Constants Part A During an action potential, Na* ions move into the cell at a rate of about 3 x 10 mol/ms How much power must be produced by the "active Nat pumping" system to produce this flow against a +30-mV potential difference? Assumo that the axon is 90 cm long and 20 um in diameter. Express your answer using two significant figures. P = Submit Request Answer

Constants Part A During an action potential, Na* ions move into the cell at a rate of about 3 x 10 mol/ms How much power must be produced by the "active Nat pumping" system to produce this flow against a +30-mV potential difference? Assumo that the axon is 90 cm long and 20 um in diameter. Express your answer using two significant figures. P = Submit Request Answer

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...

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Calculate the absolute value of the voltage across a biological membrane that has 12 mM, all other conditions [Na] outside = 140 mM and [Na+] inside being standard. A. 65.6 mV B. 0.00 mV OC. 300 mV D. 63.1 mV =
Part A Silver can be electroplated at the cathode of an electrolysis cell according to the following half-reaction: Ag*(aq) +e¯ → Ag(s) What mass of silver plates onto the cathode when a current of 6.1 A flows through the cell for 79 min ? Express the mass in grams to two significant figures. ΑΣφ g. — ш Submit Request Answer Provide Feedback
SIMULATION Voltaic Cells Under Standard Conditions Oxidizing Reducing Electrode E° cathode Eº anode = E°cell Agent Agent Potential (V) Au3+(aq) + 3 e Au(s) +1.50 Br2() + 2 e 2 Br (aq) +1.08 Cathode Anode Hg2+(aq) + 2 e Hg(t) +0.855 Ag*(aq) + e Ag(s) +0.80 Hg22+ (aq) + 2 e 2 Hа() +0.789 Cu2+(aq) + 2 e Cu(s) +0.337 Sn4+(aq) + 2 e Sn2+(s) +0.15 2 H30*(aq) + 2 e H2(g) + 2 H20(s) 0.00 Sn2+(aq) + 2 e Sn(s) -0.14 Ni2+(aq) + 2 e - Ni(s) -0.25 Cd2+(aq) + 2 e Cd(s) -0.40 Al3+(aq) + 3 e Al(s) -1.66 Mg2+(aq) + 2 e Mg(s) -2.37 Li*(aq) + e Li(s) -3.045 Click on an Oxidizing Agent and a Reducing Agent. Select Agt as the oxidizing agent. What is the value of E°cathode? Previous Next V. Gibbs Free Energy
10 Mg(s) + Cd(aq) -> Mg (aq) + Cd(s) E = +1.98 V The magnesium electrode was dipped into a 1.00 M salution of M9SO4 and the cadmium ele:trede was dipped into a solution of unknown Cd* concentration. The pctentlal of the cell was measured to be +1.54 V. What was the unknown Cd? concentration? (Answer : 1.4 x 1015 M)
Calculate the time needed to operate an electrolytic cell to plate 0.55 g of Z that is running at 138 A given the following reaction. z3 (aq) + 3e - Z(5) Atomic Mass: Z: 195.25 g/mol Round your answer to 2 decimal places.
日く A apclassroom.collegeboard.org A UPDATES AVAILABLE 29m ago Do you want to install the updates now? CollegeBoard AP'Classroom Unit 9 Progress Check: MCQ 13 14 15 16 23 24 « 0.47 V NO3 K+ Pb. Cu Submit 1.0 M Pb(NO3)2- -1.0 M Cu(NO3)2 The galvanic cell illustrated above was constructed using a salt bridge containing KNO3. A second cell is constructed from identical half-cells but uses NaNO, for the salt bridge. Which of the following best explains whether the initial potential of the second cell will be different from the initial potential of the first cell? A The initial potential of the second cell will be lower than the initial potential of the first cell because Na has a lower molar mass than K has. B The initial potential of the second cell will be higher than the initial potential of the first cell because Na is less electronegative than K is. The initial potential of the second cell will be the same as the first cell because the ions from the salt bridge are not oxidized or…
Determine the mass of Sodium (Na) produced when a 644 amp current flows for 15 hr through a cell containing molten sodium chloride (NaCI). Answer:
(Incorrect) In the electrolysis of aqueous KI, which substance is first produced at the anode? O₂ + 4H+ + 4e → 2 H₂O 1₂ +2e →21 2 H₂O +2 e → H₂ +2 OH K+ + e → K H₂ (Your answer) 12 (Correct answer) 02 K = 1.23 V = 0.53 V = -0.83 V = -2.93 V
From the initial measurements you determine the electrode slope is -56.7 mV. You measure the initial potential of an unknown to be 84.5 mV. After the addition of 0.1027 M fluoride the potential drops to 39.9 mV. What is the molarity of fluoride in the original, undiluted, unknown? Answer is 3.97. How to derive?
When measured with a F− ion-selective electrode with a Nernstian response at 25°C, the potential due to F− in unfluoridated groundwater in a nearby city was 40.7 mV more positive than the potential of tap water in a distant city. The distant city maintains its fluoridated water at the recommended level of 1.00 ± 0.05 mg F−/L. What is the concentration of F− in mg/L in groundwater in the nearby city? (Disregard the uncertainty. Assume that the Nernst potential is 0.05916 V.)
SIMULATION Voltaic Cells Under Standard Conditions Oxidizing Reducing Electrode EO °cathode -E°anode = E0 cell Agent Agent Potential (V) Au3+(aq) + 3e Au(s) +1.50 Br (t) + 2 e 2 Br (aq) +1.08 Cathode Anode Hg2+(aq) + 2 e Hg(t) +0.855 Ag*(aq) + e Ag(s) +0.80 Hg22 (aq) + 2 e 2 Hg(t) +0.789 Cu2*(aq) + 2 e Cu(s) +0.337 Sn (aq) + 2 e Sn2* (s) +0,15 2 H30*(aq) + 2 e H2(g) + 2 H20(s) 0.00 Sn2*(aq) + 2 e Sn(s) -0.14 Ni2*(aq) + 2 e Ni(s) -0.25 Cd2 (aq) + 2 e Cd(s) -0.40 Al3+ (aq) + 3 e Al(s) -1.66 Mg2+(aq) + 2 e Mg(s) -2.37 Li (aq) + e Li(s) -3.045 Click on an Oxidizing Agent and a Reducing Agent. What is the cell potential, E° cell, for the reaction? 2 Ag (aq) + Sn(s) → 2 Ag(s) + Sn²*(aq) 1 1 1 1 Gibbs Free Energy