Is there an equation? I can't help if there's no equation involved.
3.67 grams is the volume of the solution
Answer:
![[base]=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D0.28M)
Explanation:
Hello,
In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:
![pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7BpH-pKa%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7B4.9-4.76%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D1.38%5C%5C%5C%5C)
![[base]=1.38[acid]=1.38*0.20M=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D1.38%5Bacid%5D%3D1.38%2A0.20M%3D0.28M)
Regards.
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:
cesium
In particular, cesium (Cs) can give up its valence electron more easily than can lithium (Li). In fact, for the alkali metals (the elements in Group 1), the ease of giving up an electron varies as follows: Cs > Rb > K > Na > Li with Cs the most likely, and Li the least likely, to lose an electron
Explanation: