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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).
For the reaction $2\text{C}_{57}\text{H}_{110}\text{O}_6(s) + 163\ \text{O}_2(g) \to 114\ \text{CO}_2(g) + 110\ \text{H}_2\text{O}(l)$, find the mass of water produced from 445 g of C$_{57}$H$_{110}$O$_6$.
Molar mass of C$_{57}$H$_{110}$O$_6$: $57(12)+110(1)+6(16)=684+110+96=890\ \text{g mol}^{-1}$
Moles of C$_{57}$H$_{110}$O$_6 = \dfrac{445}{890} = 0.5\ \text{mol}$
From stoichiometry: 2 mol fat → 110 mol H$_2$O
So 0.5 mol fat → $\dfrac{110 \times 0.5}{2} = 27.5\ \text{mol H}_2\text{O}$
Mass of H$_2$O $= 27.5 \times 18 = \mathbf{495\ \text{g}}$
Centripetal force $F = \frac{mv^2}{R} \propto \frac{1}{R^n}$.
- $v^2 \propto \frac{1}{R^{n-1}} \implies v \propto R^{(1-n)/2}$
- $T = \frac{2\pi R}{v} \propto \frac{R}{R^{(1-n)/2}} = R^{1 - (\frac{1-n}{2})} = R^{(n+1)/2}$.
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