Did you mean the atomic mass?
Answer:
c. HF can participate in hydrogen bonding.
Explanation:
<u>The boiling points of substances often reflect the strength of the </u><u>intermolecular forces</u><u> operating among the molecules.</u>
If it takes more energy to separate molecules of HF than of the rest of the hydrogen halides because HF molecules are held together by stronger intermolecular forces, then the boiling point of HF will be higher than that of all the hydrogen halides.
A particularly strong type of intermolecular attraction is called the hydrogen bond, <em>which is a special type of dipole-dipole interaction between the hydrogen atom in a polar bond</em>, such as N-H, O-H, or F-H, and an electronegative O, N, or F atom.
The balanced chemical reaction:
C3H8 + 5O2 = 3CO2 + 4H2O
We are given the amount of the carbon dioxide to be produced. This will be the starting point of our calculations.
<span>43.62 L CO2 ( 1 mol CO2 / 22.4 L CO2 ) (5 mol O2 / 3 mol CO2 ) (
22.4 L O2 / 1 mol O2) = 72.7 L O2</span>
Answer:
Moles NH₃: 0.0593
0.104 moles of N₂ remain
Final pressure: 0.163atm
Explanation:
The reaction of nitrogen with hydrogen to produce ammonia is:
N₂ + 3 H₂ → 2 NH₃
Using PV = nRT, moles of N₂ and H₂ are:
N₂: 1atmₓ3.0L / 0.082atmL/molKₓ273K = 0.134 moles of N₂
H₂: 1atmₓ2.0L / 0.082atmL/molKₓ273K = 0.089 moles of H₂
The complete reaction of N₂ requires:
0.134 moles of N₂ × (3 moles H₂ / 1 mole N₂) = <em>0.402 moles H₂</em>
That means limiting reactant is H₂. And moles of NH₃ produced are:
0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = <em>0.0593 moles NH₃</em>
Moles of N₂ remain are:
0.134 moles of N₂ - (0.089 moles of H₂ × (1 moles N₂ / 3 mole H₂)) = <em>0.104 moles of N₂</em>
And final pressure is:
P = nRT / V
P = (0.104mol + 0.0593mol)×0.082atmL/molK×273K / 5.0L
<em>P = 0.163atm</em>
