Waste dairy whey is used to grow Kluyveromyces fragilis yeast in a continuous culture system operated at 30C and 1 atm pressure. Medium containing 4% w/w lactose (C12H22O11) and 0.15% w/w NH3 flows into a specially designed aerated bioreactor at a rate of 200 kg h21. The reactor is compartmentalised to facilitate gravity settling of the yeast; a suspension containing concentrated cells is drawn off continuously from the bottom of the reactor at a rate of 40 kg h21, while an aqueous waste stream containing 0.5 kg cells per 100 kg leaves from the top. All of the lactose provided is utilised by the culture. The biomass yield from lactose is known from preliminary experiments to be 0.25 g g21. The composition of the K. fragilis biomass is determined by elemental analysis to be CH1.63O0.54N0.16 1 7.5% ash. Required:
a. What is the RQ for this culture?
b. It is proposed to supply the reactor with air at a maximum flow rate of 305 litres min21. Will this provide sufficient oxygen? If not, what minimum level of oxygen enrichment of the air is needed if the total gas flow rate is 305 litres min21?
c. What is the concentration of residual NH3 in the aqueous waste stream?
d. What is the concentration of cells in the cell concentrate?

Less stable  hope this

  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.

2licl + 3o2 i yook the test and it was correct.