Answer: Parts per million (ppm)
Explanation:
Consider the units milligram per milliliter. This gives us one part of the solute per one million parts of solvent. That is 10^ -3/10^-3= 10^-6. This unit is commonly used in analytical chemistry to show very small concentration of analyte. A similar unit is parts per billion(ppb)
Answer:
Explanation:
K₂CrO₄ + ( COONa )₂ + 2BaCl₂ = Ba CrO₄ + ( COO ) ₂ Ba + 2 KCl + 2 NaCl
.033 M .053 M
Ksp of Ba CrO₄ is 2.10×10⁻¹⁰
Ksp of ( COO ) ₂ Ba is 1.30×10⁻⁶
A ) Ksp of Ba CrO₄ is less so it will precipitate out first .
B) Ksp = 2.10×10⁻¹⁰
Ba CrO₄ = Ba⁺² + CrO₄⁻²
C .033
C x .033 = 2.10×10⁻¹⁰
C = 63.63 x 10⁻¹⁰ M
Ba⁺² must be present in concentration = 63.63 x 10⁻¹⁰ M
C)
90% of precipitation of barium oxalate
concentration of oxalate to precipitate out = .9 x .0532 = .04788
( COO ) ₂ Ba = (COO)₂⁻² + Ba⁺²
.04788 M C
C x .04788 = 1.30×10⁻⁶
C = 27.15 x 10⁻⁶ M .
Explanation:
The given reaction at cathode will be as follows.
At cathode: 
, 
= -0.761 V
At anode: 
, = 0.761
Therefore, net reaction equation will be as follows.
Initial: 0.129 - - 0.427
Change: -0.047 - - -0.047
Equilibrium: (0.129 - 0.047) (0.427 - 0.047)
= 0.082 = 0.38
As 
for the given reaction is zero.
Hence, equation for calculating new cell potential will be as follows.
E_{cell} = ![E^{o}_{cell} - \frac{RT}{nF} ln \frac{[Zn^{2+}]_{products}}{[Zn^{2+}]_{reactants}}](https://tex.z-dn.net/?f=E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7BRT%7D%7BnF%7D%20ln%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D_%7Bproducts%7D%7D%7B%5BZn%5E%7B2%2B%7D%5D_%7Breactants%7D%7D)
= 
= 0.019
Thus, we can conclude that the cell potential of the given cell is 0.019.
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