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sergij07 [2.7K]
3 years ago
15

Pls help tyy How are NH 3 and O 2 reactants? Explain your answer:

Chemistry
2 answers:
White raven [17]3 years ago
8 0

Answer:

Here you go.

Explanation:

First of all, there are reactants and products…

Reactants are in the left side (in bold).

Products are on the right (without bold)

Example; NH3 + O2 → N2 + H2O

Vedmedyk [2.9K]3 years ago
6 0
reactants are the ones on the left side and products are the ones on the right side
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QUESTION 11
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NaCl and HCl both have chlorine but are different types of bonds.
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Answer:

In HCl ,bond between hydrogen and chlorine is covalent. But due to difference in electronegativity of hydrogen and chlorine,shared electron pair gets attracted towards chlorine. Due to this polar covalent

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What is different about most of the elements of the 7th period?
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Answer:

All elements of period 7 are radioactive. This period contains the actinides, which includes plutonium, the naturally occurring element with the heaviest nucleus; subsequent elements must be created artificially.

Explanation:

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3 years ago
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Which relationship below is an example of a predator-prey relationship? OA. A coyote eats a hen. OB. A rabbit eats grasses. O C.
Lostsunrise [7]

Answer:

A

Explanation: A wolf is a predator and a hen is prey.

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3 years ago
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A solution contains 0.021 M Cl and 0.017 M I. A solution containing copper (I) ions is added to selectively precipitate one of t
lidiya [134]

<u>Answer:</u> Copper (I) iodide will precipitate first.

<u>Explanation:</u>

We are given:

K_{sp} of CuCl = 1.0\times 10^{-6}

K_{sp} of CuI = 5.1\times 10^{-12}

Concentration of Cl^-\text{ ion}=0.021M

Concentration of I^-\text{ ion}=0.017M

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

  • <u>For CuCl:</u>

K_{sp}=[Cu^+][Cl^-]

Putting values in above equation, we get:

1.0\times 10^{-6}=[Cu^+]\times 0.021

[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M

Concentration of copper (I) ion = 4.76\times 10^{-5}M

  • <u>For CuI:</u>

K_{sp}=[Cu^+][I^-]

Putting values in above equation, we get:

5.1\times 10^{-12}=[Cu^+]\times 0.017

[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M

Concentration of copper (I) ion = 3.00\times 10^{-10}M

For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.

7 0
3 years ago
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