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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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Draw the structure for 1-nitrobutane, making sure to add all non-zero formal charges.

TiliK225 [7]

The molecular structure of 1-nitrobutane is $C_{4} H_{9} NO_{2}$ . The structure of 1-nitrobutane is shown below.

An atom's formal charge would be determined by the covalent model of chemical bonding, which assumes that almost all chemical bonds include equal sharing of electrons among all atoms, regardless their relative electronegativity.

The structure for 1-nitrobutane, making sure to add all non-zero formal charges

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It is a kind of alkane in with nitro group is attached with alkane group.

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1 year ago

The teacher tells your group to make a stock solution of sodium chloride, and then diluting it to

sergejj [24]

The idea here is that you need to figure out how many moles of magnesium chloride,

MgCl

, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.

As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.

So, how many moles of magnesium chloride must be present in the target solution?

⇒

⋅

0.158 M

⋅

250.0

⋅

−

0.0395 moles MgCl

Now determine what volume of the target solution would contain this many moles of magnesium chloride

⇒

0.0395

moles

3.15

moles

0.01254 L

Rounded to three sig figs and expressed in mililiters, the volume will be

12.5 mL

So, to prepare your target solution, use a

12.5-mL

sample of the stock solution and add enough water to make the volume of the total solution equal to

250.0 mL

This is equivalent to diluting the

12.5-mL

sample of the stock solution by a dilution factor of

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