What is the net ionic equation for the reaction 2koh(aq) + h2so4(aq) 2h2o(l) + k2so4(aq)

A net ionic equation only shows the substances that are involved in the chemical reaction. We have to expand the compounds in aqueous states, turn them into ions, and delete the duplicated ions from both sides (if possible).

So first step, write the full ionic equation. Just expand all the aqueous compounds. Other compounds/elements not in aq can be left alone and be copied directly. Make sure to write the state symbol too for each ion.

2KOH(aq) + H2SO4(aq) → 2H2O(l) + K2SO4(aq)

will be

2K+(aq) + 2OH-(aq) + 2H+(aq) + SO4 2-(aq) → 2H2O(l) + 2K+(aq) + SO4 2-(aq)

Now, just delete the duplicated ions. We see that 2K+(aq) and SO4 2-(aq) appeared in both sides of the equation, so just delete them to get the final answer!

Therefore, the net ionic equation is:

2OH-(aq) + 2H+(aq) → 2H2O(l)

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.