Chemistry, 10.11.2020 23:40, moomoofower

Can someone please tell me the answer

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Answers

answered: bryneosburn

52.068 amu

Explanation:

To calculate the average atomic mass of chromium, you have to first multiply the abundance of an isotope with its mass.

49.946 × 0.0435 = 2.173

51.941 × 0.838 = 43.527

52.941 × 0.0235 = 1.244

53.939 × 0.095 = 5.124

Next, add up all of the values, and you will get the average atomic mass.

2.173 + 45.527 + 1.244 + 5.124 = 52.068 amu

answered: Guest

the grams of barium that are in 2.3 x 10^23 atoms of barium is 52.4 g (answer c)

calculation

mass = moles of ba x molar mass of ba

molar mass of ba from periodic table= 137.3 g/mol

use the avogadro's law constant to calculate the moles of barium

that is 1 mole = 6.02 x 10^23 atoms

? moles = 2.3 x 10^23 atoms

by cross multiplication

=(1 mole x 2.3x 10^23 atoms) / 6.02 x10^23 atoms)= 0.382 moles

mass is therefore = 137.3 g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

answered: Guest

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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