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Nata [24]
3 years ago
7

What would be the voltage (Ecell) of a voltaic cell comprised of Cd(s)/Cd2+(aq) and Zr(s)/Zr4+(aq) if the concentrations of the

ions in solution were [Cd2+] = 0.5 M and [Zr4+] = 0.5 M at 298K?
Chemistry
1 answer:
lutik1710 [3]3 years ago
3 0

Answer:

1.05 V

Explanation:

Since;

E°cell= E°cathode- E°anode

E°cathode= -0.40 V

E°anode= -1.45 V

E°cell= -0.40-(-1.45) = 1.05 V

Equation of the process;

2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)

n= 8 electrons transferred

From Nernst's equation;

Ecell = E°cell - 0.0592/n log Q

Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]

Since log 1=0

Ecell= E°cell= 1.05 V

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110 degrees C

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D. They all contain carbon as an important part of their structure.
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The molar mass of hydrogen molecules is 2g/mol and the molar mass of oxygen molecules is 32g/mol. What is the ratio of the avera
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Answer:

1

Explanation:

For an ideal gas, the average kinetic energy is given by:

Ek = (3/2)*n*R*T

Where n is the number of moles, R is the gas constant (8.31 J/mol*K), and T the temperature. The gases have the same number of moles, and the same temperature, so they will have the same average kinetic energy:

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So, the ratio between then is 1.

8 0
4 years ago
(4) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2
maksim [4K]

Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:

qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ =  (V₁/(2V₁))ⁿ = (1/2)ⁿ

When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.

When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.

When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.

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3 years ago
What is the solubility of NH4CL at 50 degrees C?
Simora [160]

i believe its 50 grams.

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3 years ago
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