Living cells convert energy derived from sunlight or combustion of

Living cells convert energy derived from sunlight or combustion of food into energy-rich ATP (adenosine triphosphate) molecules. For ATP synthesis, ΔG = +34.5 kJ/mol. This energy is then made available to the cell when ATP is hydrolyzed to ADP (adenosine diphosphate). In animals, ATP is synthesized when protons pass through a complex enzyme in the mitochondrial membrane.17 Two factors account for the movement of protons through this enzyme into the mitochondrion (see the figure): (1) [H+] is higher outside the mitochondrion than inside because protons are pumped out of the mitochondrion by enzymes that catalyze the oxidation of food. (2) The inside of the mitochondrion is negatively charged with respect to the outside.

(a) The synthesis of one ATP molecule requires 2H+ to pass through the phosphorylation enzyme. The difference in free energy when a molecule travels from a region of high activity to a region of low activity is

How big must the pH difference be (at 298 K) if the passage of two protons is to provide enough energy to synthesize one ATP molecule?

(b) pH differences this large have not been observed in mitochondria. How great an electric potential difference between inside and outside is necessary for the movement of two protons to provide energy to synthesize ATP? In answering this question, neglect any contribution from the pH difference.

H2S from cigarette smoke was collected by bubbling smoke through

H2S from cigarette smoke was collected by bubbling smoke through aqueous NaOH and measured with a sulfide ion-selective electrode. Standard additions of volume VS containing Na2S at concentration +cS = 1.78 mM were then made to V0 = 25.0 mL of unknown and the electrode response, E, was measured.

From a separate calibration curve, it was found that β = 0.985 in Equation 14-14. Using T =298.15 K and n = -2 (the charge of S2-), prepare a standard addition graph with Equation 14-15 and find the concentration of sulfide in the unknown.

An ammonia gas-sensing electrode gave the following calibration points when

An ammonia gas-sensing electrode gave the following calibration points when all solutions contained 1 M NaOH.

A dry food sample weighing 312.4 mg was digested by the Kjeldahl procedure (Section 10-8) to convert all nitrogen into NH4+. The digestion solution was diluted to 1.00 L, and 20.0 mL were transferred to a 100-mL volumetric flask. The 20.0-mL aliquot was treated with 10.0 mL of 10.0 M NaOH plus enough NaI to complex the Hg catalyst from the digestion and diluted to 100.0 mL. When measured with the ammonia electrode, this solution gave a reading of 339.3 mV. Calculate the wt% nitrogen in the food sample.

The apparatus in the figure can follow the course of

The apparatus in the figure can follow the course of an EDTA titration and was used to generate the curves in Figure 11-10. The heart of the cell is a pool of liquid Hg in contact with the solution and with a Pt wire. A small amount of HgY2- added to the analyte equilibrates with a very tiny amount of Hg2+:

The redox equilibrium   is established rapidly at the surface of the Hg electrode, so the Nernst equation for the cell can be written in the form

where E- is the constant potential of the reference electrode. From Equation A, [Hg2+] = [HgY2-] / Kf[Y4-], and this can be substituted into Equation B to give

where Kf is the formation constant for HgY2-. This apparatus thus responds to the changing EDTA concentration during an EDTA titration.

Suppose that you titrate 50.0 mL of 0.010 0 M MgSO4 with 0.020 0 M EDTA at pH 10.0, using the apparatus in the figure with an S.C.E. reference electrode. Analyte contains 1.0 × 10-4 M Hg(EDTA)2- added at the beginning of the titration. Calculate the cell voltage at the following volumes of added EDTA, and draw a graph of millivolts versus milliliters: 0, 10.0, 20.0, 24.9, 25.0, and 26.0 mL.

The basal rate of consumption of O2 by a 70-kg

The basal rate of consumption of O2 by a 70-kg human is about 16 mol of O2 per day. This O2 oxidizes food and is reduced to H2O, providing energy for the organism:

(a) To what current (in amperes = C/s) does this respiration rate correspond? (Current is defined by the flow of electrons from food to O2.)

(b) Compare your answer in part (a) with the current drawn by a refrigerator using 5.00 × 102 W at 115 V. Remember that power (in watts) = work/s E ∙ I.

(c) If the electrons flow from nicotinamide adenine dinucleotide (NADH) to O2, they experience a potential drop of 1.1 V. What is the power output (in watts) of our human friend?

Chlorine has been used for decades to disinfect drinking water.

Chlorine has been used for decades to disinfect drinking water. An undesirable side effect of this treatment is reaction with organic impurities to create organochlorine compounds, some of which could be toxic. Monitoring total organic halide (designated TOX) is required for many water providers. A standard procedure for TOX is to pass water through activated charcoal, which adsorbs organic compounds. Then the charcoal is combusted to liberate hydrogen halides:

HX is absorbed into aqueous solution and measured by coulometric titration with a silver anode:

When 1.00 L of drinking water was analyzed, a current of 4.23 mA was required for 387 s. A blank prepared by oxidizing charcoal required 6 s at 4.23 mA. Express the TOX of the drinking water as µmol halogen/L. If all halogen is chlorine, express the TOX as µg Cl/L.

Cd2+ was used as an internal standard in the analysis

Cd2+ was used as an internal standard in the analysis of Pb2+by square wave polarography. Cd2+ gives a reduction wave at -0.60 (±0.02) V and Pb2+ gives a reduction wave at -0.40 (± 0.02) V. It was first verified that the ratio of peak heights is proportional to the ratio of concentrations over the whole range employed in the experiment. Here are results for known and unknown mixtures.

The unknown mixture was prepared by mixing 25.00 (±0.05) mL of unknown (containing only Pb2+) plus 10.00 (±0.05) mL of 3.23 (0.01) × 10-4 M Cd2+ and diluting to 50.00 (±0.05) mL.

(a) Disregarding uncertainties, find [Pb2+] in the undiluted unknown.

(b) Find the absolute uncertainty for the answer to part (a).

Ions that react with Ag+ can be determined electrogravimetrically by

Ions that react with Ag+ can be determined electrogravimetrically by deposition on a silver working anode:

(a) What will be the final mass of a silver anode used to electrolyze 75.00 mL of 0.023 80 M KSCN if the initial mass of the anode is 12.463 8 g?

(b) At what electrolysis voltage (versus S.C.E.) will AgBr(s) be deposited from 0.10 M Br? (Consider negligible current flow, so that there is no ohmic potential, concentration polarization, or overpotential.)

(c) Is it theoretically possible to separate 99.99% of 0.10 M KI from 0.10 M KBr by controlled potential electrolysis?

Refer to the Fourier transform infrared spectrum in Figure 19-32.(a)

Refer to the Fourier transform infrared spectrum in Figure 19-32.

(a) The interferogram was sampled at retardation intervals of 1.2660 × 10-4 cm. What is the theoretical wavenumber range (0 to ?) of the spectrum?

(b) A total of 4 096 data points were collected from δ = -Δ to δ = +Δ. Compute the value of Δ, the maximum retardation.

(c) Calculate the approximate resolution of the spectrum.

(d) The interferometer mirror velocity is given in the figure caption. How many microseconds elapse between each datum?

(e) How many seconds were required to record each interferogram once?

(f) What kind of beamsplitter is typically used for the region 400 to 4 000 cm-1? Why is the region below 400 cm-1 not observed?

Figure 19-32

(a) If a diffraction grating has a resolution of 104,

(a) If a diffraction grating has a resolution of 104, is it possible to distinguish two spectral lines with wavelengths of 10.00 and 10.01 µm?

(b) With a resolution of 104, how close in wavenumbers (cm-1) is the closest line to 1 000 cm-1 that can barely be resolved?

(c) Calculate the resolution of a 5.0-cm-long grating ruled at 250 lines/mm for first-order (n = 1) diffraction and tenth-order (n = 10) diffraction.

 (d) Find the angular dispersion (Δϕ in radians and degrees) between light rays with wavenumbers of 1 000 and 1 001 cm-1 for second-order diffraction (n = 2) from a grating with 250 lines/mm and ϕ = 30°.