Explanation:
Ouhgiuytytufryftrufurtfrthfrthfhtftryfrtfrthffthfrth
Answer:
Explanation:
Mixing Ammonia gas into a solution of Copper(II) Sulfate will give Ammonium Sulfate and a precipitate of Copper(II) Hydroxide (Cu(OH)₂). The Ksp of Cu(OH)₂ is published => 2.2 x 10⁻²². Such gives a solubility* of the Cu(OH)₂ to be ~1.77 x 10⁻⁷M => [Cu⁺²] ~1.77 x 10⁻⁷M and [OH⁻] = 2(1.77 x 10⁻⁷)M = 3.53 x 10⁻⁷M. The reaction of Ammonium Hydroxide and Copper(II) Sulfate will generate 1 x 10⁻⁴ mole Cu(OH)₂ as a precipitate but only 1.77 x 10⁻⁷ mole of the hydroxide will remain in 1 Liter of solution b/c of extreme limited solubility.
*Solubility of 1:2 ionization ratio salts = CubeRt(Ksp/4).
Ionic bonding is your answer. Hope this helps
So the question ask on what solute are present in an aqueous solution on a certain element base on the data you have given. So base on that data i came up with a chemical expression of HCIO, H+ and CIO-. I hope you are satisfied with my answer
<h3>
Answer:</h3>
0.00245 moles of HCl
<h3>
Explanation:</h3>
We are given;
Volume of KOH = 9.71 mL
Molarity of KOH = 0.252 M
Volume of HCl = 15.0 mL
We are required to determine the number of moles of HCl
<h3>Step 1: Write the balanced equation for the reaction</h3>
The reaction between HCl and KOH is given by;
KOH(aq) + HCl(aq) → KCl(aq) + H₂O(l)
<h3>Step 2: Calculating the number of moles of KOH</h3>
Molarity = Moles ÷ Volume
Rearranging the formula;
Moles = Molarity × Volume
= 0.252 M × 0.00971 L
= 0.00245 moles
<h3>Step 3: Moles of HCl </h3>
From the equation, 1 mole of KOH reacts with 1 mole of HCl
Therefore, the mole ratio of KOH : HCl is 1 : 1
Thus,
Moles of HCl will be 0.00245 moles
Therefore, there are 0.00245 moles of HCl in the solution.