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).
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Answer:
24.7 amu
Explanation:
An isotope is when an element can have different number of neutrons but they have same number of protons.
In order to calculate the average atomic mass with the given information do the following operations:
First change de percentages to fractional numbers, divide by 100.
I like to make a table, to organize all data and I believe is easier to understand.
65/100 = 0.65
35/100 = 0.35
% fraction
65.0 0.65
35.0 0.35
total100.0 1
Now multiply each mass with their corresponding fraction
24 (0.65) = 15.6
26 (0.35) = 9.1
% fraction uma uma
65.0 0.65 24 15.6
35.0 0.35 26 9.1
total100.0 1 24.7
Finally you add the resulting mass and the units will be in uma.
15.6+9.1 = 24.7
Therefore the average atomic mass of this element will be 24.7 uma.
Check the table in the document attached
Answer:
In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents. The reagent that is completely used up or reacted is called the limiting reagent, because its quantity limits the amount of products formed.
Explanation:
(a) The displacement of point P at time t=0.10s is determined as +2cm.
(b) The displacement of point P at time t=0.20s is determined as -2cm.
<h3>
What is displacement?</h3>
Displacement is the change in position of an object. It is obtained from the product of velocity and time of motion.
x = vt
<h3>Displacement of the waves after 0.1 s</h3>
x = 10 m/s x 0.1 s = 1 m
Each wave will travel 1 m to the right or to the left, depending on the initial direction.
- wave B from left will stop at point 0 m
- wave A from left will stop at point -1 m
- wave C from right will stop at point 0 m
- wave D from right will stop at point + 1 m
wave B and C superimposed and the displacement will be between A and D.
Amplitude of A = - 2cm
Amplitude of D = + 4cm
Displacement of point P = 4 cm - 2 cm = 2cm
<h3>Displacement of the waves after 0.2 s</h3>
x = 10 m/s x 0.1 s = 2 m
Each wave will travel 2 m to the right or to the left, depending on the initial direction.
- wave B from left will stop at point 1 m
- wave A from left will stop at point 0 m
- wave C from right will stop at point -1 m
- wave D from right will stop at point 0 m
Displacement of point P = (amplitude B + amplitude C) + (amplitude A + amplitude D)
Displacement of point P= (2cm - 2cm) + (2 cm - 4cm)= -2cm
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