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Lapatulllka [165]
4 years ago
8

Part C Gallium crystallizes in a primitive cubic unit cell. The length of an edge of this cube is 362 pm. What is the radius of

a gallium atom? Express your answer numerically in picometers. View Available Hint(s) radius = nothing pm p m Submit
Chemistry
1 answer:
Fudgin [204]4 years ago
5 0

Answer:

181 picometers is the radius of a gallium atom.

Explanation:

Primitive cubic cell is an arrangement in which constituent particles are present only at the corners of the cube.Also known as simple unit cell. In this arrangement number of total particles is equal to 1.

\frac{1}{8}\times 8 =1

Gallium crystallizes in a primitive cubic unit cell.

Edge length of cubic unit cell = a = 362 pm

Let the radius be gallium atom be r

Two atoms of gallium touching each other = a =2r

362 pm = 2r

r = 181 pm

181 picometers is the radius of a gallium atom.

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A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are
VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

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Explanation:

a)

The reaction at the cathode is represented as follows:

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The reaction at the anode is equal to:

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The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

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Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

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The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

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