Q: A student is solving for $x$ in the equation $4x = -2.13 \times 10^{-4} + \frac{1.75 \times 10^{-5}}{x}$. Which of the following is the correct first step to rewrite this equation in standard quadratic form?

$4x^2 + 2.13 \times 10^{-4}x - 1.75 \times 10^{-5} = 0$

Question 1

Which of the following equations is most appropriately solved using the quadratic formula?

Accepted Answer

$x^2 + 3x - 4 = 0$

Question 2

In the context of chemical equilibrium, why is the quadratic formula often used?

Accepted Answer

To solve for unknown concentrations when the equilibrium expression leads to a quadratic equation.

Question 3

Given the quadratic equation $ax^2 + bx + c = 0$, what is the correct quadratic formula for solving for $x$?

Accepted Answer

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Question 4

When using the quadratic formula in equilibrium problems, why is only one of the two possible $x$ values typically used?

Accepted Answer

Because only the value that makes physical sense (e.g., positive concentration) is valid.

Question 5

A student is solving for $x$ in the equation $4x = -2.13 	imes 10^{-4} + \frac{1.75 	imes 10^{-5}}{x}$. Which of the following is the correct first step to rewrite this equation in standard quadratic form?

Accepted Answer

$4x^2 + 2.13 	imes 10^{-4}x - 1.75 	imes 10^{-5} = 0$

Question 6

In an ICE chart for a chemical equilibrium problem, which variable is most commonly solved for using the quadratic formula?

Accepted Answer

The change in concentration, often represented as $x$

Question 7

If the discriminant $b^2 - 4ac$ in the quadratic formula is negative, what does this indicate about the solutions for $x$ in a chemical equilibrium context?

Accepted Answer

There are no real solutions, which may indicate an error in the setup or impossible equilibrium conditions.

Question 8

A reaction has an equilibrium expression that leads to the equation $x^2 + 5x + 6 = 0$. What are the possible values for $x$?

Accepted Answer

$x = -2$ and $x = -3$

Question 9

In a real-world medical scenario, a drug dissociates in the bloodstream according to the equilibrium $D \rightleftharpoons D^+ + A^-$. If the equilibrium constant expression leads to a quadratic equation for $[D^+]$, what is the best reason for discarding a negative solution for $[D^+]$?

Accepted Answer

Negative concentrations are not physically meaningful.

Question 10

Which of the following is a correct application of the quadratic formula to solve for $x$ in the equation $2x^2 - 4x - 6 = 0$?

Accepted Answer

$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2}$

Question 11

A chemical equilibrium problem leads to the equation $x^2 - 6x + 9 = 0$. What is the value of $x$ at equilibrium?

Accepted Answer

$x = 3$

Question 12

Why is it important to check the physical meaning of solutions obtained from the quadratic formula in equilibrium problems?

Accepted Answer

Because only positive, realistic concentrations are valid in chemical systems.

Question 13

If an ICE chart for a weak acid dissociation leads to the equation $x^2 + 0.1x - 0.002 = 0$, which of the following is the correct setup for the quadratic formula?

Accepted Answer

$x = \frac{-0.1 \pm \sqrt{(0.1)^2 - 4 \cdot 1 \cdot (-0.002)}}{2 \cdot 1}$

Question 14

A student finds two solutions for $x$ when solving an equilibrium problem: $x = 0.25$ and $x = -0.10$. Which value should be used and why?

Accepted Answer

$x = 0.25$, because concentrations cannot be negative.

Question 15

In a real-world scenario, a buffer solution is prepared and the equilibrium calculation leads to a quadratic equation. What is the most likely reason for this?

Accepted Answer

The concentrations of acid and conjugate base are similar and the $x$ cannot be neglected.

Using the Quadratic Formula in Chemical Equilibrium Calculations

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