The answer is 33.33 %
The explanation:
According to the reaction equation:
MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)
we can see that 1 mole of MCO3 will produce → 1 mole of CO2
-Now we need o get number of mole of CO2:
and when we have 0.22 g of CO2, so number of mole = mass / molar mass
moles = 0.22 g / 44 g/mol = 0.005 mole
∴ moles of Mg = moles of CO2 = 0.005 mole
∴ mass of Mg = moles * molar mass
= 0.005 * 84 /mol = 0.42 g
∴ Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100
= 33.33 %
Answer:
The given compound cannot be cocaine.
Explanation:
The chemist can comment on the nature of compound being cocaine or not from the depression in freezing point.
Depression in freezing point of is related to molality as:
Depression in freezing point = Kf X molality
Where
Kf = cryoscopic constant = 4.90°C/m
depression in freezing point = normal freezing point - freezing point of solution
depression in freezing point = 5.5-3.9 = 1.6°C
1.6°C = 4.90 X molality

we know that:

therefore
moles = 0.327X0.008 = 0.00261 mol


The molar mass of cocaine is 303.353
So the given compound cannot be cocaine.
Raise the boiling temperature
Answer:
Oh come on. Look at all - then look at A.
Explanation: