Answer:
Concentration of 
= 0.03101 M
Concentration of 
= 0.03721 M
Explanation:
A)
The reduction for is;
6 moles of 
= 1 mole of
number of moles of Cu reacted = 
number of moles of Cu reacted = 
number of moles of Cu reacted = 0.00747 mole
number of moles of reacted = 
number of moles of reacted = 0.001245 mole
Concentration of = 
Given that the volume = 40.15 mL = 
; we have:
Concentration of = 
Concentration of = 0.03101 M
B)
The reduction for is;
5 moles of = 1 mole of
number of moles of Cu reacted =
number of moles of Cu reacted =
number of moles of Cu reacted = 0.00747 mole
number of moles of reacted = 
number of moles of reacted = 0.001494 mole
Concentration of =
Given that the volume = 40.15 mL = ; we have:
Concentration of = 
Concentration of = 0.03721 M
Answer:
question 1: 3
question 2: the number of Valence electrons in the atom
hope it helps
The answer I'm pretty sure is a because it can't be d because ionic binding the electrons are affected and it can't be c because that's covalent bonding and it can't be b because they don't swap electrons.
1) Reaction: 3Mg + N₂ → Mg₃N₂.
m(Mg) = 0,225 g
n(Mg) = 0,225 g ÷ 24,3 g/mol = 0,009 mol.
n(Mg) : n(N₂) = 3 : 1
n₁(N₂) = 0,003 mol.
n₂(N₂) = 0,5331 ÷ 28 = 0,019 mol.
n₃(N₂) = 0,019 mol - 0,003 mol = 0,016, m(N₂) = 0,016mol·28g/mol=0,4467g.
or simpler: m(N₂) = 0,225 g + 0,5331 - 0,3114 g = 0,4467 g.
2) Answer is: 6 <span>of fluorine atoms are combined with one uranium atom.
</span>m(U) = 209 g.
m(F) = 100 g.
n(U) = m(U) ÷ M(U)
n(U) = 209 g ÷ 238 g/mol.
n(U) = 0,878 mol.
n(F) = m(F) ÷ M(F)
n(F) = 5,263 mol
n(U) : n(F) = 0,878 mol : 5,263 mol /:0,878.
n(U) : n(F) = 1 : 6.
n - amount of substance
