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.
Answer:
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Explanation:
Answer:
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Explanation:
PLEASE NOTE; I HAVE WRITTEN THE ATOMIC NUMBER IN BRACKETS, NEXT TO THE SYMBOL OF THE ELEMENT
Explanation:
Mass of fructose = 33.56 g
Mass of water = 18.88 g
Total mass of the solution = Mass of fructose + Mass of water = M
M = 33.56 g + 18.88 g =52.44 g
Volume of the solution = V = 40.00 mL
Density =
a) Density of the solution:

b) Molar mass of fructose = 180.16 g/mol
Moles of fructose = 
Molar mass of water = 18.02 g/mol
Moles of water= 
Mole fraction of fructose in this solution:


Mole fraction of water = 
c) Average molar mass of of the solution:
=

d) Mass of 1 mole of solution = 42.50 g/mol
Density of the solution = 1.311 g/mL
d) Specific molar volume of the solution:


Answer:C
Explanation: they break apart into ions as they dissolve into solution this process is known as dissociation.