A) \[x=\frac{1}{2}\]
B) \[{{K}_{P}}=\frac{{{p}_{C}}.p{{ }_{D}}}{{{p}_{A}}.{{p}_{B}}}\]
C) \[=\frac{{{\left( \frac{3x}{V} \right)}^{2}}}{{{\left( \frac{2-3x}{V} \right)}^{2}}}\]
D) \[=\frac{9\times {{\left( \frac{1}{2} \right)}^{2}}}{4-12\times \frac{1}{2}+9\times {{\left( \frac{1}{2} \right)}^{2}}}\]
Correct Answer: A
Solution :
\[Fe{{C}_{2}}{{O}_{4}}\] (where, 5 = solubility) \[C{{r}_{2}}O_{7}^{2-}\] \[\therefore \] or \[=\frac{1}{2}=0.5\] \[C{{r}_{2}}O_{7}^{2-}\] \[3U{{O}^{2+}}+C{{r}_{2}}O_{7}^{2-}+8{{H}^{+}}\xrightarrow{{}}3UO_{2}^{2+}\] For \[+2C{{r}^{3+}}+4{{H}_{2}}O\] \[{{D}_{2}}O\] \[{{H}_{2}}O\] \[N<O<S<Cl\]\[Al\] [\[Ga\] Solubility of \[_{13}Al=[Ne]3{{s}^{2}}3{{p}^{1}}\] solubility of \[_{31}Ga=[Ar]3{{d}^{10}},4{{s}^{2}}4{{p}^{1}}\]] \[Al.\]You need to login to perform this action.
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