Calculate the number of carbon atoms in 25.0 grams of isopropyl alcohol (c3h8o)

There are 7.53×10²³ atoms of carbon in 25 g of isopropyl alcohol C₃H₈O

We'll begin by calculating the number of mole in 25 g of C₃H₈O. This can be obtained as follow:

Mass of C₃H₈O = 25 g

Molar mass of C₃H₈O = (12×3) + (8×1) + 12

= 36 + 8 + 16 = 60 g/mol

Mole of C₃H₈O =?

Mole = mass / molar mass

Mole of C₃H₈O = 25 / 60

Mole of C₃H₈O = 0.417 mole

Next, we shall determine the number of mole of C in 0.417 mole of C₃H₈O.

1 mole of C₃H₈O contains 3 moles of C.

Therefore, 0.417 mole of C₃H₈O. Will contain = 0.417 × 3 = 1.251 moles of C.

Finally, we shall determine the number of atoms in 1.251 moles of C. this can be obtained as follow:

From Avogadro's hypothesis,

1 mole of C = 6.02×10²³ atoms

Therefore,

1.251 moles of C = 1.251 × 6.02×10²³

1.251 moles of C = 7.53×10²³ atoms

Thus, we can conclude that 25.0 g of isopropyl alcohol, C₃H₈O contains 7.53×10²³ atoms of carbon.

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  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.