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Keith_Richards [23]
3 years ago
11

A sample of ammonia gas at 298 K has a volume of 3.5 L at a pressure of 170.2 kPa, what is the new pressure if the gas is compre

ssed to a volume of 1.35 L at 293 K?
Chemistry
2 answers:
Maru [420]3 years ago
5 0

Answer:

430 kPa

Explanation:

Use the formula PV/T = PV/T where the left side of the equation refers to the initial values and the right side refers to the final values: (170.2 kPa)(3.5 L)/(298 K) = (P)(1.35 L)/(293 K). The P represents the unknown pressure. Solving this equation gives us a value for P of 433.855 kPa, but because of significant figures (the two from 3.5 L), we round to 430 kPa.

yarga [219]3 years ago
3 0

Answer:

The final pressure of the gas comes out to be 433.85 KPa

Explanation:

Initial volume of gas = V = 3.5 L

Initial pressure of gas =  P = 170.2 KPa

Initial temperature = T = 298 K

Final volume of gas = V' = 1.35 L

Final temperature of gas = T' = 293 K

Assuming the number of moles of gas remains constant.

The final pressure of the gas is obtained by the following equation shown below

\displaystyle \frac{PV}{T} = \displaystyle \frac{P'V'}{T'} \\\displaystyle \frac{170.2 \textrm{ KPa}\times 3.5 \textrm{ L}}{298\textrm{ K}} = \displaystyle \frac{P'\times 1.35 \textrm{ L}}{293 \textrm{ K}} \\P' = 433.85 \textrm{ KPa}

Final pressure of gas = 433.85 KPa

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The equilibrium constant of the reaction is 282. Option D

<h3>What is equilibrium constant?</h3>

The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.

Concentration of bromine = 0.600 mol /1.000-L = 0.600 M

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In this case, we must set up the ICE table as shown;

              Br2(g) + I2(g) ↔ 2IBr(g)

I          0.6            1.6           0

C      -x                -x             +2x

E    0.6 - x         1.6 - x       1.190

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Hence;

Kc = [IBr]^2/[Br2] [I2]

Kc = ( 1.190)^2/(0.005) (1.005)

Kc = 282

Learn more about equilibrium constant:brainly.com/question/15118952

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