Consider the element chlorine (CI) which has two isotopes, 3Cl andCl, where the
ID: 3150410 • Letter: C
Question
Consider the element chlorine (CI) which has two isotopes, 3Cl andCl, where the natural abundances Map each isotope are a and b (a + b = 1), respectively. If there are three Cl atoms in a molecule, the probability-or finding each combination of the two isotopes can be determined from the expansion of (a + b Using the information provided in the table, calculate the probability of finding each isotope combination of Cl in CCl3F Element Mass Mass (Da) | Abundance Probability of finding the ms Number (atom %) Number 1.007825 99.988 2.01410 0.012 10.01294 19.9 11.0093180.1 15.999491 99.757 16.99913 0.038 17.99916 0.205 18.99840 100 34.96885 75.78 36.96590 24.22 78.91834 50.69 80.91629 49.31 203.97303 1.4 205.97445 24.1 206.97588 22.1 207.97664 52.4 2 10 Probability of finding two 35Cl atoms and one 7Cl atom 16 Number 19 35 37 79 81 204 206 207 208 CI Probability of finding one 35cl atom and two 37Cl atoms Number Pb Probability of finding three 'Cl atoms NumberExplanation / Answer
Let a succcess be getting a Cl-35 atom. Note that P(Cl-35) = 0.7578, as in the table.
a)
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 3
p = the probability of a success = 0.7578
x = the number of successes = 3
Thus, the probability is
P ( 3 ) = 0.435174865 [ANSWER]
*********************
b)
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 3
p = the probability of a success = 0.7578
x = the number of successes = 2
Thus, the probability is
P ( 2 ) = 0.417257926 [ANSWER]
**********************
c)
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 3
p = the probability of a success = 0.7578
x = the number of successes = 1
Thus, the probability is
P ( 1 ) = 0.133359554 [ANSWER]
*********************
d)
That means 0 Cl35 atoms.
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 3
p = the probability of a success = 0.7578
x = the number of successes = 0
Thus, the probability is
P ( 0 ) = 0.014207655 [ANSWER]
*******************************************
Hi! Please submit the next part as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Thanks!