Answer:
what are we to do there coz I can't figure anything out oo
Answer is: A) 7.84 g.
V(Mg(NO₃)₂) = 151 mL ÷ 1000 mL/L.
V(Mg(NO₃)₂) = 0.151 L; volume of the magnesium nitrate.
c(Mg(NO₃)₂) = 0.352 M; molarity of the solution.
n(Mg(NO₃)₂) = V(Mg(NO₃)₂) · c(Mg(NO₃)₂).
n(Mg(NO₃)₂) ) = 0.151 L · 0.352 mol/L.
n(Mg(NO₃)₂) = 0.0531 mol; amount of the substance.
M(Mg(NO₃)₂) = Ar(Mg) + 2Ar(N) + 6Ar(O) · g/mol.
M(Mg(NO₃)₂) = 24.3 + 2·14 + 6·16 · g/mol.
M(Mg(NO₃)₂) = 148.3 g/mol; molar mass.
m(Mg(NO₃)₂) = n(Mg(NO₃)₂) · M(Mg(NO₃)₂).
m(Mg(NO₃)₂) = 0.0531 mol · 148.3 g/mol.
m(Mg(NO₃)₂) = 7.84; mass of magnesium nitrate.
Explanation:
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The percentage by mass of oxygen in the compound
find the total mass=( 1.900+ 0250 +0.850) = 3
the percentage mass mass of oxgyen/total mass x100
that is (0.850/3) x100=28.33%
