Given these two reactions below: Zn(OH)2 2 Zn²*(aq) + 20H-(aq) Ksp = 3.0x10-17 Zn2*(aq) + 4 OH (aq) Zn(OH),2 (aq) Kform = 2.9x1015 determine the equilibrium constant for this reaction shown: Zn(OH)2(s) + 2 OH (aq) Zn(OH),2(aq) K = ? > O 3.4x10-16 O 3.3x1016 O 8.7x10-2 O 3.8x1017

Given these two reactions below: Zn(OH)2 2 Zn²(aq) + 20H-(aq) Ksp = 3.0x10-17 Zn2(aq) + 4 OH (aq) Zn(OH),2 (aq) Kform = 2.9x1015 determine the equilibrium constant for this reaction shown: Zn(OH)2(s) + 2 OH (aq) Zn(OH),2(aq) K = ? > O 3.4x10-16 O 3.3x1016 O 8.7x10-2 O 3.8x1017

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

The following reactions all have K > 1. 1) HC₂H704 (aq) + CIO (aq)C9H7O4 (aq) + HCIO (aq) 2) C6H5COOH (aq) + CIO (aq)HCIO (aq)+ CáH5COO* (aq) 3) HC₂H704 (aq) + C6H5COO (aq) = C9H704 (aq) + C6H5COOH (aq) Arrange the substances based on their relative base strength. C9H704 C6H5COO™ C6H5COOH HCIO Clear All strongest base intermediate base weakest base
Copper (II) Phosphate is extremely insoluble, but forms a complex ion with CN- that can drive the Copper (II) solubility through Le Chatlier’s principle. Given the two reactions below, determine KT. Cu3(PO4)2 (s) -->3 Cu2+ (aq) + 2 PO43-(aq) K1=2.05x10-35 Cu(CN)42- (aq) -->Cu2+ (aq) + 4 CN- (aq) K2=1.0x10-25 Cu3(PO4)2 (s) + 12 CN- (aq) --> 3 Cu(CN)42- (aq) + 2 PO43- (aq) KT=???
How would the equilibrium be affected in regards to the value of Qc or Kc for each of the following conditions?: Fe3+ (aq) + SCN- (aq) ⇌ FeSCN2+ (aq) Adding KSCN (aq) Adding Fe(NO3) (aq) Adding NaF (s) Placing the solution into a hot water bath Placing the solution into an ice bath (To whom this may concern, Could you at least explain one as simply as possible so I'm able to understand the others enough to explain them myself? Thanks in anvance!)
[Review Topics] [References] Consider the following system at equilibrium where AH = 92.7 kJ, and K = 1.80 x 10-4, at 298 K. NH₂HS(s) → NH3 (9) + H₂S(g) When some moles of NH4HS(s) are added to the equilibrium system at constant temperature: The value of Ke Oincreases Odecreases Oremains the same 20 E The value of Qc Ois greater than Ke Ois equal to Ke O is less than Ke C The reaction must Orun in the forward direction to reestablish equilibrium. Orun in the reverse direction to feestablish equilibrium. Oremain the same. It is already at equilibrium. The number of moles of NH3 will Oincrease O decrease Oremain the same $ 4 Submit Answer 8888 R 7 F V Show Hint % 5 Retry Entire Group T G Cengage Learning | Cengage Technical Support A 6 B 9 more group attempts remaining MacBook Air Y H & 7 N C * 00 8 M ( 9 K O V O L P V 4) Previous Next> Save and Exit A
Identify the spectator ion(s) in the following reaction.Zn(OH)2(s) + 2K+(aq) + 2OH–(aq) → 2K+(aq) + Zn(OH)4–(aq) Zn(OH)2 K+ and OH- K+ K+ and Zn(OH)4 -2 Zn(OH)4– Provide explanation for answer.
You are working in the lab and your TA gives you a yellow colored solution at equilibrium. The solution contains unknown concentrations of the chemicals in the reaction below. You then add 6.00 ml of 3 M sodium hydroxide to the solution. Predict what your observation would be and select the explanation as to why this would occur. Cro,2 (aq) + H3O*(aq) Cr2O,7 (aq) + H20() yellow orange O The solution turns orange because addition of sodium hydroxide decreases the H+ concentration, thereby decreasing the pH of the solution, causing the reaction to shift toward products. O The solution turns orange because increased base concentration increases the negative charge of the solution, causing the reaction to shift toward reactants. The solution stays yellow because the addition of sodium hydroxide decreases H30+ concentration in the solution, thereby causing the pH to increase, causing the reaction to shift toward the reactants O The solution turns orange because the concentration of…
Given the equilibrium reaction, Ag2CO3(s) ⇌ 2Ag+(aq) + CO32-(aq) Which of the following is always TRUE regarding the molar solubility of Ag2CO3(s) in water? Given the equilibrium reaction, Ag2CO3(s) ⇌ 2Ag+(aq) + CO32-(aq) Which of the following is always TRUE regarding the molar solubility of Ag2CO3(s) in water? It increases in the presence of more Ag2CrO4(s). It increases in the presence of 0.10 M AgNO3. It increases as Ag+ forms a complex with 0.10 M NH3.
From the table, what overall conclusions can you make about the direction of equilibrium shift in this reaction? Are your observations consistent with Le Chatelier’s principle?
The equilibrium constant is 4.5×10-32 for the reaction Ni(CN-(aq) +→ NP* (aq) + 4 CN¯ (aq) Calculate the equilibrium concentration, expressed in units of nM (nanomolar), of cyanide (CN") for a solution that initially only contains 0.25 M of Ni(CN)42- (aq). Enter your answer in deimal notation and provide 3 significant figures. For example, enter 0.2531 as 0.253
Select the shift in the reaction that is indicated by the change in color you observed. Then explain this change. Recall the reaction is as follows: Fe3+(aq)+NCS−(aq)⇌FeNCS2+(aq)pale yellowcolorlessred-orange color Observation: When Fe3+ was added, the Fe3+–NCS–FeNCS2+ equilibrium mixture turned darker red. A. Account for the effect of adding Fe3+, including a chemical reaction. B. Account for the effect of adding NCS-, including a chemical reaction. C. Account for the effect of adding Hg(NO3)2, including a chemical reaction. D. Molarity of FeNCS2+ in solution of “known concentration”
Determine [PLUOT]eq using the following information: 2 PLUM (g) + APRICOT (g) ⇌ 2 PLUOT (g) KC = 1.7 × 108 [PLUM]eq = 0.0018 M [APRICOT]eq = 0.0034 M Group of answer choices 0.53 M 1.4 M 1.9 M 9.6 x 10–4 M 1.0 × 103 M