Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v
coal
Which of these can be mined from earth and used as an energy source?
A.Get rid of the waste and extra fluid
B.Ureters
C.Coal
D.Bladder
ans is coal
Answer:
Check the explanation
Explanation:
When,
pH = -log[H+] = 3.30
[H+] = 

![alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6](https://tex.z-dn.net/?f=alpha%5BY%5E-4%5D%20%3D%20%5BH%2B%5D%5E6%20%2B%20Ka1%5BH%2B%5D%5E5%20%2B%20Ka1Ka2%5BH%2B%5D%5E4%20%2B%20Ka1Ka2Ka3%5BH%2B%5D%5E3%20%2B%20Ka1Ka2Ka3Ka4%5BH%2B%5D%5E2%20%2B%20Ka1Ka2Ka3Ka4Ka5%5BH%2B%5D%20%2B%20Ka1Ka2Ka3Ka4Ka5Ka6)
= 
= 
When,
pH = -log[H+] = 10.15
[H+] = 
Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 =
; Ka6 = 
= 
= 
The way to working out the numbers is to increase the measure of HNO3 required by the molarity to discover what number of moles you require: 0.115. You ought to have the capacity to make sense of the recipe weight H is 1, N is 14, O is 16. The result of the quantity of moles duplicated by the recipe weight ought to give an esteem in grams. You can utilize the thickness to change over to a volume of HNO3 to add to the right volume of water.
Answer:
True
Explanation:
In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].
The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.
This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.
Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.