I’m not a 100% shure but I would personally say OIL SPILLS.
Answer:
yo whats good nebyu its nathan Imao
Explanation:
Answer:
94.325 g
Explanation:
We'll begin by converting 350 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
350 mL = 350 mL × 1 L /1000 mL
350 mL = 0.35 L
Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:
Volume = 0.35 L
Molarity of KC₂H₃O₂ = 2.75 M
Mole of KC₂H₃O₂ =?
Molarity = mole /Volume
2.75 = Mole of KC₂H₃O₂ / 0.35
Cross multiply
Mole of KC₂H₃O₂ = 2.75 × 0.35
Mole of KC₂H₃O₂ = 0.9625 mole
Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:
Mole of KC₂H₃O₂ = 0.9625 mole
Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)
= 39 + 24 + 3 + 32
= 98 g/mol
Mass of KC₂H₃O₂ =?
Mass = mole × molar mass
Mass of KC₂H₃O₂ = 0.9625 × 98
Mass of KC₂H₃O₂ = 94.325 g
Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g
Considering ideal gas:
PV= RTn
T= 25.2°C = 298.2 K
P1= 637 torr = 0.8382 atm
V1= 536 mL = 0.536 L
:. R=0.082 atm.L/K.mol
:. n= (P1V1)/(RT) = ((0.8382 atm) x (0.536 L))/
((0.082 atmL/Kmol) x (298.2K))
:. n= O.0184 mol
Then,
P2= 712 torr = 0.936842 atm
V2 = RTn/P2 = [(0.082atmL/
Kmol) x (298.2K) x (0.0184mol) ]/(0.936842atm)
:.V2 = 0.4796 L
OR
V2 = 479.6 ml
