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José S.
Tutored throughout college, Researcher
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Calculus
TutorMe
Question:

$$ \int \frac{x} {(x+2)^2} dx $$

José S.
Answer:

First we use partial fractions to simplify the integral $$ \frac{x}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2} $$ $$ x = A(x+2) + B $$ $$ x = Ax + 2A + B $$ For all x terms $$ x = Ax$$ $$A = 1$$ For all constant terms $$0 = 2A + B$$ $$ B = -2$$ So, our integral can now be expressed as $$ \int \frac{1}{x+2} - \frac{2}{(x+2)^2} dx$$ which can be solved to obtain $$ \ln{|x+2|} + \frac{2}{x+2} + C $$

Chemical Engineering
TutorMe
Question:

What would be the more preferred method of separation for proteins and cells? Explain your reasoning a. Distillation b. Absorption/Stripping Column c. Membrane separation d. Leaching e. Liquid-Liquid Extraction

José S.
Answer:

c. Membrane Separation Protein and microorganism separation is a temperature-sensitive process, which rules our the use of distillation. Furthermore, one would need to take special considerations when adding a material separating agent since these could also change the nature of the product to be separated, so absorption, leaching, and LLE are ruled out. Therefore, it is better to conduct these types of separations using membrane separation processes. Proteins can also be separated effectively using chromatography.

Physics
TutorMe
Question:

How much heat is required to raise the temperature of a kilogram of gold from 0C to room temperature (25C)? The heat capacity of Gold is 0.129 J/g/C

José S.
Answer:

In order to calculate heat, we utilize the equation Q = mc(T_f - T_i) Where m = mass c = heat capacity T_f = final temperature I_i = initial temperature Since heat capacity was given in J/g/C, we will use 1000g gold as our mass Q = 1000g X 0.129 J/g/C X (25C - 0C) Q = 1000g X 0.129 J/g/C X 25C Q = 3225J So, 3225 Joules of energy is required to raise the temperature of 1kg of gold from 0 to 25C

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