For the current reaction, 2NO2 ↔ N2O4, we have:
Kc = [tex]\frac{[N_{2} O_{4}]}{[NO_{2}]^{2} }[/tex]
Concentration of NO2 = 11.95
Concebtration of N2O4 = 6.05
Based on the current concentrations of NO2 and N2O4, what is Kc?
![For The Current Reaction, 2NO2 N2O4, We Have:Kc = [tex]\frac{[N_{2} O_{4}]}{[NO_{2}]^{2} }[/tex]Concentration](https://brainimage.s3.us-east-005.backblazeb2.com/contents/attachments/kBZEwypmHFjSk0CtoVFNObAe6SPQ3mHh.jpeg)
Answers
[tex]\displaystyle K_c \approx 0.0424[/tex]
General Formulas and Concepts:Math
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightChemistry
Equilibrium
Equilibrium Constant KConcentrations - Denoted in [Brackets]Step-by-step explanation:Step 1: Define
[RxN] 2NO₂ ⇆ N₂O₄
[Equilibrium Rate Law] [tex]\displaystyle K_c = \frac{[N_2O_4]}{[NO_2]^2}[/tex]
NO₂ = 11.95 M
N₂O₄ = 6.05 M
Step 2: Find K
Substitute [ERL]: [tex]\displaystyle K_c = \frac{[6.05]}{[11.95]^2}[/tex]Exponents: [tex]\displaystyle K_c = \frac{[6.05]}{[142.803]}[/tex]Divide: [tex]\displaystyle K_c = 0.042366[/tex]Round (Sig Figs): [tex]\displaystyle K_c \approx 0.0424[/tex]This value of K tells us that the reactants are favored in the equilibrium reaction (K < 0.1).
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Answer:
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Step-by-step explanation:
Given: The line plots shows the weights of [tex]9[/tex] grapefruits
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[tex]y[/tex]: Quantity of [tex]1\frac{1}{4}lb[/tex] grapefruit
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To check your work, simply substitute the solved value of [tex]x[/tex] into the initial equation:
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[tex] = > 3x + 10 = 4x + 54[/tex]
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The graph of y = 4 logs (x) + 2 passes through the point (5,8).
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