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3 years ago

Which of these elements would form an ionic bond between them?

Chemistry

2 answers:

Artyom0805 [142]3 years ago

7 0

Answer : The elements form an ionic bond will be, (C) Ca and Br

Explanation:

Ionic bond : It is defined as the bond that is formed by complete transfer of electrons from one atom to another atom.

That means, the atom which looses the electron is known as electropositive atom and the atom which gains the electron is known as electronegative atom.

The ionic bond is usually formed between a metal and a non-metal.

Covalent bond : It is defined as the bond which is formed when sharing of electrons takes place between the atoms.

This is usually formed between two non-metals.

(a) Chlorine and fluorine

In this pair, fluorine is a non-metal and chlorine is also non-metal. So, chlorine and fluorine will form a covalent bond.

(b) Chlorine and chlorine

In this pair, chlorine is a non-metal. So, chlorine and chlorine will form a covalent bond.

(c) Calcium and bromine

In this pair, calcium is a metal and bromine is a non-metal. So, calcium and bromine will form an ionic bond.

(d) Calcium and magnesium

In this pair, calcium and magnesium are metals. So, no bond form between calcium and magnesium.

Hence, the elements form an ionic bond will be, (C) Ca and Br

Pepsi [2]3 years ago

6 0

C. Ca and Br because they're metals and nonmetals

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Answer:

Explanation:

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Sphinxa [80]

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shutvik [7]

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3 years ago

You wish to add 5 mg/l naocl as cl2 to a solution in a disinfection test, and you have a stock solution (household bleach) that

Alecsey [184]

Answer: -

0.1 ml of bleach should be added to each liter of test solution.

Explanation:-

Let the volume of bleach to be added is B ml.

Density of stock solution = 1.0 g/ml

Mass of stock solution = Volume of stock x density of stock

= B ml x 1.0 g/ml

= B g

Amount of NaOCl in this stock solution = 5% of B g

= $\frac{5}{100}$ x B g

= 0.05 B g

Now each test solution must be added 5 mg/l NaOCl.

Thus each liter of test solution must have 5 mg.

Thus 0.05 B g = 5 mg

= 0.005 g

B = $\frac{0.005}{0.05}$

= 0.1

Thus 0.1 ml of bleach should be added to each liter of test solution.

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