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Softa [21]
3 years ago
8

Which compound is composed of oppositely charged ions? A. OCl2 B. Na2O C. NH3 D. SCl2

Chemistry
2 answers:
Roman55 [17]3 years ago
6 0

<u>Answer:</u> The correct answer is Option B.

<u>Explanation:</u>

Compound having oppositely charged ions are considered as ionic compounds.

Ionic compounds are formed when complete transfer of electrons takes place between the atoms forming a bond. This bond is formed between a metal and a non-metal or a polyatomic cation and an non-metal or a metal and a polyatomic anion. <u>For Example:</u> NaCl,NH_4Cl,BaSO_4 etc..

Covalent compounds are formed when sharing of electrons takes place between the atoms forming a bond. This bond is formed between two non-metals. <u>For Example:</u> H_2O,SO_2 etc..

For the given options:

<u>Option A:</u>  OCl_2

Oxygen and chlorine both are non-metals. Thus, it is forming a covalent compound.

<u>Option B:</u>  Na_2O

Sodium is a metal and oxygen is a non-metal. Thus, it is forming an ionic compound and contain oppositely charged ions.

<u>Option C:</u>  NH_3

Nitrogen and hydrogen both are non-metals. Thus, it is forming a covalent compound.

<u>Option D:</u>  SCl_2

Sulfur and chlorine both are non-metals. Thus, it is forming a covalent compound.

Hence, the correct answer is Option B.

vazorg [7]3 years ago
5 0
It would be B, so yeah hope that helps ya ;)

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The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

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<u>Answer:</u> The concentration of copper fluoride in the solution is 4.90\times 10^{-3}mol/L

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To calculate the molarity of solute, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

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\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L

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