Less dense than the liquid
The pH of the solution : 12
<h3>Further explanation</h3>
Reaction
HCOOH + NaOH ⇒ HCOONa + H₂O
mol HCOOH =

mol NaOH =

Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]
ICE method :
HCOOH + NaOH ⇒ HCOONa + H₂O
4 5
4 4 4 4
0 1 1 1
Concentration of [OH⁻] from NaOH :

pOH=-log[OH⁻]
pOH=-log 10⁻²=2
pH+pOH=14
pH=14-2=12
Answer:
0.022 mol O
Explanation:
Mg3(Si2O5)2(OH)2
We can see that 1 mol of this substance has 3 mol of Mg.
Oxygen altogether is 5*2 (from (Si2O5)2) + 2(from(OH)2) = 10 +2 = 12
So, 1 mol of this substance has 12 mol oxygen.
So, 1 mol of this substance contains 3 mol Mg and 12 mol O, or
ratio Mg : O = 3 : 12 = 1 : 4
1 mol Mg ----- 4 mol O
0.055 mol Mg ---x mol O
x = 0.055*4/1 = 0.220 mol O