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Inessa05 [86]
3 years ago
8

I performed an experiment and mixed copper nitrate and potassium iodide. When they reacted, they formed a precipitate, even thou

gh solubility rules would not lead one to predict so because potassium nitrate is obviously soluble and so should copper (II) iodide. One can deduce from the formation of a precipitate that copper is reduced. Write a proposed reaction for the oxidation reduction of copper (II) iodide. Justify the choice of the substance that reduces the copper based on experimental evidence. Also, justify the choice using the atomic structure of potassium ion and iodide ion.
Chemistry
1 answer:
Alexus [3.1K]3 years ago
3 0

Answer:

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

Explanation:

The reaction performed in the experiment is;

2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2

The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.

The oxidation - reduction equation is as follows;

2Cu2^+ + 2e ----> 2Cu^+ reduction half equation

2I^- ----> I2 + 2e. Oxidation half equation

Balanced redox reaction equation;

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

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Consider the synthesis of ammonia 3 H2+ N2 à 2 NH3 If a student were to react 38.5 g of nitrogen gas, how many moles of ammonia
Wittaler [7]

Answer:

2.75 mol

Explanation:

Given data:

Mass of Nitrogen = 38.5 g

Moles of ammonia produced = ?

Solution:

Chemical  equation:

N₂ + 3H₂    →     2NH₃

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 38.5 g/ 28 g/mol

Number of moles = 1.375 mol

Now we will compare the moles of ammonia and nitrogen from balance chemical equation.

           N₂            :            NH₃

            1              :             2

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4 0
3 years ago
Would an opah fish be better suited to live in cold water or warm water? explain why
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A piece of mental with a mass of 106 g is placed in a 50 mL graduated cylinder.The volume rises from 20 mL to 31 mL. What is the
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Answer:

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Explanation:

Given parameters:

Mass of the metal  = 106g

Volume of cylinder = 50mL

Volume difference  = 31mL  - 20mL  = 11mL

Unknown:

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Solution:

To find the density of the metal, we use;

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