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Daniel P.
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Chemistry
TutorMe
Question:

An aqueous solution of an unknown compound is prepared by dissolving 0.125 g of the unknown in water so that the total volume of the solution is 100.0 mL. The osmotic pressure at 298 K is measured to be $$\Pi = 58.0$$ Torr. Determine the molar mass of the unknown.

Daniel P.

The osmotic pressure is defined as $$\Pi = [B]RT$$, where is $$[B] = \cfrac{n}{V}$$. Rearranging the equation to yield moles gives $$n = \cfrac{\Pi V}{RT}$$ The molar mass $$M$$ is defined as $$\cfrac{m}{n}$$ Plugging in the mass above yields $$M = \cfrac{mRT}{\Pi V}$$, where $$R$$ is the gas constant [0.08206 (L atm)/(K mol)], $$T$$ is the temperature in Kelvin, $$V$$ is the volume, $$m$$ is the mass of the unknown and $$\Pi$$ is the osmotic pressure. It is important to note that the Osmotic pressure must be converted from Torr to atm. Plugging in all of the values returns $$M = 400$$ g/mol

Astrophysics
TutorMe
Question:

Compute the diffraction limit for a 10-meter telescope for $$\lambda = 400$$ nm. What is the FWHM of the seeing disk?

Daniel P.

The diffraction limit for any telescope is defined as $$\alpha_A = \cfrac{0.252 \lambda}{D}$$, where $$D$$ is the diameter of the telescope and $$\lambda$$ is the wavelength of light. Plugging in the various values, $$\alpha_A = 0.0101$$ arcseconds The Full-Width Half-Max (FWHM) is then computed as FWHM $$=0.9\alpha_A$$. Plugging in the value calculated above gives the value FWHM $$= 0.00907$$ arcseconds

Physics
TutorMe
Question:

If a block slides without friction down a fixed, inclined plane with $$\theta = 30$$ degrees, what is the blocks acceleration?

Daniel P.

Two forces act on the block: the gravitational force $$\textbf{F}_g$$ and the plane's normal force $$\textbf{N}$$ pushing upward on the block (no friction in this example). The block is constrained to be on the plane, and the only direction the block can move is the $$x$$-direction, up and down the plane. We take the + $$x$$-direction to be down the plane. The total force $$\textbf{F}_{\text{net}}$$ is constant. This yields $$\textbf{F}_{\text{net}} = \textbf{F}_g + \textbf{N}$$ and because $$\textbf{F}_{\text{net}}$$ is the net resultant force acting on the block, $$\textbf{F}_{\text{net}} = m\ddot{\textbf{r}}$$ or $$\textbf{F}_g + \textbf{N} = m\ddot{\textbf{r}}$$ This vector must be applied in two directions: $$x$$ and $$y$$ (perpendicularly to $$x$$). The component of force in the $$y$$-direction is zero, because no acceleration occurs in this direction. The force $$\textbf{F}_g$$ is divided vectorially into its $$x$$- and $$y$$-components. The above equation then becomes $$\textit{y-direction}$$ $$-F_g\cos \theta + N = 0$$ $$\textit{x-direction}$$ $$F_g \sin \theta = m\ddot{x}$$ with the required result $$\ddot{x} = \cfrac{F_g}{m}\sin \theta = \cfrac{mg\sin\theta}{m} = g\sin \theta$$ $$\ddot{x} = g\sin(30) = \cfrac{g}{2} = 4.9 \text{ m/s}^2$$ Therefore the acceleration of the block is a constant.

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