What mass (in grams) of mg(no3)2 is present in 151 ml of a 0.350 m solution of mg(no3)2? a) 7.84 gb) 0.0529 gc) 4.56 gd) 64.0 g

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.

*tilts head* sounds more like a statement

answer

fusion: the energy source of stars. the energy released from the collapse of the gas into a protostar causes the center of the protostar to become extremely hot. when the core is hot enough, nuclear fusion commences. fusion is the process where two hydrogen atoms combine to form a helium atom, releasing energy.

hope it