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
![molality=\frac{1.6}{4.90} = 0.327 m](https://tex.z-dn.net/?f=molality%3D%5Cfrac%7B1.6%7D%7B4.90%7D%20%3D%200.327%20m)
we know that:
![molality=\frac{moles}{mass of solvent(kg)}=\frac{moles}{0.008kg}](https://tex.z-dn.net/?f=molality%3D%5Cfrac%7Bmoles%7D%7Bmass%20of%20solvent%28kg%29%7D%3D%5Cfrac%7Bmoles%7D%7B0.008kg%7D)
therefore
moles = 0.327X0.008 = 0.00261 mol
![moles=\frac{mass}{molarmass}](https://tex.z-dn.net/?f=moles%3D%5Cfrac%7Bmass%7D%7Bmolarmass%7D)
![molarmass=\frac{mass}{moles}=\frac{0.50}{0.00261}= 191.57g/mol](https://tex.z-dn.net/?f=molarmass%3D%5Cfrac%7Bmass%7D%7Bmoles%7D%3D%5Cfrac%7B0.50%7D%7B0.00261%7D%3D%20191.57g%2Fmol)
The molar mass of cocaine is 303.353
So the given compound cannot be cocaine.
All the objects are formed from the gas and dust orbitting the sun
2.a)R b)R c)L d)L e)R f)L g)L
The answer would be 118.68 g.
Explanation for this is:4 moles of NH3 give 4 moles of NO2
so 1mole of NH3 will give 1 mole of NO2
43.9 grams of NH3 contains 2.58 moles
so 2.58 moles will be produced of NO2
which is 118.7 grams this the amount of oxygen that is used.
the answer is 360 g H2O with a 90.3 yeild percentage