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Anika [276]
3 years ago
14

Calculate the standard cell potential at 25 ∘C for the reaction X(s)+2Y+(aq)→X2+(aq)+2Y(s) where ΔH∘ = -687 kJ and ΔS∘ = -169 J/

K .
express your answer to three significant figures and include the appropriate

Chemistry
1 answer:
aivan3 [116]3 years ago
6 0

Answer:

-0.129V

Explanation:

The change in free energy is obtained from the given parameters after which the value is now applied to obtain the cell potential in volts from the formukar shown in the solution below.

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Consider a 1260-kg automobile clocked by law-enforcement radar at a speed of 85.5 km/h. If the position of the car is known to w
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The expected speed is v = 85.5 km/h
 v = 85.5 km/h = (85.5 km/h)*(0.2778 (m/s)/(km/h)) = 23.75 m/s

If there is an uncertainty of 2 meters in measuring the position, then within a 1-second time interval:
The lower measurement for the speed is v₁ = 21.75 m/s,
The upper measurement for the speed is v₂ = 25.75 m/s.
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3 years ago
Your future is affected by your choices today? I agree or disagree​
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3 years ago
Which of the following is not the same as 2.97 milligrams?
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What is the pH of a solution made by mixing 30.00 mL 0.10 M HCl with 40.00 mL of 0.10 M KOH? Assume that the volumes of the solu
nirvana33 [79]

Answer:

pH = 12.15

Explanation:

To determine the pH of the HCl and KOH mixture, we need to know that the reaction is  a neutralization type.

HCl  +  KOH  →  H₂O  +  KCl

We need to determine the moles of each compound

M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl

M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH

The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl

HCl  +  KOH  →  H₂O  +  KCl

3 m       4 m                       -

             1 m                      3 m

As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.

1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume

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- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15

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3 years ago
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