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3 years ago

Which of the following molecules would exhibit only London forces?a)CH4, BH3, and CCl4b)H2O, NH3, and CCl4c)PH3, NH3, and CCl4

Chemistry

1 answer:

Alex787 [66]3 years ago

4 0

Answer:

a)CH₄, BH₃, and CCl₄

Explanation:

London dispersion forces:-

The bond for example, in the molecule is F-F, which is non-polar in nature because the two fluorine atoms have same electronegativity values.

The intermolecular force acting in the molecule are induced dipole-dipole forces or London Dispersion forces / van der Waals forces which are the weakest intermolecular force.

Out of the given options, H₂O , NH₃ exhibits hydrogen bonding which is:-

Hydrogen bonding:-

Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.

Thus option B and C rules out.

Hence, the correct option which represents the molecules which would exhibit only London forces is:- a)CH₄, BH₃, and CCl₄

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How many milliliters of an aqueous solution of 0.170 M ammonium carbonate is needed to obtain 16.1 grams of the salt

Citrus2011 [14]

There will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.There will be needed mL of

Why?

In order to calculate how many milliliters are needed to obtain 16.1 grams of the salt given its concentration, we first need to find its chemical formula which is the following:

$(NH_{4})2CO_{3}$

Now that we know the chemical formula of the substance, we need to find its molecular mass. We can do it by the following way:

$N_{2}=14g*2=28g\\\\2H_{4}=2*1g*4=8g\\\\C=12.01g*1=12.01g\\\\O_{3}=15.99g*3=47.97g$

We have that the molecular mass of the substance will be:

$MolecularMass=\frac{28g+8g+12.01g+47.97g}{mol}=95.98\frac{g}{mol}$

Therefore, knowing the molecular mass of the substance, we need to calculate how many mols represents 16.1 grams of the same substance, we can do it by the following way:

$mol_{(NH_{4})2CO_{3}=\frac{mass_{(NH_{4})2CO_{3}}}{molarmass_{(NH_{4})2CO_{3}}}$

$mol_{(NH_{4})2CO_{3}=\frac{16.1g}{95.98\frac{g}{mol}}=0.167mol$

Finally, if we need to calculate how many milliliters are needed, we need to use the following formula:

$M=\frac{moles_{solute}}{volume_{solution}}$

$M=\frac{moles_{solute}}{volume_{solution}}\\\\volume_{solution}=\frac{moles_{solute}}{M}$

Now, substituting and calculating, we have:

$volume_{solution}=\frac{0.167mol}{0.170\frac{mol}{L}}\\\\volume_{solution}=0.982L=0.982L*1000=982.35mL$

Henc, there will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.

Have a nice day!

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