MCQ
The $pH$ of a $0.001\,M\,NaOH$ will be
- A$3$
- B$2$
- ✓$11$
- D$12$
$ = {10^{ - 3}}\,M \Rightarrow pOH = 3$
$pH + pOH = 14 \Rightarrow pH = 14 - 3 = 11$
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Correct statement is/are

$Q\,\left( {1{s^2},\,\,2{s^2}\,2{p^6},\,\,3{s^1}} \right);\,\,R\,\left( {1{s^2},\,\,2{s^2}\,2{p^2}} \right)$The element that would most readily form a diatomic molecule is
$\begin{array}{*{20}{c}}
{C{H_2} - CH - C{H_2}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,} \\
{OH\,\,\,\,\,\,\,\,OH\,\,\,\,\,\,OH}
\end{array}$
is :