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- Statement B is false because water exhibits extensive intermolecular hydrogen bonding, not intramolecular.
- Intermolecular bonding occurs between different molecules, which is responsible for water's high boiling point and surface tension.
- Molar mass of $NH_4SH = 51\text{ g/mol}$. Initial moles $= 5.1/51 = 0.1\text{ mol}$.
- At equilibrium, moles of $NH_3 = H_2S = 0.1 \times 0.3 = 0.03\text{ mol}$.
- Using $PV = nRT$: $P_{NH_3} = P_{H_2S} = \frac{0.03 \times 0.082 \times 600}{3} = 0.492\text{ atm}$.
- $K_p = P_{NH_3} \times P_{H_2S} = (0.492)^2 \approx 0.242\text{ atm}^2$.
Given \(\Delta x = \Delta p\), so \((\Delta p)^2 = \dfrac{h}{4\pi}\)
\(\Delta p = \sqrt{\dfrac{h}{4\pi}} = \dfrac{1}{2}\sqrt{\dfrac{h}{\pi}}\)
Since \(\Delta p = m \cdot \Delta v\):
\(\Delta v = \dfrac{\Delta p}{m} = \dfrac{1}{2m}\sqrt{\dfrac{h}{\pi}}\)
What mass of \(\mathrm{Al}_2\mathrm{O}_3\) is produced from 18.5 g Al? \(4\mathrm{Al} + 3\mathrm{O}_2 \rightarrow 2\mathrm{Al}_2\mathrm{O}_3\)
Molar mass Al = 27 g/mol: moles Al = 18.5/27 = 0.685 mol. From stoichiometry: 4 mol Al โ 2 mol AlโOโ, so mol AlโOโ = 0.685/2 = 0.3425 mol. Molar mass AlโOโ = 102 g/mol: mass = 0.3425ร102 = 34.94 g โ 34.9 g.
A random variable $Y$ follows a normal distribution with mean $200$ and standard deviation $10$. Compare:
Quantity A: The probability that $Y > 220$
Quantity B: $\dfrac{1}{6}$
The value $220$ is $\dfrac{220 - 200}{10} = 2$ standard deviations above the mean.
For a standard normal distribution, the probability that $Z > 2$ is approximately $0.0228$ (about $2.28\%$).
Quantity B $= \dfrac{1}{6} \approx 0.1667$.
Since $0.0228 < 0.1667$, Quantity A is less than Quantity B.
Therefore Quantity B is greater.
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