It will also be halved... but the relationship is with Kelvin not c. So 273 +273 = 546 K. half of that is 273 K, then subtract 273... you get 0 degrees c
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.
An organism that does not move on its own and makes food from its environment is a producer.
Reaction:

Magnesium is a stronger reducing agent than copper and is thus able to reduce copper(II) oxide.
Products of the reaction: Magnesium oxide and metallic copper.