Study questions platform-wide or filter by specific tests with correct answers revealed.
\(\Delta v = \dfrac{1 \times 10^{-18}}{9 \times 10^{-28}} = \dfrac{1}{9} \times 10^{10} \approx 1 \times 10^9\) cm s\(^{-1}\)
This enormously large uncertainty in velocity illustrates why classical mechanics fails for electrons — they cannot have precisely defined trajectories.
Given that $r > s > 0$, compare the two quantities below.
Column A: $\dfrac{rs}{r}$
Column B: $\dfrac{rs}{s}$
Simplify each expression:
- Column A: $\dfrac{rs}{r} = s$
- Column B: $\dfrac{rs}{s} = r$
Since we are given $r > s > 0$, it follows that $r > s$.
Therefore Column B ($r$) is greater than Column A ($s$).
- $d = 9000\text{ kg/m}^3$
- $Z = 4$ (for FCC)
- $a = 200\sqrt{2} \times 10^{-12}\text{ m}$
$M = \frac{9000 \times 6 \times 10^{23} \times 1.6 \times 10^{-29}}{4} = 0.0432\text{ kg mol}^{-1}$.
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).
Sign in to join the conversation and share your thoughts.
Log In to Comment