Answer:
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We are given that the concentration of the KCl is 2 meq /
mL. Assuming that the ampule also has exactly this concentration, therefore:
amount of KCl in ampule = (2 meq / mL) * (20 mL)
amount of KCl in ampule = 40 meq
This amount of KCl is now inside a solution of 1 Liter (also
equivalent to 1000 mL), therefore the new concentration in the resulting
solution is:
new concentration = 40 meq / 1000 mL
new concentration = 0.04 meq / mL
Since 0.04 in decimal is 4% in percentage, therefore the
strength of the resulting solution is 4% KCl.
Answer:
286 J/K
Explanation:
The molar Gibbs free energy for the vaporization (ΔGvap) is:
ΔGvap = ΔHvap - T.ΔSvap
where,
ΔHvap: molar enthalpy of vaporization
T: absolute temperature
ΔSvap: molar entropy of the vaporization
When T = Tb = 64.7 °C = 337.9 K, the reaction is at equilibrium and ΔGvap = 0.
ΔHvap - Tb . ΔSvap = 0
ΔSvap = ΔHvap/Tb = (71.8 × 10³ J/K.mol)/ 337.9 K = 212 J/K.mol
When 1.35 mol of methanol vaporizes, the change in the entropy is:
Cool liquid from 314 K to 273 K, freeze liquid at 273 K, and cool solid to 263 K.