The carbon atom(s) of pyruvate is(are) first converted to carbon dioxide by pyruvate dehydrogenase complex is the second number of carbon of pyruvate goes to oxidation and convert it to CO2 in Krebs cycle.
<h3>what is Krebs cycle ?</h3>
Krebs cycle is also known as citric acid cycle it is the conversion of sugar to the direct energy in the form of ATP which further goes to mitochondria as it is the power house of the human cell.
Pyruvate molecule release second number carbon from the chain and undergoes oxidation to form the CO2.
Therefore, second number carbon atom will converts to carbon dioxide.
Learn more about Krebs cycle , here:
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Answer:
I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.
n = 5 4th excited state
n = 4 3rd excited state
n = 3 2nd excited state
n = 2 1st excited state
n = 1 ground state
Here are the possible spectral lines.
n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.
n = 4 to 3, 4 to 2, 4 to 1 = 3 lines
n = 3 to 2, 3 to 1 = 2 lines
n = 2 to 1 = 1 line. Add 'em up. I get 10.
b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.
c.The E for any level is -21.8E-19 Joules/n^2
To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.
So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.
Explanation:
Since the molecule contains Hydrogen and is covalently bonded, it contains dipole-dipole forces and hydrogen bonds.
Answer:
4.5moles
Explanation:
First, let us balance the equation given from the question. This is illustrated below:
KClO3 —> KCl + O2
There are 2 atoms of O on the right side and 3 atoms on the left. It can be balance by putting 2 in front of KClO3 and 3 in of O2 as shown below
2KClO3 —> KCl + 3O2
Now, we have 2 atoms each of K and Cl on the left side and 1atom each of K and Cl on the right. It can be balance by putting 2 in front of KCl as shown below:
2KClO3 —> 2KCl + 3O2
Now the equation is balanced.
From the balanced equation,
2 moles of KClO3 produced 3 moles of O2.
Therefore, 3 moles of KClO3 will produce = (3 x 3) /2 = 4.5moles of O2.
Therefore 3 moles of KClO3 will produce 4.5 moles of O2
