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2 years ago

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

Chemistry

1 answer:

nikitadnepr [17]2 years 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

Reactants: 1 copper, 2 nitrate, 1 potassium, 1 iodine

Products: 1 copper, 1 nitrate, 1 potassium, 2 iodine

The balanced equation:

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

Reactants: 1 copper, 2 nitrate, 2 potassium, 2 iodine

Products: 1 copper, 2 nitrate, 2 potassium, 2 iodine

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aleksley [76]

Answer:

V ∝ abc

Explanation:

This task is a joint variation task involving only direct proportionality:

Direct variation is one in which two variables are in direct proportionality to each other. This means that as one increases, the other variable also increases and vice - versa.

Joint variation is one in which one variable is dependent on two or more variables and varies directly as each of them.

In this exercise:

If a ∝ b and a ∝ c, then a ∝ bc

Taking the above three proportionalities,

V ∝ a ∝ b ∝ c

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3 years ago

A voltaic cell is constructed from an Ni2+(aq)−Ni(s)Ni2+(aq)−Ni(s) half-cell and an Ag+(aq)−Ag(s)Ag+(aq)−Ag(s) half-cell. The in

eduard

Answer:

+1.03 V

Explanation:

The standard emf of the voltaic cell is the value of the standard potential of it, which is calculated by the standard reduction potential (E°).

The standard reduction potential is the potential needed for the reduction reaction happen, and it's determined by the reaction with the hydrogen cell (which has E° = 0.0V). The half-reactions of reduction of Ni⁺² and Ag⁺, are:

Ni⁺²(aq) + 2e⁻ → Ni(s) E° = -0.23 V

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3 years ago

What is the mass of oxygen in 3.34 g of potassium permanganate?

3241004551 [841]

Answer:

1.35 g

Explanation:

Data Given:

mass of Potassium Permagnate (KMnO₄) = 3.34 g

Mass of Oxygen: ?

Solution:

First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)

So,

Molar Mass of KMnO₄ = 39 + 55 + 4(16)

Molar Mass of KMnO₄ = 158 g/mol

Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).

Mass contributed by Oxygen (O) = 4 (16) = 64 g

Since the percentage of compound is 100

So,

Percent of Oxygen (O) = 64 / 158 x 100

Percent of Oxygen (O) = 40.5 %

It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.

So,

for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be

mass of Oxygen (O) = 0.405 x 3.34 g

mass of Oxygen (O) = 1.35 g

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4 years ago

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densk [106]

Explanation:

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Hope this helps.. 

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4 years ago

Read 2 more answers

PLZ HELP, Its The Nature of Oxidation and Reduction for chemistry

elixir [45]

Answer:

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So, from the given equation:

a. It is an oxidation reaction as Rb loses one elctron.

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c. It is a reduction reaction as H atom gains electrons.

d. It is an oxidation reaction as P loses 3 electrons.

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3 years ago