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3 years ago

WILL MARK BRAINIEST

Chemistry

2 answers:

lions [1.4K]3 years ago

4 0

Answer:

The correct option is B. The nonpolar substance has a lower boiling point than the polar substance.

Explanation:

Substances with hydrogen bonding, an intermolecular force, will have much higher boiling points than those that have ordinary dipole-dipole intramolecular forces. Non-polar molecules have lower boiling points, because they are held together by the weak van der Waals forces.

Note

However, if size of non polar increases, then its boiling point increases due to increases in van der walls forces with large size.

Alinara [238K]3 years ago

3 0

The correct answer is letter B: the non polar substance has a lower boiling point than the polar substance.

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C.
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2 years ago

How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo

riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

$(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)$

The expression used will be:

$\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = heat released by the reaction = ?

m = mass of benzene = 125 g

$c_{p,g}$ = specific heat of gaseous benzene = $1.06J/g^oC$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC$

$\Delta H_{vap}$ = enthalpy change for vaporization = $33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g$

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

$\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]$

$\Delta H=-67682.5J=-67.7kJ$

Therefore, the energy removed must be, -67.7 kJ

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If 20.0 g LIOH react with excess KCl, giving a 17.0

adell [148]

The yield of lithium chloride is 1.92 grams.

Option D.

<h3>Explanation:</h3>

In this reaction, we can see that 1 mole of lithium hydroxide reacts with 1 mole of potassium chloride to produce 1 mole of lithium chloride and 1 mole of potassium hydroxide.

Molecular weight of lithium hydroxide is 24.

Molecular weight of lithium chloride is 42.5.

So 24 grams of lithium hydroxide produces 42.5 grams of lithium chloride.

So, 20 grams of lithium hydroxide produces $20 * 24 / 42.5$ grams =11. 29 grams of lithium chloride.

But this is when the yield is 100%.

But yield is 17%.

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