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Answer: The answer is A. A conductor that allows electricity to flow easily
corrected question: A chemist adds 135mL of a 0.21M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Round your answer to significant digits.
Answer:
5.37g
Explanation:
0.21M means ; 0.21mol/dm³
1dm³=1L , so we can say 0.21mol/L
if 0.21mol of Zinc nitrate is contained in 1L of water
x will be contained in 135mL of water
x= 0.21*135*10³/1
=0.02835moles
number of moles= mass/ molar mass
mass= number of moles *molar mas
molar mass of Zn(NO₃)₂=189.36 g/mol
mass= 0.02835 *189.36
mass=5.37g
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