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Nataly_w [17]
1 year ago
14

Imagine that a researcher has developed a way to isolate thylakoids from chloroplasts and to make the ph of the thylakoid space

acidic. If these isolated thylakoids are transferred to a ph8 solution in the dark, what might the researcher expect to observe?.
Chemistry
1 answer:
PilotLPTM [1.2K]1 year ago
8 0

In the given condition where the thylakoids are isolated from the chloroplasts to make the pH of the thylakoid space acidic, the isolated thylakoid will form the ATP.

ATP is also referred to as Adenosine Triphosphate and is correctly called as the energy currency of the cell as it is responsible molecule for carrying the energy units in the cell. When the isolated thylakoids are made to transfer in the pH = 8 basic solution, then the acidic lumen of the chloroplast will be having a higher concentration of H⁺ ions. When they are placed in the basic solution then the H⁺ ions will move out because of the occurrence of concentration gradient which is higher in the lumen than the surrounding. The chloroplasts will now bear synthesis of ATP. Therefore, the isolated thylakoid is able to make ATP when the lumen is acidic, and the surroundings are made basic in nature.

Thus, the thylakoid which is isolated will bear the synthesis of ATP.

Learn more about ATP at:

brainly.com/question/252380

#SPJ9

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8 0
3 years ago
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Be sure to answer all parts. what are the concentrations of hso4−, so42−, and h in a 0. 31 m khso4 solution? (hint: h2so4 is a s
disa [49]

At equilibrium the concentrations of:

[HSO₄⁻] = 0.10 M;

[SO₄²⁻] = 0.037 M;

[H⁺] = 0.037 M;

There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.

HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid.  HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.

R   HSO_4^-    ⇄  H^+ + SO_4^2^-

I    0.14

C   - x               +x       +x

E   0.14-x        x         x

K_a = 1.3 × 10^-^2 for HSO^-_4 . As a result,

\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a

K_a is large. It is no longer valid to approximate that [HSO^-_4] at equilibrium is the same as its initial value.

\frac{x^2}{0.14-y} = 1.3 * 10^-^2

x^2+1.3*10^-^2x - 0.14 × 1.3 × 10^-^2= 0

Solving the quadratic equation for x , x \geq 0 since x represents a concentration;

                             x=0.0366538

Then, round the results to 2 significant figure;

  • [SO_4^2^-] = x = 0.037 mol. L ^-^1
  • [H^+] = x = 0.037 mol. L ^-^1
  • [HSO_4^-] = 0.14 - x = 0.10 mol. L ^-^1

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brainly.com/question/14469428

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Hitman42 [59]

Answer:

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The electrophile now is the isopropyl group which attacks the benzene to yield the product.

4 0
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Maksim231197 [3]

Answer:

Explanation:

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When it moves over hot location then bottom layer gets hot and lighter. Then it moves towards poles.

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The sodium-potassium pump does not run out of ions since ion exchange is essential for the action potential to take place and to maintain homeostasis.

The cell has variable concentrations of different substances compared to the environment that surrounds it, with significant differences with sodium and potassium.

  • The main function of the sodium-potassium pump is to maintain homeostasis of the intracellular medium, controlling the concentrations of these two ions.

  • In order to carry out the adequate exchange of sodium and potassium ions in the extra and intracellular medium, the cells need an active transport process that is carried out thanks to the sodium potassium pump.

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