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A gaseous hydrocarbon produces 0.72 g of water and 3.08 g of CO$_2$ upon complete combustion. What is the empirical formula of the hydrocarbon?
Moles of H$_2$O $= \dfrac{0.72}{18} = 0.04\ \text{mol}$ â H atoms $= 0.08\ \text{mol}$
Moles of CO$_2 = \dfrac{3.08}{44} = 0.07\ \text{mol}$ â C atoms $= 0.07\ \text{mol}$
Ratio C : H $= 0.07 : 0.08 = 7 : 8$
Empirical formula: $\mathbf{C_7H_8}$ (like toluene)
A particle is projected vertically upward with $u = 10\text{ m/s}$. Air exerts a resistive force $F = -0.2v^2$ on it ($m = 2\text{ kg}$, $g = 10\text{ m/s}^2$). The maximum height attained is:
Net force (taking up as positive): $F_{net} = -mg - 0.2v^2 = -20 - 0.2v^2$
Using $ma = v\dfrac{dv}{dx}$: $2v\dfrac{dv}{dx} = -20 - 0.2v^2$
Separating variables: $\dfrac{v\,dv}{20+0.2v^2} = -\dfrac{dx}{2}$
Integrating from $v = 10$ to $v = 0$: let $k = 20 + 0.2v^2$, $dk = 0.4v\,dv$
$\dfrac{1}{0.4}\big[\ln(20) - \ln(40)\big] = -\dfrac{H}{2} \Rightarrow H = \dfrac{2\ln 2}{0.4} = \mathbf{5\ln 2}$
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