A) \[2.8\times {{10}^{-3}}\]
B) \[2.8\times {{10}^{-4}}\]
C) \[2\times {{10}^{-7}}\]
D) \[{{10}^{-7}}\]
Correct Answer: B
Solution :
\[b(ac+b)+a(bc+a)-(1-{{c}^{2}})=0\] M moles 10 20 0 0 0 0 20 0 i.e., almost all \[b(ac+b)+a(bc+a)-(1-{{c}^{2}})=0\] gets precipitated, Hence, the remaining \[\Rightarrow \] and \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc=1\] concentration can be determined from \[(2\pi ct/\lambda )\] value of \[(2\pi x/\lambda )\]. \[x\] \[\lambda \] \[\left[ \frac{2\pi Ct}{\lambda } \right]=\,\left[ \frac{2\pi x}{\lambda } \right]=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\] \[\frac{2\pi C}{\lambda }=\frac{2\pi x}{\lambda t}\] \[\frac{x}{\lambda }\]You need to login to perform this action.
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