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Dovator [93]
3 years ago
5

calculate the standard potential arising from the reaction in which NADH is oxidized to NAD and the corresponding biological sta

ndard reaction Gibbs energy.
Chemistry
1 answer:
melisa1 [442]3 years ago
7 0

Answer:

-219.99kJ

Explanation:

The acronym '' NADH'' simply stands for what is known as coenzyme 1 with full meaning of Nicotinamide Adenine Dinucleotide Hydride. This substance is useful in the production of energy. The oxidation reaction of NADH causes it to produce NADP⁺ and the oxygen produces water when it is in the reduction process. The balanced equation for the oxidation reaction is given below as:

NADPH ---------------------------------------------------------------------> NADP⁺H⁺ + 2e⁻.

Also, the balanced equation for the reduction reaction is given below as:

\frac{1}{2} O₂ + 2H⁺ + 2e⁻ --------------------------------------------------------------> H₂O.

It can be shown from the above REDOX reaction that the total number of electrons getting transferred is 2.

The  Gibbs energy = -nFE. where n = 2, F = faraday's constant = 96485.3329 C and E = overall cell potential.

The overall cell potential = E[ reduction reaction] - E[oxidation reaction] = 0.82 - (- 0.32 ) = 1.14 V.

Hence, the Gibbs energy = - 2 × 96485.3329 × 1.14 = -219.99kJ

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Answer:

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Explanation:

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3 years ago
Lab report enthalpy
Mice21 [21]

Answer:

The purpose of this experiment was to verify Hess’s law associated with three reactions. Our experiment confirmed it by combining the enthalpy changes of two reactions to equal of that another single reaction (Reaction 2 in our case). Styrofoam cup was used as a calorimeter to measure the change in temperature, which is a necessary variable for further calculations. We discovered that sum of enthalpy changes for reactions one and three equaled the enthalpy change in reaction two. This experiment was possible because if one of the reactions is the same as the combination of other two, the heat of the reaction of the single reaction should be equal to the heat of other two reaction combined.

Explanation: when you see the reactions that the Hess's law to find the enthalpy, which we did from this lab.

8 0
3 years ago
Read 2 more answers
A water sample contains 50Mg of ferric sulfate. determine the normality of the solution
Lesechka [4]

Answer:

Assuming Volume of dolution = 1 liter . The Normality is:

N=3.75\times 10^{-4}N

Explanation:

<u>Normality : </u>It is the gram equivalent of solute per liter of solution. It is represented by N.

Normality=\frac{Number\ of\ equivalent\ weight}{Volume(L)}

Number of  equivalents weight =

\frac{Molar\ mass}{n}

n= acidity of base or basicity of acid

for salts , n = charge present on cation

The relation between Normality and Molarity is :

Normality =Molarity\times n

n = charge present on cation (<u>not moles)</u>

Fe_{2}(SO_{4})_{3} = ferric sulphate

Here <u><em>Fe is cation and its oxidation state is = + 3 so n= 3</em></u>

1. First , calculate the molarity(M)

Molar Mass of  ferric sulfate = Fe2(SO4)3

2(mass of Fe)+3 (mass of S) + 12(mass of O)

= 2(56)+3(32)+12(16)

= 400 grams

Molarity =\frac{Moles}{Volume}

Moles=\frac{mass}{Molar\ mass}

Molar mass = 400 gram

mass = 50 mg = 0.05 grams

moles=\frac{0.05}{400}

moles=1.25\times 10^{-4}

Molarity =\frac{1.25\times 10^{-4}}{Volume}

let volume = 1 liter

Molarity =\frac{1.25\times 10^{-4}}{1}

molarity=1.25\times 10^{-4}M

<u><em>2. Multiply the molarity with n </em></u>

Normality =Molarity\times n

Normality =1.25\times 10^{-4}M\times n

n = 3

Normality =1.25\times 10^{-4}M\times 3

Normality =3.75\times 10^{-4}N

7 0
4 years ago
Given the net ionic equation, Ba2+ + SO42- =&gt; BaSO4, how many grams of barium chloride must be present to react with 200 gram
zheka24 [161]

Answer: 208g/mol

Explanation:

First of all we have to write the balance equation for the reaction.

BaCl2 + Fe2(SO4)3____> BaSO4 + FeCl3

After balancing we have.

3BaCl2 +Fe2(SO4)3_____> 3BaSO4 +2FeCl3

Looking at the equation, we find out that 3 moles of barium chloride reacts with 1 mole of iron iii sulfate

Therefore we have

3moles of BaCl2 _____> 400g/mole of iron iii sulfate

Xmole of BaCl2 _____> 200g/mole of iron iii sulfate

X = 2 * 200g/mol divide by 400g/mol

X = 1mole

1 mole of BaCl2 will be need to react with 200g/mol of iron iii sulfate.

This 1 mole of BaCl2 is equivalent to 208g/mol of BaCl2.

Therefore the gram of barium chloride that must be present is = 208g/mol//

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Marta_Voda [28]
It is C that is the most testable hypotheses
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