The correct answer should be answer choice B
Answer:
0.294 M
Explanation:
The computation of the final molarity of acetate anion is shown below:-
Lead acetate = Pb(OAc)2
Lead acetate involves two acetate ion.
14.3 gm lead acetate = Mass ÷ Molar mass
= 14.3 g ÷ 325.29 g/mol
= 0.044 mole
Volume of solution = 300 ml.
then
Molarity of lead is
= 0.044 × 1,000 ÷ 300
= 0.147 M
Therefore the molarity of acetate anion is
= 2 × 0.147
= 0.294 M
Because noble gases already have full shells of electrons therefore meaning they do not need to react to form anything.Whereas if they had 1 electron short of full shells , they would become VERY reactive.
Answer:
34 g of NH₃ were produced in the reaction
Explanation:
N₂ (g) + 3H₂ (g) → 2NH₃ (g)
Moles of N₂ → 28 g / 28 g/m = 1 mol
Moles of H₂ → 25g / 2 g/m = 12.5 moles
Clearly, the limiting is the nitrogen.
1 mol of N₂ produced 2 moles of ammonia
So, If I have 1 mol, I'll produce the same amount
2 moles of NH₃ = Mol . Molar mass
2 m . 17 g/m = 34 g