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.
Answer:
Mercury
Explanation:
Mercury is a liquid element that contains 80 electrons when not charged.
A) The limiting reactant is Al
b) Br2 is the excess reactant
c) The amount moles of AlBr3 that get produced will be equal to the number of moles of Al to begin with.
d) 0
amino group
carboxyl group
R-group
single Hydrogen arom
Explanation:
The structures of both acetone and propanal are shown below:
In the formula of propanal there is -CHO functional group at the end.
In acetone -CO- group is present in the middle that is on the second carbon.
The molecular formula is C3H6O.
Both have same molecular formula but different structural formulas.
