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kirill115 [55]
4 years ago
9

Oxygen and chlorine gas are mixed in a container with partial pressure of 401 mmHg and 136 mmHg Respectively. What is the total

pressure inside the container?
Chemistry
2 answers:
Alexandra [31]4 years ago
5 0

Answer:

The answer is 53.2L

Explanation:

Avogadro’s Law

Avogadro’s Law, also known as the Mole-Volume Law or Volume Amount Law, states that volume (V) and moles (n) are directly proportional as long as pressure (P) and temperature (T) are held constant. This mole-volume relationship is depicted with the addition or removal of gas molecules from a closed container with a moveable piston.

Avogadro's Law (Moles & Volume)

As more and more gas molecules are pumped into the container they push up against the moveable piston and thereby increase the volume inside the container. This direct mole-volume relationship can be plotted onto a chart and provide the following:

Moles-Volume-Plot

Moles-Volume Plot

The direct relationship between the number of moles and volume at constant temperature and pressure is illustrated by the expression:

Volume-moles-Direct-Relationship

Volume & moles (Direct Relationship)

Avogadro’s Law Formula

By rearranging the Ideal Gas Law we can isolate V and n:

Avogadro-Law-Derived-Formula

Derived Formula (Avogadro's Law)

If temperature (T) and pressure (P) are held constant then the formula simplifies into:

Avogadro-Law-V/n-constant

Avogadro's Law (V/n = Constant)

Incorporating the two sets of data (2 volumes, 2 moles) produces the Avogadro’s Law formula as:

Avogadro-Law-V1n1-V2n2

Avogadro's Law (V1/n1 = V2/n2)

lukranit [14]4 years ago
3 0

Answer:

537 mmHg is the total pressure

Explanation:

In a mixture of gases, the total pressure from the mixture is the sum of the partial pressure from each gas in the mixture.

Our mixture contains O₂ and Cl₂

Partial pressure O₂ + Partial pressure Cl₂ = Total pressure

401 mmHg + 136 mmHg = 537 mmHg

Total pressure → 537 mmHg

This is the Dalton's Law,  the total pressure of a mixture of gases is equal to the sum of the partial pressures of the gases that are contained in the mixture.

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6 0
4 years ago
0.25 mol of nitric acid is used to make a 0.10 M nitric acid solution. How many grams of solute was used?
Yanka [14]

Answer:

15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution

Explanation:

Step 1: Data given

Nitric acid = HNO3

Molar mass of H = 1.01 g/mol

Molar mass of N = 14.0 g/mol

Molar mass O = 16.0 g/mol

Number of moles nitric acid (HNO3) = 0.25 moles

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Step 2: Calculate molar mass of nitric acid

Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)

Molar mass HNO3 = 1.01 + 14.0 + 3*16.0

Molar mass HNO3 = 63.01 g/mol

Step 3: Calculate mass of solute use

Mass HNO3 = moles HNO3 * molar mass HNO3

Mass HNO3 = 0.25 moles * 63.01 g/mol

Mass HNO3 = 15.75 grams

15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution

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