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Andru [333]
2 years ago
12

Calculate the average charge on arginine when ph=9.20. (hint : find the average charge for each ionizable group and sum these to

gether.)
Physics
1 answer:
weqwewe [10]2 years ago
4 0
Arginine is a basic aminoacid, because it has two amino groups and one acid group. At a low pH, every ionizable group is protoned. At a little higher pH, the acid group looses its proton. A little higher pH, one amino group looses its proton. At a very high pH, all ionizable groups are not protoned.

Pkas

<span> <span><span> <span> pka1 = 1.82 </span> <span> pka2 = 8.99 </span> <span> pka3 = 12.48 </span> </span> </span></span> So 9.20 is higher tan the second pKa and lower than the third pka. This means the acid has already lost its proton, and one of the aminos too, but the second amino hasn’t. When an acid is not protoned, it has a negative charge. When an amino is not protoned, it’s neutral. When an amino is protoned, it has a positive charge. So this amnino acid has one positive charge (one of the aminos) and one negative charge (the acid), what makes it neutral.
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2 years ago
A student measures the pH of a solution to be 6.8. Which should the student add if she wants to decrease the pH of the solution?
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Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

hence, B_o = 1.013μT

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