Answer:
the answer is 6
Explanation:
there is 3 hydrogen molecules in NH3 and there's 2 molecules of NH3 so in total, there are 6 hydrogen molecules on the products side.
Moles of Hydrogen present: 100 / 2 = 50 moles
Moles of Nitrogen present: 200 / 28 = 7.14 moles
Hydrogen required by given amount of nitrogen = 7.14 x 3 = 21.42 moles
Hydrogen is excess so we will calculate the Ammonia produced using Nitrogen.
Molar ratio of Nitrogen : Ammonia = 1 : 2
Moles of ammonia = 7.14 x 2 = 14.28 moles
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:
Explanation:
The nitrates of Bi,Sn and Cd is ruled out because their sulfides are insoluble in acidic medium.
Nitrates of Ni or Co may be present because their sulfides are insoluble in basic medium. The presence of other nitrates are ruled out.
486 kPa will be the pressure of gas at 27 degrees.
Explanation:
Data given :
Initial pressure of the gas P1 = 810 kPa
initial temperature of the gas T1 = 227 degrees OR 500.15 K
final pressure of the gas =?
final temperature of the gas = 27 degrees OR 300.15 K
Gay Lussac's law is used to calculate the pressure of the gas at 27 degrees or 300.15 K
= 
P2 = 
Putting the values in the equation:
P2 = 
P2 = 486 KPa
486 Kpa is the pressure of the gas when temperature was reduced to 27 degrees.
