Answer is: A. 1.1 3 1023 NiCl2 formula units.
m(NiCl₂) = 24.6 g; mass of nickel(II) chloride.
M(NiCl₂) = 129.6 g/mol; molar mass of nickel(II) chloride.
n(NiCl₂) = m(NiCl₂) ÷ M(NiCl₂).
n(NiCl₂) = 24.6 g ÷ 129.6 g/mol.
n(NiCl₂) = 0.19 mol; amount of nickel(II) chloride.
Na = 6.022·10²³ 1/mol; Avogadro constant.
N(NiCl₂) = n(NiCl₂) · Na.
N(NiCl₂) = 0.19 mol · 6.022·10²³ 1/mol.
N(NiCl₂) = 1.13·10²³; number of formula units.
It is C 2 1 2 1
You have to 1st balance the nitrates then balance the silvers to get the coefficients
Answer:
The answer to your question is: letter c
Explanation:
Data
V1 = 612 ml n1 = 9.11 mol
V2 = 123 ml n2 = ?
Formula


n2 = 1.83 mol