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The system $x+ky+3z=0$, $3x+ky-2z=0$, $2x+4y-3z=0$ has a non-zero solution $(x,y,z)$. Find the value of $\dfrac{xz}{y^2}$.
For non-zero solution, $\Delta=0$:
$\begin{vmatrix}1&k&3\\3&k&-2\\2&4&-3\end{vmatrix}=1(β3k+8)βk(β9+4)+3(12β2k)=β3k+8+5k+36β6k=44β4k=0$
So $k=11$.
With $k=11$: from equations 1 and 2: $x+11y+3z=0$ and $3x+11y-2z=0$. Subtracting: $2x-5z=0\Rightarrow x=\frac{5z}{2}$.
Substituting back: $\frac{5z}{2}+11y+3z=0\Rightarrow 11y=-\frac{11z}{2}\Rightarrow y=-\frac{z}{2}$.
$\frac{xz}{y^2}=\frac{\frac{5z}{2}\cdot z}{\frac{z^2}{4}}=\frac{\frac{5z^2}{2}}{\frac{z^2}{4}}=\frac{5}{2}\times4=\mathbf{10}$
Bulk modulus $K = \rho \frac{dP}{d\rho}$.
- Rearranging for change in density $d\rho$: $d\rho = \frac{\rho dP}{K}$
- Given $dP = P$, the increase is $\frac{\rho P}{K}$.
Reduction: $Mn^{2+} + 2e^{-} \rightarrow Mn$ ($E^{0}_{red} = -1.18\text{ V}$)
Oxidation: $2Mn^{2+} \rightarrow 2Mn^{3+} + 2e^{-}$ ($E^{0}_{ox} = -1.51\text{ V}$)
$E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} = -1.18 - 1.51 = -2.69\text{ V}$. Since $E^{0}_{cell}$ is negative, the reaction is non-spontaneous.
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