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nevsk [136]
3 years ago
10

A mixture of nitrogen and neon gas is compressed from a volume of to a volume of , while the pressure is held constant at . Calc

ulate the work done on the gas mixture. Round your answer to significant digits, and be sure it has the correct sign (positive or negative).
Chemistry
1 answer:
Mamont248 [21]3 years ago
7 0

Answer:

-266 kJ

Explanation:

There is some info missing. I think this is the original question.

<em>A mixture of nitrogen and neon gas is expanded from a volume of 53.0 L to a volume of 90.0 L, while the pressure is held constant at 71.0 atm. Calculate the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits.</em>

<em />

Given data

Initial volume = 53.0 L

Final volume = 90.0 L

Pressure = 71.0 atm

We can find the work (w) associated with the expansion of the gaseous mixture using the following expression.

w = - P × ΔV = - 71.0 atm × (90.0 L - 53.0 L) = -2.63 × 10³ atm . L

w = -2.63 × 10³ atm . L × (101.3 J/ 1 atm . L) = -2.66 × 10⁵ J = -266 kJ

The negative sign means that the system does work on the surroundings.

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The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

7 0
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How many atoms of oxygen are in 0.100 miles of silicon dioxide, SiO^2
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Answer:

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Goryan [66]

Answer:

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Explanation:

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