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Bianca W.
Chemical engineer with strong math and science background, Tutor for 3 years
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Chemistry
TutorMe
Question:

Staurosporine (FW: 466.53 g/mol) is a potent reagent used to inhibit various PKC isoforms in human cell lines, and is supplied in a vial containing 2 mg of lyophilized powder. You want to treat 12 ml of HL-60 cells with a final concentration of 5 nM staurosporine, and you only want to add 6 $$\mu$$L of staurosporine solution to the cells. What concentration should you make your stock solution of staurosporine in order to make this 6 $$\mu$$L addition?

Bianca W.
Answer:

To get the answer for the concentration of the stock solution, I'll use the equation M1V1=M2V2. This equation is a representation of the law of conservation of mass and tells you that the mass of the solvent (staurosporine) in the original stock solution is the same as the mass in the final 12 mL solution. The mass is equal to the concentration times the volume. M1=x, V1= 6µL, M2= 5 nM, and V2= 12 mL + 6µL First, I'll convert the 12 mL to µL: 12 ml*(106 µL/ 103 mL)=12,000µL. So, V2 becomes 12,006 µL M1= M2*V2/V1 = 5nM*(12,000 µL)/6µL=10005 nM*(106 µM/109 nM)=10 µM The stock solution needs to have a concentration of 10 µM.

Calculus
TutorMe
Question:

Find y if y'(2)=0, y(2)=1, and y''-4=0

Bianca W.
Answer:

I'll start with the equation y''-4=0. Adding 4 to both sides of the equation gives y''=4. Integrate both sides of the equation: $ \int {y}''dy = \int 4 dx $ ${y}'=\int 4 dx=4x+C$ Now we can use y'(2)=0 to solve for C by plugging in 2 and 0 for x and y' respectively. Giving 0=(4)(2) +c. Solving that equation gives C=-8. The equation is now y'=4x -8. Integrate both sides of that equation: $\int {y}'dy=\int 4x-8 dx$ $y=\int 4x-8 dx = 2x^{2}-8x+C% Now use y(2)=1 to solve for c by plugging in 2 and 1 for x and y, respectively, giving 1=2*(2^2)-8*2+c. Solving that equation gives c=9. That gives you a final answer of: $y=2x^{2}-8x+9$

Chemical Engineering
TutorMe
Question:

Explain the concept of accumulation. (Do not merely cite a general mass or energy balance equation, but rather clearly explain the meaning of the term “accumulation”.)

Bianca W.
Answer:

Accumulation is the time rate of change of an extensive property, such as mass or total energy, within a control volume (fixed region in space). For processes operating in a steady state, the accumulation is zero.

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