Answer:
Explanation:
Hello,
In this case, the dissociation reaction is:
For which the equilibrium expression is:
![Ksp=[Pb^{2+}][I^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5BI%5E-%5D%5E2)
Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)
In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:
![[Pb^{2+}]=1.39x10^{-3}M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3D1.39x10%5E%7B-3%7DM)
![[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M](https://tex.z-dn.net/?f=%5BI%5E-%5D%3D1.39x10%5E%7B-3%7DM%2A2%3D2.78x10%5E%7B-3%7DM)
Thereby, the solubility product results:
Regards.
I think it may be “past climate of Earth seen in glaciers”/ I learned about this and that one doesn’t really ring a bell, plus it doesn’t make sense to me. I four sure know it’s not the fitting like a puzzle one. Sorry if I get this wrong.
The mass in grams of NH₃ produced from the reaction is 3.4 g
<h3>Balanced equation</h3>
We'll begin by writing the balanced equation for the reaction. This illustrated below:
N₂ + 3H₂ -> 2NH₃
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
<h3>How to determine the volume of NH₃ produced</h3>
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
Therefore,
2.24 dm³ of N₂ will react to produce = 2.24 × 2 = 4.48 dm³ of NH₃
<h3>How to determine the mass of NH₃ produced</h3>
We'll begin by obtained the mole of 4.48 dm³ of NH₃. Details below:
22.4 dm³ = 1 mole NH₃
Therefore,
4.48 dm³ = 4.48 / 22.4
4.48 dm³ = 0.2 mole of NH₃
Finally, we shall determine the mass of NH₃ as follow:
- Molar mass of NH₃ = 17 g/mol
- Mole of NH₃ = 0.2 mole
- Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 0.2 × 17
Mass of NH₃ = 3.4 g
Learn more about stoichiometry:
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