Answer:
[N2] = 0.3633M
[H2] = 1.090M
[NH3] = 0.2734M
Explanation:
Based on the reaction of the problem, Kc is defined as:
Kc = 0.159 = [NH3]² / [N2] [H2]³
<em>Where [] are the equilibrium concentrations.</em>
The initial concentrations of the reactants is:
N2 = 1.00mol / 2.00L = 0.500M
H2 = 3.00mol / 2.00L = 1.50M
When the equilibrium is reached, the concentrations are:
[N2] = 0.500M - X
[H2] = 1.50M - 3X
[NH3] = 2X
<em>Where X is reaction quotient</em>
Replacing in the Kc equation:
0.159 = [2X]² / [0.500 - X] [1.50 - 3X]³
0.159 = 4X² / 1.6875 - 13.5 X + 40.5 X² - 54 X³ + 27 X⁴
0.268313 - 2.1465 X + 6.4395 X² - 8.586 X³ + 4.293 X⁴ = 4X²
0.268313 - 2.1465 X + 2.4395 X² - 8.586 X³ + 4.293 X⁴ = 0
Solving for X:
X = 0.1367. Right solution.
X = 1.8286. False solution. Produce negative concentrations
Replacing:
[N2] = 0.500M - 0.1367M
[H2] = 1.50M - 3*0.1367M
[NH3] = 2*0.1367M
The equilibrium concentrations are:
<h3>[N2] = 0.3633M</h3><h3>[H2] = 1.090M</h3><h3>[NH3] = 0.2734M</h3>
Answer:
The correct answer is - n and l.
Explanation:
The size of an orbital is determined by the principal number of shell which is represented by n. The larger the energy level (n) bigger the size of the orbital. N can be any integer value: 1, 2, 3 . . . . and so on.
l represents the angular momentum or subshell number provides the overall shape of an orbital in this subshell only integer values between 0 and n-1 are permitted.
Thus, n & l are two quantum numbers that determine the energy level of an orbital in a multielectron atom.
The answer is B.
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Answer:
2C2H6 + 7O2 → 4 CO2 + 6H20
Explanation:
