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 
Answer:
D
Explanation:
Because if I said its 35 degrees outside that has nothing to do with climate maybe if I said its usually 35 degrees in the month of December but just saying what temperature it doesn't tell you anything about climate
Well, the definition of climate is the weather conditions prevailing in an area in general or over a long period so I don't really know
So maybe it is d
Answer:
pOH = 5.961
Explanation:
To find the pH of a weak base we can use Henderson-Hasselbalch equation for weak bases:
pOH = pKb + log [(CH₃)₃NHCl] / [(CH₃)₃N]
<em>Where pKb is -log Kb = 4.187 and [] could be taken as moles of each specie.</em>
<em />
<em>Moles (CH₃)₃NHCl:</em>
0.0441L * (0.15mol/L) = 6.615x10⁻³moles
<em>Moles (CH₃)₃N:</em>
0.0233L * (0.16mol/L) = 3.728x10⁻³moles
And pOH is:
pOH = pKb + log [(CH₃)₃NHCl] / [(CH₃)₃N]
pOH = 4.187 + log [6.615x10⁻³moles] / [3.728x10⁻³moles]
<h3>pOH = 5.961</h3>
<em />
Answer:
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Explanation:
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