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AleksAgata [21]
3 years ago
8

A voltaic cell consists of a nickel electrode in a solution containing Ni2+ ions, and a copper electrode in a solution containin

g Cu2+ ions. Which is the cell potential? a. E subscript c e l l space end subscript superscript 0 space end superscript equals minus 0.59 V b. E subscript c e l l space end subscript superscript 0 space end superscript equals plus 0.59 V c. E subscript c e l l space end subscript superscript 0 space end superscript equals minus 0.09 V d. E subscript c e l l space end subscript superscript 0 space end superscript equals plus 0.09 V
Chemistry
1 answer:
Marysya12 [62]3 years ago
4 0
First, we write the half equations for the reduction of the chemical species present:

Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V

In order to determine the potential of the cell, we find the difference between the two values. For this:

E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V

The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
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2 years ago
Enter your answer in the provided box.
Lina20 [59]

Answer:

5.06atm

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (Litres)

V2 = final volume (Litres)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 1.34 atm

P2 = ?

V1 = 5.48 L

V2 = 1.32 L

T1 = 61 °C = 61 + 273 = 334K

T2 = 31 °C = 31 + 273 = 304K

Using P1V1/T1 = P2V2/T2

1.34 × 5.48/334 = P2 × 1.32/304

7.34/334 = 1.32P2/304

Cross multiply

334 × 1.32P2 = 304 × 7.34

440.88P2 = 2231.36

P2 = 2231.36/440.88

P2 = 5.06

The final pressure is 5.06atm

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3 years ago
A student attempts to separate 4.656 g of a sand/salt mixture just like you did in this lab. After carrying out the experiment,
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Answer:

Explanation:

a ) Total mixture = 4.656 g

Sand recovered = 2.775 g

percent composition of sand in the mixture

= (2.775 g / 4.656 g ) x 100

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b )

Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .

Total mixture = 4.656 g

percent recovery = (3.627 / 4.656 ) x 100

= 77.9 % .

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2 years ago
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