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.
<span>orbital shell is the</span><span> the circular paths around the nucleus of an atom along which the electrons traverse.</span>
Explanation:
The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion. The oxidation number of oxygen in a compound is usually –2. The oxidation state of hydrogen in a compound is usually +1.
The oxidation state of Al in Al(OH)
4
−
x+4(+1−2)=−1
∴x=+3
The oxidation state of Mn in MnO
2
y+2(−2)=0
∴y=+4
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