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Sedaia [141]
3 years ago
13

Succinate dehydrogenase catalyzes the conversion of succinate to fumarate. The reaction is inhibited by malonic acid, which rese

mbles succinate but cannot be acted upon by succinate dehydrogenase. Increasing the ratio of succinate to malonic acid reduces the inhibitory effect of malonic acid. What is malonic acid's role with respect to succinate dehydrogenase? Malonic acid
Chemistry
1 answer:
allsm [11]3 years ago
3 0

Answer:

Here malonic acid acts as a competitive inhibitor of succinate dehydrogenase

Explanation:

Malonic acid structurally resembles succinic acid as a result the enzyme succinate dehydrogenase cannot distinguish between malonic acid and succinic acid.

      That"s why malonic acid interact with succinate dehydrogenase thereby blocking the catalytic activity of the later.

      As this mechanism is a type of competitive inhibition that"s why increasing the concentration of substrate succinic acid can reduce the inhibitory effect of malonic acid.

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LenKa [72]
In a non-flowering plant, the embryo is in spores found in the stem, and in a flowering plant, the embryo is in seeds found in the flower.
(don’t count on my answer but I think it might be this one and I apologize if you get it wrong)
6 0
3 years ago
A 55.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a
balu736 [363]

Answer:

0.14 M

Explanation:

To determinate the concentration of a new solution, we can use the equation below:

C1xV1 = C2xV2

Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL

1.50x55.0 = C2x278

C2 = 0.30 M

The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL

Then,

0.30x139 = C2x294

C2 = 0.14 M

4 0
3 years ago
Read 2 more answers
"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical</em>

<em />

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

  • <em>pKa NH₃/NH₄⁺</em>

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = <em>4.74</em>

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

  • <em>Moles NH₃</em>

<em>2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃</em>

  • <em>H-H equation:</em>

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

<em />

3 0
4 years ago
Which two formulas represent compounds
SVEN [57.7K]

CO2 ; H20- They are the only ones that, on both sides, combined with another element and bonding of atoms

4 0
3 years ago
The iron content of foods can be determined by dissolving them in acid (forming Fe3+), reducing the iron(III) to iron(II), and t
____ [38]

Answer:

  • <em>Oxidation half-reaction</em>:

Fe²⁺(aq) →  Fe³⁺(aq) + 1e⁻

  • <em>Reduction half-reaction</em>:

Ce⁴⁺(aq) + 1e⁻ →  Ce³⁺(aq)

Explanation:

The reaction that takes place is:

  • Fe²⁺(aq) + Ce⁴⁺(aq) → Fe³⁺(aq) + Ce³⁺(aq)

The <em>oxidation half-reaction</em> is:

  • Fe²⁺(aq) →  Fe³⁺(aq) + 1e⁻

It is an oxidation because the oxidation state of Fe increases from 2+ to 3+.

The <em>reduction half-reaction</em> is:

  • Ce⁴⁺(aq) + 1e⁻ →  Ce³⁺(aq)

It is a reduction because the oxidation state of Ce decreases from 4+ to 3+.

4 0
3 years ago
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