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lina2011 [118]
3 years ago
8

What is the temperature of 0.80 mol of a gas stored in a 275 mL cylinder at 175 kPa?

Chemistry
1 answer:
vitfil [10]3 years ago
5 0

Answer:

The answer to your question is   T = 7.25 °K  

Explanation:

Data

Temperature = ?

moles = 0.80

volume = 275 ml

pressure = 175 kPa

constant of ideal gases = 0.082 atm l/ mol °K

Process

1.- Use the ideal gas law to solve this problem

                  PV = nRT

-Solve for Temperature

                 T = PV / nR

2.- Convert Pressure to atm

                  1 atm --------------- 101.325 kPa

                    x     --------------- 175 kPa

                    x = (175 x 1) / 101.325

                    x = 1.73 atm

3.- Convert volume to liters

                   1000 ml ---------------- 1 l

                    275 ml ----------------- x

                      x = (275 x 1) / 1000

                      x = 0.275 l

4.- Find the temperature

                   T = (0.275 x 1.73) / (0.8 x 0.082)

-Simplification

                    T = 0.476 / 0.0656

-Result

                     T = 7.25 °K  

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\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}

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\text{reaction fills a} \ \mathbf{100 \  L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \  \text{of oxygen gas, }

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Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

From the above reaction; the ICE table can be represented as:

                    \mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

I (mol/L)     0.086            0.28                 0              0

C                   -4x                -3x               +2x           +6x

E                 0.086 - 4x     0.28 - 3x      +2x             +6x

At equilibrium;

The water vapor = \dfrac{2.6 \ mol}{100 \ L} = 6x

x = \dfrac{2.6}{100} \times \dfrac{1}{6}

x = 0.00433

\text{equilibrium constant}  ({k_c}) =  \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }

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Replacing the value of x, we have:

K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}

K_c = \mathbf{5.5 \times 10^{-8} \ to  \ 2 \ significant \ figures}

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Explanation:

The deuterium is:

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Then, the mass defect is:

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I hope it helps you!  

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