Urgent
fluorine-20 has a half-life of 11 seconds. identify the amount of fluorine-20 left from a 25 gram sample after 55 seconds.

a 0.78125 g
b 78125 mg
c 0.6352 g
d 7.8124 g

A. 0.78 g  

Step-by-step explanation:

The half-life of F-20 (11 s) is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

  No. of                  Fraction           Mass

half-lives     t/s   Remaining   Remaining/g

      0             0          1                      25

      1             11           ½                     12.5

      2           22          ¼                      6.2

      3           33          ⅛                       3.1

      4           44          ¹/₁₆                      1.6

      5           55          ¹/₃₂                   0.78

      6           66          ¹/₆₄                    0.39

We see that 0.78 g of F-20 remain after five half-lives (55 s).

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step-by-step explanation:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.