What amount of heat (in kJ) is required to convert 14.0 g of an unknown liquid (MM = 67.44 g/mol) at 43.5 °C to a gas at 128.2 °C? (specific heat capacity of liquid = 1.18 J/g･°C; specific heat capacity of gas = 0.792 J/g･°C; ∆Hvap = 30.1 kJ/mol; normal boiling point, Tb = 97.4°C)

1.24 kJ is required to convert 14 g of liquid from 43.5°C to 128.2°C

Explanation:

This is a typical calorimetry problem:

We have to assume, no heat is lost to sourrounding.

First of all, we need to go from 43.5°C to 97.4°C, the boiling point.

Q = Ce . m . ΔT

We replace data, 1.18° J/g . 14 g . (97.4°C - 43.5°C)

Heat for the first stage is: 890.4 Joules

Now we have to change the state, and we need the ΔH. As we do not have latent heat, we can proceed like this:

1 mol release 30.1 kJ at vaporization.

We convert the mass to moles → 14 g.  1mol/ 67.44g = 0.207 mol

0.207 mol will release (0.207 . 30.1 kJ) = 6.25 kJ

Now, we are at gaseous phase.

Q = Ce . m . ΔT → 0.792 J/g°C . 14g . (128.2°C - 97.4°C)

Q = 341.5 Joules

To determine the amount of heat, we sum all the obtained values:

890.4 Joules + 6250 Joules + 341.5 Joules = 1238.2 J

We convert to kJ →  1238.2 J . 1kJ / 1000J = 1.24 kJ

electron microscopy and scanning tunneling microscopes gives us a greater understanding of atoms by being able to see the atom itself, and lets us see some atomic reactions.

explanation:

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.