I'LL GIVE YOU BRAINLIEST ANSWER IF I GET A REAL CORRECT ANSWER When solving for the answer to the problem, the following pattern is used. The missing
information is given letter designations A through F, for labeling purposes.
3.45x1021 fu Cr(ClO3)3
LO2
E
x 7/8 x
mol C
1
mol D
x
F
• You need type in the requested information for the letter sections, please read what is being
asked since it could be a single number, or word or more.
. Please remember to type a space between number and unit of measurement and chemical, for
example - 3 g Knot 3gK or 3g K

  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.

  balanced chemical equations follow the law of conservation of mass and charges. the masses and charges on either sides of the equation should be balanced.in conservation of mass, the number of atoms of each element on left side should be equal to the number of atoms of each element on the right side of the equation.1)k₂so₄ + 2ca(no₃)₂ --> 2kno₃ + 2caso₄number of ions        left side             right side  k⁺         2                         2so₄²⁻   1                         2ca²⁺     2                         2no₃⁻     4                         2in this equation the masses of so₄²⁻ and no₃ are not balanced. therefore this equation violates the law of conservation of mass2)ti + 2 cl₂ > ticl₄        left side       right side  ti           1               1cl           4               4this equation follows the law of conservation of mass3)2c₂h₂ + 5o₂ > 4co₂   + 2h₂o              left side     right side  c               4               4h               4               4o             10             10this equation follows the law of conservation of mass4)baso₄ + 2naoh --> ba(oh)₂   + na₂so₄          left side       right side  ba       1                   1so₄²⁻     1                   1na         2                   2  oh⁻       2                   2this equation follows the law of conservation of massanswer is 1st equation