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dimulka [17.4K]
3 years ago
6

The molecular mass of propane-1,2-diol is 76.1 amuamu . Calculate the molecular mass of propane-1,3-diol, an isomer of propane-1

,2-diol
Chemistry
2 answers:
sashaice [31]3 years ago
8 0

Answer:

76.1 amu

Explanation:

Let us recall that isomers refer to two different compounds with the same molecular formula but different atom to atom connectivity and different chemical properties. When two compounds are isomers, we can essentially represent them with exactly the same molecular formula.

Now propane-1,2-diol and propane-1,3-diol are both represented by the molecular formula C3H8O2 since they are isomers of each other. When two compounds have the same molecular formula, they must essentially have the same molecular mass. Hence the molecular mass of propane-1,3-diol is also 76.1 amu.

galben [10]3 years ago
4 0

Answer:

molecular mass of propane-1,3-diol, an isomer of propane-1,2-diol = 76.1 amu

Explanation:

An isomer of any compound will also have the same molecular mass of the comlund

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barxatty [35]

Answer:

The reaction will be  non spontaneous at these concentrations.

Explanation:

AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)

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K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}

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\Delta G^o=-2.303\times R\times T\times \log K_c

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\Delta G^o=69,117.84 J/mol=69.117 kJ/mol

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Reaction quotient of an equilibrium = Q

Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}

\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)

\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])

\Delta G=40.588 kJ/mol

  • For reaction to spontaneous reaction:  \Delta G.
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Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

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The given sentence is part of a longer question.

I found this question with the same sentence. So, I will help you using this question:

For the reaction N2O4<span>(g) ⇄ 2NO</span>2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm<span>).  If </span>Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?


(a) Q = K<span>;   The reaction </span>is at equilibrium.
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Q is the reaction quotient. It is a measure of the progress of a chemical reaction.

The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.

At equilibrium both Kp and Q are equal. Q = Kp

If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,

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Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.

Since, Q = 1 and Kp = 0.15, Q > Kp and the reaction will proceed to the left.
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