In reaction 1 of the Krebs cycle, acetyl‑CoA formed in the pyruvate dehydrogenase reaction condenses with the four‑carbon compound to form <em>citrate </em>with the elimination of coenzyme A. Since the product has three carboxyl groups, this pathway is referred to as the cycle. In reaction 2 of the Krebs cycle, this product then undergoes to form<em> isocitrate. </em>The enzyme is called aconitase because the compound cis‑aconitate is the <em>intermediate product</em> of the reaction. Reaction 3 eliminates CO2 to form the five‑carbon dicarboxylic acid <em>α-cetoglutarate. </em>Oxidation also occurs, with electrons transferred from the substrate to <em>COO-</em> . Consequently, this reaction is an oxidative decarboxylation.
In the image, you can see the reaction 2 in Krebs cycle is a two steps reaction with an intermediate cis-aconitase and a product called isocitrate.
Answer:
lesser the molar mass of the gas higher the no. of moles included in a certain mass sample. ie at STP more volume is required for the gas having less molar mass.
He has the smallest molar mass.
Therefore bag of He is the biggest.
Answer:
A) oxidizing agent is SO2
B) NaClO is the oxidizing agent
Explanation:
A) This is a redox reaction in which oxidation and reduction occur simultaneously.
Thus, in 2H2S(g) + SO2(g) -> 2H2O(l) + 3S(s);
H2S is reduced as follows;
H2S → S + 2H+ + 2e−
We can see that SO2 has been reduced while H2S gets oxidized since it has changed state from - 2 to 0 . Thus sulphur dioxide is the oxidizing agent.
B) SO2(g) + H2O(l) + NaClO(aq) -> NaCl(aq) + H2SO4(aq)
In this, SO2 undergoes oxidation and NaClO is the oxidizing agent
Burette is a very accurate measuring instrument when adding solutions and has a measurement error of 0.05 mL.
Small volumes of solutions can be transferred from the burette at a controllable rate.
In this instance NaOH is in the burette.
Initial reading of NaOH is 0.20 mL
end point is the point at which the chemical reaction reaches completion. In acid base reactions, end point is when all the H⁺ ions have reacted with OH⁻ ions.
final reading of NaOH is 24.10 mL
to find the volume of NaOH dispensed we have to find the difference between final reading and initial reading
volume of NaOH added = 24.10 mL - 0.20 mL = 23.90 mL
volume of NaOH dispensed is 23.90 mL
