Answer is: 4.826 grams of anhydrous nickel(II) sulfate could be obtained.
m(NiSO₄×7H₂O) = 8.753 g; mass of heptahydrate.
m(NiSO₄×6H₂O) = 8.192 g; mass of hexahydrate.
m(H₂O) = m(NiSO₄×7H₂O) - m(NiSO₄×6H₂O).
m(H₂O) = 8.753 g - 8.192 g.
m(H₂O) = 0.561 g.
m(NiSO₄) = m(NiSO₄×7H₂O) - 7 · m(H₂O).
m(NiSO₄) = 8.753 g - 7 · 0.561 g.
m(NiSO₄) = 4.826 g.
1. Ionic Bond
2.It is a chemical reaction because the reactants react to create products and it is endothermic because it takes in heat from the sunlight which helps it to photosynthesise
Hope that helps ☻
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
