<span>Their orbitals are completely filled</span>
LiOH is soluble. Na2CO3 is soluble. Cu(OH)2 is insoluble.
Answer:
if you will decrease the HCO3- so the less H+ ion will be form and reaction will more likely shift to product and when HCO3- decreases pH value increases and vica versa for the increasing HCO3 the more H+ ion will be form and reaction will shift to product and the pH value will decreases!!
I've doubt in reaction shift coz whatever is the amount of HCO3- this is completely gonna form number of H+ so reaction shift may or may not be same!!
Answer:
9.47 mL
Explanation:
The reaction that takes place is:
- 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
First we <u>calculate how many KOH moles reacted</u>, using <em>the given concentration and volume of KOH solution</em>:
- 0.061 mol/L = 0.061 mmol/mL
- 0.061 mmol/mL * 26.7 mL = 1.6287 mmol KOH
Then we <u>convert KOH moles into H₂SO₄ moles</u>, using the <em>stoichiometric coefficients</em>:
- 1.6287 mmol KOH *
= 0.8144 mmol H₂SO₄
Finally we <u>calculate the required volume of the H₂SO₄ solution</u>, using<em> the number of moles and given concentration</em>:
- 0.8144 mmol ÷ 0.086 mmol/mL = 9.47 mL