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Airida [17]
3 years ago
13

Onsider the following elementary reaction:

Chemistry
1 answer:
Brilliant_brown [7]3 years ago
3 0

Answer:

C_{O_2}=\frac{k_2C_{NO_2}C_{O}}{k_1C_{NO}}

Explanation:

Hello,

In this case, for the given elementary reaction, the rate law takes the form:

r=-k_1C_{NO}C_{O_2}+k_2C_{NO_2}C_{O}

Nevertheless, at equilibrium, the rate becomes zero:

0=-k_1C_{NO}C_{O_2}+k_2C_{NO_2}C_{O}

Thus, we can solve for the concentration of O₂ in terms of the rate constants as shown below:

k_1C_{NO}C_{O_2}=k_2C_{NO_2}C_{O}\\\\C_{O_2}=\frac{k_2C_{NO_2}C_{O}}{k_1C_{NO}}

Best regards.

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What reproductive process do bacteria use by forming two identical cells from one parent cell?
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This is a kind of an asexual reproduction called binary fission. This involves division of the parent cell into two identical cells. This is common for prokaryotic reproduction like archaea and bacteria and it also happens in some single-celled eukaryotes such as protists and unicellular fungi.
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What is it called when energy is lost or removed?
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I think it would be ionization
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Predict the ground-state electron configuration of the following ions. Write the answer in abbreviated form beginning with noble
Tanzania [10]

Answer: The ground-state electronic configuration of

1) Ru^{2+}=[Kr]4d^6

2) W^{3+]=4f^{14}5d^3

Explanation: The electronic configuration of elements is defined as the distribution of electrons around the nucleus of that element. It depends on the atomic number of the element.

Atomic number of the element = Number of electrons.

  • For Ru-element

Atomic number = 44

Number of electrons = 44

Electronic configuration of Ru-element = [Kr]4d^75s^1

To form Ru^{2+}, 2 electrons are released from the neutral Ru-element.

So, the electronic configuration of Ru^{2+}=[Kr]4d^6

  • For W-element

Atomic number = 74

Number of electrons = 74

Electronic configuration of W-element = [Xe]4f^{14}5d^46s^2

To form W^{3+}, 3 electrons are released from the neutral W-element.

So, the electronic configuration of W^{3+}=[Xe]4f^{14}5d^3

5 0
4 years ago
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6) What types of atomic orbitals are in the third principal energy
Bumek [7]

Answer:

Types of atomic orbitals present in the third principal energy are <u>s, p and d only .</u>

Explanation:

  • <u>OPTION A-:  s and p atomic orbitals -</u> these two orbitals are present in second principal energy level. Therefore , the option is incorrect.
  • <u> OPTION B-: p and d only -</u> This option is wrong as there is no such principal level energy where , s atomic orbital is absent .

  • <u>OPTION C-: s , p and d only -</u>these orbitals are present in<u> third principal energy level</u>. The third major level of energy has one orbital, three orbitals of p, and five orbitals of d, each of which can contain up to 10 electrons. The third stage thus holds a maximum of 18 electrons. This option is correct .
  • <u>OPTION D-: s , p, d and f only -</u>There is also a f sublevel at the <u>fourth and higher stages,</u> containing seven f orbitals, which can accommodate up to 14 electrons at most. Therefore, up to 32 electrons will hold the fourth level: 2 in the s orbital, 6 in the three p orbitals, 10 in the five d orbitals, and 14 in the seven f orbitals. This option is incorrect .

<u>Thus , the correct option is C (s , p and d only .)</u>

6 0
4 years ago
Consider the malate dehydrogenase reaction from the citric acid cycle. Given the listed concentrations, calculate the free energ
Ahat [919]

<u>Answer:</u> The Gibbs free energy of the reaction is 21.32 kJ/mol

<u>Explanation:</u>

The chemical equation follows:

\text{Malate }+NAD^+\rightleftharpoons \text{Oxaloacetate }+NADH

The equation used to Gibbs free energy of the reaction follows:

\Delta G=\Delta G^o+RT\ln K_{eq}

where,

\Delta G = free energy of the reaction

\Delta G^o = standard Gibbs free energy = 29.7 kJ/mol = 29700 J/mol  (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314J/K mol

T = Temperature = 37^oC=[273+37]K=310K

K_{eq} = Ratio of concentration of products and reactants = \frac{\text{[Oxaloacetate]}[NADH]}{\text{[Malate]}[NAD^+]}

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[NAD^+]=490mM

Putting values in above expression, we get:

\Delta G=29700J/mol+(8.314J/K.mol\times 310K\times \ln (\frac{0.130\times 2.0\times 10^2}{1.37\times 490}))\\\\\Delta G=21320.7J/mol=21.32kJ/mol

Hence, the Gibbs free energy of the reaction is 21.32 kJ/mol

4 0
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