Answer:
Yes, yield.
Explanation:
N2(g) + 3 H2(g) → 2 NH3 (g) balanced equation
First, find limiting reactant:
Moles H2 = 1.83 g x 1 mole/2 g = 0.915 moles H2
Moles N2 = 9.84 g N2 x 1 mole/28 g = 0.351 moles N2
The mole ratio of H2: N2 is 3:1, so H2 is limiting (0.915 is less than 3 x 0.351)
Theoretical yield of NH3 = 0.915 mol H2 x 2 mol NH3/3 mol H2 = 0.61 moles NH3
The answer is: To see how fast hydrogen peroxide decomposes into water and oxygen.
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Answer:
Explanation:
From the given information:
The equation for the reaction can be represented as:
The I.C.E table can be represented as:
2SO₂ O₂ 2SO₃
Initial: 14 2.6 0
Change: -2x -x +2x
Equilibrium: 14 - 2x 2.6 - x 2x
However, Since the amount of sulfur trioxide gas to be 1.6 mol.
SO₃ = 2x,
then x = 1.6/2
x = 0.8 mol
For 2SO₂; we have 14 - 2x
= 14 - 2(0.8)
= 14 - 1.6
= 12.4 mol
For O₂; we have 2.6 - x
= 2.6 - 1.6
= 1.0 mol
Thus;
[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,
[O₂] = 1/50 = 0.02 M ,
[SO₃] = 1.6/50 = 0.032 M
Kc = [SO₃]² / [SO₂]² [O₂]
= ( 0.032²) / ( 0.248² x 0.02)
= 0.8325
Recall that; the equilibrium constant for the reaction = 0.8325;
If we want to find:
Then:
Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.
<span>The solid lines between N and Mg are actually ionic bonds. N has 5 valence electrons (2 of which are paired). Of the 3 that are unpaired, 2 are part of covalent bonds with adjacent carbon atoms. N accepts an extra electron to complete its octet, but gets a formal charge of -1. This allows for formation of an ionic bond with Mg, which is +2. Two of these charged N atoms therefore neutralize the charge of the central Mg. As for the coordinate (dative) covalent bonds, Mg has empty orbitals - the ionic bonds with the charged N atoms give it only 4/8 possible valence electrons.
The other two N atoms (dotted lines) have a formal charge of 0 since they form three covalent bonds with adjacent carbon atoms, but they still have a lone pair. Therefore, just to improve stability, each of these N atoms can "donate" its lone pair to Mg in order to complete its octet.
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