Answer:
The answer to your question is 280 g of Mg(NO₃)₂
Explanation:
Data
Efficiency = 30.80 %
Mg(NO₃)₂ = ?
Magnesium = 147.4 g
Copper (II) nitrate = excess
Balanced Reaction
Mg + Cu(NO₃)₂ ⇒ Mg(NO₃)₂ + Cu
Reactants Elements Products
1 Mg 1
1 Cu 1
2 N 2
6 O 6
Process
1.- Calculate the theoretical yield
Molecular weight Mg = 24
Molecular weight Mg(NO₃)₂ = 24 + (14 x 2) + (16 x 6)
= 24 + 28 + 96
= 148 g
24 g of Mg -------------------- 148 g of Mg(NO₃)₂
147.4 g of Mg ------------------- x
x = (147.4 x 148) / 24
x = 908.96 g of Mg(NO₃)₂
2.- Calculate the Actual yield
yield percent = 
Solve for actual yield
Actual yield = Yield percent x Theoretical yield
Substitution
Actual yield =
x 908.96
Actual yield = 279.95 ≈ 280g
The molarity of a Ba(OH)2 solution required to prepare a1.0 OH- solution is calculated as follows
write the equation for dissociation of Ba(Oh)2
that is,
Ba(Oh)2 -----> Ba^2+ + 2Oh-
by use of reacting ratio between Ba(Oh)2 to Oh which is 1:2 the molarity of Ba(oh)2 = 1.0/2 = 0.5 M
It's B, because a negative charged Ion is an anion. And since it's 1- ,one valence electron is been added to H. So it has 2 valence electrons.
Using the general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
(105*220)/275 = (P₂*95)/310
(105*220)*310/275 = P₂*95
(105*220*310)/(275*95) = P₂
274.11 = P₂
P₂ = 274.11 kPa
New Pressure = 274.11 kPa