4 years ago

Match the cofactors below with their roles in the pyruvate dehydrogenase complex reaction.

1 answer:

8 0

Cofactors:
A. Coenzyme A (CoA-SH)
B. NAD+
C. Thiamine pyrophosphate (TPP)
D. FAD
E. Lipoic acid in oxidized form

Roles:
E... Attacks and attaches to the central carbon in pyruvate.
A...Oxidizes FADH2.
C...Accepts the acetyl group from reduced lipoic acid.
D... Oxidizes the reduced form of lipoic acid.
B... Initial electron acceptor in oxidation of pyruvate.

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Inside the room the air is often warmer near the ceiling than near the floor. Which of the following accounts for the difference

Rudiy27

The correct answer would be C. Convection, which describes that heat rises.

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3 years ago

Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of SO2 by O2 to give SO3

SIZIF [17.4K]

Answer:

See explanation

Explanation:

We know that the process of the oxidation of SO2 to SO3 is catalysed by NO2 gas. It occurs in two stages and i will show the balanced reaction equation of the both stages below;

Step 1

2NO2(g) + 2SO2(g) ------> 2NO(g) + 2SO3(g)

Step 2

2NO(g) + O2(g) -------> 2NO2(g)

So, the overall reaction equation is;

2SO2(g) + O2(g) ------> 2SO3(g)

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3 years ago

A resting adult requires about 240 ml of pure oxygen/mind and breathes about 12 times every minute. if inhaled air contains 20 p

Dmitry_Shevchenko [17]

12 times breathe give 240 ml of pure $O_{2}$ . Each breathe gives 20 ml of $O_{2}$ .

Let us consider, volume of air per breathe= x ml.

Pure $O_{2}$ from inhaled air= $\frac{20}{100}x$ ml and Pure $O_{2}$ from exhaled air= $\frac{16}{100}x$ ml.

Pure $O_{2}$ from inhaled and exhaled air= 20 ml

So, $\frac{20}{100}x$ + $\frac{16}{100}x$ = 20

Therefore, x = 55.5 ml

So, volume of air per breath= 55.5 ml.

7 0

3 years ago

An oil tanker spills a large amount of oil near the ocean shoreline. Which application of chemistry would best solve this proble

DochEvi [55]

I have heard they can use hair is stocking or nets to absorb the oil out of the water

3 0

3 years ago

What mass of copper is deposited when a current of 10.0a is passed through a solution of copper(ii) nitrate for 30.6 seconds?

asambeis [7]

Data Given:

Time = t = 30.6 s

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 63.54/2 = 31.77 g

Amount Deposited = W = ?

Solution:
According to Faraday's Law,

W = I t e / F

Putting Values,

W = (10 A × 30.6 s × 31.77 g) ÷ 96500

W = 0.100 g

Result:
0.100 g of Cu²⁺ is deposited.

3 0

4 years ago