Answer:
In the third step of the citric acid cycle, the oxidation of isocitrate takes place and one molecule of carbon dioxide is released.
Explanation:
In the first step of citric acid cycle, acetylCoA combines with a four-carbon molecule, oxaloacetate, forming a six-carbon molecule, citrate.
In the second step, the citrate in the presence of enzyme anicotase is converted into isocitrate.
<u>In the third step, the oxidation of isocitrate takes place and one molecule of carbon dioxide is released leaving behind one five-carbon molecule called as α-ketoglutarate. During this step, NAD⁺ is reduced to form NADH. </u>
<u>This is first round of the citric acid cycle that could possibly release a carbon atom originating from this acetyl CoA.</u>
On series of reaction, another carbon dioxide molecule also being relased and oxaloacetate is regenerated again.
4 x 1000 so 4 kilometers are equal to 4000 meters
when the thermal energy is the energy contained within a system that is responsible for its temperature.
and when the thermal energy is can be determined by this formula:
q = M * C *ΔT
when q is the thermal energy
and M is the mass of water = 100 g
and C is the specific heat capacity of water = 4.18 joules/gram.°C
and T is the difference in Temperature = 50 °C
So by substitution:
∴ q = 100 g * 4.18 J/g.°C * 50
= 20900 J = 20.9 KJ
It's the first option choice on Plato
Answer:
a) Unsaturated
b) Supersaturated
c) Unsaturated
Explanation:
A saturated solution contains the <u>maximum amount of a solute that will dissolve in a given solvent at a specific temperature</u>.
An unsaturated solution contains <u>less solute than it has the capacity to dissolve. </u>
A supersaturated solution, <u>contains more solute than is present in a saturated solution</u>. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.
According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:
a) 30 g in 100 mL of H₂O at 20 °C ⇒ unsaturated
b) 65 g in 100 mL of H₂O at 50 °C ⇒ supersaturated
c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)
