Study questions platform-wide or filter by specific tests with correct answers revealed.
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 $= 27.5 \times 18 = \mathbf{495\ \text{g}}$
A force $\vec{F} = 6t^2\,\hat{i} + 4t\,\hat{j}$ acts on a particle of mass $3\text{ kg}$, initially at rest. Find its velocity at $t = 3\text{ s}$.
Acceleration: $\vec{a} = \dfrac{\vec{F}}{m} = 2t^2\,\hat{i} + \dfrac{4t}{3}\,\hat{j}$
Velocity: $\vec{v} = \displaystyle\int_0^3 \vec{a}\,dt = \left[\dfrac{2t^3}{3}\,\hat{i} + \dfrac{2t^2}{3}\,\hat{j}\right]_0^3 = \dfrac{2(27)}{3}\,\hat{i} + \dfrac{2(9)}{3}\,\hat{j} = \mathbf{18\hat{i} + 6\hat{j}}$
Sign in to join the conversation and share your thoughts.
Log In to Comment