Answer:
The molar solubility of lead bromide at 298K is 0.010 mol/L.
Explanation:
In order to solve this problem, we need to use the Nernst Equaiton:
![E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}](https://tex.z-dn.net/?f=E%20%3D%20E%5E%7Bo%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D)
E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.
At equilibrium, E = 0, therefore:
![E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\](https://tex.z-dn.net/?f=E%5E%7Bo%7D%20%20%3D%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D%20%5C%5C%5C%5Clog%20%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D%20%3D%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%20%5C%5C%5C%5Clog%5Bred%5D%20%3D%20%20log%5Box%5D%20-%20%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%5C%5C%5C%5C%5Bred%5D%20%3D%2010%5E%7B%20log%5Box%5D%20-%20%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%7D%20%5C%5C%5C%5C%5Bred%5D%20%3D%2010%5E%7B%20log0.733%20-%20%20%5Cfrac%7B2x5.45x10%5E%7B-2%7D%20%20%7D%7B0.0591%7D%7D%5C%5C%5C%5C)
[red] = 0.010 M
The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.
I believe the answer is carbon atoms
12 moles of water H₂O are produced from the combustion of pentane.
Explanation:
We have the following combustion of pentane (C₅H₁₂):
C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O
Knowing the chemical reaction we devise the following reasoning:
if 1 moles of pentane C₅H₁₂ produces 6 moles of water H₂O
then 2 moles of pentane C₅H₁₂ produces X moles of water H₂O
X = (2 × 6) / 1 = 12 moles of water H₂O
Learn more about:
combustion of organic compounds
brainly.com/question/7295137
brainly.com/question/884053
#learnwithBrainly
Every organic molecules/compound contains carbon (c).
Some other very abundant are hydrogen, nitrogen, oxygen, phosphorus, and sulfur.
I learned this with the acronym CHNOPS.
C - Carbon
H - Hydrogen
N - Nitrogen
O - Oxygen
P - Phosphorus
S - Sulfur
Hope this helps!
Answer:
The answer to your question is 41.6 g of AgCl
Explanation:
Data
mass of NH₄Cl = 15.5 g
mass of AgNO₃ = excess
mass of AgCl = 35.5 g
theoretical yield = ?
Process
1.- Write the balanced chemical reaction.
NH₄Cl + AgNO₃ ⇒ AgCl + NH₄NO₃
2.- Calculate the molar mass of NH₄Cl and AgCl
NH₄Cl = 14 + 4 + 35.5 = 53.5 g
AgCl = 108 + 35.5 = 143.5 g
3.- Calculate the theoretical yield
53.5 g of NH₄Cl -------------------- 143.5 g of AgCl
15.5 g of NH₄Cl ------------------- x
x = (15.5 x 143.5) / 53.5
x = 2224.25 / 53.5
x = 41.6 g of AgCl
