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Ivan
1 year ago
7

Aqueous Copper (II) nitrate reacts with aqueous potassium iodide to form Copper (II) iodide solid and potassium nitrate

Chemistry
1 answer:
nikitadnepr [17]1 year ago
7 0

Answer:

Cu(NO₃)₂ (aq) + 2 KI (aq) ---> CuI₂ (s) + 2 KNO₃

Explanation:

When writing the reaction with the symbols, you need to take into account the charges of the ions. If he charges on the ions do not balance out in a molecule, they need to be made up for in the form of subscripts. For example, copper (+2) and iodine (-1) have charges which do not balance. Thus, to make the molecule neutral, you need to have two iodine atoms (CuI₂).

The unbalanced equation:

Cu(NO₃)₂ (aq) + KI (aq) ---> CuI₂ (s) + KNO₃

<u>Reactants</u>: 1 copper, 2 nitrate, 1 potassium, 1 iodine

<u>Products</u>: 1 copper, 1 nitrate, 1 potassium, 2 iodine

The balanced equation:

1 Cu(NO₃)₂ (aq) + 2 KI (aq) ---> 1 CuI₂ (s) + 2 KNO₃

<u>Reactants</u>: 1 copper, 2 nitrate, 2 potassium, 2 iodine

<u>Products</u>: 1 copper, 2 nitrate, 2 potassium, 2 iodine

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Data provided:

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A particle with a charge of 9.40 nC is in a uniform electric field directed to the left. Another force, in addition to the elect
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Answer:

a. Work done by the electric force = -2.85 * ×10⁻⁵ J

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c. The magnitude of the electric field is 33.7kV/m

Explanation:

Given.

Charge = Q = 9.40 nC

Distance = d = 9.00 cm = 0.09m

Amount of work = 7.10×10⁻⁵ J

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This is calculated by; change in Kinetic Energy i.e. ∆KE

∆KE = ∆K2 - ∆Kæ

Where K2 = 4.25×10⁻⁵ J

The body is released at rest, so the initial velocity is 0.

So, K1 = 0

Also, total work done = W1 + W2

Where W2 = 7.10×10⁻⁵J

So, W1 + W2 = W = K2

W1 + 7.10×10⁻⁵ = 4.25×10⁻⁵

W1 = 4.25×10⁻⁵ - 7.10×10⁻⁵

W = -2.85 * ×10⁻⁵ J

Work done by the electric force = -2.85 * ×10⁻⁵ J

b. What is the potential of the starting point with respect to the end point?

The change in potential energy is given as

W = ∆U

W = Q|V2 - V1| where V1 = 0 because the body starts from rest

So, W = QV2

Make V the Subject of the formula

V2 = W/Q

V2 = -2.85 * ×10⁻⁵ J / 9.40 nC

V2 = -2.85 * ×10⁻⁵ J / 9.40 * 10^-9C

V2 = −3031.9148936170212765957V

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The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. What is the magnitude of the electric field?

The magnitude of the electric field is calculated as follows;

W = -Fd = -QEd

And E = V/d

E = -3.03 * 10³ V / 0.09 m

E = −33687.943262411347517730 V/m

E = -33.7kV/m

The magnitude of the electric field is 33.7kV/m

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