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geniusboy [140]
3 years ago
12

In the liquid and solid states, molecules are held together by attractions called intermolecular forces. there are several types

of intermolecular forces. london dispersion forces, found in all substances, result from the motion of electrons. these work to attract both polar and nonpolar molecules to one another via instantaneous dipole moments. dipole-dipole forces arise from molecular dipole moments. ion-dipole forces result from the interaction of an ion and a molecular dipole. hydrogen-bond forces result from the attraction of a hydrogen atom bonded to a small highly electronegative atom (n, o, and f) and the unshared electron pairs of another electronegative atom physical properties such as boiling point, melting point, vapor pressure, viscosity, and surface tension are all affected by the strength of the intermolecular forces within a substance. part a what happens to these physical properties as the strength of intermolecular forces increases?
Chemistry
1 answer:
BaLLatris [955]3 years ago
3 0
<span>Since these molecules are all non-polar, the only intermolecular force of attraction will be London dispersion forces. Since these increase by the size of the molecule, the boiling points will decrease in the same order: Parafin > Heptadecane > hexane > 2,2-dimethylbutane > propane For these two, hexane > 2,2-dimethylbutane, dispersion forces are greater in a molecule which is longer and unbranched compared to one which is branched and more compact.</span>
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Answer: -15.4^00C

Explanation:

T^0_f-T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_f = boiling point of solution = ?

T^o_f = boiling point of solvent (X) = -10.1^oC

k_f = freezing point constant  = 5.32^oC/kgmol

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte like urea)

= mass of solute (urea) = 29.82 g

= mass of solvent (X) = 500.0 g

M_2 = molar mass of solute (urea) = 60 g/mol

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

(-10.1-(T_f))^oC=1\times (5.32^oC/m)\times \frac{(29.82g)\times 1000}{60\times (500.0g)}

T_f=-15.4^0C

Therefore, the freezing point of  solution is -15.4^0C

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3 years ago
If 17. 6 g of hcl are used to produce a 14. 5 l solution, what is the ph of the solution?.
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This problem is providing us with the mass of hydrochloric acid and the volume of solution and asks for the pH of the resulting solution, which turns out to be 1.477.

<h3>pH calculations</h3>

In chemistry, one can calculate the pH of a solution by firstly obtaining its molarity as the division of the moles of solute by the liters of solution, so in this case for HCl we have:

M=\frac{17.6g*\frac{1mol}{36.46g} }{14.5L} \\\\M=0.0333 M

Next, due to the fact that hydrochloric acid is a strong acid, we realize its concentration is nearly the same to the released hydrogen ions to the solution upon ionization. Thereby, the resulting pH is:

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Learn more about pH calculations: brainly.com/question/1195974

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