There are several process in the nature, which keeps water cycling between them. By Precipitation, Earth's sphere gets water which we use in several ways, and by evaporation & sublimation, it goes back to upper spheres of the Earth's atmosphere and cycle happens again & again.
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Grams ethanol = 33 ml times .789 gms/ml = 26.037 gms
<span>Moles ethanol = 26.037 gms / 46 gms/mole = .57 moles </span>
<span>Moles water = 67 ml or 67 grams/18 gms/mole = 3.22 moles </span>
<span>total moles = .57 + 3.72 = 4.29 moles </span>
<span>Mole fraction ethanol = .57 moles ethanol / 4.29 moles total = 0.13</span>
<span>Moles fraction water = 3.72 moles water / 4.29 moles total = 0.87</span>
<span>Partial pressure of ethanol = mole fraction ethanol (.13) _ times VP ethanol 43.9 torr) = 5.707 torr </span>
<span>partial pressure water = mole fraction water .87) times VP water (l7.5 torr) = 15.23 torr </span>
<span>Total vapor pressure over solution = 5.71 torr + 15.23 torr = 20.94 torr</span>
Answer:
Answer: 882mmHg
Explanation:
In a closed-end manometer, the gas added is under a pressure that is measured against atmospheric pressure. The difference in pressure between both gases is equal to the difference in mercury levels. Thus, the pressure of the argon can be:
783 mmHg - 96.0mmHg = 687mmHg
783 mmHg - 96.0mmHg = 879mmHg
Based on the options, the possible answer (The nearest to the second value) is:
<h3>Answer: 882mmHg</h3>
Answer:
0.500 moles of CO2 has a volume of 11.2 L at STP (option B)
Explanation:
Step 1: Data given
Volume of a gas at STP = 11.2 L
STP: Pressure = 1 atm and temperature = 273 K
Step 2: Calculate volume
p*V= n*R*T
V = (n*R*T)/p
⇒with V = the volume of the gas = TO BE DETERMINED
⇒with n = the number of moles of the gas
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 273 K
⇒with p = the pressure of the gas = 1 atm
A
) 0.250 mole of NH3
V = (0.250 * 0.08206 * 273) / 1
V = 5.6 L
B
) 0.500 mole of CO2
V = (0.500 * 0.08206 * 273) / 1
V = 11.2 L
C
) 0.750 mole of NH3
V = (0.750 * 0.08206 * 273) / 1
V = 16.8 L
D) 1.00 mole of CO2
V = (1.00 * 0.08206* 273) / 1
V = 22.4 L
0.500 moles of CO2 has a volume of 11.2 L at STP (option B)
