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![Ky z lokOn loxioly oxioly Soitto = 1.786*105 pkb =-109 kb -10g (1-786x103) 34-75 E K0 H3 = LOH33 770.003 50.603M Lw #] = bois](http://img.homeworklib.com/questions/e74df760-33de-11eb-b82b-e9a0d8c8c37c.png?x-oss-process=image/resize,w_560)
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15. Suppose 0.003 moles of KOH are added to a 1.0 L solution...
A) How many milliliters of 0.15 M HCl would have to be added to 1.0 L...
A) How many milliliters of 0.15 M HCl would have to be added to 1.0 L of a buffer containing 0.25 M NH3 and 0.45 M NH4 to make the pH decrease by 0.05 pH unit? The Ka value = 5.6 x 10^-10. B) How many milliliters of the same HCl solution would, if added to 0.204 L of pure water, make the pH decrease by 0.05 pH unit?
Which of the following cannot be a buffer solution? Group of answer choices 1.0-L solution, in...
Which of the following cannot be a buffer solution? Group of answer choices 1.0-L solution, in which 0.10 mol acetic acid (CH3COOH) and 0.050 mol potassium hydroxide (KOH) are dissolved. 0.50-L solution, in which 0.25 mol ammonia (NH3) and 0.18 mol hydrochloric acid (HCl) are dissolved. 0.75-L solution, in which 1.0 mol hydrofluoric acid (HF) and 2.0 mol sodium hydroxide (NaOH) are dissolved. 1.0-L solution, in which 3.0 mol lithium hydroxide (LiOH) and 5.0 mol formic acid (HCOOH) are dissolved....
A 1.0 L solution initially contains 0.3 M CH3COOH. Approximate the number of moles of CH3COOH...
A 1.0 L solution initially contains 0.3 M CH3COOH. Approximate the number of moles of CH3COOH and CH3COO− in the solution after 0.1 mol of solid NaOH is added, and calculate the pH of the resulting solution. Ka of CH3COOH is 1.8 ×10−5. 1) The number of moles of CH3COOH present is [ Select ] ["0.2", "0.3", "0.1"] mol. 2) The number of...
What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00? Ka =...
What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00? Ka = 1.8×10-5
1. Calculate the equilibrium concentrations of all species present and the pH of a solution obtained...
1. Calculate the equilibrium concentrations of all species present and the pH of a solution obtained by adding 0.100 moles of solid NaOH to 1.00 L of 15.0 M NH3. Kb = 1.8 × 10–5 2. One mole of a weak acid HA was dissolved in 2.0 L of water. After the system had come to equilibrium, the concentration of HA was found to be 0.45 M. Calculate the Ka for this weak acid. 3. Calculate the pH of a...
28. A 0.50 L buffer solution is 0.20 M in HF and 0.40 M in NaF. The Ka for HF is 1.5 x 10*. Show all work for full...
28. A 0.50 L buffer solution is 0.20 M in HF and 0.40 M in NaF. The Ka for HF is 1.5 x 10*. Show all work for full credit. (a) Calculate the pH of the solution after the addition of 0.025 moles of solid KOH. (3 pts) (b) Calculate the pH of the solution after the addition of 0.050 moles of HCI. (3 pts)
Chem II Common Assignment: Equilibrium and Buffers 1. Calculate the equilibrium concentrations of all species present...
Chem II Common Assignment: Equilibrium and Buffers 1. Calculate the equilibrium concentrations of all species present and the pH of a solution obtained by adding 0.100 moles of solid NaOH to 1.00 L of 15.0 M NH3. Kb = 1.8 × 10–5 2. One mole of a weak acid HA was dissolved in 2.0 L of water. After the system had come to equilibrium, the concentration of HA was found to be 0.45 M. Calculate the Ka for this weak...
Question 10 0/5 points A buffer solution is made such that the initial concentrations of lactic...
Question 10 0/5 points A buffer solution is made such that the initial concentrations of lactic acid (HC3H5O3) and the lactate ion (C3H503) are both 0.600 M. What is the resulting pH if 0.020 mol of hydrochloric acid is added to 0.500 L of the buffer solution? (The Ky of HC3H5O3 is 1.4 x 10" 4.) HINT: Use your RICE chart! [A-] Note: pH = pka + log [HA] Ka x Ky = Kw = 1.0 x 10-14 2.92 3.80...
Show all work neatly please l mol Cu was added to a solution containing 4 moles...
Show all work neatly please
l mol Cu was added to a solution containing 4 moles of HNO_3 which react according to the following Chemical equation 3Cu_(s) + 8HNO_3(aq) rightarrow 3Cu(NO_3)_2(aq) + 4H_2O_(l) + 2NO_(g) What is the limiting reactant (LR) and how many moles of No could be produced? LR = Cu, 1.5 moles of No. LR = 4HO_3, 2 moles of No. LR = C4, 213 moles of No. LR = HNO_3, 1/2 moles of No. LR =...
A solution is prepared by dissolving 0.23 mol of formic acid and 0.27 mol of sodium...
A solution is prepared by dissolving 0.23 mol of formic acid and 0.27 mol of sodium formate in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of NaOH to this buffer solution causes the pH to increase slightly. The pH does not increase drastically because the NaOH reacts with the present in the buffer solution. The K, of formic acid is 1.8 x 10+ A formic acid B. sodium formate C water D. sodium E....
28. A 0.50 L buffer solution is 0.20 M in HF and 0.40 M in NaF. The Ka for HF is 1.5 x 10*. Show all work for full credit. (a) Calculate the pH of the solution after the addition of 0.025 moles of solid KOH. (3 pts) (b) Calculate the pH of the solution after the addition of 0.050 moles of HCI. (3 pts)
Question 10 0/5 points A buffer solution is made such that the initial concentrations of lactic acid (HC3H5O3) and the lactate ion (C3H503) are both 0.600 M. What is the resulting pH if 0.020 mol of hydrochloric acid is added to 0.500 L of the buffer solution? (The Ky of HC3H5O3 is 1.4 x 10" 4.) HINT: Use your RICE chart! [A-] Note: pH = pka + log [HA] Ka x Ky = Kw = 1.0 x 10-14 2.92 3.80...
Show all work neatly please
l mol Cu was added to a solution containing 4 moles of HNO_3 which react according to the following Chemical equation 3Cu_(s) + 8HNO_3(aq) rightarrow 3Cu(NO_3)_2(aq) + 4H_2O_(l) + 2NO_(g) What is the limiting reactant (LR) and how many moles of No could be produced? LR = Cu, 1.5 moles of No. LR = 4HO_3, 2 moles of No. LR = C4, 213 moles of No. LR = HNO_3, 1/2 moles of No. LR =...
A solution is prepared by dissolving 0.23 mol of formic acid and 0.27 mol of sodium formate in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of NaOH to this buffer solution causes the pH to increase slightly. The pH does not increase drastically because the NaOH reacts with the present in the buffer solution. The K, of formic acid is 1.8 x 10+ A formic acid B. sodium formate C water D. sodium E....
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