1 year ago

Calculate the percent covalent character of a chemical bond between iodine and lithium.

Chemistry

1 answer:

raketka [301]1 year ago

3 0

Answer:

27966692

Explanation:

I don't know if this is right or not

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Answer:

$\boxed {\boxed {\sf 2.8 \ m }}$

Explanation:

The formula for molality is:

$m=\frac{moles \ of \ solute}{kg \ of \ solvent}$

There are 0.210 moles of KBr and 0.075 kilograms of pure water.

$moles= 0.210 \ mol \\kilograms = 0.075 \ kg$

Substitute the values into the formula.

$m= \frac{ 0.210 \ mol }{0.075 \ kg}$

Divide.

$m= 2.8 \ mol/kg= 2.8 \ m$

The molality is <u>2.8 moles per kilogram</u>

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