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 .
Answer:
82.5 metres
Explanation:
speed=330m|s
initial time taken =0.5÷2=0.25
distance=speed×time
330m|s × 0.25s = 82.5m
P₁ = 0.90 atm
V₁ = 50.0 mL
T₁ = 298 K
P₂ = 1 atm
T₂ = 273 K
V₂ = P₁ x V₁ x T₂ / T₁ x P₂
V₂ = 0.90 x 50.0 x 273 / 298 x 1
V₂ = 12285 / 298
V₂ = 41 mL
Answer (1)
hope this helps!
Answer:
VH2SO4 = 145.3 mL
Explanation:
Mw BaO2 = 169.33 g/mol
⇒ mol BaO2 = 53.5g * ( mol BaO2 / 169.33 g BaO2) = 0.545 mol BaO2
⇒according to the reaction:
mol BaO2 = mol H2SO4 = 0.545 mol
⇒ V H2SO4 = 0.545 mol H2SO4 * ( L H2SO4 / 3.75 mol H2SO4 )
⇒V H2SO4 = 0.1453 L (145.3 mL)
