Answer:
a) Keq = 4.5x10^-6
b) [oxaloacetate] = 9x10^-9 M
c) 23 oxaloacetate molecules
Explanation:
a) In the standard state we have to:
ΔGo = -R*T*ln(Keq) (eq.1)
ΔGo = 30.5 kJ/moles = 30500 J/moles
R = 8.314 J*K^-1*moles^-1
Clearing Keq:
Keq = e^(ΔGo/-R*T) = e^(30500/(-8.314*298)) = 4.5x10^-6
b) Keq = ([oxaloacetate]*[NADH])/([L-malate]*[NAD+])
4.5x10^-6 = ([oxaloacetate]/(0.20*10)
Clearing [oxaloacetate]:
[oxaloacetate] = 9x10^-9 M
c) the radius of the mitochondria is equal to:
r = 10^-5 dm
The volume of the mitochondria is:
V = (4/3)*pi*r^3 = (4/3)*pi*(10^-15)^3 = 4.18x10^-42 L
1 L of mitochondria contains 9x10^-9 M of oxaloacetate
Thus, 4.18x10^-42 L of mitochondria contains:
molecules of oxaloacetate = 4.18x10^-42 * 9x10^-9 * 6.023x10^23 = 2.27x10^-26 = 23 oxaloacetate molecules
Answer: Three samples that can be used to test for poisons are blood samples, hair samples, and gastric contents samples.
Advantages of hair samples area is that it can indicate past exposure to toxins or poisons and also provide information on how long the exposure has been going on. Poisons can sometimes be detected in hair for long periods of time. The advantages of gastric contents samples area is it shows whether the poison was digested or not. An advantage of blood samples is that it’s the most useful toxicological samples. The disadvantage of hair samples is that the results can be tainted by external sources. The disadvantage of gastric content samples is that it doesn’t always show which toxin was digested. The disadvantage of blood samples is if taken near or from the heart it may have artificially high levels of toxins that wouldn’t reflect the amount distributed throughout the body at the time of death.
Explanation:
Answer:
92.93 g
Explanation:
Number of half lives that have elapsed in eight days =8/14.3 = 0.559
Fraction of the radioactive nuclide that remains after 0.559 half lives is given by
N/No=(1/2)^0.559
Where N= mass of radioactive nuclides remaining after a time t
No= mass of radioactive nuclides originally present
N/No=(1/2)^0.559= 0.679
Mass of nuclides present eight days before= 63.1g/0.679
Mass of nuclides present eight days before=92.93 g
Answer:
right
Explanation:
the air that circulates to the right is deflected, because of the earths gravitational rotation, and how its tilted.
The total pressure 1566 mm Hg.
Pressure is the force applied perpendicular to the floor of an object per unit location over which that pressure is shipped. Gauge strain additionally spelled gage pressure is the pressure relative to the ambient stress. pressure.
The SI unit of stress is pascal which is identical to one newton in keeping with rectangular meter. apparently, this name was given in 1971. earlier than that pressure in SI became measured in newtons in line with square meter.
Partial pressure of Argon = 429 mm Hg
Partial pressure of Neon = 1.2 atm
∵ 1 atm = 760 mm Hg
=> 1.2 atm = 1.2 * 760 mm Hg
= 912 mm Hg
Partial presser of He = 225 mm Hg
S0, Total pressure = Par + P ne + P he
= ( 429 + 912 + 225 ) mm Hg
= 1566 mm Hg
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