During aerobic respiration, electrons travel downhill in which sequence? A) food → citric acid cycle → ATP → NAD+
1 answer:
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Answer : The partial pressure of at equilibrium is, 1.0 × 10⁻⁶
Explanation :
The partial pressure of =
The partial pressure of =
The partial pressure of =
The balanced equilibrium reaction is,
Initial pressure 1.0×10⁻² 2.0×10⁻⁴ 2.0×10⁻⁴
At eqm. (1.0×10⁻²-2p) (2.0×10⁻⁴+p) (2.0×10⁻⁴+p)
The expression of equilibrium constant for the reaction will be:
Now put all the values in this expression, we get :
The partial pressure of at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶
Therefore, the partial pressure of at equilibrium is, 1.0 × 10⁻⁶
Answer:
1) 1,1,1-trichloropropane
2) 1,1,2-trichloropropane
3) 1,2,2-trichloropropane
4) 1,2,3-trichloropropane
Explanation:
For this question, we must remember that isomers are molecules that have the <em>same formula but different structure</em>s. For the formula we can draw a <u>linear chain of three carbons</u> and change the position of the chlorine atoms in the carbon chain.
With this in mind, if we put all the chlorine atoms on the same carbon we will get <u>1,1,1-trichloropropane</u>. If we change an atom from chlorine to carbon 2 we will obtain <u>1,1,2-trichloropropane</u>. If we move another chlorine atom to carbon two we will get <u>1,2,2-trichloropropane</u>. Finally, if we put a chlorine atom in each carbon we will obtain <u>1,2,3-trichloropropane</u>.
See figure one for further explanations
I hope it helps!
