MCQ
If the solubility product of $BaS{O_4}$ is $1.5 \times {10^{ - 9}} $ in water, its solubility in moles per litre, is
- A$1.5 \times {10^{ - 9}}$
- ✓$3.9 \times {10^{ - 5}}$
- C$7.5 \times {10^{ - 5}}$
- D$1.5 \times {10^{ - 5}}$
Solubility constant $ = S \times S$
$1.5 \times {10^{ - 19}} = {S^2}$;$S = \sqrt {1.5 \times {{10}^{ - 19}}} $;$S = 3.9 \times {10^{ - 5}}$
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$\mathop C\limits_6 {H_3} - \mathop C\limits_5 H = \mathop C\limits_4 H - \mathop C\limits_3 {H_2} - \mathop C\limits_2 \equiv \mathop C\limits_1 H$
The state of hybridization of carbons $1, 3$ and $5$ are in the following sequence