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3 years ago

If the electronnegavity difference between elements A and X is 0.8, the bond in AX will most likely be what?

Chemistry

1 answer:

siniylev [52]3 years ago

8 0

Answer:

Polar covalent bond

Explanation:

When the electronegativity difference between two elements A and X is 0.8, the bond AX formed will most likely be a polar covalent bond.

A polar covalent bond is one whose electronegativity difference is between 0.5 and 2.1.

In such a bond type, we have heteronuclear species with one of the species having a higher electronegativity value.

When electronegativity difference is less than 0.5, a non-polar covalent bond forms.
If the value is greater than 2.1, an ionic bond will form.

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Calculate the mole fraction of kbr (molar mass 119.00 g/mol) in a solution made by dissolving 0.30 g kbr in 0.400 l of H2O (d =

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The mole fraction of KBr in the solution is 0.0001

<h3>How to determine the mole of water</h3>

We'll begin by calculating the mass of the water. This can be obtained as follow:

Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
Density of water = 1 g/mL
Mass of water =?

Density = mass / volume

1 = Mass of water / 400

Croiss multiply

Mass of water = 1 × 400

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Finally, we shall determine the mole of the water

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Mole of water = 400 / 18.02

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<h3>How to de terminethe mole of KBr</h3>

Mass of KBr = 0.3 g
Molar mass of KBr = 119 g/mol
Mole of KBr = ?

Mole = mass / molar mass

Mole of KBr = 0.3 / 119

Mole of KBr = 0.0025 mole

<h3>How to determine the mole fraction of KBr</h3>

Mole of KBr = 0.0025 mole
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Mole fraction = mole / total mole

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