Answer:
1/3
Explanation:
Pyruvate is produced by the glycolysis in cytoplasm. The oxidation of pyruvate takes place in mitochondrial matrix.
Pyruvate is converted to acetyl-CoA in the reaction given below:
Pyruvate + NAD⁺ + CoA-SH ⇒ acetyl-CoA + NADH + CO₂
1 molecule of carbon dioxide is eliminated from 1 molecule of pyruvate.
Also,
2 molecules of carbon dioxide is eliminated from 2 molecules of pyruvate (as glucose on glycolysis yields 2 molecules of pyruvate).
Also, acetyl-CoA further goes into the citric acid cycle and produces 2 molecules of carbon dioxide.
Thus pyruvate produces total 3 molecules of CO₂ and hence glucose produces 6 molecules of CO₂ (as glucose on glycolysis yields 2 molecules of pyruvate)
Thus,
<u>Fraction = 2/6 = 1/3</u>
Answer:
Explanation:
Due to Coulomb´s law electric force can be described by the formula 
, where K is the Coulomb´s constant (
), 
= Charge 1 (Na+ in this case), 
is the charge 2 (Cl-) and r is the distance between both charges.
Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is 
.
so we have ![W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]](https://tex.z-dn.net/?f=W%3DW_%7Bf%7D%20-W_%7Bi%7D%20%3D%28K%5Cfrac%7Bq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%7Drf%7D%7Br_%7Bf%7D%20%5E%7B2%7D%7D%29-%28K%5Cfrac%7Bq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%7Dri%7D%7Br_%7Bi%7D%20%5E%7B2%7D%7D%29%3DKq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%5B%5Cfrac%7B1%7D%7B%7Br_%7Bf%7D%7D%7D%20-%5Cfrac%7B1%7D%7B%7Br_%7Bi%7D%7D%7D%5D)
Given that ri= 1.1nm= 
and rf= infinite distance
![W=(9x10^{9})(1.6x10^{-19})(-1.6x10^{-19})[\frac{1}{\alpha }-\frac{1}{(1.1x10^{-9})}]=2.1x10^{-19}J](https://tex.z-dn.net/?f=W%3D%289x10%5E%7B9%7D%29%281.6x10%5E%7B-19%7D%29%28-1.6x10%5E%7B-19%7D%29%5B%5Cfrac%7B1%7D%7B%5Calpha%20%7D-%5Cfrac%7B1%7D%7B%281.1x10%5E%7B-9%7D%29%7D%5D%3D2.1x10%5E%7B-19%7DJ)
Answer:
(1) cathode: Y
(2) anode X
(3) electrons in the wire flow toward: Y
(4) electrons in the wire flow away from: X
(5) anions from the salt bridge flow toward X
(6) cations from the salt bridge flow toward Y
(7) gains mass: Y
(8) looses mass X
Explanation:
The voltaic cell uses two different metal electrodes, each in an electrolyte solution. The anode will undergo oxidation and the cathode will undergo reduction. The metal of the anode will oxidize, going from an oxidation state of 0 (in the solid form) to a positive oxidation state, and it will become an ion. At the cathode, the metal ion in the solution will accept one or more electrons from the cathode, and the ion’s oxidation state will reduce to 0. This forms a solid metal that deposits on the cathode. The two electrodes must be electrically connected to each other, allowing for a flow of electrons that leave the metal of the anode and flow through this connection to the ions at the surface of the cathode. This flow of electrons is an electrical current that can be used to do work, such as turn a motor or power a light.
