MCQ
$60 \,g$ of a compound on analysis gave $C = 24\,{\rm{ }}g,$ $H = 4\,{\rm{ }}g$ and $O = 32\, g$. Its Empirical formula is
- A${C_2}{H_4}{O_2}$
- B${C_2}{H_2}O$
- C$C{H_2}{O_2}$
- ✓$C{H_2}O$
Element No. of Moles Simple Ratio
|
$C = 24$ |
$24/12 = 2$ |
$1$ |
|
$H=4$ |
$4/1 = 4$ |
$ 2$ |
|
$O = 32$ |
$32/16 = 2$ |
$1$ |
Therefore $C{H_2}O$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Column $I$ | Column $II$ |
| $(A)$ Silica gel | $(i)$ Transistor |
| $(B)$ Silicon | $(ii)$ Ion-exchanger |
| $(C)$ Silicone | $(iii)$ Drying agent |
| $(D)$ Silicate | $(iv)$ Sealant |
$\Delta {H^o}\, = \, +234.1\,\,kJ$
$C(s)\, + {O_2}(g)\, \to \,\,C{O_2}(g)$
Use these equation and $\Delta H^o$ value to calculate $\Delta H^o$ for this reaction:
$4Fe(s) + 3O_2(g) \to 2Fe_2O_3(s)$
.....$kJ$