Answer: 5 is the molarity
Explanation:
The molarity formula is moles over liters and that in your case is 2.50 moles divided by .500 L which results in 5 which is your answear hope this helped god bless
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
Answer:
Following are the solution to these question:
Explanation:
Calculating the mean:
![\bar{x}=\frac{175+104+164+193+131+189+155+133+151+176}{10}\\\\](https://tex.z-dn.net/?f=%5Cbar%7Bx%7D%3D%5Cfrac%7B175%2B104%2B164%2B193%2B131%2B189%2B155%2B133%2B151%2B176%7D%7B10%7D%5C%5C%5C%5C)
![=\frac{1571}{10}\\\\=157.1](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1571%7D%7B10%7D%5C%5C%5C%5C%3D157.1)
Calculating the standardn:
![\sigma=\sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}}\\\\](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7B%5Cfrac%7B%5CSigma%28x_i-%5Cbar%7Bx%7D%29%5E2%7D%7Bn-1%7D%7D%5C%5C%5C%5C)
Please find the correct equation in the attached file.
![=28.195](https://tex.z-dn.net/?f=%3D28.195)
For point a:
![=3s+yblank \\\\=3 \times 28.195+50\\\\=84.585+50\\\\=134.585\\](https://tex.z-dn.net/?f=%3D3s%2Byblank%20%5C%5C%5C%5C%3D3%20%5Ctimes%2028.195%2B50%5C%5C%5C%5C%3D84.585%2B50%5C%5C%5C%5C%3D134.585%5C%5C)
For point b:
![=3 \ \frac{s}{m}\\\\ = \frac{(3 \times 28.195)}{1.75 \times 10^9 \ M^{-1}}\\\\= 4.833 \times 10^{-8} \ M](https://tex.z-dn.net/?f=%3D3%20%5C%20%5Cfrac%7Bs%7D%7Bm%7D%5C%5C%5C%5C%20%3D%20%5Cfrac%7B%283%20%5Ctimes%2028.195%29%7D%7B1.75%20%5Ctimes%2010%5E9%20%5C%20M%5E%7B-1%7D%7D%5C%5C%5C%5C%3D%204.833%20%5Ctimes%2010%5E%7B-8%7D%20%5C%20M)
For point c:
![= 10 \frac{s}{m} \\\\= \frac{(10 \times 28.195)}{1.75 x 10^9 \ M^{-1}}\\\\ = 1.611 \times 10^{-7}\ M](https://tex.z-dn.net/?f=%3D%2010%20%5Cfrac%7Bs%7D%7Bm%7D%20%5C%5C%5C%5C%3D%20%5Cfrac%7B%2810%20%5Ctimes%2028.195%29%7D%7B1.75%20x%2010%5E9%20%5C%20M%5E%7B-1%7D%7D%5C%5C%5C%5C%20%3D%201.611%20%5Ctimes%2010%5E%7B-7%7D%5C%20%20M)
It is calculated by using the slope value that is
. The slope value
is ambiguous.