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Joy S.
UC Berkeley undergrad
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Writing
TutorMe
Question:

How do I accomplish clarity in my essay?

Joy S.
Answer:

Clarity has a lot to do with your choice of words and sentence syntax. I know too many students who believe using bigger words in their essays will help them score a higher grade. But this is not completely true. While showcasing a more advanced and versatile vocabulary does show your aptitude for the English language, big complex words can clutter and slow down your writing. For the reader, it becomes annoying if you feel the need to look up every 3rd word in the dictionary! When you want to achieve clarity in your writing, try to avoid overly gaudy or extravagant word choice and limit the length of your sentences. Even better, your essay should have a good mix of long, short, and medium sentences as this aids in keeping the pace of your writing consistent.

Chemistry
TutorMe
Question:

3NO2 (g) + CO (g) <--> NO (g) + CO2 (g) + N2O4 (g) a) What's the K expression for the reaction above? Use pressure for the concentrations of gaseous compounds. b) If K = 70, find delta G naught? (R = 8.314 J/molK, T = 298K) c) Assume the reaction begins at equilibrium, and then later the reaction vessel doubles in volume. Calculate Q. d) Now that our volume is larger, will the reaction proceed int he forward or reverse direction to reach equilibrium again?

Joy S.
Answer:

a) The first thing we do is check if the reaction is balanced. Number of moles of each element on the products side equal the number of moles of each element on the reactants side; thus, the reaction is already balanced. If we are given: aA + bB <--> cC + dD Then, K = ([C]^c [D]^d) / ([A]^a [B]^b) Using this formula, we get: K = (Pressure of N2O4 * Pressure of CO2 * Pressure of NO) / (Pressure of CO * (Pressure of NO2)^3) b) To solve this, we need to use a formula to calculate delta G naught. We are given R, K, and temperature. Thus we should use: delta G naught = -RTInK Check to make sure all values are in the correct units and then solve for delta G naught. = -(8.314 J/molK)(298K)(In70) = -10,525.95 J/mol c) If we double the volume of the vessel, this will decrease the pressure of the gasses. To find Q, we need to use the same formula we used to find K. While K is the equilibrium constant at exactly 298K, Q is the equilibrium constant at any other decided temperature we want. Q = ([C]^c [D]^d) / ([A]^a [B]^b) By doubling the volume, this will half the pressure because pressure and volume are inversely proportional to one another. This is stated in Boyle's Law: P = 1/V We can apply this concept to help us solve for Q. Because the problem tells us we started at equilibrium, this means that all of our gasses have a pressure equal to 1. Q = (1/2*1/2*1/2) / (1/2*(1/2)^3) = 2 Thus, 2 * K = 2 * 70 = 140 d) Our reaction will progress in the reverse direction, because Q is greater than K

Biology
TutorMe
Question:

You are given a list containing: CO2, Cl-, sucrose and glycerol. Rank the listed molecules from least diffusible to most diffusible according to how easily they diffuse across the plasma membrane.

Joy S.
Answer:

Answer: Cl-, sucrose, glycerol, CO2 Cl- is the least diffusible because it is a charged ion. Charged compounds have a difficult time diffusing across a plasma membrane because of the amphiphile (containing both a hydrophilic head and hydrophobic tails) quality of the lipid bilayer. Cl- would be transported via a voltage-gated ion channel such as those found in neurons. Sucrose is less diffusible than glycerol because its molecular structure is larger. Larger molecules have a harder time diffusing across membranes compared to smaller molecules such as CO2.

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