Answer:
B: Unit conversions express an amount in a different unit.
Explanation:
The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.
A(t) = A(o) (1/2)^(t/d)
where t is the certain period of time. Substituting the known values,
A(t) = (20 mg)(1/2)^(85.80/14.30)
Solving,
A(t) = 0.3125 mg
Hence, the answer is 0.3125 mg.
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.
The surrounding ecosystem in and around the water
Answer:
a food pyramid.
Explanation:
the food pyramid shows how autotrophs are at the bottom and how the top predators are at the top, in this case the eagle.
