Explanation:
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Formula units in 450 g of
is 1.93 × 10²⁴ formula units.
<u>Explanation:</u>
First we have to find the number of moles in the given mass by dividing the mass by its molar mass as,

Now, we have to multiply the number of moles of Na₂SO₄ by the Avogadro's number, 6.022 × 10²³ formula units/mol, so we will get the number of formula units present in the given mass of the compound.
3.2 mol × 6.022 × 10²³ = 1.93 × 10²⁴ formula units.
So, 1.93 × 10²⁴ formula units is present in 450g of Na₂SO₄.
Answer:
Explanation:
We will need a balanced equation with moles, so let's gather all the information in one place.
CH₃C₆H₄NH₂·HCl + (CH₃CO)₂O ⟶ CH₃C₆H₄NHCOCH₃ + junk
V/mL: 70.
c/mol·L⁻¹: 0.167
For simplicity in writing , let's call p-toluidine hydrochloride A and N-acetyl-<em>p</em>-toluidine B.
The equation is then
A + Ac₂O ⟶ B + junk
1. Moles of A

2. Moles of B
The molar ratio is 1 mol B:1 mol A
Moles of B = moles of A = 12 mmol = 0.012 mol

The given question is incomplete. The complete question is as follows.
A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 212 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)
Explanation:
When
and
are added then white precipitate forms. And, reaction equation for this is as follows.
It is given that mass (m) is 212 mg or 0.212 g (as 1 g = 1000 mg). Molecular weight of
is 233.43.
Now, we will calculate the number of moles as follows.
No. of moles = mass × M.W
= 
= 0.00091 mol of
Hence, it means that 0.00091 mol of
. Now, we will calculate the mass as follows.
Mass = moles × MW
=
= 0.124 grams or 124 mg of barium
Thus, we can conclude that mass of barium into the original solution is 124 mg.