Answer:
The percent composition of fluorine is 65.67%
Explanation:
Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.
That is, the percentage composition is the percentage by mass of each of the elements present in a compound.
The calculation of the percentage composition of an element is made by:

In this case, the percent composition of fluorine is:

percent composition of fluorine= 65.67%
<u><em>The percent composition of fluorine is 65.67%</em></u>
Answer:
Part a: <em>Units of k is </em>
<em> where reaction is first order in A and second order in B</em>
Part b: <em>Units of k is </em>
<em> where reaction is first order in A and second order overall.</em>
Part c: <em>Units of k is </em>
<em> where reaction is independent of the concentration of A and second order overall.</em>
Part d: <em>Units of k is </em>
<em> where reaction reaction is second order in both A and B.</em>
Explanation:
As the reaction is given as

where as the rate is given as
![r=k[A]^x[B]^y](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5Ex%5BB%5D%5Ey)
where x is the order wrt A and y is the order wrt B.
Part a:
x=1 and y=2 now the reaction rate equation is given as
![r=k[A]^1[B]^2](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E1%5BB%5D%5E2)
Now the units are given as
![r=k[A]^1[B]^2\\M/s =k[M]^1[M]^2\\M/s =k[M]^{1+2}\\M/s =k[M]^{3}\\M^{1-3}/s =k\\M^{-2}s^{-1} =k](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E1%5BB%5D%5E2%5C%5CM%2Fs%20%3Dk%5BM%5D%5E1%5BM%5D%5E2%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B1%2B2%7D%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B3%7D%5C%5CM%5E%7B1-3%7D%2Fs%20%3Dk%5C%5CM%5E%7B-2%7Ds%5E%7B-1%7D%20%3Dk)
The units of k is 
Part b:
x=1 and o=2
x+y=o
1+y=2
y=2-1
y=1
Now the reaction rate equation is given as
![r=k[A]^1[B]^1](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E1%5BB%5D%5E1)
Now the units are given as
![r=k[A]^1[B]^1\\M/s =k[M]^1[M]^1\\M/s =k[M]^{1+1}\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E1%5BB%5D%5E1%5C%5CM%2Fs%20%3Dk%5BM%5D%5E1%5BM%5D%5E1%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B1%2B1%7D%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B2%7D%5C%5CM%5E%7B1-2%7D%2Fs%20%3Dk%5C%5CM%5E%7B-1%7Ds%5E%7B-1%7D%20%3Dk)
The units of k is 
Part c:
x=0 and o=2
x+y=o
0+y=2
y=2
y=2
Now the reaction rate equation is given as
![r=k[A]^0[B]^2](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E0%5BB%5D%5E2)
Now the units are given as
![r=k[B]^2\\M/s =k[M]^2\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k](https://tex.z-dn.net/?f=r%3Dk%5BB%5D%5E2%5C%5CM%2Fs%20%3Dk%5BM%5D%5E2%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B2%7D%5C%5CM%5E%7B1-2%7D%2Fs%20%3Dk%5C%5CM%5E%7B-1%7Ds%5E%7B-1%7D%20%3Dk)
The units of k is 
Part d:
x=2 and y=2
Now the reaction rate equation is given as
![r=k[A]^2[B]^2](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E2%5BB%5D%5E2)
Now the units are given as
![r=k[A]^2[B]^2\\M/s =k[M]^2[M]^2\\M/s =k[M]^{2+2}\\M/s =k[M]^{4}\\M^{1-4}/s =k\\M^{-3}s^{-1} =k](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E2%5BB%5D%5E2%5C%5CM%2Fs%20%3Dk%5BM%5D%5E2%5BM%5D%5E2%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B2%2B2%7D%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B4%7D%5C%5CM%5E%7B1-4%7D%2Fs%20%3Dk%5C%5CM%5E%7B-3%7Ds%5E%7B-1%7D%20%3Dk)
The units of k is 
1) START counting for sig. figs. On the FIRST non-zero digit.
2) STOP counting for sig. figs. On the LAST non-zero digit.
3) Non-zero digits are ALWAYS significant.
4) Zeroes in between two non-zero digits are significant. All other zeroes are insignificant.
I this is really hard who can help me complete all of these because this is acknowledged
Answer: 100kPa
Explanation:
P1 = 3.00 x 10² kPa , P2 =?
T1 = 30°C = 30 +273 = 303k
T2 = —172°C = —172 + 273 = 101k
P1/T1 = P2/T2
3 x 10² / 303 = P2 / 101
P2 = (3 x 10² / 303) x 101
P2 = 100kPa