A) \[{{\tan }^{-1}}\frac{2}{\sqrt{3}}\]
B) \[\pi /6\]
C) \[{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]
D) \[\pi /2\]
Correct Answer: C
Solution :
On homogenising \[{{C}_{2}}{{H}_{6}}+\frac{3}{2}{{O}_{2}}\to 2C{{O}_{2}}+3{{H}_{2}}O\] with the help of the line \[\Delta \text{x}=2\left( -94.1 \right)+3\left( -68.3 \right)-\left( -21.1 \right)\] we get \[\underset{s}{\mathop{A{{g}_{2}}Cr{{O}_{4}}}}\,\to \underset{2s}{\mathop{2A{{g}^{+}}}}\,+\underset{s}{\mathop{CrO_{4}^{2-}}}\,\] \[{{K}_{sp}}={{\left( 2s \right)}^{2}}s=4{{s}^{3}}\] \[s={{\left( \frac{{{K}_{sp}}}{4} \right)}^{1/3}}={{\left( \frac{32\times {{10}^{-12}}}{4} \right)}^{1/3}}=2\times {{10}^{-4}}M\] \[k={{\wedge }_{eq}}.C\] \[=\left( 91{{\Omega }^{-1}}c{{m}^{2}}e{{q}^{-1}} \right)\left( \frac{2.54}{159/2\times 1000}eq.c{{m}^{-3}} \right)\] On comparing with \[=2.9\times {{10}^{-3}}{{\Omega }^{-1}}c{{m}^{-1}}\] we get \[\frac{{{\left( {{t}_{1/2}} \right)}_{1}}}{{{\left( {{t}_{1/2}} \right)}_{2}}}={{\left[ \frac{{{a}_{2}}}{{{a}_{1}}} \right]}^{n-1}}\] and b = 0 \[\left( {{\text{t}}_{\text{1/2}}} \right)\text{=0}\text{.1s}\text{. }{{\text{a}}_{1}}=400\] The angle between the lines is \[{{\left( {{\text{t}}_{\text{1/2}}} \right)}_{2}}=0.8s,{{a}_{2}}=50\] \[\frac{0.1}{0.8}={{\left[ \frac{50}{400} \right]}^{n-1}}\]
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