Answer:

Explanation:
We are given that 25 mL of 0.10 M
is titrated with 0.10 M NaOH(aq).
We have to find the pH of solution
Volume of 
Volume of NaoH=0.01 L
Volume of solution =25 +10=35 mL=
Because 1 L=1000 mL
Molarity of NaOH=Concentration OH-=0.10M
Concentration of H+= Molarity of
=0.10 M
Number of moles of H+=Molarity multiply by volume of given acid
Number of moles of H+=
=0.0025 moles
Number of moles of
=0.001mole
Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles
Concentration of H+=
pH=-log [H+]=-log [4.28
]=-log4.28+2 log 10=-0.631+2

Seas are smaller than oceans. Oceans are where land and water meet, and are usually bigger.
Answer:
Filtration
Explanation:
Metal carbonate is insoluble, it is possible to filter off the unreacted substances leaving the desired salt solution
Neutralization reactions are the reactions type which form salts.
Explanation:
Salts are formed by ionic bonds when the oxidation states of anions and cations are equal and have opposite signs. So one should be highly electronegative in nature and another should be highly electropositive in nature. So the electropositive element will be ready to give electrons and the electronegative element will be ready to accept all the electrons given by the electropositive element. As a whole the compound will be neutrally charged by adding of equal number of positively charged and negatively charged ions.
The reduction or addition of electrons will be occurring in cations and the oxidation or removal of electrons will be occurring in anions.
So the salt formation is based on neutralization reactions.
Potassium oxide has the antifluorite structure. The antifluorite structure have compounds with the stoichiometry X₂Y, where X is the cation and Y is the anion. In the antifluorite structure <span>positions of the </span>cations<span> and </span>anions<span> are reversed relative to their positions in calcium fluoride.</span>
Potassium ions coordinated to 4 oxide ions, <span>potassium ions are all in the tetrahedral holes.</span>