The formation constant, Kf, of Fe(CN)63-(aq), Fe3+(aq) + 6 CN-(aq) ⇌ Fe(CN)63-(aq), is 1.0 x 1042....
The formation constant, Kf, of
Fe(CN)63-(aq), Fe3+(aq) + 6
CN-(aq) ⇌ Fe(CN)63-(aq), is 1.0 x
1042. What are the equilibrium concentrations of
Fe3+(aq), CN-(aq), and
Fe(CN)63-(aq) if we add 0.113 mol of
Fe(NO3)3 to 1.000 L of a 0.965 M aqueous
solution of NaCN? Assume the volume remains fixed at 1.000
L.
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The formation constant, Kf, of
Fe(CN)63-(aq), Fe3+(aq) + 6
CN-(aq) ⇌ Fe(CN)63-(aq), is 1.0 x
1042....
Question 4 The formation constant, Kf, of PbC142 (aq), Pb2+(aq) + 4 Cl(aq) = PbC1 +...
Question 4 The formation constant, Kf, of PbC142 (aq), Pb2+(aq) + 4 Cl(aq) = PbC1 + (aq), is 2.5 x 1015. What are the equilibrium concentrations of Pb2+(aq), Cl"(aq), and PbC1,2 (aq) if we add 0.105 mol of Pb(NO3)2 to 1.000 L of a 0.865 M aqueous solution of NaCl? Assume the volume remains fixed at 1.000 L.
A solution is prepared by adding 0.075 mole of K3 [Fe(CN)6] to 0.72 L of 2.2...
A solution is prepared by adding 0.075 mole of K3 [Fe(CN)6] to 0.72 L of 2.2 M NaCN. Assuming no volume change, calculate the concentrations of Fe(CN)63- and Fe3+ in this solution. The K (overall) for the formation of Fe(CN)63- is 1.0 x 1042. [Fe(CN)6*] = [Fe3+] = Submit Answer Try Another Version 2 item attempts remaining
A solution is prepared by adding 0.063 mole of K3 [Fe(CN)6] to 0.59 L of 2.1...
A solution is prepared by adding 0.063 mole of K3 [Fe(CN)6] to 0.59 L of 2.1 M NaCN. Assuming no volume change, calculate the concentrations of Fe(CN)63- and Fe3+ in this solution. The K (overall) for the formation of Fe(CN)63- is 1.0 x 1042. M [Fe(CN)6*] = C [Fe3+]= C M Submit Anchor Another Voreion 2 itom attomnte romaining
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Q1 The solubility product, Kps, of Al (OH) 3 (s) is 1.0 x 10-33. What is its solubility (in g / L) in an aqueous solution of NaOH with a pH of 12.21? Q2 The formation constant, Kf, of PbCl42- (aq), Pb2 + (aq) + 4 Cl- (aq) ⇌ PbCl42- (aq), is 2.5 x 1015. What are the equilibrium concentrations of Pb2 + (aq) , Cl- (aq), and PbCl42- (aq) if we add 0.103 mol of Pb (NO3) 2 to...
Calculate the overall formation constant for Fe(CN) given that the overall formation constant for Fe(CN)6)-is ~1032,...
Calculate the overall formation constant for Fe(CN) given that the overall formation constant for Fe(CN)6)-is ~1032, and that Fe3+ (aq) + e-= Fe2+ (aq) [Fe(CN)s! (aq) +e Fe(CN) +0.77v +0.36 (aq) E This the wite 39 answer λ 10
3. Iron(II) nitrate and hydrocyanic acid react to form the hexacyanoferrate(II) ion, [Fe(CN).]*(aq), with a formation...
3. Iron(II) nitrate and hydrocyanic acid react to form the hexacyanoferrate(II) ion, [Fe(CN).]*(aq), with a formation constant of 1.438x10's. This complex ion has an intense blue color in solution. a. If you add 182.734g of iron(II) nitrate solid to 750.0mL of 0.8324M CN'(aq) solution, what is the concentration of CN'(aq) still in solution when the reaction reaches equilibrium? b. Copper(II) also forms a complex ion with cyanide ions, tetracyanocuprate(II)ion, [Cu(CN).] (aq), with a formation constant of 1.028x10”. This complex ion...
Equilibrium Concentrations for a Simple Addition Reaction At a certain temperature K = 1.10x103 for the reaction: Fe3+(...
Equilibrium Concentrations for a Simple Addition Reaction At a certain temperature K = 1.10x103 for the reaction: Fe3+(aq)SCN"(aq) FeSCN2+(aq) 5.00x102 mol of Fe(NO3)3 is added to 8.70x 10-1 L of 1.84 x 10-1 M KSCN. Neglecting any volume change and assuming that all species remain in solution--: Calculate the equilibrium concentration of Fe3+ (in mol/L) mol/L 1 pts 提交答案 Tries 0/8 Calculate the equilibrium concentration of SCN (in mol/L). mol/L 1 pts [提交答案 Tries 0/8 Calculate the equilibrium concentration of...
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The formation constant of M(CN),1'-is 2.50 x 101, where M is a generic metal A 0.130...
The formation constant of M(CN),1'-is 2.50 x 101, where M is a generic metal A 0.130 wole quantity of M(NO3)> is alled to a liter of' 1.010 M NaCN solution. What is the concentration of Mº ions at equilibrium? M? 1- 3.513 x10-14
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The formation constant* of [M(CN) 6 ]4− is 2.50 × 10 17 , where M is a generic metal. A 0.140 mole quantity of M(NO3)2 is added to a liter of 1.210 M NaCN solution. What is the concentration of M2+ ions at equilibrium?
Question 4 The formation constant, Kf, of PbC142 (aq), Pb2+(aq) + 4 Cl(aq) = PbC1 + (aq), is 2.5 x 1015. What are the equilibrium concentrations of Pb2+(aq), Cl"(aq), and PbC1,2 (aq) if we add 0.105 mol of Pb(NO3)2 to 1.000 L of a 0.865 M aqueous solution of NaCl? Assume the volume remains fixed at 1.000 L.
A solution is prepared by adding 0.075 mole of K3 [Fe(CN)6] to 0.72 L of 2.2 M NaCN. Assuming no volume change, calculate the concentrations of Fe(CN)63- and Fe3+ in this solution. The K (overall) for the formation of Fe(CN)63- is 1.0 x 1042. [Fe(CN)6*] = [Fe3+] = Submit Answer Try Another Version 2 item attempts remaining
A solution is prepared by adding 0.063 mole of K3 [Fe(CN)6] to 0.59 L of 2.1 M NaCN. Assuming no volume change, calculate the concentrations of Fe(CN)63- and Fe3+ in this solution. The K (overall) for the formation of Fe(CN)63- is 1.0 x 1042. M [Fe(CN)6*] = C [Fe3+]= C M Submit Anchor Another Voreion 2 itom attomnte romaining
Calculate the overall formation constant for Fe(CN) given that the overall formation constant for Fe(CN)6)-is ~1032, and that Fe3+ (aq) + e-= Fe2+ (aq) [Fe(CN)s! (aq) +e Fe(CN) +0.77v +0.36 (aq) E This the wite 39 answer λ 10
3. Iron(II) nitrate and hydrocyanic acid react to form the hexacyanoferrate(II) ion, [Fe(CN).]*(aq), with a formation constant of 1.438x10's. This complex ion has an intense blue color in solution. a. If you add 182.734g of iron(II) nitrate solid to 750.0mL of 0.8324M CN'(aq) solution, what is the concentration of CN'(aq) still in solution when the reaction reaches equilibrium? b. Copper(II) also forms a complex ion with cyanide ions, tetracyanocuprate(II)ion, [Cu(CN).] (aq), with a formation constant of 1.028x10”. This complex ion...
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The formation constant of M(CN),1'-is 2.50 x 101, where M is a generic metal A 0.130 wole quantity of M(NO3)> is alled to a liter of' 1.010 M NaCN solution. What is the concentration of Mº ions at equilibrium? M? 1- 3.513 x10-14
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