The element of bromine has entropy of vaporization as -93.1 J/(mol∙K) which is calculated as S=-ΔH/T.
What is an element?It is defined as a substance which cannot be broken down further into any other substance. Each element is made up of its own type of atom. Due to this reason all elements are different from one another.
Elements can be classified as metals and non-metals. Metals are shiny and conduct electricity and are all solids at room temperature except mercury. Non-metals do not conduct electricity and are mostly gases at room temperature except carbon and sulfur.
The entropy of vaporization for bromine is calculated by the formula S=-ΔH/T,substitution in given formula gives,
S=-30.91/332=-93.1 J/(mol∙K)
Thus, the element of bromine has entropy of vaporization as -93.1 J/(mol∙K) .
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Final answer:
The entropy of vaporization for bromine is 93.1 J/(mol·K), which means the correct answer is D. 93.1 J/(mol·K).
Explanation:
To find the entropy of vaporization (ΔSvap) for bromine, we can use the formula ΔSvap = ΔHvap / Tb, where ΔHvap is the heat of vaporization and Tb is the boiling point in Kelvin. Given that the heat of vaporization (ΔHvap) for bromine is 30.91 kJ/mol and its boiling point is 59 °C, we first need to convert these units appropriately. The boiling point in Kelvin is 59 °C + 273.15 = 332.15 K.
Converting the heat of vaporization to J/mol (since 1 kJ = 1000 J), we have 30.91 kJ/mol = 30910 J/mol.
Now, we can calculate the entropy of vaporization as follows: ΔSvap = 30910 J/mol / 332.15 K = 93.1 J/(mol·K).
Use tabulated electrode potentials to calculate ∆G° rxn for each reaction at 25 °C in kJ.
(a) Pb2+(aq) + Mg(s) ➝ Pb(s) + Mg2+(aq)
(b) Br2(l) + 2 Cl-(aq) ➝ 2 Br-(aq) + Cl2( g)
(c) MnO2(s) + 4 H+(aq) + Cu(s) ➝ Mn2+(aq) + 2 H2O(l) + Cu2+(aq)
Answer: (a) [tex]\Delta G^0=-432.25kJ[/tex] , (b) [tex]\Delta G^0=55.96kJ[/tex] and (c) [tex]\Delta G^0=-171.74kJ[/tex]
Explanation: (a) Oxidation half reaction for the given equation is:
[tex]Mg(s)\rightarrow Mg^2^+(aq)+2e^-E^0=2.37V[/tex]
The reduction half equation is:
[tex]Pb^2^+(aq)+2e^-\rightarrow Pb(s)E^0=-0.13V[/tex]
[tex]E^0_c_e_l_l=E^0_r_e_d_u_c_t_i_o_n+E^0_o_x_i_d_a_t_i_o_n[/tex]
[tex]E^0_c_e_l_l=-0.13V+2.37V[/tex]
[tex]E^0_c_e_l_l=2.24V[/tex]
[tex]\Delta G^0=-nFE^0_c_e_l_l[/tex]
where n is the number of moles of electrons transferred and F is faraday constant.
2 moles of electrons are transferred in the cell reaction which is also clear from both the half equations.
[tex]\Delta G^0=-2mol*96485\frac{C}{mol}*2.44V[/tex]
[tex]\Delta G^0=-432252.8J[/tex]
or [tex]\Delta G^0=-432.25kJ[/tex]
(b) Oxidation half reaction for the given equation is:
[tex]2Cl^-(aq)\rightarrow Cl_2(g)+2e^-E^0=-1.36V[/tex]
Reduction half equation is:
[tex]Br_2(l)+2e^-\rightarrow 2Br^-E^0=1.07V[/tex]
[tex]E^0_c_e_l_l=1.07V+(-1.36V)[/tex]
[tex]E^0_c_e_l_l=1.07V-1.36V[/tex]
[tex]E^0_c_e_l_l=-0.29V[/tex]
Now, we can calculate the [tex]\Delta G^0[/tex] same as we did for part a.
[tex]\Delta G^0=-2mol*96485\frac{C}{mol}*(-0.29V)[/tex]
[tex]\Delta G^0=55961.3J[/tex]
or [tex]\Delta G^0=55.96kJ[/tex]
(c) Oxidation half reaction for the given equation is:
[tex]Cu(s)\rightarrow Cu^2^+(aq)+2e^-E^0=-0.34V[/tex]
reduction half equation is:
[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^2^+(aq)+2H_2O(l)E^0=1.23V[/tex]
[tex]E^0_c_e_l_l=1.23V+(-0.34V)[/tex]
[tex]E^0_c_e_l_l=0.89V[/tex]
[tex]\Delta G^0=-2mol*96485\frac{C}{mol}*0.89V[/tex]
[tex]\Delta G^0=-171743.3J[/tex]
or [tex]\Delta G^0=-171.74kJ[/tex]
To calculate the standard Gibbs free energy change (∆G°) for each reaction at 25°C in kJ, we can use tabulated electrode potentials. By writing half-cell reactions and summing the electrode potentials, we can determine the overall reaction's standard potential (E°). Then, using the formula ∆G° = -nFE°, we can calculate the standard Gibbs free energy change.
Explanation:Use tabulated electrode potentials to calculate ∆G° rxn for each reaction at 25 °C in kJ.
a) Pb2+(aq) + Mg(s) ➝ Pb(s) + Mg2+(aq)First, we need to write half-cell reactions for the given equation:Pb2+ + 2e- ➝ Pb (E° = -0.126 V)Mg2+ + 2e- ➝ Mg (E° = -2.37 V)Next, we can sum up the electrode potentials to calculate the overall reaction's standard potential:E° rxn = E°(Pb) - E°(Mg)E° rxn = -0.126 V - (-2.37 V) = 2.244 VFinally, we can use the formula ∆G° = -nFE° to calculate the standard Gibbs free energy change:∆G° = -2F(2.244 V)b) Br2(l) + 2 Cl-(aq) ➝ 2 Br-(aq) + Cl2( g)First, we need to write half-cell reactions for the given equation:Br2 + 2e- ➝ 2Br- (E° = 1.07 V)Cl2 + 2e- ➝ 2Cl- (E° = 1.36 V)Next, we can sum up the electrode potentials to calculate the overall reaction's standard potential:E° rxn = E°(2Br-) - E°(2Cl-)E° rxn = 2(1.07 V) - 2(1.36 V) = -0.640 VFinally, we can use the formula ∆G° = -nFE° to calculate the standard Gibbs free energy change:∆G° = -2F(-0.640 V)c) MnO2(s) + 4 H+(aq) + Cu(s) ➝ Mn2+(aq) + 2 H2O(l) + Cu2+(aq)First, we need to write half-cell reactions for the given equation:MnO2 + 4H+ + 2e- ➝ Mn2+ + 2H2O (E° = 1.23 V)Cu2+ + 2e- ➝ Cu (E° = 0.34 V)Next, we can sum up the electrode potentials to calculate the overall reaction's standard potential:E° rxn = E°(Mn2+) - E°(Cu)E° rxn = 1.23 V - 0.34 V = 0.89 VFinally, we can use the formula ∆G° = -nFE° to calculate the standard Gibbs free energy change:∆G° = -2F(0.89 V)Learn more about Calculating standard Gibbs free energy change here:https://brainly.com/question/34263086
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The solubility of Cd(OH)2 can be increased through formation of the complex ion CdBr2−4 (Kf=5×103). If solid Cd(OH)2 is added to a NaBr solution, what would the initial concentration of NaBr need to be in order to increase the molar solubility of Cd(OH)2 to 1.0×10−3 moles per liter?
Answer:
Concentration of sodium bromide required = 2.38 M (around 2.4 M)
Explanation:
The equilibrium representing the complex ion formation is:
[tex]Cd^{2+} + 4Br^{-}\rightleftharpoons [CdBr_{4}]^{2-} .....Kf =5*10^{3}[/tex]-----(1)
where K(f) = formation equilibrium
The equilibrium representing the dissolution of Ca(OH)2 is:
[tex]Cd(OH)_{2}\rightleftharpoons Cd^{2+}+2OH^{-}.....Ksp = 2.5*10^{-14}[/tex]---(2)
where Ksp = solubility product
adding Equation (1) and equation(2) gives the net reaction:
[tex]Cd(OH)_{2} + 4Br^{-}\rightleftharpoons [CdBr_{4}]^{2-}+2OH^{-}[/tex]
[tex]K = K_{f}*K_{sp} = 5*10^{3}*2.5*10^{-14}=\frac{[CdBr_{4}^{2-}][OH^{-}]^{2}}{[Br{-}]^{4}}[/tex]
[tex]12.5*10^{-11} =\frac{1*10^{-3} *[2*10^{-3}]^{2}}{[Br-]^{4} }\\[/tex]
[tex][Br-] = 2.38 M[/tex]
The study of chemicals is called chemistry. when the amount of reactant or product gets equal is said to be an equilibrium state.
The correct answer is 2.38M.
What is equilibrium constant?The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium. A state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.The data is given in the question is as follows:-
[tex]Kf =5*10^3\\Ksp =2.5*10^{-14}[/tex]
The reaction in the given question is as follows:-
[tex]Cd(OH)_2 +4Br^- +[CdbBr_4]^{2-} +2OH^-[/tex]
The formula we used to solve the question is as follows:-
[tex]K =\frac{{[cdbr_4^2-]}[OH]^2}{{Br^-}^2}[/tex]After placing the value, the correct answer for the bromine is 2.38M.
Hence, the correct answer is 2.38M
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Which statement about van der Waals forces is true?
a)When the forces are weaker, a substance will have higher volatility.
b)When the forces are stronger, a substance will have lower viscosity.
c)When the forces are weaker, the boiling point of a substance will be higher.
d)When the forces are stronger, the melting point of a substance will be lower.
Answer:
A
Explanation:
Van der Waals forces are the weak electric forces of attraction between molecules and their strength is dependent on the distance between the molecules. The longer the distance between the molecules the weaker the forces. Weaker Van der Waals forces mean that molecules can easily escape from the liquid - hence meaning higher volatility.
Answer:
A!!!
Explanation: