The number of C atoms in 0.524 moles of C is 3.15 atoms.
The number of
molecules in 9.87 moles
is 59.43 molecules.
The moles of Fe in 1.40 x
atoms of Fe is 0.23 x 
The moles of
in 2.30x
molecules of
is 3.81.
<h3>What are moles?</h3>
A mole is defined as 6.02214076 ×
of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
A. The number of C atoms in 0.524 mole of C:
6.02214076 ×
x 0.524 mole
3.155601758 atoms =3.155 atoms
B. The number of
molecules in 9.87 moles of
:
6.02214076 ×
x 9.87
59.4385293 molecules= 59.43 molecules
C. The moles of Fe in 1.40 x
atoms of Fe:
1.40 x
÷ 6.02214076 × 
0.2324754694 x
moles.
0.23 x
moles.
D. The moles of
in 2.30x
molecules of
:
2.30x
÷ 6.02214076 × 
3.819239854 moles=3.81 moles
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Answer:
The three-point test
Explanation:
The three-point test refers to a ground test utilizing an auxiliary current electrode and an auxiliary potential electrode.
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. More NO2 and SO2 will form
Explanation:
Le Chatelier's Principle : It predicts the behavior of equilibrium due to change in pressure , temperature , volume , concentration etc
It states that When external changes are introduced in the equilibrium then it will shift the equilibrium in a direction to reduce the change.
In given Reaction SO3 is introduced(increased) .
So equilibrium will shift in the direction where SO3 should be consumed(decreased)
Hence the equilibrium will go in backward direction , i.e

So more and more Of NO2 and SO2 will form