The correct answer is c.
answer b is incorrect <span />
Answer:
The final electron acceptor of the electron transport chain is oxygen
Explanation:
Four electrons gotten from cytochrome c are involved in the conversion of a molecule of oxygen (O2) to two molecules of water (H2O). This final electron transfer occurs in complex IV. Complex IV, also known as cytochrome c oxidase, facilitates the the use of four protons from the matrix of the mitochondrion, in the production of water molecules while pumping four protons to the intermembrane space of the mitochondrion.
Answer:
Eqv Pt pH = 8.73
Explanation:
HOAc + NaOH => NaOAc + H₂O
50ml(0.10M HOAc) + 50ml(0.10M NaOH) => 100ml(0.05M NaOAc) + H₂O
For neutralized system, 100ml of 0.05M NaOAc remains
NaOAc => Na⁺ + OAc⁻
Na⁺ + H₂O => No Rxn
OAc⁻ + H₂O => HOAc + OH⁻
C(i) 0.05M ----- 0M 0M
ΔC -x ----- +x +x
C(f) 0.05-x
≅ 0.05M ----- x x
Kb = Kw/Ka = [HOAc][OH⁻]/[OAc⁻] = 1 X 10⁻¹⁴/1.7 X 10⁻⁵ = (x)(x)/(0.05M)
=> x = [OH⁻] = SqrRt(0.05 x 10⁻¹⁴/1.7 x 10⁻⁵) = 5.42 x 10⁻⁶M
=> pOH = -log[OH⁻] = -log(5.42 x 10⁻⁶) = 5.27
pH + pOH = 14 => pH = 14 - pOH = 14 - 5.27 = 8.73 Eqv Pt pH
Answer:
See explanation
Explanation:
In this case, we have to remember that if we want to remove water from the reaction vessel we have to heat the vessel. So, we can convert the liquid water into <u>gas water</u> and we can remove it from the vessel. In this case, the products of dehydration for both molecules are <u>(E)-4-methylpent-2-ene</u> and <u>cyclohexene</u> with boiling points of <u>59.2 ºC</u> and <u>89 ºC</u> respectively. The boiling point of water is <u>100 ºC</u>, therefore if we heat the vessel the products and water would leave the system, and the products would be lost.
See figure 1
I hope it helps!
