500 cc of 2N Na2CO3 are mixed with 400 cc of 3N H2SO4 and volume was diluted to one litre. Will the resulting solutiomacidic, basic or neutral?Also calculate the molarity of the dilute solution.

Given info : 500 cc of 2N Na2CO3 are mixed with 400 cc of 3N H2SO4 and volume was diluted to one litre. To find : will the resulting solution is acidic , basic or neutral ? Calculate the molarity of the dilute solution. solution : no of moles of Na2CO3 = normality/n %3D - factor x volume 2/2 x 500/1000 = 0.5 mol %D no of moles of H2SO4 = 3/2 x 400/100O = 0.6 mol %3D We see, Na2CO3 + H2S04 => Na2S04 + CO2 + H2O Here one mol of Na2C03 reacts with one mole of H2SO4. So, 0.5 mol of Na2CO3 reacts with 0.5 mol of H2SO4. so, remaining 0.1 mol of H2SO4 makes solution acidic. Now molarity of solution = remaining no of moles of H2SO4/volume of solution= 0.1/1 = %3D 0.1M

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

2b2t

explanation:

2b2t