How many ml of 0.100 m naoh is needed to titrate 20.0 ml of 0.100 m h2so4

So the first thing we must do is write a balanced equation for the reaction and we know the equation is balnced when all the species on the RHS is equal to the species on the LHS
                      2NaOH  +  H₂SO₄   →  Na₂SO₄  +  2H₂O

So now it's time to identify what you know from the question (volume & conc. of H₂SO₄) and use that info to find the unknown (volume of NaOH)

If 1000 ml  of H₂SO₄ contain 0.100 mol   [0.100 M is the amount of moles in
                                                                 1 L (1000 ml)]
then let 20 ml of H₂SO₄ contain x mol     [20 ml is the amount of the acid that                                                                   took place in the reaction]
    
⇒  x  =  
       
         = 0.002 mol

Mole ratio of NaOH  to  H₂SO₄  can be obtained from the balanced equation
        2NaOH  +  1H₂SO₄   →  Na₂SO₄  +  2H₂O

  thus mole ratio of   NaOH  to  H₂SO₄  is 2 : 1

∴ if mole of of H₂SO₄  =  0.002 mol
  then moles of NaOH = (0.002 mol) × 2
                                    = 0.004 mol

Now molarity = 
    ∴ by transposition

       Volume = 
                    = 
                    = 0.04 L
                    = 40 mL

Answer : option a) 84explanation :

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.

Remember that rust prevented by placing a sacrificial electrode in electrical contact with the iron. the sacrificial electrode must be composed of a metal that  oxidizes  more easily than iron (that is, it must be below iron in table 18.1).