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![→ CAnewes ) a) Cata(s) = Ca 2+ + 2F - o [F + H+ = HF (aq)] x2 – adding ② and ③ we get Cat(s) + 2HT = Ca ²+ + 2HF le laf (s)](http://img.homeworklib.com/questions/65d1fa60-e00c-11eb-8858-a72fe14e9ff0.png?x-oss-process=image/resize,w_560)
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Problem 3: a) Write the mass balance for CaF2 in H2O if the Rx's are: CaF2(s)...
1. Write the equilibrium constant expression for the reaction N2 (g) + 3 H2 (g) D ...
1. Write the equilibrium constant expression for the reaction N2 (g) + 3 H2 (g) D 2 NH3 (g) 2. For the reaction in question 1, delta H = -92 kj*mol^-1 Predict the direction of shift of the equilibrium of the above reaction caused by adding N2 at constant volume and temperature removing H2 at constant volume and temperature adding N2 at constant volume and temperature raising the temperature at constant total pressure 3. Hydrogen fluoride (HF) is a weak...
1. Solid calcium fluoride (CaF2) establishes the following equilibrium in solution: CaF2(s) = Ca2+(aq) + 2F-(aq)...
1. Solid calcium fluoride (CaF2) establishes the following equilibrium in solution: CaF2(s) = Ca2+(aq) + 2F-(aq) Ke = 1.5 x 10-10 A solution initially contains 2.45 g of CaF2. a. In which direction will the reaction move to reach equilibrium? Explain your reasoning b. Calculate the equilibrium concentrations of Ca2+ and F. How much CaF2 (in mg) dissolves in solution? c. Another solution initially contains 2.45 g of CaF2 and 0.0025 M NaF. What are the equilibrium concentrations of Ca2+...
Consider these two equilibrium processes. (i) CaF2 (s) ↔ Ca2+ + 2 F−
Consider these two equilibrium processes. (i) CaF2 (s) ↔ Ca2+ + 2 F− Kosp = 3.2 x 10−11 (ii) HF ↔ H+ + F− Koa = 6.8 x 10−4 (a) (0.6 pt) Use equations (i) and (ii), and their K’s to find Kooverall for this process: (iii) CaF2 (s) + 2 H+ ↔ Ca2+ + 2 HF Keq = ?? (b) (0.6 pt) Find Ksp, conditional for reaction (iii) when the pH is such that the fraction...
If 1.2 moles of CaF2 (s) are added to 2.7 L of water what will the...
If 1.2 moles of CaF2 (s) are added to 2.7 L of water what will the concentration of F- be at equilibrium? The Kop - 3,9 x 10-11 CaF, (s) = Ca (aq) + 2F (aq) O(A) 5.38 x 10 M (B) 3. 12 x 10-6 M (C) 2.14 x 10- M (D) 7.91 x 10-M (E) 4.27 x 10-4 M
Problem 1: Balance the following reaction in (a) acidic solution and (b) basic solution MnO2 (s)...
Problem 1: Balance the following reaction in (a) acidic solution and (b) basic solution MnO2 (s) + H Asos (aq) → Mn2+ (aq) + HASO (aq) (a) Acidic solution 1. Split the reaction into two half-reactions. Mn q + ac Mnat HAs Oy > Hj As Oy + ae" 2. Consider the following half reaction: MnO2 (s) → Mn2+ (aq) Balance everything but oxygen and hydrogen atoms. MnOade > Moi 3. Take your answer to (2) and balance the oxygen...
10.60 Write and balance a chemical equation show occurs when solid barium is combined with each...
10.60 Write and balance a chemical equation show occurs when solid barium is combined with each of the following: a. Ca(CN(ag) ing the single-displacement reaction that b. AgCIO (ag) c. Ni(C2H O2)(ag) 10.64 Identify the substance being oxidized in each of the following unbalanced reactions: a. F2 (g) + H2O (l) ? HF (aq) + O2 (g) c. Mg (I) + TiCla (g) ? MgCl2 (s) + Ti (s)
2) Write the equilibrium constant expressions for each of the following reactions. (For gas-phase reactions, write...
2) Write the equilibrium constant expressions for each of the following reactions. (For gas-phase reactions, write the Ke expression.) a) 2 NO(g) + O2(g) = 2 NO2(g) b) 4 Ag(s) + O2(g) + 2 Ag2O(S) c) CaCO3(s) + CO2(aq) + H2O(l) = Ca2+(aq) + 2 HCO3(aq) 3) a) Write the K, expressions for reactions a and b in problem 2. b) If the value of K for reaction a in problem 2 is 2.8 x 1011 at 200°C, what is...
2. The equilibrium constant for the following reaction is called the "solubility product" of calcium fluoride:...
2. The equilibrium constant for the following reaction is called the "solubility product" of calcium fluoride: CaF2(s) - Cal(aq) + 2F-(ay) K = Kp = 3.2 x 10-11 (a) Write an expression for the equilibrium constant of this reaction in terms of concentrations. Why do you suppose we call this a solubility product instead of a solubility quotient or ratio? (b) Calculate the equilibrium concentrations of Ca2+ and F if excess solid CaF is placed in water. (c) in which...
HF(aq) + H2O() H0"(aq) + F (aq) EC. 2 H2O() H3O* (aq) + OH(aq) if [HFJinitial 0.05M, then K,=[H,O" ][A]/[HA...
HF(aq) + H2O() H0"(aq) + F (aq) EC. 2 H2O() H3O* (aq) + OH(aq) if [HFJinitial 0.05M, then K,=[H,O" ][A]/[HA] Kw=[HO"][OH] 0.05M [HF] +[F] equilibria (1) (2) mass balance (3) charge balance al+LHol L0H] (4) Given the setup above, manipulate the equations so you are left with this equation: Ka [H3O'1{[H,o'j-K[H3O']} / {0.05-[H20*1-K.//H2O]} ( you do not have to distribute the negative in the denominator)
HF(aq) + H2O() H0"(aq) + F (aq) EC. 2 H2O() H3O* (aq) + OH(aq) if...
HF(aq) + H2O() H0"(aq) + F (aq) EC. 2 H2O() H3O* (aq) + OH(aq) if [HFJinitial 0.05M, then K,=[H,O" ][A]/[HA...
HF(aq) + H2O() H0"(aq) + F (aq) EC. 2 H2O() H3O* (aq) + OH(aq) if [HFJinitial 0.05M, then K,=[H,O" ][A]/[HA] Kw=[HO"][OH] 0.05M [HF] +[F] equilibria (1) (2) mass balance (3) charge balance al+LHol L0H] (4) Given the setup above, manipulate the equations so you are left with this equation: Ka [H3O'1{[H,o'j-K[H3O']} / {0.05-[H20*1-K.//H2O]} ( you do not have to distribute the negative in the denominator)
HF(aq) + H2O() H0"(aq) + F (aq) EC. 2 H2O() H3O* (aq) + OH(aq) if...
1. Solid calcium fluoride (CaF2) establishes the following equilibrium in solution: CaF2(s) = Ca2+(aq) + 2F-(aq) Ke = 1.5 x 10-10 A solution initially contains 2.45 g of CaF2. a. In which direction will the reaction move to reach equilibrium? Explain your reasoning b. Calculate the equilibrium concentrations of Ca2+ and F. How much CaF2 (in mg) dissolves in solution? c. Another solution initially contains 2.45 g of CaF2 and 0.0025 M NaF. What are the equilibrium concentrations of Ca2+...
If 1.2 moles of CaF2 (s) are added to 2.7 L of water what will the concentration of F- be at equilibrium? The Kop - 3,9 x 10-11 CaF, (s) = Ca (aq) + 2F (aq) O(A) 5.38 x 10 M (B) 3. 12 x 10-6 M (C) 2.14 x 10- M (D) 7.91 x 10-M (E) 4.27 x 10-4 M
Problem 1: Balance the following reaction in (a) acidic solution and (b) basic solution MnO2 (s) + H Asos (aq) → Mn2+ (aq) + HASO (aq) (a) Acidic solution 1. Split the reaction into two half-reactions. Mn q + ac Mnat HAs Oy > Hj As Oy + ae" 2. Consider the following half reaction: MnO2 (s) → Mn2+ (aq) Balance everything but oxygen and hydrogen atoms. MnOade > Moi 3. Take your answer to (2) and balance the oxygen...
10.60 Write and balance a chemical equation show occurs when solid barium is combined with each of the following: a. Ca(CN(ag) ing the single-displacement reaction that b. AgCIO (ag) c. Ni(C2H O2)(ag) 10.64 Identify the substance being oxidized in each of the following unbalanced reactions: a. F2 (g) + H2O (l) ? HF (aq) + O2 (g) c. Mg (I) + TiCla (g) ? MgCl2 (s) + Ti (s)
2) Write the equilibrium constant expressions for each of the following reactions. (For gas-phase reactions, write the Ke expression.) a) 2 NO(g) + O2(g) = 2 NO2(g) b) 4 Ag(s) + O2(g) + 2 Ag2O(S) c) CaCO3(s) + CO2(aq) + H2O(l) = Ca2+(aq) + 2 HCO3(aq) 3) a) Write the K, expressions for reactions a and b in problem 2. b) If the value of K for reaction a in problem 2 is 2.8 x 1011 at 200°C, what is...
2. The equilibrium constant for the following reaction is called the "solubility product" of calcium fluoride: CaF2(s) - Cal(aq) + 2F-(ay) K = Kp = 3.2 x 10-11 (a) Write an expression for the equilibrium constant of this reaction in terms of concentrations. Why do you suppose we call this a solubility product instead of a solubility quotient or ratio? (b) Calculate the equilibrium concentrations of Ca2+ and F if excess solid CaF is placed in water. (c) in which...
HF(aq) + H2O() H0"(aq) + F (aq) EC. 2 H2O() H3O* (aq) + OH(aq) if [HFJinitial 0.05M, then K,=[H,O" ][A]/[HA] Kw=[HO"][OH] 0.05M [HF] +[F] equilibria (1) (2) mass balance (3) charge balance al+LHol L0H] (4) Given the setup above, manipulate the equations so you are left with this equation: Ka [H3O'1{[H,o'j-K[H3O']} / {0.05-[H20*1-K.//H2O]} ( you do not have to distribute the negative in the denominator)
HF(aq) + H2O() H0"(aq) + F (aq) EC. 2 H2O() H3O* (aq) + OH(aq) if...
HF(aq) + H2O() H0"(aq) + F (aq) EC. 2 H2O() H3O* (aq) + OH(aq) if [HFJinitial 0.05M, then K,=[H,O" ][A]/[HA] Kw=[HO"][OH] 0.05M [HF] +[F] equilibria (1) (2) mass balance (3) charge balance al+LHol L0H] (4) Given the setup above, manipulate the equations so you are left with this equation: Ka [H3O'1{[H,o'j-K[H3O']} / {0.05-[H20*1-K.//H2O]} ( you do not have to distribute the negative in the denominator)
HF(aq) + H2O() H0"(aq) + F (aq) EC. 2 H2O() H3O* (aq) + OH(aq) if...
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