1) Zn(CH₃COO)₂(s) + 2KOH(aq) = Zn(OH)₂(s) + 2CH₃COOK(aq)
Ksp{Zn(OH)₂}=1.2*10⁻¹⁷
2) Zn(CH₃COO)₂(s) + 2NaCN(aq) = Zn(CN)₂(s) + 2CH₃COONa(aq)
Ksp{Zn(CN)₂}=2.6*10⁻¹³
Ksp{Zn(OH)₂}<Ksp{Zn(CN)₂}
Zn(OH)₂ precipitates first
D) + ΔH and +ΔE
Given this is one of the answer choices
First, calculating the amount of zinc carbonate by multiplying the given mass of the ore by the percent zinc carbonate.
mass of zinc carbonate = (976 tons)(0.78)
= 761.28 tons
Then, calculate for %Zn in zinc carbonate
%Zn = ((65.38) / (125.39)) x 100%
% Zn = 52.141%
tons of Zn from the sample,
mass of Zn = (761.28 tons)(0.52141)
= 396.94 tons Zn
<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C
14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O
molar mass of C=12 g/mole
molar mass of H=1 g/mole
0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C
0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H
mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O
molar mass of O=16 g/mole
9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602
C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602
C2.66H2.66O1 is the empirical formula;
to obtain whole numbers multiply by 3
3[C2.66H2.66O1] = C8H8O3
above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu
The empirical formula weight and the molecular formula weight are the same .
Molecular formula is C8H8O3.</span>
