Answer:
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Explanation:
Step 1: Data given
Molarity of Na2CrO4 = 0.010 M
Molarity of NaBr = 2.5 M
Ksp(PbCrO4) = 1.8 * 10^–14
Ksp(PbBr2) = 6.3 * 10^–6
Step 2: The balanced equation
PbCrO4 →Pb^2+ + CrO4^2-
PbBr2 → Pb^2+ + 2Br-
Step 3: Define Ksp
Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]
1.8*10^-14 = [Pb^2+] * 0.010 M
[Pb^2+] = 1.8*10^-14 /0.010
[Pb^2+] = 1.8*10^-12 M
The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M
Ksp PbBr2 = [Pb^2+][Br-]²
6.3 * 10^–6 = [Pb^2+] (2.5)²
[Pb^2+] = 1*10^-6 M
The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Answer:
the measured amount of product that is made from a given amount of reactant
Explanation:
google & quizlet
We will use boiling point formula:
ΔT = i Kb m
when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35
and Kb is the boiling point constant =5.03
and m = molality
i = vant's Hoff factor
so by substitution, we can get the molality:
1.35 = 1 * 5.03 * m
∴ m = 0.27
when molality = moles / mass Kg
0.27 = moles / 0.015Kg
∴ moles = 0.00405 moles
∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol
Answer:
Explanation:
(a) Balanced equation
Cu + 2AgNO₃ ⟶ Cu(NO₃)₂+ 2Ag
(b) Calculation
You want to convert atoms of Cu to atoms of Ag.
The atomic ratio is ratio is 2 atoms Ag:1 atom Cu
Its d its the only one that makes sense
