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aev [14]
1 year ago
6

In the reaction catalyzed by the pyruvate dehydrogenase complex, the two carbons constituting the acetyl group are

Chemistry
1 answer:
lara [203]1 year ago
4 0

Transferred to the lipoamide by an earlier intermediate in the process.

The pyruvate dehydrogenase complex (PDC) is a mitochondrial multienzyme complex composed of three different enzymes

<h3>What reaction is catalyzed by enzyme 2 of the pyruvate dehydrogenase complex ?</h3>

the pyruvate dehydrogenase complex is the bridge between glycolysis and the citric acid cycle

  • Five coenzymes are used in the pyruvate dehydrogenase complex reactions: thiamine pyrophosphate or TPP, flavin adenine dinucleotide or FAD, coenzyme A or CoA, nicotinamide adenine dinucleotide or NAD, and lipoic acid.
  • during the reactions catalyzed by the pyruvate dehydrogenase complex, it is first reduced to dihydrolipoamide, a dithiol or the reduced form of the prosthetic group, and then, reoxidized to the cyclic form.

Learn more about Pyruvate dehydrogenase complex here:

brainly.com/question/16346028

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Hclo is a weak acid (ka = 4.0 × 10–8) and so the salt naclo acts as a weak base. what is the ph of a solution that is 0.089 m in
Anuta_ua [19.1K]

Answer: 14

Explanation:

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How does the composition of a mixture of hydrogen and oxygen differ from the composition of a compound containing hydrogen and o
IrinaK [193]
Chemical compounds have chemically bonded molecules so that they exhibit different properties (i.e. chemical) compared to the individual molecules comprising the compound. Mixtures are simply the combinations of different molecules and compounds that are not chemically bonded together, and can therefore be separated by physical means. Mixtures usually retain the properties of its components. 

The hydrogen and oxygen molecules in a mixture do not form strong bonds between each other. The molecules of both gases are only contained in the same space or volume and the individual molecules retain their chemical properties. 

A compound containing hydrogen and oxygen molecules exhibit different chemical (and even physical) properties compared to the individual molecules themselves.

Water for example, is a compound with 2 hydrogen atoms and 1 oxygen atom, chemically-bonded together. Hydrogen gas is highly flammable, water is not. Oxygen gas is an essential reagent for combustion (or burning) reactions, water is not. 
 
Thus, throwing a lighted match to a gaseous mixture of hydrogen (H_{2}) and oxygen (O_{2}) would create fire, or even an explosion (since hydrogen is flammable and oxygen feeds the reaction). Throwing a match to water vapor (H_{2}O) would not create fire. 
5 0
3 years ago
What does it mean if an elemet has mo subscript in a chemical formula​
kogti [31]
If no subscript appears, one atom of that element is present.
3 0
3 years ago
Read 2 more answers
Determine the rate law and the value of k for the following reaction using the data provided.
blsea [12.9K]

<u>Answer:</u> The rate law expression is \text{Rate}=k[NO]^2[O_2]^1 and value of 'k' is 1.727\times 10^3M^{-2}s^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

Rate law expression for the reaction:

\text{Rate}=k[NO]^a[O_2]^b

where,

a = order with respect to nitrogen monoxide

b = order with respect to oxygen

  • <u>Expression for rate law for first observation:</u>

8.55\times 10^{-3}=k(0.030)^a(0.0055)^b       ....(1)

  • <u>Expression for rate law for second observation:</u>

1.71\times 10^{-2}=k(0.030)^a(0.0110)^b       ....(2)

  • <u>Expression for rate law for third observation:</u>

3.42\times 10^{-2}=k(0.060)^a(0.0055)^b      ....(3)

Dividing 1 from 2, we get:

\frac{1.71\times 10^{-2}}{8.55\times 10^{-3}}=\frac{(0.030)^a(0.0110)^b}{(0.030)^a(0.0055)^b}\\\\2=2^b\\b=1

Dividing 1 from 3, we get:

\frac{3.42\times 10^{-2}}{8.55\times 10^{-3}}=\frac{(0.060)^a(0.0055)^b}{(0.030)^a(0.0055)^b}\\\\2^2=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[NO]^2[O_2]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

8.55\times 10^{-3}=k[0.030]^2[0.0055]^1\\\\k=1.727\times 10^3M^{-2}s^{-1}

Hence, the rate law expression is \text{Rate}=k[NO]^2[O_2]^1 and value of 'k' is 1.727\times 10^3M^{-2}s^{-1}

4 0
3 years ago
0 3
Alik [6]
To find heat, you need the following formula

heat= specific heat x mass x ΔT

specific heat (for water)= 4.184--> according to the table
mass= 100 grams
ΔT= 75 -10= 65 °C

let's plug the values.

specific heat= 4.184 x 100 x 65= 27200 joules or 27.2 Kilojoules
6 0
3 years ago
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