Answer:
The molality of isoborneol in camphor is 0.53 mol/kg.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample = 
= 165°C
Depression in freezing point = 
Depression in freezing point is also given by formula:
= The freezing point depression constant
m = molality of the sample
i = van't Hoff factor
We have: = 40°C kg/mol
i = 1 ( organic compounds)
The molality of isoborneol in camphor is 0.53 mol/kg.
Solids are usually more dense than liquids and gases.
Well depending on how many people on one side of the family, you would inherite their trait more commonly.
Answer:
6L
Explanation:
<em>if it's 3L per 200kPa</em>
then it would be;
4L per 300kPa
5L per 400kPa
6L per 500kPa
that's how i'd work it out in my head, hope it helps, but not sure though!
Answer:
1670 ml
Explanation:
molarity x Volume (Liters) = moles => Volume (Liters) = moles/Molarity
Volume needed = 2.50mol/1.50M = 1.67 Liters = 1670 ml.
