Answer:
Natural selection is a process in which organisms are fit to adapt and live in the environment. Natural selection was an idea developed by Charles Darwin, Darwin contributed this idea when he was studying birds and tortoises at the Galapagos Islands. He then wrote his famous book, <em><u>O</u></em><em><u>r</u></em><em><u>i</u></em><em><u>g</u></em><em><u>i</u></em><em><u>n</u></em><em><u> </u></em><em><u>o</u></em><em><u>f</u></em><em><u> </u></em><em><u>S</u></em><em><u>p</u></em><em><u>e</u></em><em><u>c</u></em><em><u>i</u></em><em><u>e</u></em><em><u>s</u></em><em><u>.</u></em> Evolution is also very similar, evolution is a process in which organisms reproduce and mutate in order to survive and adapt. As you can see, natural selection and evolution are alike in many ways.
☆anvipatel77☆
•Expert•
Brainly Community Contributor
Answer:
M = 184.1 g/mol
Explanation:
Given data:
Density = 5.95 g/L
Temperature = 100°C (100+273 = 373 K)
Pressure = 755 mmHg = (755/760 = 0.99 atm )
Molar mass of gas = ?
Formula:
d = PM/RT
Solution:
M = dRT / P
M = 5.95 g/L × 0.0821 atm.L/mol.K × 373 K/ 0.99 atm
M = 182.21atm.g/mol / 0.99 atm
M = 184.1 g/mol
Explanation:
We know that relation between
and
is as follows.
= 14
As it is given that
is 8.18. Therefore, calculate the value of
as follows.
= 14
= 14
= 14 - 8.18
= 5.82
Similarly, as value of pH is given as 7.18. Therefore, value of pOH will be as follows.
pH + pOH = 14
7.18 + pOH = 14
pOH = 6.82
Let us take that B represents the enzyme. Hence, its reaction with proton will be as follows.
(protonated active enzyme)
Hence, pOH = ![pK_{b} + log\frac{[BH^{+}]}{[B]}](https://tex.z-dn.net/?f=pK_%7Bb%7D%20%2B%20log%5Cfrac%7B%5BBH%5E%7B%2B%7D%5D%7D%7B%5BB%5D%7D)
6.82 = 5.82 + ![log_{10} \frac{[BH^{+}]}{[B]}](https://tex.z-dn.net/?f=log_%7B10%7D%20%5Cfrac%7B%5BBH%5E%7B%2B%7D%5D%7D%7B%5BB%5D%7D)
= 10
Therefore, percentage of active enzyme = %
= ![\frac{10}{10 + 1} \rimes 100](https://tex.z-dn.net/?f=%5Cfrac%7B10%7D%7B10%20%2B%201%7D%20%5Crimes%20100)
%
= 90.9%
Thus, we can conclude that 90.9% is the percentage of the enzyme which is active in a buffer at pH 7.18.
4.33898305 is the density
Divide mass/volume