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Could someone please help with the equation, have put the numbers into my calculator but do not get the right answer, unsure where I am going wrong?

the \( \log \mathrm{K}^{p 0} \mathrm{Na}_{\mathrm{a}} \cdot \mathrm{Mg}_{\mathrm{p}}{ }^{2+} \) for measurements at \( 21.5^{\circ} \mathrm{C} \) at activities of \( 10^{3} \). \[ \begin{array}{l} \mathrm{F}=9.648 \times 10^{4} \mathrm{C} / \mathrm{mol} \\ \mathrm{R}= 8.314 \mathrm{VC} / \mathrm{moIK} \\ \log K_{\mathrm{A}, \mathrm{X}}^{\mathrm{Pot}}=\frac{z_{\mathrm{A}} F\left(E_{\mathrm{X}}-E_{\mathrm{A}}\right)}{R T \ln 10}+\log \left(\frac{\mathcal{A}_{\mathrm{A}}}{\left(\mathcal{A}_{\mathrm{X}}\right)^{z_{N} / z_{\mathrm{X}}}}\right) \end{array} \] [5 marks]

The electron affinity, Eea, is the energy released when an el

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