Answer:
[OH⁻] = 3.54 × 10⁻¹⁰ M
Solution:
pOH is related to [OH⁻] as,
pOH = - log [OH⁻]
Putting value of pOH,
9.45 = -log [OH⁻]
Solving for [OH⁻],
[OH⁻] = 10⁻⁹·⁴⁵ ∴ 10 = Antilog
[OH⁻] = 3.54 × 10⁻¹⁰ M
Answer:
Since with LiBr no precipitation takes place. So, Ag+ is absent
When we add Li2SO4 to it, precipitation takes place.
Ca2+(aq) + SO42-(aq) ----> CaSO4(s) ...Precipitate
Thus, Ca2+ is present.
When Li3PO4 is added, again precipitation takes place.Reaction is:
Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate
A. Ca2+ and Co2+ are present in solution
B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)
C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)
Answer: 24 moles of 
are produced.
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
is accompanied with = 1 mole of
Thus 24 moles of is accompanied with =
of
Thus 24 moles of are produced.
⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️2 one.
