Study questions platform-wide or filter by specific tests with correct answers revealed.
Work done (in atmยทdmยณ) when 1 mol ideal gas expands from 15 dmยณ to 20 dmยณ against constant external pressure 2 atm:
\(W = -P_{\text{ext}} \Delta V = -2 \times (20-15) = -2 \times 5 = -10\ \mathrm{atmยทdmยณ}\). Negative sign indicates work done by the system (expansion).
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.
Evaluate $\displaystyle\int_{\pi/4}^{3\pi/4}\dfrac{dx}{1+\cos x}$
$1+\cos x = 2\cos^2(x/2)$
$\displaystyle\int_{\pi/4}^{3\pi/4}\dfrac{dx}{2\cos^2(x/2)}=\dfrac{1}{2}\int_{\pi/4}^{3\pi/4}\sec^2\!\left(\dfrac{x}{2}\right)dx=\left[\tan\!\left(\dfrac{x}{2}\right)\right]_{\pi/4}^{3\pi/4}$
$=\tan\!\left(\dfrac{3\pi}{8}\right)-\tan\!\left(\dfrac{\pi}{8}\right)$
Using $\tan(3\pi/8)=\cot(\pi/8)$ and $\cot\theta-\tan\theta=2\cot2\theta$:
$=\cot(\pi/8)-\tan(\pi/8)=2\cot(\pi/4)=2\cdot1=\mathbf{2}$
According to the text, approximately what proportion of jobs eliminated in the 1990s were supervisory/middle management positions?
The text explicitly states that it is estimated that two-thirds ($\frac{2}{3}$) of the jobs eliminated in the 1990s were supervisory/middle management jobs, reflecting the flattening of organizational structures.
Sign in to join the conversation and share your thoughts.
Log In to Comment