Calculate the concentration of oxalate ion (c2o42–) in a 0.175 m solution of oxalic acid (c2h2o4). [for oxalic acid, ka1 = 6.5 × 10–2, ka2 = 6.1 × 10–5.]

C2H2O4 C2HO4- + H+

0.175 - x x x + y

C2HO4- c2o42– + H+

x - y y x + y

K2 = (y) (x +y) / (x-y)

As, y << x

So, K2 = (y) (x) / (x)

K2 = y =6.1 × 10^–5

Hence, concentration of (c2o42–) 6.1 × 10^–5 M

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

mark brainliest and have a great day!

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.