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Answer:
The minimum concentration of Cl⁻ that produces precipitation is 12.6M
Explanation:
The Ksp of PbCl₂ is expressed as:
PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq)
The Ksp is:
Ksp = 1.6 = [Pb²⁺] [Cl⁻]²
When Ksp = [Pb²⁺] [Cl⁻]² the solution begind precipiration.
A 0.010M Pb(NO₃)₂ is 0.010M Pb²⁺, thus:
1.6 = [0.010M] [Cl⁻]²
160 = [Cl⁻]²
12.6M = [Cl⁻]
<h3>The minimum concentration of Cl⁻ that produces precipitation is 12.6M</h3>
A) NOPE!
B)Maybe
C)YUP this is the answer
D)meh...
Answer:
Explanation:
Step 1: Data given
The equilibrium constant, Kc, for the following reaction is 4.76 * 10^-4 at 431 K
The equilibrium concentration of Cl2(g) is 0.233 M
Step 2: The balanced equation
PCl5(g) ⇄ PCl3(g) + Cl2(g)
Step 3: The initial concentration
[PCl5]= Y M
[PCl] = 0M
[Cl2] = 0M
Step 4: Calculate the concentration at equilibrium
[PCl5] = Y + X M = Y - 0.233 M
[PCl]= XM = 0.233 M
[Cl2]= XM = 0.233 M
Step 5: Define Kc
Kc = [Cl2]* [PCl3] / [PCl5]
4.76 * 10^-4 = 0.233² / (Y -0.233)
0.000476 = 0.05429 / (Y - 0.233)
Y - 0.233 = 0.05429 / 0.000476
Y - 0.233 = 114.05 M
Y = 114.283 M = the initial concentration
The concentration of PCl5 at the equilibrium is 114.05 M
