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determine the enthalpy change for the reaction: Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)
q (enthalpy change) = (mass of solution) x (specific heat capacity) x (temperature change)q = 50g (from the 50cm cubed) x 4.18 x 6.3 = 1316.7 Jalso, number of moles...
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The equation below represents an equilibrium reaction HNO3(aq) + NH3(aq) ↔ NO3- (aq) + NH4+(aq). Which chemical in the forward reaction is a Bronsted-Lowry acid? * A. HNO3 B. NH3 C. NO3- D. NH4+
A. HNO3
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An ore sample contains 11% impurity in additon to a mix of CuS and Cu2S. Heating 100g of the mix produces 75.4g of copper metal with a purity of 89.5%. What is weight 5 of CuS and Cu2S in the ore?
75.4* 0.895=67.483 - it is the mass of pure copper mass of S in sample is 100-67.483-100*0.11=21.517 n of S= 21.517/32=0.6724 moles n of Cu =67.483/63.5=1.06 moles 64* (1.06 -...
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Which species are present in aqueous solution of 0.12 M NH3 solution except for H2O? Lütfen birini seçin: a. NH4+, OH- b. NH3, NH4+ c. NH3, NH4+, OH- d. NH3, NH4+, H3O+, OH- e. NH3, OH-
c. NH3, NH4+, OH-
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obtain the ΔH° of the following: <ol><li>Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)</li></ol>
ΔH=ΔE+PΔVor ΔE=ΔH−PΔVPΔV=1 atm×22.4 L=22.4 L atm=22.4×101.3 J=2307 J=2.31 kJ∴ΔE=−154.4−2.31=−156.71 kJ
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Calculate how much (as grams) the following elements are needed to produce a charge equal to the Faraday’s constant. Electron charge is 1.602*10^–19 As
a) Li (using Li/Li+)
b) Zn (using Zn/Zn2+)
m = (Q/F)*(M/z) m is the mass of the substance liberated at an electrode in grams Q is the total electric charge passed through the substance F = 96,485...
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Calculate ∆G° for the reaction occurring in the following cell :
Al (s) | Al³⁺ 1(M) || Cu²⁺ 1(M) Cu (s)
Standard electrode potentials :
E°= −1.66 V and E°= + 0.34 V
Al³⁺/Al Cu²⁺/Cu
E°=0.34-(-1.66)=2 V∆G°=-nFE°/1000=-6*96485*2=-1157 kJ/mol
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If 2.40g of CuSO4 is dissolved in 9.95*10^2 ml of 0.350 M NH3, calculate the concentrations of the following species at equilibrium. (a) Cu2+ (b)NH3 (c)Cu(NH3)4^2+
C0(NH3)=0.35Mn(CuSO4)=2.40/159.6=0.015 molC0(Cu2+)=0.015/0.995=0.0151 MCu2++4NH3=Cu(NH3)42+k1234 = [Cu2+]*[NH3]4/[Cu(NH3)42+] = 9.33*10-13 (instability constant of this complex)(0.0151-x)*(0.35-4x)4/x=9.33*10-13x=0.0151 = [Cu(NH3)42+] (ammost all Cu2+ react with NH3)[NH3] = 0.35-0.0151*4=0.2896M(0.0151-x)*0.294/0.0151=9.33*10-13(0.0151-x)=1.99*10-12 = [Cu2+]Answer: [Cu2+]=1.99*10-12M; [NH3] =0.2896M; [Cu(NH3)42+]=0.0151M
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Calculate the reduction potential of the Cu2+/Cu electrode when [Cu2+] = 1.0 ´ 10-8 M.
a.
+0.10 V
b.
+0.37 V
c.
+0.33 V
d.
+0.34 V
e.
+0.35 V
It should be used the Nernst equation: E = E0 + (0.05916V )/z log[Cu2+] For Cu2+/Cu electrode: E0 = +0.34 V and z = 2 Therefore: E = 0.34 +...
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Balance the half-reactions assuming that they occur in acidic solution.Zn(s)⟶Zn2+(aq)
Zn(s) - 2e ⟶ Zn2+(aq) Oxidation half-reaction 2H+(aq) + 2e ⟶ H2(g) Reduction half-reaction Zn(s) + 2H+(aq) ⟶ Zn2+(aq) + H2(g) total equation
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