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pychu [463]
3 years ago
14

An analogy makes a comparison between objects based on their similar qualities. Cassidy wanted to create an analogy for the moti

on of atoms in solids, liquids, and gases, so she compared them to marbles in a tray. Which best compares the phases of matter to marbles in a tray?
A. A solid is like the tray being shaken and the marbles moving around it, and a liquid is like the tray being shaken slowly and all the marbles moving in their positions.
B. A solid is like the tray being shaken slowly and all the marbles moving in their positions, a liquid is like the tray being shaken and the marbles moving around it, and a gas is like the tray being shaken hard and the marbles moving vigorously around it.
C. A gas is like the tray being shaken slowly and all the marbles moving in their positions, and a solid is like the tray being shaken hard and the marbles moving vigorously around it.
D. A liquid is like the tray being shaken hard and the marbles moving vigorously around it, and a gas is like the tray being shaken slowly and all the marbles moving in their positions.
Chemistry
2 answers:
cricket20 [7]3 years ago
7 0
B
atoms in a solid can move very little because of how compact they are
in a liquid atoms move a little more freely
in a gas atoms are bouncing fast as they are very spaced out
lukranit [14]3 years ago
3 0
B. a solid is the most dense of all the phases of matter, so the atoms are the closest and move the slowest. a liquid is in the middle, and a gas has the most movement and space between the atoms.
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Answer : The actual cell potential of the cell is 0.47 V

Explanation:

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The given redox reaction is :

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Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

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Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

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Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

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