Answer:
94.2 g/mol
Explanation:
Ideal Gases Law can useful to solve this
P . V = n . R . T
We need to make some conversions
740 Torr . 1 atm/ 760 Torr = 0.974 atm
100°C + 273 = 373K
Let's replace the values
0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K
n will determine the number of moles
(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)
n = 0.032 moles
This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?
Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol
Answer:
1. CBr4
2. C5H12
3. CF4
4. C8H17NH2
Explanation:
1. The boiling point of a molecule depends on its structure. Because the boiling point of molecules of similar size, depends on the differences in functional groups of the molecule.
For example CBr4 have a higher boiling point than other compounds because they have both London dispersion forces and dipole-dipole interactions.
2. Non-polar molecules like C5H12 will have low melting and boiling points, because they are held together by the weak van der Waals forces.
The deeper the diver takes the helium balloon, the more it reduces in size. This is due to the pressure of the water column above pressing on the balloon. According to Boyle’s law (P= k*1/V.), as the volume of the balloon decreases, the pressure of the helium inside increases.
