From the equation above the reacting ratio of KClO3 to O2 is 2:3 therefore the number of moles of oxygen produced is ( 4 x3)/2 = 6 moles since four moles of KClO3 was consumed
mass=relative formula mass x number of moles
That is 32g/mol x 6 moles =192grams
Answer:
In 1 mol of Pb₃(PO₄)₄ occupies 1001.48 grams
Explanation:
This compound is the lead (IV) phosphate.
Grams that occupy 1 mole, means the molar mass of the compound
Pb = 207.2 .3 = 621.6 g/m
P = 30.97 .4 = 123.88 g/m
O = (16 . 4) . 4 = 256 g/m
621.6 g/m + 123.88 g/m + 256 g/m = 1001.48 g/m
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
