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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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The Lyman series comprises a set of spectral lines. All of these lines involve a hydrogen atom whose electron undergoes a change
mihalych1998 [28]

Answer:

a) 1.2*10^-7 m

b) 1.0*10^-7 m

c) 9.7*10^-8 m

d) ultraviolet region

Explanation:

To find the different wavelengths you use the following formula:

\frac{1}{\lambda}=R_H(1-\frac{1}{n^2})

RH: Rydberg constant = 1.097 x 10^7 m^−1.

(a) n=2

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(2)^2})=8227500m^{-1}\\\\\lambda=1.2*10^{-7}m

(b)

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(3)^2})=9751111,1m^{-1}\\\\\lambda=1.0*10^{-7}m

(c)

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(4)^2})=10284375m^{-1}\\\\\lambda=9.7*10^{-8}m

(d) The three lines belong to the ultraviolet region.

8 0
2 years ago
A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnet
Mumz [18]
<h3>Question:</h3>

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.

<h3>Answer:</h3>

1.6nT [in the negative z direction]

<h2>Explanation:</h2>

The magnetic field, B, due to a distance of finite value b, is given by;

B = (μ₀IL) / (4πb\sqrt{b^2 + L^2})                -----------(i)

Where;

I = current on the wire

L = length of the wire

μ₀ = magnetic constant = 4π × 10⁻⁷ H/m

From the question,

I = 20A

L = 2.0cm = 0.02m

b = 5.0m

Substitute the necessary values into equation (i)

B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 \sqrt{5.0^2 + 0.02^2})

B = (10⁻⁷ x 20 x 0.02) / (5.0 \sqrt{5.0^2 + 0.02^2})

B = (10⁻⁷ x 20 x 0.02) / (5.0 \sqrt{25.0004})

B = (10⁻⁷ x 20 x 0.02) / (25.0)

B = 1.6 x 10⁻⁹T

B = 1.6nT

Therefore, the magnetic field at the point x = 5.0m  on the x-axis is 1.6nT.

PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.

5 0
2 years ago
Para proteger un computador de sobrecargas eléctricas, Juan coloca un filamento delgado de cobre llamado fusible en su circuito,
Lynna [10]

Answer:

Los fusibles están diseñados de tal forma que estos se "rompen" o se funden, cuando la demanda eléctrica supera un dado valor (cuando demasiada electricidad pasa a través de el).

Una vez el filamento se rompe, la corriente ya no puede circular por el (podes pensar en esta situación como un cable roto, la electricidad no puede circular por este cable)

Entonces, al romperse el filamento, en caso de una sobrecarga eléctrica, el flujo de electricidad se corta, y de esta forma se protege al computador de posibles sobrecargas.

7 0
1 year ago
Two students, sitting on frictionless carts, push against each other. Both are initially at rest and the mass of student 1 and t
Zepler [3.9K]

Answer:

  v₂ = v/1.5= 0.667 v

Explanation:

For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.

Initial moment before pushing

    p₀ = 0

Final moment after they have been pushed

    p_{f} = m₁ v₁ + m₂ v₂

   p₀ =  p_{f}

   0 = m₁ v₁ + m₂ v₂

   m₁ v₁ = - m₂ v₂

Let's replace

   M (-v) = -1.5M v₂

   v₂ = v / 1.5

  v₂ = 0.667 v

6 0
2 years ago
Which combination of initial horizontal velocity, (vh) and initial vertical velocity, (vv) results in the greatest horizontal ra
Naya [18.7K]

If an object is projected with vertical speed given as

v_v

now the time of flight of the object that time in which it comes back on ground

\Delta y = v_v t + \frac{1}{2}(-g)t^2

now here we will have

0 = v_v t - \frac{1}{2}gt^2

t = \frac{2v_v}{g}

now the range of projectile is given as

R = horizontal\: speed \times time

R = v_h(\frac{2v_v}{g}

now here we know that

v_v = v_0 sin\theta

v_h = v_0 cos\theta

now the range is given as

R = \frac{2(vsin\theta)(vcos\theta)}{g}

R = \frac{v^2sin2\theta}{g}

now in order to have maximum range we can say

sin2\theta = 1

2\theta = 90^0

so we will have

\theta = 45 ^0

so now we can say

v_v = v_h = \frac{v_0}{\sqrt2}

so both speed must be same to have maximum horizontal range

8 0
2 years ago
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