A) \[6.67\times {{10}^{-4}}wb\]
B) \[3.2\times {{10}^{-5}}wb\]
C) \[5.9\times {{10}^{-4}}wb\]
D) \[8.65\times {{10}^{-4}}wb\]
Correct Answer: D
Solution :
Given that N = 20 \[\therefore \] \[56u=\frac{56}{14}Natom=4Natom\] \[\begin{align} & CaC{{O}_{3}}\xrightarrow{Heat}CaO+C{{O}_{2}}\uparrow \\ & \text{XColourless} \\ & \text{gas} \\ \end{align}\] \[\begin{align} & CaO+{{H}_{2}}O\to Ca{{\left( OH \right)}_{2}} \\ & \text{ResidueY} \\ \end{align}\] \[\begin{align} & Ca{{\left( OH \right)}_{20}}+2C{{O}_{2}}\to Ca{{\left( HC{{O}_{3}} \right)}_{2}} \\ & \text{YExcessZ} \\ \end{align}\] the flux through the coil is given by \[\begin{align} & Ca{{\left( HC{{O}_{3}} \right)}_{2}}\xrightarrow{Heat}CaC{{O}_{3}}+C{{O}_{2}}\uparrow +{{H}_{2}}O \\ & ZX \\ \end{align}\] \[\begin{align} & Zn{{\left( OH \right)}_{2}}+2\overset{-}{\mathop{OH}}\,\to ZnO_{2}^{2-}+2{{H}_{2}}O \\ & \text{AcidBaseSaltWater} \\ \end{align}\] \[\begin{align} & Zn{{\left( OH \right)}_{2}}+2{{H}^{+}}\to Z{{n}^{2+}}+2{{H}_{2}}O \\ & \text{BaseAcidSaltWater} \\ \end{align}\] \[{{\left[ \text{Cu}{{\left( \text{N}{{\text{H}}_{\text{3}}} \right)}_{\text{4}}} \right]}^{\text{2+}}}\text{}{{\left[ \text{Cu}{{\left( \text{en} \right)}_{\text{2}}} \right]}^{\text{2+}}}\text{}{{\left[ \text{Cu}\left( \text{trien} \right) \right]}^{\text{2+}}}\text{.}\]
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