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Tanzania [10]
3 years ago
6

Which intermolecular force would affect melting point the least?

Chemistry
1 answer:
scoray [572]3 years ago
7 0

Answer: B Is the answer

JupiterJ
3 years ago
what is B bro, they switch the answers
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3 years ago
The heat of vaporization Δ1, of toluene (C6H5CH3) is 38.1 kJ/mol. Calculate the change in entropy AS when 207. g of toluene boil
AveGali [126]

Answer: The change in entropy change when 207 g of toluene boils at 110.6^{o}C is 223 J/K.

Explanation:

It is given that mass of toulene is 207 g and its molar mass is 92.14 g/mol. So, its moles will be calculated as follows.

     No. of moles = \frac{mass}{\text{molar mass}}

                           = \frac{207 g}{92.14 g/mol}

                           = 2.25 moles

As we are given that heat of vaporization for 1 mol is 38.1 kJ/mol. So, heat of vaporization for 2.25 moles will be calculated as follows.

            2.25 moles \times 38.1 KJ/mol

            = 85.725 KJ

Now, we know that the relation between enthalpy change and entropy change is as follows.

           \Delta S = \frac{\Delta H}{\Delta T}

                        = \frac{85.725 kJ}{(110.6 + 273) K}

                        = 0.223 kJ/K

or,                     = 223 J/K         (as 1 kJ = 1000 J)

Thus, we can conclude that the change in entropy change when 207 g of toluene boils at 110.6^{o}C is 223 J/K.

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4 years ago
The quantity, q grams, of the radioactive isotope strontium-90 present in a sample decreases by half over a period of about t =
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