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Natalija [7]
2 years ago
14

On a clear day at sea level, with a temperature of 25 °C, the partial pressure of N2 in air is 0.78 atm and the concentration of

nitrogen in water is 5.3⋅10−4 M. When the partial pressure of N2 is ________ atm, the concentration in water is 2.0⋅10−3 M.
Chemistry
1 answer:
joja [24]2 years ago
5 0

Answer : The partial pressure of nitrogen gas is, 2.94 atm

Explanation:

According top the Henry's Law, the concentration of a gas in a liquid is directly proportional to the partial pressure of the gas.

C\propto P

C=K_H\times P

K_H is Henry's constant.

or,

\frac{C_1}{C_2}=\frac{P_1}{P_2}

where,

C_1 = initial concentration of gas = 5.3\times 10^{-4}M

C_2 = final concentration of gas = 2.0\times 10^{-3}M

P_1 = initial partial pressure of gas = 0.78 atm

P_2 = final partial pressure of gas = ?

Now put all the given values in the above formula, we get the final partial pressure of the gas.

\frac{5.3\times 10^{-4}M}{2.0\times 10^{-3}M}=\frac{0.78atm}{P_2}

P_2=2.94atm

Therefore, the partial pressure of nitrogen gas is, 2.94 atm

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A 24.00 g sample contains 14.60 g Cl and 9.400 g B. What is the percent composition (by mass) of boron in this sample?
vaieri [72.5K]

Answer:

39.2 %

Explanation:

The following data were obtained from the question:

Mass of sample = 24 g

Mass of Cl = 14.6 g

Mass of B = 9.4 g

Percentage composition of boron =?

The percentage composition (by mass) of boron in the sample can be obtained as illustrated below:

Percentage composition of boron = mass of B /mass of sample × 100

Percentage composition of boron = 9.4/24 × 100

Percentage composition of boron = 39.2 %

Therefore, the percentage composition (by mass) of boron in the sample is 39.2 %

8 0
2 years ago
You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain
IrinaVladis [17]

Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of O_2:-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

Moles= \frac{49.8\ g}{32\ g/mol}

Moles_{O_2}= 1.55625\ mol

Since pressure and volume are constant, we can use the Avogadro's law  as:-

\frac {V_1}{n_1}=\frac {V_2}{n_2}

Given ,  

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

\frac {V_1}{n_1}=\frac {V_2}{n_2}

\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}

n₁ = 0.778125 moles

Moles of carbon dioxide = 0.778125 moles

Molar mass of CO_2 = 44.0 g/mol

Mass of CO_2 = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>

5 0
2 years ago
The concentration of ozone in a sample of air that has a partial pressure of O3 of 0.33 torr and a total pressure of air of 695
Goryan [66]

Answer:

0.047 %

Explanation:

Step 1: Given data

  • Partial pressure of ozone (pO₃): 0.33 torr
  • Total pressure of air (P): 695 torr

Step 2: Calculate the %v/v of ozone in the air

Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.

<em>%v/v = pO₃/P × 100%</em>

%v/v = 0.33 torr/695 torr × 100%

%v/v = 0.047 %

8 0
2 years ago
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