Answer:
Explanation:
Please, find the image with the pictured molecule for this question attached.
The molecule has one oxygen atom (red) covalently bonded to one hydrogen atom (light grey), one nitrogen atom (blue) covalently bonded to two hydrogen atoms (light grey), and two carbon atoms (dark grey) bonded each to two hydrogen atoms (light grey).
<em>Hydrogen bondings</em> are intermolecular bonds (bonds between atoms of two different molecules not between atoms of the same molecule). The hydrogen bonds are attractions between the positive end of one hydrogen atom and the negative end of a small atom of other molecule (N, O, or F).
Since, nitrogen and oxygen are much more electronegative than hydrogen atoms, you conclude that:
- The two hydrogen atoms covalently bonded to the nitrogen atoms have considerably partial positive charge.
- The hydrogen atom covalently bonded to the oxygen atom also has a a relative large partial positive charge.
So, those are three ends of the molecule that can form hydrogen bonds with water molecules.
The hydrogen bondings are only possible when hydrogen is covalently bonded to N, O or F atoms.
You will observe the the project that your trying to do an watch the change
Complete dominance.... Hope this helps
<span>CH3OH and NH3
I HOPE THIS HELPS :D</span>
39.25 g of water (H₂O)
Explanation:
We have the following chemical reaction:
2 H₂ + O₂ → 2 H₂O
Now we calculate the number of moles of each reactant:
number of moles = mass / molar weight
number of moles of H₂ = 14.8 / 2 = 7.4 moles
number of moles of O₂ = 34.8 / 32 = 1.09 moles
We see from the chemical reaction that 2 moles of H₂ will react with 1 mole of O₂ so 7.4 moles of H₂ will react with 3.7 moles of O₂ but we only have 1.09 moles of O₂ available. The O₂ will be the limiting reactant. Knowing this we devise the following reasoning:
if 1 moles of O₂ produces 2 moles of H₂O
then 1.09 moles of O₂ produces X moles of H₂O
X = (1.09 × 2) / 1 = 2.18 moles of H₂O
mass = number of moles × molar weight
mass of H₂O = 2.18 × 18 = 39.25 g
Learn more about:
limiting reactant
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