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Inga [223]
2 years ago
9

When sulfate (SO42-) serves as the electron acceptor at the end of a respiratory electron transport chain, the product is?

Chemistry
1 answer:
o-na [289]2 years ago
3 0

When sulfate (SO₄²⁻) serves as the electron acceptor at the end of a respiratory electron transport chain, the product is hydrogen sulfide (H₂S).

How sulfate acts as electon acceptor and electron donor?

  • Sulfate (SO₄²⁻) is used as the electron acceptor in sulfate reduction, which results in the production of hydrogen sulfide (H2S) as a metabolic byproduct.
  • Many Gram negative bacteria identified in the -Proteobacteria use sulfate reduction, which is a rather energy-poor process.
  • Gram-positive organisms connected to Desulfotomaculum or the archaeon Archaeoglobus also utilise it.
  • Electron donors are needed for sulfate reduction, such as hydrogen gas or the carbon molecules lactate and pyruvate (organotrophic reducers) (lithotrophic reducers).

Learn more about the Electron transport chain with the help of the given link:

brainly.com/question/24372542

#SPJ4

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Answer: The last electron will be filled in first orbital of 3p sub-shell.

Explanation: Filling of electrons in orbitals is done by using Hund's Rule.

Hund's rule states that the electron will be singly occupied in the orbital of the sub-shell before any orbital is doubly occupied.

For filling up of the electrons in Sulfur atom having 16 electrons. First 10 electrons will completely fill according to Aufbau's Rule in 1s, 2s and 2p sub-shells and last 6 electrons are the valence electrons which will be filled in the order of 3s and then 3p.

3s sub-shell will be fully filled and the orbitals of 3p sub-shell will be first singly occupied and then pairing will take place. Hence, the last electron will be filled in the first orbital of 3p-sub-shell.

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Consider the equation:
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Answer:1. Rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}

2. The rate constant (k) for the reaction is 3.50M^\frac{-1}{2}s^{-1}

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

rate=k[CHCl_3]^x[Cl_2]^y

k= rate constant

x = order with respect to CHCl_3

y = order with respect to Cl_2

n = x+y= Total order

1. a) From trial 1: 0.0035=k[0.010]^x[0.010]^y  (1)

From trial 2: 0.0069=k[0.020]^x[0.010]^y   (2)

Dividing 2 by 1 :\frac{0.0069}{0.035}=\frac{k[0.020]^x[0.010]^y}{k[0.010]^x[0.010]^y}

2=2^x,2^1=2^x therefore x=1.

b)  From trial 2: 0.0069=k[0.020]^x[0.010]^y   (3)

From trial 3: 0.0098=k[0.020]^x[0.020]^y   (4)

Dividing 4 by 3:\frac{0.0098}{0.0069}=\frac{k[0.020]^x[0.020]^y}{k[0.020]^x[0.010]^y}

1.4=2^y,2^{\frac{1}{2}}=2^y therefore y=\frac{1}{2}

rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}

2. to find rate constant using trial 1:

0.0035=k[0.010]^1[0.010]^\frac{1}{2}  

k=3.50M^\frac{-1}{2}s^{-1}

5 0
3 years ago
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