it has less tightly bound electrons, is able to lose electron easily as compare to metal B at it has 4 unpaired electron in 3d sub-shell.
When HCl is added to metal ions, metal chlorides are produced. In this problem, it is asked whether the given ions precipitate or not when added to HCl. According to the rule, all chlorides except Ag+, Pb 2+, Hg2 2+ are soluble. Hence the ion that would precipitate is only lead (II) ion.
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
2.1 x 10 ^23
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Answer:
=154.8 J
Explanation:
The rise in temperature is contributed by the change in temperature.
Change in enthalpy = MC∅, where M is the mass of the substance, C is the specific heat capacity and ∅ is the change in temperature.
Change in temperature = 100.0°C-20.0°C=80°C
ΔH=MC∅
The specific heat capacity of gold= 0.129 J/g°C
ΔH= 15.0g×0.129J/g°C×80°C
=154.8 J
