A) \[x\,\in \,\left( -\infty ,\,\,{{\log }_{2}}\,\frac{4}{3} \right)\,\cup \,(1,\,\,\infty )\]
B) \[x\,\in \,\left( {{\log }_{2}}\,\left( \frac{4}{3} \right),\,\,0 \right)\,\]
C) \[x\,\in \,\left( {{\log }_{2}}\,\left( \frac{4}{3} \right),\,\,1 \right)\,\]
D) Cannot say
Correct Answer: C
Solution :
Put \[\alpha =\beta =\gamma \ne {{90}^{o}}\] \[{{p}_{1}}=\frac{nRT}{{{V}_{1}}}=\frac{2\times 0.0821\times 243.6}{20}=2\,\text{atm}\] \[\Delta S=nR\,\,In\,\left( \frac{{{p}_{1}}}{{{p}_{2}}} \right)\] \[=2\times 2=\,In\,2\] \[=2\times 2\times 0.693\] \[[\because \,I{{n}^{2}}\,or\,{{\log }_{e}}2=0.693]\]\[FeC{{l}_{3}}+3NaOH\xrightarrow{\,}\,Fe{{(OH)}_{3}}+3NaCl\] For f(t) to be decreasing, \[NaOH\xrightarrow{\,}N{{a}^{\oplus }}+O{{H}^{\odot -}}\] \[O{{H}^{\odot -}}\] \[Fe{{(OH)}_{3}}\] \[Fe{{(OH)}_{3}}\]You need to login to perform this action.
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