Explanation:
by considering that ch4 is limiting, we get that answer using mole ratios.
In which situation would hydrogen bonding be present?
2 answers:
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Answer:
Yes, precipitation of barium iodate will occur.
Explanation:
Molarity of barium nitrate solution = 0.050 M
Volume of barium nitrate solution =15.0 mL = 0.0150 L
1 mL = 0.001 L
Moles of barium nitrate = n
Moles of barium ions:
Molarity of potassium iodate solution = 0.10 M
Volume of potassium iodate solution =100.0 mL = 0.1000 L
1 mL = 0.001 L
Moles of potassium iodate = n'
Moles of iodate ions =
After mixing of both solution in 250 mL in erlenmeyer flask
Volume of the final solution = 250 mL = 0.250 L
Concentration of barium ions in 250 mL solution :
Concentration of iodate ions:
Solubility product of barium iodate,
Ionic product of the barium iodate in solution :
( precipitation)
As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.
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.