For ksp of silver sulfate (ag2so4) at 1. 2 ✕ 10-5, the solubility of silver sulfate is mathematically given as
x = 5.26*10^{-4}
<h3>What is the
solubility of silver
sulfate in 0.15 M AgNO3?</h3>
Generally, the equation for the Chemical reaction is mathematically given as
Ag2SO4 ⇄ 2Ag+ + SO4 2-
Therefore
Ksp = [Ag+ ]2 *[SO4 2-]
1.2*10-5 = ( 0.15 +2x)2*(x)
x = 5.26*10^{-4}
In conclusion, the solubility of silver sulfate is
x = 5.26*10^{-4}
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Answer:
Es la tercera por qué cuando el material varía va cambiando de estado
Explanation:
Answer:
Part C: P2 = 0.30 atm
Part D: V1 = 16.22 L.
Explanation:
Part C:
Initial pressure (P1) = 2.67 atm
Initial volume (V1) = 5.54 mL
Final pressure (P2) =.?
Final volume (V2) = 49 mL
The final pressure (P2) can be obtained as follow:
P1V1 = P2V2
2.67 x 5.54 = P2 x 49
Divide both side by 49
P2 = (2.67 x 5.54)/49
P2 = 0.30 atm
Therefore, the final pressure (P2) is 0.30 atm
Part D:
Initial pressure (P1) = 348 Torr
Initial volume (V1) =?
Final pressure (P2) = 684 Torr
Final volume (V2) = 8.25 L
The initial volume (V1) can be obtained as follow:
P1V1 = P2V2
348 x V1 = 684 x 8.25
Divide both side by 348
V1 = (684 x 8.25)/348
V1 = 16.22 L
Therefore, the initial volume (V1) is 16.22 L
Answer:
2.1 kg of water
Explanation:
Step 1: Given data
- Moles of lithium bromide (solute): 4.3 moles
- Molality of the solution (m): 2.05 m (2.05 mol/kg)
- Mass of water (solvent): ?
Step 2: Calculate the mass of water required
Molality is equal to the moles of solute divided by the kilograms of solvent.
m = moles of solute/kilograms of solvent
kilograms of solvent = moles of solute/m
kilograms of solvent = 4.3 mol /(2.05 mol/kg) = 2.1 kg
