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belka [17]
2 years ago
14

How many moles of methane occupy a volume of 2.00 l at 50.0°c and 0.500 atm answer?

Chemistry
1 answer:
Sauron [17]2 years ago
5 0
Data Given:
                  Pressure  =  P  =  0.5 atm
 
                  Volume  =  V  =  2.0 L

                  Temperature  =  T  =  50 °C + 273  =  323 K

                  Moles  =  n  =  ?

Solution:
              Let suppose the gas is acting Ideally, Then According to Ideal Gas Equation.

                              P V  =  n R T
Solving for n,
                              n  =  P V / R T

Putting Values,
                             n  =  (0.5 atm × 2.0 L) ÷ (0.0821 atm.L.mol⁻¹.K⁻¹ × 323 K)

                             n  =  0.0377 mol
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Answer:

[Ni²⁺] = 1.33 M

Explanation:

To do this, we need to use the Nernst equation which (in standard conditions of temperature)

E = E° - RT/nF lnQ

However R and F are constant, and the reaction is taking place in 25 °C so we can assume the nernst equation like this:

E = E° - 0.05916/n logQ

As the nickel is in the cathode, this means that this element is being reducted while Cadmium is being oxidized, therefore the REDOX reaction would be:

Cd(s) + Ni²⁺(aq) --------> Cd²⁺(aq) + Ni(s)

With this, Q:

Q = [Cd²⁺] / [Ni²⁺]

Now, we need to know the value of the standard reduction potentials, which can be calculated with the semi equations of reduction and oxidation:

Cd(s) ------------> Cd²⁺ + 2e⁻       E°₁ = 0.40 V

Ni²⁺ + 2e⁻ -------------> Ni(s)        E°₂ = -0.25 V

E° = E°₁ + E°₂

E° = 0.40 - 0.25 = 0.15 V

Now that we have all the data, we can solve for the [Ni²⁺]:

0.133 = 0.15 - 0.05916/2 log(5/[Ni²⁺])

0.133 - 0.15 = -0.05916/2 log(5/[Ni²⁺])

-0.017 = -0.02958 log(5/[Ni²⁺])

-0.017/-0.02958 = log(5/[Ni²⁺])

0.5747 = log(5/[Ni²⁺])

10^(0.5747) = 5/[Ni²⁺]

[Ni²⁺] = 5/3.7558

<h2>[Ni²⁺] = 1.33 M</h2>
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Explanation:

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