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3 years ago

For the current reaction, 2NO2 ↔ N2O4, we have:

Mathematics

1 answer:

vivado [14]3 years ago

4 0

<h3>Answer:</h3>

$\displaystyle K_c \approx 0.0424$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Equilibrium</u>

Equilibrium Constant K
Concentrations - Denoted in [Brackets]

<h3>Step-by-step explanation:</h3>

<u>Step 1: Define</u>

[RxN] 2NO₂ ⇆ N₂O₄

[Equilibrium Rate Law] $\displaystyle K_c = \frac{[N_2O_4]}{[NO_2]^2}$

NO₂ = 11.95 M

N₂O₄ = 6.05 M

Substitute [ERL]: $\displaystyle K_c = \frac{[6.05]}{[11.95]^2}$
Exponents: $\displaystyle K_c = \frac{[6.05]}{[142.803]}$
Divide: $\displaystyle K_c = 0.042366$
Round (Sig Figs): $\displaystyle K_c \approx 0.0424$

This value of K tells us that the reactants are favored in the equilibrium reaction (K < 0.1).

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vichka [17]

Rearrange the ODE as

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Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
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So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago

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Otrada [13]

Answer:

adjacent, supplementary

Step-by-step explanation:

"Two lines intersecting in a right angle with a square indicating a right angle in quadrant one and a ray splitting quadrant two with a two labeling the left angle and one labeling the angle on the right"

Based on your description, it seems like they are right next to each other and equal 180 (straight line)

Angles directly next to each other = adjacent

Angles that add up to 180 = supplementary

This indicates adjacent and supplementary angles

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2 years ago

Simple Algebra problem. Includes the x and y axis. Also need help finding the perimeter.

Lina20 [59]

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2 years ago

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zheka24 [161]

Answer:

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