i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:
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ii. Given the dissolution of some substance
,
the Ksp, or the solubility product constant, of the preceding equation takes the general form
![K_{sp} = [B]^y [C]^z](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BB%5D%5Ey%20%5BC%5D%5Ez)
.
The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.
So, given our dissociation equation in question i., our Ksp expression would be written as:
![K_{sp} = \mathrm{[Pb^{2+}] [SO_4^{2-}]}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5Cmathrm%7B%5BPb%5E%7B2%2B%7D%5D%20%5BSO_4%5E%7B2-%7D%5D%7D)
.
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iii. Presumably, what we're being asked for here is the <em>molar </em>solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).
We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:
So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.
Answer:
a) Molarity KCl = 0.755 M
b) molality HNO = 5.09 m
Explanation:
- Formality (F) = moles sto / L sln
- Molarity (M) = # dissolved specie / L sln
- molality (m) = moles sto / Kg ste
- %p/p = ( g sto / g ste ) * 100
a) KCl ↔ K+ + Cl-
moles KCL:
⇒ 20 g KCl * ( mol / 74.6 g ) = 0.268 mol KCl
⇒ F = 0.268 mol KCl / 0.355 L = 0.755 F
⇒ M [ K+ ] = 1 * ( 0.755) = 0.755 M
b) 24% HNO:
calculation base: 1 g solution:
⇒24 = ( g sto / g sln) * 100
⇒ 0.24 = g sto / 1
⇒ g sto = 0.24g
⇒g sln = 1 - 0.24 = 076g sln
⇒Kg ste = 0.76 g * ( Kg / 1000g ) = 7.6 E-4 Kg ste
moles sto (HNO):
⇒ 0.24g * ( mol / 62.03g) = 3.869 E-3 moles HNO
⇒ m = 3.869 E-3 moles HNO / 7.6 E-4 Kg sln
⇒ m = 5.09 mol/Kg
The number of mole of S₂O₃ produced from the reaction is 13.33 moles
<h3>Balanced equation </h3>
4Sb + 3O₂ —> 2S₂O₃
From the balanced equation above,
3 moles of O₂ reacted to produce 2 moles of S₂O₃
<h3>How to determine the mole of S₂O₃ produced </h3>
From the balanced equation above,
3 moles of O₂ reacted to produce 2 moles of S₂O₃.
Therefore,
20 moles of O₂ will react to to produce = (20 × 2) / 3 = 13.33 moles of S₂O₃.
Thus, 13.33 moles of S₂O₃ were obtained from the reaction.
Learn more about stoichiometry:
brainly.com/question/14735801
D because it’s not going into then air or to the clouds
I Think The answer is c I hope it helps My friend Message Me if I’m wrong and I’ll change My answer and fix it for you
