Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Answer:3844 cm^2
Explanation:
62 squared since it is a square, and area is base times height
Answer:
(c) The retention time would be higher (d) The retention time would be lower.
Explanation:
For the polar solutes which were separated using the hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase, the retention time would be higher if eluent were changed from 80 vol% to 90 vol% acetonitrile in water.
However, for the polar solutes which were separated using the normal-phase chromatography on bare silica with methyl t=butyl ether and 2-propanol solvent, the retention time would be lower if the eluent were changed from 40 vol% to 60 vol% 2-propanol.
Answer:1 mol of Mg(NO3)2 contains 6.022*10^23 molecules
3 mol Mg(NO3)2 contains 3*6.022*10^23 = 1.81*10^24 molecules
Each Mg(NO3)2 molecule contains 2 N atoms
Number of N atoms = 2*1.81*10^24 = 3.62*10^24 N atoms.
Density is defined as mass/volume (or m/v).
So,
(126.0 g)/(12.5 cm^3)= 10.08 g/cm^3
If your teacher requires correct significant figures, the answer is 10.1 g/cm^3.
If not, the first answer is fine.
