- A$CaC_2 + CaCN_2$
- ✓$CaC_2 + Ca_3P_2$
- C$CaC_2 + CaCO_3$
- D$Ca_3P_2 + CaCN_2$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\begin{array}{*{20}{l}}
{(u){H_2}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{C} CHC{H_3}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (v){\mkern 1mu} C{H_2}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{C} CHCl{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (w){\mkern 1mu} C{H_3}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{C} H_2^ + {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (x){\mkern 1mu} H - C \equiv C - H} \\
{(y){\mkern 1mu} C{H_3}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{C} N{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (z){\mkern 1mu} {{(C{H_3})}_2}C\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{N} N{H_2}}
\end{array}$

|
$A$ $(mol/l)$ |
$B$ $(mol/l)$ |
Rate |
| $0.05$ | $0.05$ | $1.2\times 10^{-3}$ |
| $0.10$ | $0.05$ | $2.4\times 10^{-3}$ |
| $0.05$ | $0.10$ | $1.2\times 10^{-3}$ |