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.
Answer:
80 liters
Explanation:
At STP, 1 mole of ideal gas has a volume of 22.4 liters.
Therefore, since liters and moles are directly proportional, we can use stoichiometry directly.
40L CH₄ × (2L H₂ / 1L CH₄) = 80L H₂
When a atom loses electrons it becomes a ion
