The minimum volume of 0.2M sodium sulfide that will precipitate out aluminum from 50.0 mL, 0.25 M aluminum nitrate would be 0.094 L or 94 mL
<h3>Stoichiometric calculation</h3>From the equation of the reaction:
Mole ratio of Na2S and Al(NO3)3 = 3:2
Mole of 50.0 mL, 0.25 M Al(NO3)3 = 50/1000 x 0.25
= 0.0125 mole
Equivalent mole of Na2S = 3/2 x 0.0125
= 0.0188 mole
Volume of 0.2M, 0.0188 mole Na2S = 0.0188/0.2
= 0.094 L or 94 mL
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Answer:
The answer is
<h2>19.31 g/cm³</h2>Explanation:
The density of a substance can be found by using the formula
From the question
mass of gold = 475.09g
volume = 24.6 cm³
The density of the gold is
We have the final answer as
<h3>19.31 g/cm³</h3>Hope this helps you
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.