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Stella [2.4K]
3 years ago
15

Consider the conversion of succinate to fumarate in the Citric Acid Cycle (reaction below). This reaction is endergonic under st

andard conditions (ΔGo’≈ 6 kJ/mol). How might this reaction be made favorable under equilibrium conditions? Your answer should include the relationship of this reaction to a canonical electron transport chain (i.e. an electron transport chain that uses oxygen as a terminal electron acceptor).
Chemistry
1 answer:
balu736 [363]3 years ago
5 0

Answer:

Succinate oxidation to fumarate The following reactions transform succinate to regenerate oxalacetate. The first of these reactions is carried out by an oxidation catalyzed by succinate dehydrogenase. The hydrogen acceptor is FAD, since the free energy change is insufficient to allow NAD to interact. The final product is fumarate.

Explanation:

The condensation reaction of GDP + Pi and the hydrolysis of Succinyl-CoA involve the H2O necessary to balance the equation.

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The valences of metal x,y and z are 1,2 and 3 respectively. What are the formulae of their;a) hydroxides, b) sulphates, c) hydro
Rina8888 [55]

Answer:

See answer below

Explanation:

AS we know that the valence for those metals X, Y, and Z are 1, 2 and 3, we can determine the formula of each compound.

1. Hydroxides.

An hydroxide is formed when an oxyde of a metal reacts with water. When this happens, the general molecular formula is:

Meₐ(OH)ₙ

Where:

a: valence or charge of the hydroxide (Which is -1)

n: valence of the metal.

Following this, the formula for X, Y and Z would be:

XOH

Y(OH)₂

Z(OH)₃

2. Sulphates

Sulphates follow a similar rule of hydroxide in the general molecular formula, but instead of having a charge of -1, it has a charge of -2 so:

Mₐ(SO₄)ₙ

So, following the rule:

X₂SO₄

Y₂(SO₄)₂ ------> YSO₄

Z₂(SO₄)₃

3. Hydrogens

Following the same rule as the previous, hydrogens works with a charge of -1, so:

MₐHₙ

Then:

XH

YH₂

ZH₃

4. Carbonates.

This follows the same rule as sulphates, with the same charge so:

Mₐ(CO₃)ₙ

Then:

X₂CO₃

YCO₃

Z₂(CO₃)₃

5. Nitrates

Follow the same rule as the hydroxides, with the same charge of -1.

Mₐ(NO₃)ₙ

Then:

XNO₃

Y(NO₃)₂

Z(NO₃)₂

6. Phosphates

In the case of phosphates, these have a charge of -3 so:

Mₐ(PO₄)ₙ

Then:

X₃PO₄

Y₃(PO₄)₂

Z₃(PO₄)₃ ----> ZPO₄

Hope this helps

6 0
2 years ago
Be sure to answer all parts. identify and label the species in each reaction. (a) nh4+(aq) + h2o(l) ⇌ nh3(aq) + h3o+(aq) acid ba
quester [9]
A)

NH⁴⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃0⁺<span>(aq)

- acid </span>a species that able to donate (H+): NH⁴⁺
- base a species that is able to accept a proton (H+): H₂O
- conjugate base a species formed when acid donates a proton (H+): NH₃
- conjugate acid a species formed by a base accepts a proton (H+): H₃0⁺

b)

CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq)

- base a species that is able to accept a proton (H+): CN⁻
- acid a species that able to donate (H+): H₂O
- conjugate acid a species formed by a base accepts a proton (H+): HCN
- conjugate base a species formed when acid donates a proton (H+): OH⁻

5 0
3 years ago
nadira is using her laptop for about 6 hours straight,yet her laptops barley hot or warm. Does her laptop have high or low speci
makkiz [27]

Her computer will start to heat up, and the temp. would be 60 degrees celcius.

I hope this helps you ᕕ( ᐛ )ᕗ

8 0
3 years ago
Which sub-atomic particle is positively charged and found in the nucleus?
Anarel [89]
The proton has a positive charge.
4 0
3 years ago
Nh3 is a weak base (kb = 1.8 × 10–5 m) and so the salt nh4cl acts as a weak acid. what is the ph of a solution that is 0.013 m i
ankoles [38]
First, we need to get the value of Ka:

when Ka = Kw / Kb 

we have Kb = 1.8 x 10^-5 

and Kw = 3.99 x 10^-16  so, by substitution:

Ka = (3.99 x 10^-16) /  (1.8 x 10^-5) = 2.2 x 10^-11

by using the ICE table :

                 NH4+  + H2O →NH3 +  H+
intial          0.013                       0         0 

change       -X                          +X      +X

Equ        (0.013-X)                      X         X

when Ka = [NH3][H+] / [NH4+] 

by substitution:

2.2 x 10^-11 = X^2 / (0.013 - X)  by solving this equation for X

∴X = 5.35 x 10^-7

∴[H+] = X = 5.35 x 10^-7

∴PH = - ㏒[H+]
 
        = -㏒(5.35 x 10^-7)
        = 6.27
6 0
2 years ago
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