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Delicious77 [7]
3 years ago
8

Four (4.0) liters of helium gas are stored at 125 atm pressure. If the temperature and number of particles do not change, what w

ill be the new volume when the pressure is decreased to 50 atm? Show your work.
Please answer ASAP!!
Chemistry
1 answer:
seropon [69]3 years ago
5 0

Answer:

10 L

Explanation:

The only variables are pressure and volume, so we can use Boyle's Law:

p1V1 = p2V2

Data:

p1 = 125 atm; V1  = 4.0 L  

p2 =  50 atm; V2 = ?

Calculation:

125 × 4.0 = 50V2

       500 = 50 V2

        V2 = 500/50 = 10 L

The new volume will be 10 L.

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Whenever a number or a variable is divided<br> by itself, it is always equal to
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Duh

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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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5 0
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