Balance the equation below for the reaction between MnO4− and C2O42−. 2MnO4− + 16H+ + 5C2O42− → 2Mn2+ + 10CO2 + 8H2O You are correct. Your receipt no. is 155-7521 Help: Receipt Previous Tries Calculate the corrected volumes for KMnO4: Experiment Number #1 #2 #3 #4 Corrrected Volume (mL) 15.35 16.75 12.05 13.35 You are correct. Your receipt no. is 155-2845 Help: Receipt Previous Tries Calculate the moles of KMnO4 added : Experiment Number #1 #2 #3 #4 Moles KMnO4 (moles) 3.07×10-4 3.35×10-4 2.41×10-4 2.67×10-4 You are correct. Your receipt no. is 155-7075 Help: Receipt Previous Tries Calculate the moles of C2O42− in the sample : Experiment Number #1 #2 #3 #4 Moles C2O42− (moles) 7.68e-4 8.38e-4 6.03e-4 6.68e-4 Answer Submitted: Your final submission will be graded after the due date. Tries 1/3 Previous Tries Calculate the mass of C2O42− in the sample : Experiment Number #1 #2 #3 #4 Mass of C2O42− (g) 6.76e-2 7.38e-2 5.31e-2 5.88e-2 Answer Submitted: Your final submission will be graded after the due date. Tries 1/3 Previous Tries Calculate the percent of C2O42− in the sample : Experiment Number #1 #2 #3 #4 Percent C2O42− % 54.5 54.3 34.3 42.0 Answer Submitted: Your final submission will be graded after the due date. Tries 1/3 Previous Tries Now, using the statistical treatment of data, calculate the percent of C2O42− in your sample : 46.28

Iwould you figure it out but you don't have a picture or in your question as to what the physical property you're looking for is

  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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