MCQ
When $Cl_2$ is passed through cold dil. $NaOH$ the products are
- ✓$NaCl, \,NaOCl$
- B$NaCl, \,NaClO_2$
- C$ NaCl,\, NaClO_3$
- D$ NaCl,\, NaClO_4$
Therefore, the product of the reaction of $I _2 I _2$ with $H _2 O _2. H _2 O _2$ in the basic medium is Iodine.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Column $I$ | Column $II$ |
| $(A)$ $CH_3COONa$ | $(i)$ Almost neutral, $pH > 7$ or $< 7$ |
| $(B)$ $NH_4Cl$ | $(ii)$ Acidic, $pH < 7$ |
| $(C)$ $NaNO_3$ | $(iii)$ Alkaline, $pH > 7$ |
| $(D)$ $CH_3COONH_4$ | $(iv)$ Neutral, $pH = 7$ |
$Fe(OH)_{3(s)} \rightleftharpoons Fe^{3+}_{(aq)} + 3OH^-_{(aq)}$
is decreased by $1/4$ times, then equilibrium concentration of $Fe^{3+}$ will increase by.....$ times$
$(1)$ $BeCl _2, CO _2, BCl _3, CHCl _3$
$(2)$ $SO _2, C _6 H _5 Cl , H _2 Se , BrF _5$
$(3)$ $BF _3, O _3, SF _6, XeF _6$
$(4)$ $NO _2, NH _3, POCl _3, CH _3 Cl$