Ammonia can be determined spectrophotometrically by reaction with phenol in the presence of hypochlorite. A 4.37 mg sample of a protein was chemically digested to
convert all of its nitrogen to ammonia in a one-to-one reaction. After treatment, the total
volume of the sample was 100 mL. Then 10 mL of the solution was placed in a 50 mL
volumetric flask and treated with 7 mL of a phenol and hypochlorite solution. The
sample was allowed to react and then diluted to 50 mL. Its absorbance in a 1 cm cuvette
was measured. For reference, a 1.00 x 10-2 g of NH4Cl was dissolved in 1 L of water.
As with the sample, 10 mL was transferred to 50 mL volumetric flask, treated and
diluted, and analyzed in the same manner as the unknown sample to convert all of its
nitrogen to ammonia in a one-to-one reaction. A blank was prepared using distilled water in place of the unknown. The following data was measured:
SAMPLE ABSORBANCE (AT 625 NM)

BLANK 0.140

REFERENCE 0.308

UNKNOWN 0.592
find:
Correct Absorbance of the Reference. _
Ammonia Complex Concentration in the Reference Cuvette _
e of the ammonia complex. _
Correct Absorbance of the Unknown. _
Ammonia Complex of the Unknown in the Cuvette. _
Concentration of the ammonia the 100 mL sample. _
Mass percent of nitrogen in the protein. _
(hint: assume all mol of ammonia can from nitrogen in the protein)

Ithink this is what you meant. the temperature at which a liquid begins to boil is called the boiling point. the boiling point of water is 212°f or 100°c.  and at that point will start to turn into steam (gas)

  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.