The answer is; acetyl-CoA
This organic compound is an intermediate of the TCA/Citric/Krebs cycle. To make fats, the coenzyme is carboxylated to manolyCoA. This becomes a precursor fo palmitate and other lipids in the body by the addition of more manolyCoA.
The coenzyme is also a precursor to the formation of alpha-ketoglutarate, another intermediate of the TCA cycle, which is also a precursor in the formation of proteins.
Answer: He theorized that all material bodies are made up of indivisibly small “atoms.” Aristotle famously rejected atomism in On Generation and Corruption. Aristotle refused to believe that the whole of reality is reducible to a system of atoms.
Explanation:
Explanation:
The given data is as follows.
T = = (120 + 273.15)K = 393.15 K,
As it is given that it is an equimolar mixture of n-pentane and isopentane.
So, = 0.5 and = 0.5
According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.
(393.15 K) = 9.2 bar
(393.15 K) = 10.5 bar
Hence, we will calculate the partial pressure of each component as follows.
=
= 4.6 bar
and,
=
= 5.25 bar
Therefore, the bubble pressure will be as follows.
P =
= 4.6 bar + 5.25 bar
= 9.85 bar
Now, we will calculate the vapor composition as follows.
=
= 0.467
and,
=
= 0.527
Calculate the dew point as follows.
= 0.5, = 0.5
= 0.101966
P = 9.807
Composition of the liquid phase is and its formula is as follows.
=
= 0.5329
=
= 0.467
Answer:
Volume is 1.065L
Explanation:
Hello,
We can easily solve this problem by using general gas equation.
PV / T = K
P1V1/T1 = P2V2/T2
Data;
P1 = 3.0atm
P2 = 1.0atm
T1 = 20°C = (20 + 273.15)K = 293.15K
T2 = 20°C = (20 + 273.15)K = 293.15K
V1 = 355mL = 0.355L
V2 = ?
From the data given, we can substitute it into the equation,
(P1 × V1) / T1 = (P2 × V2) / T2
(3.0 × 0.355) / 293.15 = (1.0 × V2) / 293.15
1.065 = 1.0V2
Divide both sides by 1.0
V2 = 1.065L
The volume of CO₂ released is 1.065L
Answer:
Kp for the reverse reaction is 76.9
Explanation:
Step 1: Kp at 1000 K = 0.013
Step 2: The balanced equation
2NO (g) + Br2 (g) ⇆ 2NOBr (g)
2NOBr (g) ⇆ 2NO (g) + Br2 (g)
Step 3: Calculate Kp
Kp for the reaction 2NO (g) + Br2 (g) ⇆ 2NOBr (g)
Kp = p(NOBr)²/(pBr2)*(pNO)²
Kp = 0.013
Kp for the reverse reaction 2NOBr (g) ⇆ 2NO (g) + Br2 (g)
Kp' = (pBr2)*(pNO)² /p(NOBr)²
Kp' = 1/Kp = 1/0.013 = 76.9
Kp for the reverse reaction is 76.9