Answer:
Explanation:
K₂CrO₄ + ( COONa )₂ + 2BaCl₂ = Ba CrO₄ + ( COO ) ₂ Ba + 2 KCl + 2 NaCl
.033 M .053 M
Ksp of Ba CrO₄ is 2.10×10⁻¹⁰
Ksp of ( COO ) ₂ Ba is 1.30×10⁻⁶
A ) Ksp of Ba CrO₄ is less so it will precipitate out first .
B) Ksp = 2.10×10⁻¹⁰
Ba CrO₄ = Ba⁺² + CrO₄⁻²
C .033
C x .033 = 2.10×10⁻¹⁰
C = 63.63 x 10⁻¹⁰ M
Ba⁺² must be present in concentration = 63.63 x 10⁻¹⁰ M
C)
90% of precipitation of barium oxalate
concentration of oxalate to precipitate out = .9 x .0532 = .04788
( COO ) ₂ Ba = (COO)₂⁻² + Ba⁺²
.04788 M C
C x .04788 = 1.30×10⁻⁶
C = 27.15 x 10⁻⁶ M .
Am so sorry I don’t know how to help you hope you find the answer
Answer:
0.13 M ( 2 s.f)
Explanation:
2Cl2O5 (g)-->2Cl2(g) +5O2 (g)
rate= (17.4 M -1 .s -1 ) [Cl2O5]2
From the rte above, we can tell that our rate constant (k) = 17.4 M -1 .s -1
The units of k tells us this is a second order reaction.
Initial Concentration [A]o = 1.46M
Final Concentration [A] = ?
Time = 0.400s
The integrated rate law for second order reactions is given as;
1 / [A] = (1 / [A]o) + kt
1 / [A] = [ (1/ 1.46) + (17.4 * 0.4) ]
1 / [A] = 0.6849 + 6.96
1 / [A] = 7.6496
[A] = 1 / 7.6496
[A] = 0.13073 M ≈ 0.13 M ( 2 s.f)
it would be b hope i helped
