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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).
The displacement of a particle is given by $s = 3t^3 + 7t^2 + 5t + 8$ metres, where $t$ is in seconds. The acceleration at $t = 1\text{ s}$ is:
Velocity: $v = \dfrac{ds}{dt} = 9t^2 + 14t + 5$
Acceleration: $a = \dfrac{dv}{dt} = 18t + 14$
At $t = 1\text{ s}$: $a = 18(1) + 14 = \mathbf{32\text{ m/s}^2}$
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