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.
Answer:
Lead(III) Fluoride PbF3 Molecular Weight -- EndMemo.
Explanation:
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Answer:
mass CaI2 = 23.424 Kg
Explanation:
From the periodic table we obtain for CaI2:
⇒ molecular mass CaI2: 40.078 + ((2)(126.90)) = 293.878 g/mol
∴ mol CaI2 = (4.80 E25 units )×(mol/6.022 E23 units) = 79.708 mol CaI2
⇒ mass CaI2 = (79.708 mol CaI2)×(293.878 g/mol) = 23424.43 g
⇒ mass CaI2 = 23.424 Kg
Answer:
6.23 KOH 90% son necesarios
Explanation:
Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.
Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:
<em>Equivalentes KOH:</em>
0.100L * (1eq / L) = 0.100eq = 0.100moles
<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>
0.100moles * (56.1056g/mol) = 5.61 KOH se requieren
<em>KOH 90%:</em>
5.61g KOH * (100g KOH 90% / 90g KOH) =
<h3>6.23 KOH 90% son necesarios</h3>
Answer:
The volume of water in water bath is 1,011 Liters.
Explanation:
Length of the water bath, L = 1.85 m
Width of the water bath, W= 0.810 m
Height of the water bath ,H= 0.740 m
Height of the water in water bath, h= 0.740 m - 2.57 inches
1 m = 39.37 inch
Volume of the water in bath = L × W × h
The volume of water in water bath is 1,011 Liters.
