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emmasim [6.3K]
2 years ago
11

Question

Chemistry
1 answer:
AlekseyPX2 years ago
5 0
Largest

K
Ca
Ga
Ge
As
Br
Kr
Smallest
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2.50 liter of a gas has a pressure of 165. kPa at 25.0°C. If the pressure increases to 600. kPa and the temperature to 100.0°C,
tiny-mole [99]

Answer:

0.861 L

Explanation:

We are given pressure, volume, and temperature, so let's apply the Combined Gas Law:

(P₁V₁)/T₁ = (P₂V₂)/T₂

Convert the temperatures to degrees Kelvin.

25.0°C -> 298 K, 100.0°C -> 373 K

Plug in the initial conditions on the left, then the final/new on the right, and solve for the unknown:

(165(2.5))/298 = (600(V₂))/373

V₂ = (165(2.5)(373))/(298(600))

V₂ = 0.861 L

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Read 2 more answers
For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [
sveta [45]

Explanation:

The given data is as follows.

     E^{o} = 0.483,     [Co^{3+}] = 0.173 M,

     [Co^{2+}] = 0.433 M,     [Cl^{-}] = 0.306 M,

     P_{Cl_{2}} = 9.0 atm

According to the ideal gas equation, PV = nRT

or,             P = \frac{n}{V}RT    

Also, we know that

                Density = \frac{mass}{volume}

So,         P = MRT

and,          M = \frac{P}{RT}

                    = \frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}

                    = \frac{9.0}{24.436}

                    = 0.368 mol/L

Now, we will calculate the cell potential as follows.

          E = E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}

             = 0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}

             = 0.483 - 0.02955 log \frac{0.0689}{0.0162}

             = 0.483 - 0.02955 \times 0.628

             =  0.483 - 0.0185

             = 0.4645 V

Thus, we can conclude that the cell potential of given cell at 25^{o}C is 0.4645 V.

4 0
3 years ago
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