [517]2 years ago

4 0

Answer:

Pressure of the Oxygen gas collected over water at 23.5°C = 728.35 mmHg = 0.958 atm

Explanation:

The gas collected over water will mix with water vapour.

The total pressure of the setup is then a sum of the partial pressure of the gas collected over the water and the vapour pressure of water at the temperature given.

Total Pressure = (Partial Pressure of Oxygen) + (Vapour Pressure of water at 23.5°C)

Total Pressure = 750 mmHg

Partial Pressure of Oxygen = ?

Vapour Pressure of water at 23.5°C = 21.6472 mmHg from literature

750 = (Partial Pressure of Oxygen) + 21.6472

Partial Pressure of Oxygen = 750 - 21.6472 = 728.3528 mmHg = 0.958 atm

Hope this Helps!!!

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Read 2 more answers

If you have 30.O g of hydrogen gas burned in excess oxygen how many moles of water can you make

vladimir1956 [14]

Answer:

15 moles

Explanation:

Data given:

mass of hydrogen (H₂) = 30.0 g

amount of oxygen (O₂) = excess

moles of water = ?

Solution:

First we look to the reaction in which hydrogen react with oxygen and make (H₂O)

Reaction:

2H₂ + O₂ -----------> 2H₂O

Now look at the reaction for mole ratio

2H₂ + O₂ -----------> 2H₂O

2 mole 2 mole

So it is 2:2 mole ratio of hydrogen to water

As we Know

molar mass of H₂ = 2(1) = 2 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now convert moles to gram

2H₂ + O₂ -----------> 2H₂O

2 mole (2 g/mol) 2 mole (18 g/mol)

4 g 36 g

So,

we come to know that 4 g of hydrogen gives 36 g of water then how many grams of water will be produce by 30 grams of hydrogen.

Apply unity formula

4 g of H₂ ≅ 36 g of H₂O

30 g of H₂ ≅ X of H₂O

Do cross multiplication

X of H₂O = 30 g x 36 g / 4 g

X of H₂O = 270 g

Now convert grams of H₂O into moles

No. of moles = mass in grams/molar mass

Put values in above formula

No. of moles = 270 g / 18 (g/mol)

No. of moles = 15 mol

so 30 gram of hydrogen produce 15 mol of water.

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3 years ago

The pressure on a 200 milliliter sample of CO2 (g) at constant temperature is increased from

balu736 [363]

The answer for the following problem is mentioned below.

Therefore the final volume of the gas is 100 ml.

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 600 mm of Hg

Final pressure ( $P_{2}$ ) = 1200 mm of Hg

Initial volume ( $V_{1}$ ) = 200 ml

To find:

Final volume ( $V_{2}$ )

We know;

According to the ideal gas equation,

P × V = n × R × T

Where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So,

From the above mentioned equation,

P × V = constant

$\frac{P_{1} }{P_{2} }$ = $\frac{V_{1} }{V_{2} }$

Where,

( $P_{1}$ ) represents the initial pressure of the gas

( $P_{2}$ ) represents the final pressure of the gas

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

So;

$\frac{600}{1200}$ = $\frac{V_{2} }{200}$

$V_{2}$ = 100 ml

Therefore the final volume of the gas is 100 ml.

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