See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

0 hv hv KE     0 15 0 15 0 KE h v v 2 10 v KE 4KE 5 10 v          15 15 0 0 15 0 8 10 4v 5 10 v 3v 3 10           15 0 v 1 10 