Answer:
3 mol
Explanation:
Given data:
Volume of ethane = 15.0 L
Temperature = 30°C
Pressure = 5.0 atm
Number of moles of ethane = ?
Solution:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
30+273 = 303 K
5.0 atm × 15.0 L = n×0.0821 atm.L/ mol.K × 303 K
75 atm.L = n× 24.87 atm.L/ mol
n = 75 atm.L / 24.87 atm.L/ mol
n = 3 mol
Answer:Methyl benzoate has a very high boiling point i.e. 199°C.Therefore a water-cooled condenser is not advisable as it might crack the condenser tube.So in such cases air condenser tube is normally used for such distillation procedure.
Answer:
227.78g of the precipitate are produced
Explanation:
Based on the reaction, 3 moles of CuCl2 produce 1 mole of Cu3(PO4)2 (The precipitate).
To solve this question we need to find the moles of CuCl2 added. With these moles and the reactio we can find the moles of Cu3(PO4)2 and its mass as follows:
<em>Moles CuCl2:</em>
285mL = 0.285L * (6.3mol / L) = 1.7955 moles CuCl2
<em>Moles Cu3(PO4)2:</em>
1.7955 moles CuCl2 * (1mol Cu3(PO4)2 / 3mol CuCl2) = 0.5985 moles Cu3(PO4)2
<em>Mass Cu3(PO4)2 -380.58g/mol-</em>
0.5985 moles Cu3(PO4)2 * (380.58g/mol) =
227.78g of the precipitate are produced
Answer:

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Answer: A. Cilla Is Correct.