Answer:
24e⁻ are transferred by the reaction of respiration.
Explanation:
C₆H₁₂O₆ + 6O₂ → 6 H₂O + 6CO₂
This is the reaction for the respiration process.
In this redox, oxygen acts with 0 in the oxidation state on the reactant side, and -2 in the product side - REDUCTION
Carbon acts with 0 in the glucose (cause it is neutral), on the reactant side and it has +4, on the product side - OXIDATION
6C → 6C⁴⁺ + 24e⁻
In reactant side we have a neutral carbon, so as in the product side we have a carbon with +4, it had to lose 4e⁻ to get oxidized, but we have 6 carbons, so finally carbon has lost 24 e⁻
6O⁻² + 6O₂ + 24e⁻ → 6O₂²⁻ + 6O⁻²
In reactant side, we have 6 oxygen from the glucose (oxidation state of -2) and the diatomic molecule, with no charge (ground state), so in the product side, we have the oxygen from the dioxide with -2 and the oxygen from the water, also with -2 at the oxidation state. Finally the global charge for the product side is -36, and in reactant side is -12, so it has to win 24 e⁻ (those that were released by the C) to be reduced.
Answer:
pCH₄ = 105.1 - 0.42 = 104.68 torr
pI₂ = 7.96 -0.42 = 7.54 torr
pCH₃I = 0.42 torr
pHI = 0.42 torr
Explanation:
Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:
aA(g) + bB(g) ⇄ cC(g) + dD(g)
, where p is the partial pressure in the equilibrium. By the reaction given:
CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)
105.1 torr 7.96 torr 0 0 <em> initial partial pressure</em>
-x -x +x +x <em> react</em>
105.1-x 7.96-x x x <em>equilibrium</em>
Then:
x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²
0.9997x² + 0.0255x - 0.1891 = 0
Using Bhaskara's rule:
Δ = (0.0255)² - 4x(0.9997)x(-0.1891)
Δ = 0.7568
Using only the positive term, x = 0.42 torr.
So,
pCH₄ = 105.1 - 0.42 = 104.68 torr
pI₂ = 7.96 -0.42 = 7.54 torr
pCH₃I = 0.42 torr
pHI = 0.42 torr
Answer:
127.0665 amu
Explanation:
Firstly, to answer the question correctly, we need to access the percentage compositions of the iodine and the contaminant iodine. We can do this by placing their individual masses over the total and multiplying by 100%.
We do this as follows. Since the mass of the contaminant iodine is 1.00070g, the mass of the 129I in that particular sample will be 12.3849 - 1.00070 = 11.3842g
The percentage abundances is as follows:
Synthetic radioisotope % = 1.0007/12.3849 * 100% = 8.1%
Since there are only two constituents, the percentage abundance of the 129I would be 100 - 8.1 = 91.9%
Now, we can use these percentages to get the apparent atomic mass. We get this by multiplying the percentage abundance’s by the atomic masses of both and adding together.
That is :
[8.1/100 * 128.9050] + [91.9/100 * 126.9045] = 10.441305 + 116.6252355 = 127.0665 amu
A significant figure is every symbol that made the number itself.
In this case, the number 40.00 has four figures but only two of them are significant 40, this is because you haven't got any more decimals than the first zero.
If you have a case with zeros in front, you take to the first non zero digits.
For example, 0.071004 you wold express as 0.071 and those 7, and 1 are the significant ones.
Answer:
Gd → Gd⁺ + 1e⁻, Gd⁺ → Gd⁺² + 1e⁻, Gd⁺² → Gd⁺³ + 1e⁻
Explanation:
The ionization energy is the energy necessary to remove one electron of the atom, transforming it in a cation. The first ionization energy is the energy necessary to remove the first electron, the second energy, to remove the second electron, and then successively.
Thus, for gadolinium (Gd)
Fisrt ionization:
Gd → Gd⁺ + 1e⁻
Second ionization:
Gd⁺ → Gd⁺² + 1e⁻
Third ionization:
Gd⁺² → Gd⁺³ + 1e⁻
