You can put a known amount sodium into some sort of time release mechanism such as a pill made from soluble material. Then you can place the sodium into a calorimeter with a known mass of water and record the temperature change the water undergoes during the reaction. Then you can use the equation q(water)=m(water)c(water)ΔT to find the amount of heat absorbed by the water. since the amount of heat absorbed by the water is the amount of heat released from the sodium, q(sodium)=-q(water). Than you can use the equation q(sodium)=m(sodium)c(sodium)ΔT and solve for c(sodium)
I hope this helps and feel free to ask about anything that was unclear in the comments.
150/30 = 5
HF1 20/2 = 10
HF2 10/2 = 5
HF3 5/2 = 2.5
HF4 2.5/2 = 1.25
HF5 1.25/2 = 0.625
Answer: 0.63g
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.
Answer:
in my opinion find some stuff to help u with it test if you are aloud to bring in a revision boom into your test and get a good night sleep if you can and try not to get stressed about it
