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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 $= 27.5 \times 18 = \mathbf{495\ \text{g}}$
Using the Relativistic Doppler Effect formula for an approaching source:
$f' = f \sqrt{\frac{1 + v/c}{1 - v/c}}$
Given $v = 0.5c$:
$f' = 10 \sqrt{\frac{1 + 0.5}{1 - 0.5}} = 10 \sqrt{\frac{1.5}{0.5}} = 10 \sqrt{3} \approx 10 \times 1.732 = 17.32 \text{ GHz}$.
A projectile is fired at $45°$ to the horizontal. The elevation angle of the projectile at its highest point, as seen from the point of projection, is:
At max height $H = \dfrac{u^2}{4g}$; horizontal distance $= \dfrac{R}{2} = \dfrac{u^2}{2g}$
Elevation angle: $\tan\phi = \dfrac{H}{R/2} = \dfrac{u^2/(4g)}{u^2/(2g)} = \dfrac{1}{2}$
$\phi = \mathbf{\tan^{-1}\!\left(\dfrac{1}{2}\right)}$
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