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Arte-miy333 [17]
3 years ago
6

The standard free energy change for a reaction can be calculated using the equation ΔG∘′=−nFΔE∘′ ΔG∘′=−nFΔE∘′ where nn is the nu

mber of electrons transferred, FF is Faraday's constant, 96.5 kJ·mol−1·V−1, and ΔE∘′ΔE∘′ is the difference in reduction potential. For each of the given reactions, determine the number of electrons transferred (n)(n) and calculate standard free energy (ΔG∘′)(ΔG∘′) . Consider the half-reactions and overall reaction for reaction 1. half-reactions:fumarate 2−+2H+CoQH2↽−−⇀succinate−↽−−⇀CoQ+2H+ half-reactions:fumarate−+2H+↽−−⇀succinate2−CoQH2↽−−⇀CoQ+2H+ overall reaction:fumarate2−+CoQH2↽−−⇀succinate2−+CoQΔE∘′=−0.009 V
Chemistry
1 answer:
Lelechka [254]3 years ago
6 0

Answer:

ΔG°′ = 1.737 KJ/mol

Explanation:

The reaction involves the transfer of two electrons in the form of hydride ions from reduced coenzyme Q, CoQH₂ to fumarae to form succinate and oxidized coenzyme Q, CoQ.

The overall equation of reaction is as follows:

fumarate²⁻ + CoQH₂ ↽⇀ succinate²⁻ + CoQ ;    ΔE∘′=−0.009 V

Using the equation  for standard free energy change; ΔG°′ = −nFΔE°′

where n = 2; F = 96.5 KJ.V⁻¹.mol⁻¹; ΔE°′ = 0.009 V

ΔG°′ = - 2 * 96.5 KJ.V⁻¹.mol⁻¹ * 0.009 V

ΔG°′ = 1.737 KJ/mol

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What is the mass of a magnesium block that measures 2.00 cm
skad [1K]

Answer:

m = 31.284 grams

Explanation:

Given that,

The dimension of a magnesium block is 2.00 cm  x 3.00 cm x 3.00 cm.

The density of magnesium is, d = 1.738 g/cm³

We need to find the mass of the magnesium block. We know that the density of an object is given by its mass per unit its volume. So,

d=\dfrac{m}{V}\\\\\text{Where m=mass}\\\\m=d\times V\\\\m=1.738\ g/cm^3\times (2\times 3\times 3)\ cm^3\\\\m=31.284\ \text{grams}

So, the mass of the block is 31.284 grams.

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3 years ago
In the chemical reaction below, what is the product?<br> C + O2 + CO2
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You’ll need to be sure to count all the atoms in each side of the chemical equation.
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Which of the following was not a big ideal of chemistry
ikadub [295]

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3 0
2 years ago
How many mL of a 4% mass/volume Mg(NO3)2 solution would contain 1.2 grams of magnesium nitrate?
Phoenix [80]
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7 0
3 years ago
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid. HCl(aq), as described by
melisa1 [442]

Answer:

0.88 g

Explanation:

Using ideal gas equation to calculate the moles of chlorine gas produced as:-

PV=nRT

where,

P = pressure of the gas = 805 Torr

V = Volume of the gas = 235 mL = 0.235 L

T = Temperature of the gas = 25^oC=[25+273]K=298K

R = Gas constant = 62.3637\text{torr}mol^{-1}K^{-1}

n = number of moles of chlorine gas = ?

Putting values in above equation, we get:

805torr\times 0.235L=n\times 62.3637\text{ torrHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{805\times 0.235}{62.3637\times 298}=0.01017\ mol

According to the reaction:-

MnO_2+4HCl\rightarrow MnCl_2+2H_2O+Cl_2

1 mole of chlorine gas is produced when 1 mole of manganese dioxide undergoes reaction.

So,

0.01017 mole of chlorine gas is produced when 0.01017 mole of manganese dioxide undergoes reaction.

Moles of MnO_2 = 0.01017 moles

Molar mass of MnO_2 = 86.93685 g/mol

So,

Mass=Moles\times Molar\ mass

Applying values, we get that:-

Mass=0.01017moles \times 86.93685\ g/mol=0.88\ g

<u>0.88 g of MnO_2(s) should be added to excess HCl (aq) to obtain 235 mL of Cl_2(g) at 25 degrees C and 805 Torr.</u>

6 0
3 years ago
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