Answer:
Most of free energy available from oxidation of the glucose remains in pyruvate.
Explanation:
The overall reaction of the process glycolysis is:
Glucose + 2 NAD⁺ + 2 ADP + 2 Pi ⇒ 2 Pyruvate + 2 NADH + 2 H⁺ + 2ATP
Glucose is oxidized to give 2 molecules of pyruvate and 2 molecules of NADH and ATP (Energy currency).
<u>Though the free energy of oxidation of glucose is high but only 2 NADH is formed because the most of the free energy that is being released from the oxidation of glucose remains in the pyruvate which is produced in the reaction and thus only 2 molecules are formed.</u>
Answer:
439.7nm
Explanation:
Energy of a quantum can be calculated using below formula
E=hv...........eqn(1)
But v=λ/ c .........eqn(2)
If we substitute eqn(2) into eqn(1) we have
E= hc/(λ)
Where E= energy
h= Plank's constant= 6.62607004 × 10-34 m2 kg / s
c= speed of light
c= 2.998 × 10^8 m/s
λ= wavelength= ?
But the energy was given in Kj , it must be converted to Kj/ photon for unit consistency.
Energy E= 272 kJ/mol × 1mol/6.02× 10^23
Energy= 451.83× 10^-24 Kj/ photon
E= hc/(λ)...........eqn(1)
If we make λ subject of the formula
λ= hc/E
Then substitute the values we have
λ= [(6.626 × 10^-34) × (2.998 × 10^8)]/451.83× 10^-24
λ=(0.00043965) × (1Kj/1000J) × (10^9nm/1m)
λ=439.7nm
Hence, the longest wavelength of radiation with enough energy to break carbon-sulfur bonds is 439.7nm
Answer:
100 g
Explanation:
From the question given above, the following data were obtained:
Original amount (N₀) = 400 g
Time (t) = 4 years
Half-life (t½) = 2 years
Amount remaining (N) =?
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Time (t) = 4 years
Half-life (t½) = 2 years
Number of half-lives (n) =?
n = t / t½
n = 4 / 2
n = 2
Thus, 2 half-lives has elapsed.
Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:
Original amount (N₀) = 400 g
Number of half-lives (n) = 2
Amount remaining (N) =?
N = 1/2ⁿ × N₀
N = 1/2² × 400
N = 1/4 × 400
N = 0.25 × 400
N = 100 g
Thus, the amount of the radioactive isotope remaing is the 100 g.
Answer:
b. The shorter the half-life, the more dangerous the radioisotope.
S + O2 → SO2
<span>z / (32.0655 g S/mol) x (1 mol SO2 / 1 mol S) x (64.0638 g SO2/mol) = (1.9979 z) g SO2 </span>
<span>C + O2 → CO2 </span>
<span>(9.0-z) / (12.01078 g C/mol) x (1 mol CO2 / 1 mol C) x (44.00964 g CO2/mol) = (32.9776 - 3.66418 z) g CO2 </span>
<span>Add the two masses of SO2 and CO2 and set them equal to the amount given in the problem: </span>
<span>(1.9979 z) + (32.9776 - 3.66418 z) = 27.9 </span>
<span>Solve for z algebraically: </span>
<span>z = 3.0 g S</span>
