EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 8220103151757
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 10, Problem 10.54AP
Interpretation Introduction
Interpretation:
The "percent blood alcohol content" (BAC) of Bobbin Weaver is to be calculated. Whether the Officer Order arrest Bobbin Weaver or not is to be stated.
Concept introduction:
The stoichiometry of a chemical species involved in a
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Consider the following reactions.
2CH4 (g) ⇌ C2H6 (g) + H2 (g) Kc = 9.5x10–13
CH4 (g) + H2O (g) ⇌ CH3OH (g) + H2 (g) Kc = 2.8x10–21
Calculate the equilibrium constant, Kc value for the following reaction.
2CH3OH (g) + H2 (g) ⇌ C2H6(g) + 2H2O (g)
Please help answer these questions, I'm stuck on them
4. Explain using equilibrium principles, how the Carbonic Acid/Bicarbonate
system regulates breathing.
carbonie
anhydrase
co, + H,0S H,co,S HCO, + H*
carbon dioxide +
carbonic acid
bicarbonate +
water
hydrogen lon
Chapter 10 Solutions
EBK ORGANIC CHEMISTRY
Ch. 10 - Prob. 10.1PCh. 10 - Prob. 10.2PCh. 10 - Prob. 10.3PCh. 10 - Prob. 10.4PCh. 10 - Prob. 10.5PCh. 10 - Prob. 10.6PCh. 10 - Prob. 10.7PCh. 10 - Prob. 10.8PCh. 10 - Prob. 10.9PCh. 10 - Prob. 10.10P
Ch. 10 - Prob. 10.11PCh. 10 - Prob. 10.12PCh. 10 - Prob. 10.13PCh. 10 - Prob. 10.14PCh. 10 - Prob. 10.15PCh. 10 - Prob. 10.16PCh. 10 - Prob. 10.17PCh. 10 - Prob. 10.18PCh. 10 - Prob. 10.19PCh. 10 - Prob. 10.20PCh. 10 - Prob. 10.21PCh. 10 - Prob. 10.22PCh. 10 - Prob. 10.23PCh. 10 - Prob. 10.24PCh. 10 - Prob. 10.25PCh. 10 - Prob. 10.26PCh. 10 - Prob. 10.27PCh. 10 - Prob. 10.28PCh. 10 - Prob. 10.29PCh. 10 - Prob. 10.30PCh. 10 - Prob. 10.31PCh. 10 - Prob. 10.32PCh. 10 - Prob. 10.33PCh. 10 - Prob. 10.34PCh. 10 - Prob. 10.35PCh. 10 - Prob. 10.36PCh. 10 - Prob. 10.37PCh. 10 - Prob. 10.38PCh. 10 - Prob. 10.39APCh. 10 - Prob. 10.40APCh. 10 - Prob. 10.41APCh. 10 - Prob. 10.42APCh. 10 - Prob. 10.43APCh. 10 - Prob. 10.44APCh. 10 - Prob. 10.45APCh. 10 - Prob. 10.46APCh. 10 - Prob. 10.47APCh. 10 - Prob. 10.48APCh. 10 - Prob. 10.49APCh. 10 - Prob. 10.50APCh. 10 - Prob. 10.51APCh. 10 - Prob. 10.52APCh. 10 - Prob. 10.53APCh. 10 - Prob. 10.54APCh. 10 - Prob. 10.55APCh. 10 - Prob. 10.56APCh. 10 - Prob. 10.57APCh. 10 - Prob. 10.58APCh. 10 - Prob. 10.59APCh. 10 - Prob. 10.60APCh. 10 - Prob. 10.61APCh. 10 - Prob. 10.62APCh. 10 - Prob. 10.63APCh. 10 - Prob. 10.64APCh. 10 - Prob. 10.65APCh. 10 - Prob. 10.66APCh. 10 - Prob. 10.67APCh. 10 - Prob. 10.68APCh. 10 - Prob. 10.69APCh. 10 - Prob. 10.70AP
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- Write the balanced chemical reaction for this equilibrium expression: K= [H2O]P[O2] [H2O2]?arrow_forwardApart from the identity of the central atom, what is the major difference between NH4+ and CH4? How does this difference affect the relative solubility of the compounds in water?arrow_forward(CH3)2O →BF3(g)⇋BF3(g)+(CH3)2O(g) If you place 1.2 g of the complex in a 61.5 mL flask at 125 ℃, what is the total pressure in the flask at equilibrium? What are the partial pressures of the Lewis Acid, Lewis Base, and the complex? (Given: The value of K (here Kp) for the reaction is 0.17 at 125 ℃.) P(BF3)= P((CH3)2O)= P((CH3)2O→BF3)=arrow_forward
- Write the expression for the equilibrium constant for the reaction before you can calculate for the Keq or Kc: 2 P₂O5(g) 4 PO₂(g) + O₂(g) Calculate the equilibrium constant if the equilibrium concentrations are: [P₂O5] = 0.63 mol/L. [PO₂] = 0.74 mol/L, [O₂]= 0.21 mol/L.arrow_forwardWhen you add HCl to a solution containing aqueous sodium acetylsalicylate (structure shown below), a precipitate formed. Which is the structure of this precipitate?arrow_forwardWhen the reversible reaction HC2H;O2 (aq) E> H(aq) + C2H3O2 (aq) is at equilibrium at room temperature, pH of the reaction mixture is 5. What will be the change in pH when you add a large amount of NaOH (aq) to the reaction mixture? a) pH will not change because NaOH is not a part of the reaction equation. b) pH will not change because acetic acid and NaOH form a buffer. c) pH will increase because NaOH will completely neutralize acetic acid. d) pH will decrease because NaOH will completely neutralize acetic acid. In the reversible reaction A (aq) + B(aq) E > C(aq), reactant A is very expensive. What are two ways to get it to react as fully as possible to form as much C as possible? a) remove C as it forms, and use an excess of A b) add another reactant that will form a precipitate with B c) remove C as it forms, and use an excess of B d) add another reactant that will form a precipitate with Aarrow_forward
- CH4arrow_forwardCalculate the mass of PCl5, that must be heated in a 1.0L flask at 250 oC in order to produce enough chlorine to give an equilibrium concentration,[Cl2]eq. of 0.20 mol/L? PCl5(g) <——> PCl3(g) + Cl2(g) Kc = 0.480arrow_forward1. Which of the following solutions are acidic? Which are basic? Distilled H2O NaCl NH4Cl NaCH3COO (NH4)2CO3arrow_forward
- M11.arrow_forwardConsider the reaction below. Co(H,O),* (aq) + 4 Cl¯ (aq) = COC1," (aq) + 6 H,O (1I) The equilibrium constant expression (K) of the reaction does not include the concentration of H,O even though the reaction produces six moles of water. Explain.arrow_forwardWhen only one concentration at equilibrium is known the concentrations of the others can be calculated using stoichiometry. This is possible because all relationships in a chemical reaction are direct proportions. Calculate the the missing concentrations, then calculate the K, when a 0.050 M solution of CH3NH₂ is made, and at equilibrium [CH3NH₂] = 0.040 M. CH3NH2 + H₂O → CH³NH₂+¹ + OH-¹ 0 0 start: 0.050 M equil: 0.045 M 0.005 M The amount of CH3NH₂ that was converted into product was 0.050 M -0.045 M = For each 1 mol CH3NH3+¹ (0.005 M CH3NH₂ X- Kea 1 mol CH3NH₂ (0.005 M CH3NH2)(- 1 mol OH-1 1 mol CH3NH₂ [CH3NH3 +¹][OH-¹] [ of CH3NH₂ converted into products, there will be 0.005 M of CH3NH3*¹ and -) = ][ CH3NH3+1 OH-1 [CH3NH₂] [ ] (notice that the information to calculate both the pOH and pk, are provided) A. HF B. H₂O C. F-1 D. H30+1 J. 0.005 M K. 0.040 M L. 0.045 M E. CH3NH₂ F. CH3NH₂+¹ M. 0.00056 M N. 0.0056 M G. OH-1 H. 0.030 M I. 0.050 M O. 0.11 M P. 0.50 M of OH-1, orarrow_forward
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