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asambeis [7]
3 years ago
5

Two sealed, rigid 5.0L containers each contain a gas at the same temperature but at a different pressure, as shown above. Also s

hown are the results of transferring the entire contents of container 1 to container 2. No gases escape during the transfer. Assuming ideal behavior, which statement is correct regarding the total pressure of the gases after they are combined
Chemistry
1 answer:
Elodia [21]3 years ago
7 0

Answer:

The question with options are

Two sealed, rigid 5.0L containers each contain a gas at the same temperature but at a different pressure, as shown above. Also shown are the results of transferring the entire contents of container 1 to container 2. No gases escape during the transfer. Assuming ideal behavior, which statement is correct regarding the total pressure of the gases after they are combined?

A) The total pressure of the gases in the mixture is the sum of the initial pressures of oxygen gas and nitrogen gas because pressure only depends on the total amount of gas when volume and temperature are held constant.

B) The total pressure of the gases in the mixture is lower than the sum of the initial pressures of oxygen and nitrogen because some of the energy of the particles will be lost due to an increase in the number of collisions.

C) The total pressure of the gases in the mixture is higher than the sum of the initial pressures of oxygen and nitrogen because of the inter molecular forces that develop between oxygen and nitrogen molecules.

D) The total pressure of the gases in the mixture cannot be determined because the actual value of the temperature is not given.

The correct answer to the question is (A), a representation of Dalton's Law of Partial pressure

A) The total pressure of the gases in the mixture is the sum of the initial pressures of oxygen gas and nitrogen gas because pressure only depends on the total amount of gas when volume and temperature are held constant.

Explanation:

The above solution can be explained by Dalton's law of partial pressures  which states that at constant temperature and pressure, the pressure of a given mass gas is equal to the sum of the partial pressures of the individual gases that make up the mixture :

P_{Total}​ = P_{gas in container 1} + P_{gas in container 2}

The partial pressure of a gas is the pressure exerted by the gas if the gas is the only constituent of the container

Dalton's Law is used to

Calculate the gas partial pressure given the mole ratio and total pressure

Determine the number of moles of a gas present in a mixture given the partial pressure and total pressure

Calculate the pressure exerted by a mixture of gases given the partial pressures of the individual gases at constant temperature and pressure

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3 years ago
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Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
Katen [24]

<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

Evaluating the value of 'x'

\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

K_c=\frac{[HI]^2}{[H_2][I_2]}

[HI]_{eq}=2x=(2\times 0.079)=0.158M

[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M

[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

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3 years ago
1. Baking soda and vinegar combined *
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Answer:

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Explanation:

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Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to
morpeh [17]

Explanation:

1.) 175 km to μm

1 km=10^9 \mu m

175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m

3.) 385 nm to dm

1 nm=10^{-8} dm

385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm

5.) 492 μm  to m

1 μm =  10^{-6} m

492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m

7.) 52\times 10^3 dm to mm

1 dm = 100 mm

52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm

9.) 321\times 10^{35} mm to km

1 mm = 10^{-6} km

321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km

11.) 456\times 10^3 m to km

m = 0.001 km

456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km

13.) 422\times 10^3 m to nm

1 m = 10^{9} nm

422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm

15.) 4.87\times 10^{30} m to pm

1 m = 10^{12} pm

4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm

17.) 5.26\times 10^3 m to um

1 m =  10^{6} \mu m

5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m

19.) 1.25\times 10^{35}m to Mm

1 m =  10^{-6} Mm

1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm

21.) 4.22\times 10^3 Tm to nm

1 Tm = 10^{21} nm

4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm

6 0
3 years ago
what is the most appropriate unit to measure the following; the length of a football field. b) the diameter of a tennis ball​
hjlf

Explanation:

a) m (metre)

b) mm (millimeter)

6 0
3 years ago
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