A lime-soda softening plant treats a flow of 100 MGD, and the water has a hardness of 230 mg/L as CaCO3 and alkalinity of 185 mg/L as CaCO3, a Mg2 ion content of 42 mg/L, and 4 mg/L of CO2. Assume the system requires the addition of 1.25 meq/L of quicklime to raise the pH to a sufficient level that any precipitation will take place; this quicklime is in addition to any other quicklime needed. The commercial grade of quicklime has a purity of 88%, and the soda ash has a purity of 93%. Determine: The pounds of quicklime and soda ash required per million gallons. The tons of quicklime and soda ash required per month if a month is considered to be 30 days.

  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.

Using dilution equationm1v1 =m2v2 ; where m1 and m1 are the intiial and final concentration, and v1 and v2 are the initial and final volume.5  × 5 m = 8 l  × m2m2 = 25/8 = 3.125 mtherefore; the resulting molarity is 3.125 m

The chlorine is the chemical property