Answer: I think the answer is C)
Explanation:
Answer:
14.7838
Explanation:
Given that;
A student uses a solution that contains 16 g of water to conduct an experiment and at the end of an hour ; the amount have decreased by 3.5%
The mathematically illustration for this is given as
= 16 g × 3.5%
= 
= 0.56
After an hour; the amount have decreased by 16 - 0.56
= 15.44
after 2 hours the water decreased by another 4.25%
i.e 
⇒ 0.6562
after two hours , the amount would have decreased by:
= 15.44 - 0.6562
= 14.7838
A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the
solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.
Separate the components of a mixture.
Answer:
It’s twice as much as 9+10
Explanation:
