The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>
Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):
mass O₂ = 120 g
mol O₂(MW=32 g/mol) :
Mol ratio of reactants(to find limiting reatants) :
mol of H₂O based on O₂ as limiting reactants :
mol H₂O :
mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :
Each column is called a group<span>. The elements in each </span>group have<span> the same number of electrons in the outer orbital. Those outer electrons are also called valence electrons.</span>
Answer:
Explanation:
Well, obviously a molecule with polar bonds can be polar in itself. It's like saying I am an atheltic person who can just reach the basketball rim with my head and also I can dunk.
But if the question is how can a molecule that in non-polar have polar bonds, well, its because the polar bonds' dipole cancels each other out. It's like a tight rope. If a person pulls in one direction, it intuitively, the rope would go in that direction. However, if a person pulls in the other direction with the same amount of force, the rope stays still. This is the same case. Although molecules can have different electronegativities, the pull of electrons in one direction is cancelled out by a pull in the opposite direction, making the net dipole 0.
This is common for main VSERP shaped molecules like linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral.
Answer:
b. E = 2,28V
Explanation:
The maximum work is the same than ΔG. As ΔG could be written as:
ΔG = nFE <em>(1)</em>
Where n is moles of electrons transferred, F is faraday constant (96485 J/Vmol) and E is the voltage of the cell.
For the reaction:
CH₃OH(l) + ³/₂O₂(g) → CO₂(g) + 2H₂O(l)
The oxidation state of C in CH₃OH is -2 but in CO₂ is +4, that means transferred electrons are +4 - -2 = <em>6e⁻</em>
Replacing in (1):
1320x10³ J = 6mol e⁻×96485J/Vmol×E
<em>E = 2,28V</em>
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I hope it helps!
