The reaction between NaOH and H₂SO₄ is as follows;
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
NaOH is a strong acid and H₂SO₄ is a strong acid, therefore complete ionization into their respective ions takes place.
number of acid moles reacted - 0.112 M / 1000 mL/L x 39.1 mL = 0.0044 mol
the number of base moles required for neutralisation = 0.0044 x 2 = 0.0088 mol
Number of NaOH moles in 25.0 mL - 0.0088 mol
Therefore in 1000 mL - 0.0088 mol/ 25.0 mL x 1000 mL/L = 0.352 mol/L
Therefore molarity of NaOH - 0.352 M
Answer:1 mol of Mg(NO3)2 contains 6.022*10^23 molecules
3 mol Mg(NO3)2 contains 3*6.022*10^23 = 1.81*10^24 molecules
Each Mg(NO3)2 molecule contains 2 N atoms
Number of N atoms = 2*1.81*10^24 = 3.62*10^24 N atoms.
Cs -137 has a half life of about 30 years. If 60 years pass, there is two half lives passed so 1/2 * 1/2= 1/4. Take 1/4 and multiply it with the mass given:
(1/4)*20mg=5mg left
C protons, elctrons 093939393