Answer:
The statements that correctly describes pyruvate dehydrogenase includes:
- Several copies each of E 1 and E 3 surround E 2.
-A regulatory kinase and phosphatase are part of the mammalian PDH complex.
-E 2 contains three domains.
Explanation:
Pyruvate dehydrogenase is a hydrolase key enzyme in glucose metabolism which converts pyruvate to acetyl- ChoA. It also forms a complex that catalyzes an irreversible reaction that is the entry point of pyruvate into the TCA cycle. Pyruvate dehydrogenase complex contains E1, E2 and E3 enzymes that transform pyruvate, NAD+, coenzyme A into acetyl-CoA, CO2, and NADH. Also, A regulatory kinase and phosphatase are part of the mammalian PDH complex and E 2 contains three domains.
Answer:
Frequency = 6.16 ×10¹⁴ Hz
λ = 4.87×10² nm
Explanation:
In case of hydrogen atom energy associated with nth state is,
En = -13.6/n²
For n = 2
E₂ = -13.6 / 2²
E₂ = -13.6/4
E₂ = -3.4 ev
Kinetic energy of electron = -E₂ = 3.4 ev
For n = 4
E₄ = -13.6 / 4²
E₄ = -13.6/16
E₄ = -0.85 ev
Kinetic energy of electron = -E₄ = 0.85 ev
Wavelength of radiation emitted:
E = hc/λ = E₄ - E₂
hc/λ = E₄ - E₂
by putting values,
6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev - (-3.4ev )
6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev
λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55ev
λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55× 1.6×10⁻¹⁹ J
λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J
λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J
λ = 4.87×10⁻⁷ m
m to nm:
4.87×10⁻⁷ m ×10⁹nm/1 m
4.87×10² nm
Frequency:
Frequency = speed of electron / wavelength
by putting values,
Frequency = 3×10⁸m/s /4.87×10⁻⁷ m
Frequency = 6.16 ×10¹⁴ s⁻¹
s⁻¹ = Hz
Frequency = 6.16 ×10¹⁴ Hz
Cholesterol is an example of a lipid.
Answer:
The Third Law of thermodynamics states that the entropy of a pure substance in a perfect crystalline state at zero temperature is zero.
About 8.0 moles of methane.Number of moles = MassMolar mass.
And thus we get the quotient:
128.3⋅g16.04⋅g⋅mol−1=8.0⋅moles of methane.
Note that the expression is dimensionally consistent, we wanted an answer in moles, and the quotients gives, 1mol−1=11mol=mol as required.
