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Arisa [49]
3 years ago
6

If you only wanted to increase the particle motion of a gas without increasing any of its other properties, which would the most

correct situation?
a. Keep the gas at a constant pressure and keep the temperature constant, but increase the volume of the gas

b. Keep the gas in a fixed container at constant pressure and increase the temperature

c. Keep the gas in a fixed container at constant pressure and decrease the temperature

d. Keep the gas at a constant volume and keep the temperature constant, but decrease the pressure of the gas
Chemistry
1 answer:
andrey2020 [161]3 years ago
5 0

Answer:

c

Explanation:

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What would the pOH of a solution be if the pH is 9.4
Alika [10]
<h2>Answer:</h2>

pOH = 4.6

<h2>Explanations:</h2>

The sum of pH and pOH of a solution is 14as shown:

pH+pOH=14

Given the following parameter

pH = 9.4

Substitute the given parameters into the formula to have:

\begin{gathered} 9.4+pOH=14 \\ pOH=14-9.4 \\ pOH=4.6 \end{gathered}

Hence the pOH of the solution of pH of 9.4 is 4.6

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1 year ago
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Andre45 [30]
I believe the answer is B.
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3 years ago
How to solve for K when given your anode and cathode equations and voltage
Aleks04 [339]

Answer:

See Explanation

Explanation:

In thermodynamics theory the Free Energy (ΔG) of a chemical system is described by the expression ΔG = ΔG° + RTlnQ. When chemical system is at equilibrium ΔG = 0. Substituting into the system expression gives ...

0 = ΔG° + RTlnKc, which rearranges to ΔG° = - RTlnKc.  ΔG° in electrochemical terms gives ΔG° = - nFE°, where n = charge transfer, F = Faraday Constant = 96,500 amp·sec and E° = Standard Reduction Potential of the electrochemical system of interest.

Substituting into the ΔG° expression above gives

-nFE°(cell) = -RTlnKc => E°(cell) = (-RT/-nF)lnKc = (2.303·R·T/n·F)logKc

=> E°(cell) = (0.0592/n)logKc = E°(Reduction) - E°(Oxidation)

Application example:

Calculate the Kc value for a Zinc/Copper electrochemical cell.

Zn° => Zn⁺² + 2e⁻  ;    E°(Zn) = -0.76 volt  

Cu° => Cu⁺² + 2e⁻ ;    E°(Cu) =  0.34 volt

By natural process, charge transfer occurs from the more negative reduction potential to the more positive reduction potential.

That is,

           Zn° => Zn⁺² + 2e⁻ (Oxidation Rxn)

Cu⁺² + 2e⁻ => Cu°             (Reduction Rxn)

E°(Zn/Cu) = (0.0592/n)logKc

= (0.0592/2)logKc = E°(Cu) - E°(Zn) = 0.34v - (-0.76v) = 1.10v

=> logKc = 2(1.10)/0.0592 = 37.2

=> Kc = 10³⁷°² = 1.45 x 10³⁷

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3 years ago
What determines the amount of chemical energy a substance has
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This would probably be based on the number and type of chemical bonds the substance has.
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a substance that has no fixed shape and yields easily to external pressure; a gas or a liquid.

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