Explanation:
Salts are the solutes in an aqueous solution. An aqueous solution is solution whose solvent water.
- To form a solution,a solute must be dissolved in a solvent.
- For aqueous solutions, the solvent which is the dissolving medium is made up of water.
- The solute is the substance that is dissolved in it.
- Salts for example can be a solute in an aqueous solution.
- A salt is generally an ionic compound consisting of positive ions such as metallic, ammonium ans complex ions and negative ions such as acid radicals and complex ions.
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Answer:
% of n-propyl chloride = 43.48 %
Explanation:
There are 2 secondary hydrogens and 6 primary hydrogens
The rate of abstraction of seondary hydrogen = 3.9 X rate of abstraction of primary hydrogen
probability of formation of isopropyl chloride = 3.9 X 1 (relative rate X relative number of secondary hydrogens)
Probability of formation of n-propyl chloride = 1 X 3 (relative rate X relative number of primary hydrogens)
Total probability = 3.9
% of n-propyl chloride = 3 X 100 / 6.9 = 43.48 %
1) The forward reaction is N2 (g) + O2 (g) → 2NO
(that reaction requires special contitions because at normal pressures and temperatures N2 and O2 do not react to form another compound.
2) The equiblibrium equation is
N2 (g) + O2 (g) ⇄ 2NO
3) Then, the reverse reaction is
2NO → N2(g) + O2(g)
Answer: 2NO → N2(g) + O2(g)
Answer:
15.70mg would remain
Explanation:
Partition coefficient is used to extract or purify a solute from a solvent selectively to avoid interference from other substances. For the problem, formula is:
Kp = Concentration 9-fluorenone in ether / Concentration of solute in H₂O
After the solute, 9-fluorenone, is extracted with water, the mass that remains in ether is:
(19mg - X)
<em>Where X is the mass that now is in the aqueous phase</em>
Replacing in Kp formula:
9.5 = (19mg - X) / 5mL / (X /10mL)
0.95X = 19mg - X / 5mL
4.75X = 19 - X
5.75X = 19
X = 19 / 5.75
X = 3.30mg
That means 9-fluorenone that remain in the ether layer is:
19mg - 3.30mg =
<h3>15.70mg would remain</h3>
In the presence of heat, copper (II) hydroxide decomposes in to copper (II) oxide.
Cu(OH)₂ (s) ----> CuO (s) + H₂O (l)
upon decomposition, water is removed from Cu(OH)₂
the amount of Cu(OH)₂ decomposed - 3.67 g
number of moles of Cu(OH)₂ - 3.67 g / 97.5 g/mol = 0.038 mol
stoichiometry of Cu(OH)₂ to CuO is 1:1
therefore number of CuO moles formed are - 0.038 mol
CuO reacts with sulfuric acid to form CuSO₄
CuO + H₂SO₄ ---> CuSO₄ + H₂O
stoichiometry of CuO to H₂SO₄ is 1:1
therefore number of H₂SO₄ moles that should react is 0.038 mol
the molarity of H₂SO₄ is 3M
this means that in 1000 ml - 3 mol of H₂SO₄ present
so if 3 mol are present in 1000 ml
then volume for 0.038 mol = 1000/3 * 0.038
= 12.67 ml
