The chemical name begins with Hydro The Chemical name ends with hydroxide and the ide ending is changed to ic also the polyatomic ate ending changed to ic
1 g/mL ------------ 8.3454 lbs/gal
1.09 g/mL --------- ?
lbs/gal = 1.09 x 8.3454 / 1
ibs/gal = 9.096 / 1
= <span>9.096 lbs/gal</span>
The width of the cuvette is important in the spectrophotometric analysis since wider cuvette<span> translates to more absorbing species present in the path where light passes, hence more absorbance is read in the analysis. other factors include the concentration of the sample and the species itself present.</span>
What's the answer. tell me please please please please please please please please please please please please
<u>Answer:</u> The cell potential of the cell is +0.118 V
<u>Explanation:</u>
The half reactions for the cell is:
<u>Oxidation half reaction (anode):</u> 
<u>Reduction half reaction (cathode):</u> 
In this case, the cathode and anode both are same. So,
will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BNi%5E%7B2%2B%7D_%7Bdiluted%7D%5D%7D%7B%5BNi%5E%7B2%2B%7D_%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= ?
= 
= 1.0 M
Putting values in above equation, we get:


Hence, the cell potential of the cell is +0.118 V