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Elden
1 month 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:
Sav [3.1K]1 month ago
4 0
Arginine is classified as a basic amino acid since it has two amino groups alongside a single acid group. At a low pH level, all ionizable groups are protonated. As the pH rises slightly, the acid group loses its proton. When the pH increases further, one of the amino groups also loses a proton. At considerably high pH levels, none of the ionizable groups remain protonated.

Pkas

<span> <span><span> <span> pka1 = 1.82 </span> <span> pka2 = 8.99 </span> <span> pka3 = 12.48 </span> </span> </span></span> Thus, 9.20 is above the second pKa and below the third pKa. This indicates that the acid has already lost its proton, as has one of the amino groups, while the second amino group remains protonated. When an acid is not protonated, it carries a negative charge. An unprotonated amino group is neutral, whereas when protonated, the amino group bears a positive charge. Therefore, this amino acid exhibits one positive charge (from one of the amino groups) and one negative charge (from the acid), resulting in an overall neutral charge.
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Considering the activity series given for nonmetals, what is the result of the below reaction? Use the activity series provided.
ValentinkaMS [3465]

Response: The result would be no reaction.

Clarification:

7 0
2 months ago
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What mass needs to be attached to a spring with a force constant of 7N/m in order to make a simple harmonic oscillator oscillate
Sav [3153]

Answer:

The mass will be 4.437 kg

Explanation:

The force constant k is given as 7 N/m

The time period of oscillation T is 5 sec

Thus, angular frequency \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{5}=1.256rad/sec

It is known that angular frequency is computed via

\omega =\sqrt{\frac{k}{m}}

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Squaring both sides gives us

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6 0
1 month ago
A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s
ValentinkaMS [3465]
This question is incomplete. The query is regarding a 3.00 cm diameter coin rolling up a 30.0° incline. With an initial angular speed of 60.0 rad/s, it rolls without slipping. Given that the moment of inertia of the coin is (1/2) MR², the distance the coin travels up the incline is calculated as 0.124 m.
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Two charges of 15 pC and −40 pC are inside a cube with sides that are of 0.40-m length. Determine the net electric flux through
Keith_Richards [3271]

To address this issue, we will utilize the principles related to Gauss' law, which states that the electric flux across a surface corresponds to the object's charge divided by the permittivity of vacuum. In mathematical terms, this can be expressed as

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It's crucial to remember that the net charge equals the difference between the two specified charges, so upon substitution,

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The negative sign indicates that the flux is directed into the surface

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2 months ago
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