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).
Checking options:
- Hโบ (Z=1): 0 electrons
- Liโบ (Z=3): 2 electrons โ โ isoelectronic with Beยฒโบ
- Naโบ (Z=11): 10 electrons
- Mgยฒโบ (Z=12): 10 electrons
The coordinates of a particle at any time $t$ are $x = 5t - 2t^2$ and $y = 10t$ (metres, seconds). The acceleration of the particle at $t = 2\text{ s}$ is:
$a_x = \dfrac{d^2x}{dt^2} = -4\text{ m/s}^2$ (constant)
$a_y = \dfrac{d^2y}{dt^2} = 0$
Acceleration vector $= -4\hat{i}\text{ m/s}^2$ at all times, including $t = 2\text{ s}$. Magnitude $= \mathbf{4\text{ m/s}^2}$, directed in $-x$ direction.
Sign in to join the conversation and share your thoughts.
Log In to Comment