Question

Solubility product of radium sulphate is $N_{2} + 3H_{2}$what will be the solubility of $2NH_{3;}$in $\Delta H = - 22.08kcal$?

• A. $N_{2}$
• B. $N_{2}$
• C. $N_{2}$
• D. $HCO_{3}^{-}$

Answer 1

Answer: (A)

$N_{2}$

Answer 2

: $RaSO_{4}$ $Ra^{2 +} + SO_{4}^{2 -}$

$K_{sp} = \lbrack Ra^{2 +}\rbrack\lbrack SO_{4}^{2 -}\rbrack$

Concentration of $SO_{4}^{2 -}$ from $Na_{2}SO_{4} = 0.10M$

$Ra^{2 +} = \frac{4 \times 10^{- 11}}{0.10} = 4 \times 10^{- 10}M$