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Mariulka [41]
3 years ago
15

The pKa of an acid depends partly on its environment. Predict the effect of each of the following environmental changes on the p

Ka of a glutamic acid side chain. (a) A lysine side chain is brought into proximity. (b) The terminal carboxyl group of the protein is brought into close proximity.
Chemistry
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

A. Decrease

B. Increase

Explanation:

A. A lysine side chain is brought into proximity:

lysine is basic in its nature, it would have to stabilize the electrostatic interaction and also weak interactions existing among the acidic amino acids and hydrogen bond. This would then reduce or decrease the pH and thus pKa of the protein.

B. The effect of this on pKA is an increase on the pKA. The carboxyl group is acidic. When it is removed, it would then cause the pKA to be increased. There would be accumulation of more negative change by basic amino acids increasing the pKA.

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Zina [86]

Answer:

Explanation:

n CaCO3 = mass / m.wt

             = 500  /( 40 + 12 + 16x 3)

           =   5 mole

n CaO = 5 moles  ( from the balanced equation we have 1:1 moles )

mass of CaO = nCaO X m.wt

                       5 x(  40 +16 )

                 =   280 grams

5 0
3 years ago
Which subatomic part of the atom (a-c) are involved in chemical bonds and reactions?
Marianna [84]
B because it can stabilize
8 0
3 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
The metalloid that has three valence electrons is?
bezimeni [28]
The metalloid that has three valence electrons is Boron~
8 0
3 years ago
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Which element has a larger atomic radius than sulfur?
Novay_Z [31]

Cadium

Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period.

6 0
3 years ago
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