Answer:
0.3 moles of aluminum
Explanation:
The reaction of Aluminium with copper sulfate reacts to give aluminium sulfate and copper metal ,
The balanced chemical reaction is as follows -
2Al + 2 CuSO4 ----> Al2(SO4)3 + 3CuFrom the above balanced equation ,considering the stoichiometry , 3 mole of Copper is produced , by using 2 moles of Aluminium,using unitary method ,hence, 1 mole of copper is produced , by using 2/3 moles of Aluminum ,(from the question , 0.45 moles of copper )Therefore , 0.45 mole of copper is produced , by using 2/3 * 0.45 mole Aluminium,Solving, 2/3 * 0.45 mole = 0.3 molHence, 0.3 moles of aluminum will be required to produce, 0.45 moles of copper metal.To produce 0.45 moles of copper metal, 0.30 moles of aluminum are required.
To determine how many moles of aluminum (Al) are needed to produce 0.45 moles of copper (Cu) metal, we need to refer to the balanced chemical reaction between aluminum and copper (II) chloride (CuCl₂).
The balanced equation is:
2 Al + 3 CuCl₂ → 2 AlCl₃ + 3 Cu
This means that 2 moles of aluminum produce 3 moles of copper. We can set up a ratio to find out how many moles of aluminum are needed to produce 0.45 moles of copper:
(2 moles Al / 3 moles Cu) = (x moles Al / 0.45 moles Cu)
Solving for x,
x = (2/3) x 0.45 = 0.30 moles of Al
Therefore, 0.30 moles of aluminum will be required to produce 0.45 moles of copper metal.
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:
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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