Calculate the heat required for the following two processes, and compare the results. a. A 100.0 g sample of solid ethanol melts at its melting point. Hfus = 4.94 kJ/mol

b. A 100.0 g sample of liquid ethanol vaporizes at its boiling point. Hvap = 38.6 kJ/mol

A. 10.74 KJ

B. 83.92 KJ

Explanation:

A. Data obtained from the question include the following:

Mass (m) of ethanol = 100g

Heat of fusion, Hfus = 4.94 kJ/mol

Heat (Q) =..?

Next, we shall determine the number of mole in 100g of ethanol, C2H5OH. This is illustrated below:

Mass of C2H5OH = 100g

Molar mass of C2H5OH = (2x12)+ (5x1) + 16 + 1 = 46g/mol

Number of mole (n) of C2H5OH =..?

Mole = Mass /Molar Mass

Number of mole (n) of C2H5OH = 100/46 = 2.174 moles.

Now, we can obtain the heat required to melt the ethanol as follow:

Q = n x Hfus

Q = 2.174 mol x 4.94 kJ/mol

Q = 10.74KJ

Therefore, 10.74 KJ of heat is required to melt the ethanol.

B. Data obtained from the question include the following:

Mass of C2H5OH = 100g

Heat of vaporisation, Hvap = 38.6 kJ/mol

Heat (Q) =..?

As calculated above, the number of mole in 100g of ethanol, C2H5OH is 2.174 moles.

The heat required to vaporise the ethanol can be obtained as follow:

Q = n x Hvap

Q = 2.174 mol x 38.6 kJ/mol

Q = 83.92 KJ

Therefore, 83.92 KJ of heat is required to vaporise the ethanol.

C. From the above calculations, a higher amount of heat energy i.e 83.92 KJ is required to vaporise the ethanol and a lesser amount of heat energy i.e 10.74 KJ is needed to melt the ethanol.

that's what i got you can count the number of atoms

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3