Chemistry, 25.09.2019 01:30 jandrew2052

The lowa regional utilities association (ilrua) is contracted to receive up to 3.0 million gallons per day of water from the newton and marshalltown water works, and is able to treat an additional 3.0 million gallons per day at the waverly treatment plant. the treated water from all three sources has an average temperature of 55 °f (12.8 °c), and is distributed to 27 water towers by 31 pumping stations. the irua distribution system consists of over 4,656 miles of pipeline ranging in diameter from 2 in. to 20 in.3 a. calculate the volumetric flow rate [gal/min] of water to each pumping station, assuming that the water is distributed equally amongst the 31 pumping stations. b. density (p) is the reciprocal of specific volume (v). use table b.5 from elementary principles of chemical processes to calculate the density [lbm/ft] of the distributed water. c. use the provided nomograph to determine the mass flow rate [lbm/h] of water to each pumping station. d. use the provided nomograph to determine the range of the velocity [ft/s] of water to each pumping station for the provided range of pipeline diameters, and compare this range to the stated range of reasonable velocities for water flow through city pipe. e. the 27 water towers have a range of capacities from 150,000 gallons to 2,000,000 gallons.3 based upon this information comment on the validity of the assumption that the water is distributed equally to the 31 pumping stations, and discuss how this may impact the velocities determined in part d.

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Iread all the questions and answers, that person done a lot of work so go ahead and give them brainliest!

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c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.

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Sorry i dont understand the question can you rephase it?

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remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.

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