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
Sediments move one place to another in the process called “ erosion”
Answer:
yaeh
Explanation:
a)Ca(OH)
2
+CO
2
⟶CaCO
3
+H
2
O
No. of atoms:Ca−1;O−4;H−2;C−1
b)Zn+AgNO
3
⟶ZnNO
3
+Ag
No. of atoms:Zn−1;Ag−1;N−1;O−3.
Answer:
2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.
Explanation:
Generally, moles of solute in solution before dilution must equal moles of solute after dilution.
By definition Molarity = moles solute/volume of solution in Liters
=> moles solute = Molarity x Volume (L)
Apply moles before dilution = moles after dilution ...
=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution
=> (M)(2.5L)before = (1.2M)(10.0L)after
=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate