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⁺
Answer:
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Answer:
308 g
Explanation:
Data given:
mass of Fluorine (F₂) = 225 g
amount of N₂F₄ = ?
Solution:
First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄
Reaction:
2F₂ + N₂ -----------> N₂F₄
Now look at the reaction for mole ratio
2F₂ + N₂ -----------> N₂F₄
2 mole 1 mole
So it is 2:1 mole ratio of Fluorine to N₂F₄
As we Know
molar mass of F₂ = 2(19) = 38 g/mol
molar mass of N₂F₄ = 2(14) + 4(19) =
molar mass of N₂F₄ = 28 + 76 =104 g/mol
Now convert moles to gram
2F₂ + N₂ -----------> N₂F₄
2 mole (38 g/mol) 1 mole (104 g/mol)
76 g 104 g
So,
we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.
Apply unity formula
76 g of F₂ ≅ 104 g of N₂F₄
225 g of F₂ ≅ X of N₂F₄
Do cross multiplication
X of N₂F₄ = 104 g x 225 g / 76 g
X of N₂F₄ = 308 g
So,
308 g N₂F₄ can be produced from 225 g F₂
44. (a) N2O3 (b) SF4 (c) AlCl3 (d) Li2CO3
46. H Br
δ+ δ−
48. The metallic potassium atoms lose one electron and form +1 cations,
and the nonmetallic fluorine atoms gain one electron and form –1 anions.
K → K+
+ e–
19p/19e–
19p/18e–
F + e–
→ F–
9p/9e–
9p/10e–
The ionic bonds are the attractions between K+
cations and F–
anions.
50. See Figure 3.6.
52. (a) covalent…nonmetal-nonmetal (b) ionic…metal-nonmetal
54. (a) all nonmetallic atoms - molecular (b) metal-nonmetal - ionic
56. (a) 7 (b) 4
58. Each of the following answers is based on the assumption that nonmetallic
atoms tend to form covalent bonds in order to get an octet (8) of
electrons around each atom, like the very stable noble gases (other than
helium). Covalent bonds (represented by lines in Lewis structures) and lone
pairs each contribute two electrons to the octet.
(a) oxygen, O
If oxygen atoms form two covalent bonds, they will have an octet of electrons
around them. Water is an example:
H O H
(b) fluorine, F
If fluorine atoms form one covalent bond, they will have an octet of electrons
around them. Hydrogen fluoride, HF, is an example:
H F
(c) carbon, C
If carbon atoms form four covalent bonds, they will have an octet of electrons
around them. Methane, CH4, is an example:
H H
H
H
C
(d) phosphorus, P
If phosphorus atoms form three covalent bonds, they will have an octet
