Answer:
2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓
Ksp = [2s]² . [s] → 4s³
Explanation:
Ag₂CrO₄ → 2Ag⁺ + CrO₄⁻²
Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.
2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓ Ksp
That's the expression for the precipitation equilibrium.
To determine the solubility product expression, we work with the Ksp
Ag₂CrO₄ (s) ⇄ 2Ag⁺ (aq) + CrO₄⁻² (aq) Ksp
2 s s
Look the stoichiometry is 1:2, between the salt and the silver.
Ksp = [2s]² . [s] → 4s³
Answer:
6Fe^2+(aq) -------> 6Fe^3+(aq) + 6e
Explanation:
The balanced oxidation half equation is;
6Fe^2+(aq) -------> 6Fe^3+(aq) + 6e
A redox reaction is actually an acronym for oxidation-reducation reaction. Since the both reactions are complementary, there can't be oxidation without reduction and there can't be reduction without oxidation.
The main characteristic of redox reactions is that electrons are transferred in the process. The number of electrons transferred is usually deduced from the balanced reaction equation. For this reaction, the balanced overall reaction equation is;
Cr2O7^2–(aq) + 6Fe^2+(aq) +14H^+(aq)→ 2Cr^3+(aq) + 6Fe^3+ (aq) + 7H2O(l)
It is clear from the equation above that six electrons were transferred. Thus six Fe^2+ ions lost one electron each in the oxidation half equation as shown in the balanced oxidation half equation above.
Answer:
Here's what I've found
Explanation:
Refer to the attachment !
If it is incomplete or complete combustion. Normally the problems question with tell you.
EX:
"In the presence of plenty of oxygen" is complete combustion
"Not enough oxygen present" is incomplete combustion
Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag
∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5
There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
