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

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Answer 1

Answer:

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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The standard entropy of Pb(s) at 298.15 K is 64.80 J K–1 mol–1. Assume that the heat capacity of Pb(s) is given by: CP,m(Pb, s) J K−1mol−1 = 22.13 + 0.01172 T K + 1.00 x 10−5 T 2 K2 The melting point is 327.4 ℃ and the heat of fusion is 4770 J mol-1. Assume that the heat capacity of Pb(l) is given by: CP,m(Pb, l) J K−1mol−1 = 32.51 − 0.00301 T K Calculate the standard entropy of Pb(l) at 500 ℃

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Answer:

$s_(Pb(l),500\°C)=100.83(J)/(mol*K)$

Explanation:

Hello,

In this case, for the calculation of the standard entropy of liquid lead at 500 °C (773.15 K), starting by solid lead 298.15 K we need to consider three processes:

1. Heating of solid lead at 298.15 K to 600.55 K (melting point).

2. Melting of solid lead to liquid lead.

3. Heating of liquid lead at 600.55 K (melting point) to 773.15 K.

Which can be written in terms of entropy by:

$s_(Pb(l),500\°C)=s_(Pb(s),298.15K)+s_1+s_2+s_3$

Whereas each entropy is computed as follows:

$s_1=\int\limits^(600.55K)_(298.15K) {(22.13 + 0.01172 T + 1.00 x 10^(-5) T^2)/(T) } \, dT =20.4(J)/(mol*K)\n\n\ns_2=(4770(J)/(mol) )/(600.55K)= 7.94(J)/(mol*K)\n\n\ns_3=\int\limits^(773.15K)_(600.55K) {(32.51-0.00301T)/(T) } \, dT=7.69(J)/(mol*K)$

Therefore, the standard entropy of liquid lead at 500 °C turns out:

$s_(Pb(l),500\°C)=64.80+20.4+7.94+7.69\n\ns_(Pb(l),500\°C)=100.83(J)/(mol*K)$

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Answer:

164 g

Explanation:

Identify which of the following equations are balanced.A. 6Li(s)+N2(g)→2Li3N(s)6Li(s)+N2(g)→2Li3N(s).B. 2KMnO4(s)→K2MnO4(s)+MnO2(s)+O2(g). C. 2KClO3(s)→2KCl(s)+O2(g)2KClO3(s)→2KCl(s)+O2(g). D. Mg(s)+N2(g)→Mg3N2(s)Mg(s)+N2(g)→Mg3N2(s).

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Answer:

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What is the polarity of a Na-Br bond?nonpolar covalent

definitely polar covalent

likely ionic

slightly polar covalent

slightly ionic

Answers

Answer:

likely ionic

Explanation:

The bond between , Sodium and Bromine , in sodium bromide salt , is ionic in nature .

Since ,

The difference in the electronegativity of the two atom is high .

According to Pauling scale ,

The Electronegativity of Bromine = 2.96

The Electronegativity of sodium = 0.93

Hence ,

The Difference in electronegativity = ( 2.96 - 0.93 ) = 2.03

Hence ,

The electronegativity difference value is more than 1.8 ,

Therefore,

The bond is ionic in nature .

Which is not a product of the fractional distillation of petroleum?A) Gasoline
B) Jet Fuel
C) steel

Answers

Answer:
Option-C, STEEL is not a product of the fractional distillation of petroleum.

Explanation:
Petroleum is the mixture of Hydrocarbons *carbon and hydrogen containing compounds) present beneath the Earth's surface. Petroleum is formed from the remains of animals and plants beneath earth's surface in an anaerobic conditions.
Petroleum contains from small hydrocarbons (gases) to medium (liquids) and long chain hydrocarbons (Solids). These hydrocarbons are separated from each other by Fractional Distillation method (separation due to difference in boiling points)
Gasoline is a derivative of one of the fraction of petroleum used in internal combustion engines.
Jet Fuel is also derived from Kerosene and Naphtha fractions of petroleum.
While, Steel is inorganic Alloy (mixture of metals) composed of mainly Iron, Carbon and other elements.

The molecule XeF2 does not obey the octet rule. Draw its Lewis structure and state the type of octet-rule exception. Include lone pair electrons in the Lewis structure.

Answers

Answer: See attachment.

Explanation:

Exceptions to the octet rule fall into three categories:

An incomplete octet.
An odd number of electrons.
More than eight valence electrons around the central atom.

In addition to the 3s and 3p orbitals, xenon also has 3d orbitals that can be used in bonding. These orbitals enable xenon to form an expanded octet.

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