Answer:
1. NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂
2. The reactant that is reduced is Q
3. The charge on iron on the right side is +2, Fe²⁺
Explanation:
NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂
The reaction above is catalysed by NADH:ubiquinone oxidoreductase (complex 1), which transfers a hydride ion from NADH to FMN, from which two electrons pass through a series of of Fe-S centers to the iron-sulfur protein N-2. Electron transfer from N-2 to Ubiquinone forms QH₂
The species in a reaction which gains hydrogen irons is reduced, Therefore, the reactant that is reduced is Q, ubiquinone to form QH₂, ubiquinol.
To determine the oxidation number of iron on the right side of the reaction below,
QH2 + 2cyt c ( Fe3+) ⟶ Q + 2cyt c(Fex) + 2H^+
Sum of charges on the left side = Sum of charges on the right side
Sum of charges on the left side = 2 *+3 = +6
Therefore 2 * x + 2= 6
2x = 6 -2 = 4
x = 4/
x = 2
Therefore the charge on iron on the right side is +2, Fe²⁺
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Answer:
Advantakes= it is a renewable source, you can but it anywhere with sunlight
disatvantage= it cost a lot to place/replace, it uses a lot of different materials
Answer:- C. Hafnium.
Solution:- Mass of the sample is 46.0 g and it's volume is
.
From mass and volume, we can calculate it's density using the formula:



On the basis of the density, this substance could either be mercury or hafnium. Since the substance is a solid at room temperature where as mercury is liquid. So, it can't be mercury.
The right choice is C) Hafnium.