Study questions platform-wide or filter by specific tests with correct answers revealed.
Acidic medium (oxidising agent):
$[\text{Fe(CN)}_6]^{4-} + H_2O_2 + 2H^+ \to [\text{Fe(CN)}_6]^{3-} + \text{H}_2\text{O}$
H$_2$O$_2$ is reduced to H$_2$O. Other product: H$_2$O.
Alkaline medium (reducing agent):
$H_2O_2 \to H_2O + \frac{1}{2}O_2$ (H$_2$O$_2$ is oxidized)
$2[\text{Fe(CN)}_6]^{3-} + H_2O_2 + 2OH^- \to 2[\text{Fe(CN)}_6]^{4-} + O_2 + 2H_2O$
Other products: H$_2$O and O$_2$.
So the pair is: H$_2$O (acidic) and (H$_2$O + O$_2$) (alkaline).
Areal velocity is defined as the area swept by the position vector per unit time: $\frac{dA}{dt}$.
In polar coordinates, $dA = \frac{1}{2} r^2 d\theta$.
So, $\frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt} = \frac{1}{2} r^2 \omega$.
Since angular momentum $L = m r^2 \omega$, we can substitute $r^2 \omega = \frac{L}{m}$.
Therefore, $\text{Areal Velocity} = \frac{1}{2} \left( \frac{L}{m} \right) = \frac{L}{2m}$.
Sign in to join the conversation and share your thoughts.
Log In to Comment