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## How many grams of lead chromate are formed from the reaction of 15.0mL of 0.40M potassium chromate with 15mL of lead nitrate? What was the limitng reagent from the equation?

Stefan V.
Featured 1 month ago

Here's what I got.

#### Explanation:

Since you didn't provide the concentration of the lead(II) nitrate solution, we're going to have to improvise a bit here.

You're dealing with a double replacement reaction in which aqueous potassium chromate, ${\text{K"_2"CrO}}_{4}$, and aqueous lead(II) nitrate, #"Pb"("NO"_3)_2#, will react to form the insoluble lead(II) chromate, ${\text{PbCrO}}_{4}$, and aqueous potassium nitrate, ${\text{KNO}}_{3}$.

The balanced chemical equation that describes this reaction looks like this

${\text{K"_ 2"CrO"_ (4(aq)) + "Pb"("NO"_ 3)_ (2(aq)) -> "PbCrO"_ (4(s)) darr + 2"KNO}}_{3 \left(a q\right)}$

Notice that the two reactants react in a $1 : 1$ mole ratio. This tells you that the reaction will *always consume equal numbers of moles of potassium chromate and of lead(II) nitrate.

Moreover, lead(II) chromate is produced in the same $1 : 1$ mole ratio with both reactants, which means that the number of moles of lead(II) chromate produced by the reaction will match the number of moles of the reactants that take part in the reaction.

Use the molarity and volume of the potassium chromate solution to find how many moles of this reactant you have present

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

#n_(K_2CrO_4) = "0.40 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(15.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#

${n}_{{K}_{2} C r {O}_{4}} = {\text{0.00600 moles K"_2"CrO}}_{4}$

Now, you have three possible scenarios to account for here

$\textcolor{w h i t e}{a}$

• If you have fewer moles of lead(II) nitrate than moles of potassium chromate

In this case, lead(II) nitrate will be the limiting reagent because it will be completely consumed by the reaction before all the moles of potassium chromate will get the chance to react.

${n}_{P b {\left(N {O}_{3}\right)}_{2}} < \text{0.00600 moles}$

Since you have equal volumes of both solutions, you can say that if the concentration of the lead(II) nitrate is lower than that of potassium chromate, the former will act as a limiting reagent.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{c}_{P b {\left(N {O}_{3}\right)}_{2}} \textcolor{red}{<} {\text{0.40 mol L}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ lead(II) nitrate will be the limiting reagent

$\textcolor{w h i t e}{a}$

• if you have equal numbers of moles of each reactant

In this case, both of the reactants will be completely consumed by the reaction, i.e. you won't be dealing with a limiting reagent.

Since the lead(II) chromate is produced in a $1 : 1$ mole ratio with the two reactants, it follows that the reaction will produce $0.00600$ moles of ${\text{K"_2"CrO}}_{4}$.

Use the compound's molar mass to find how many Grams would contain this many moles

$0.00600 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles K"_2"CrO"_4))) * "184.19 g"/(1color(red)(cancel(color(black)("mole K"_2"CrO"_4)))) = color(green)(|bar(ul(color(white)(a/a)"1.1 g } \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\textcolor{w h i t e}{a}$

• if you have more moles of lead(II) nitrate than moles of potassium chromate

This time, the potassium chromate will be completely consumed by the reaction before all the moles of lead(II) nitrate get the chance to react.

Potassium chromate will act as a limiting reagent. The reaction will once again produce $0.00600$ moles of lead(II) chromate. So, for

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{c}_{P b {\left(N {O}_{3}\right)}_{2}} \textcolor{red}{>} {\text{0.40 mol L}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ potassium chromate will be the limiting reagent

You will have

$0.00600 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles K"_2"CrO"_4))) * "184.19 g"/(1color(red)(cancel(color(black)("mole K"_2"CrO"_4)))) = color(green)(|bar(ul(color(white)(a/a)"1.1 g } \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answers are rounded to two sig figs.

It's worth noting that lead(II) chromate is a yellow insoluble solid that precipitates out of solution as the reaction proceeds.

## How would you determine the ionization energy of a hydrogen atom (in kJ/mol) if the electron is in its ground state?

Michael
Featured 1 month ago

I will suggest two ways you could do this.

#### Explanation:

$\textcolor{b l u e}{\left(1\right)}$

Use the Rydberg expression:

The wavelength $\lambda$ of the emission line in the hydrogen spectrum is given by:

$\frac{1}{\lambda} = R \left[\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2}\right]$

$R$ is the Rydberg Constant and has the value $1.097 \times {10}^{7} {\text{m}}^{- 1}$

${n}_{1}$ is the principle quantum number of the lower energy level

${n}_{2}$ is the principle quantum number of the higher energy level.

The energy levels in hydrogen converge and coalesce:

Since the electron is in the ${n}_{1} = 1$ ground state we need to consider series 1. These transitions occur in the u.v part of the spectrum and is known as The Lyman Series.

You can see that as the value of ${n}_{2}$ increases then the value of $\frac{1}{n} _ {2}^{2}$ decreases. At higher and higher values the expression tends to zero until at $n = \infty$ we can consider the electron to have left the atom resulting in an ion.

The Rydberg expression now becomes:

$\frac{1}{\lambda} = R \left[\frac{1}{n} _ {1}^{2} - 0\right] = \frac{R}{n} _ {1}^{2}$

Since ${n}_{1} = 1$ this becomes:

$\frac{1}{\lambda} = R$

$\therefore \frac{1}{\lambda} = 1.097 \times {10}^{7}$

$\therefore \lambda = 9.116 \times {10}^{- 8} \text{m}$

We can now find the frequency and hence the corresponding energy:

$c = \nu \lambda$

$\therefore \nu = \frac{c}{\lambda} = \frac{3 \times {10}^{8}}{9.116 \times {10}^{-} 8} = 3.291 \times {10}^{15} {\text{s}}^{- 1}$

Now we can use the Planck expression:

$E = h \nu$

$\therefore E = 6.626 \times {10}^{- 34} \times 3.291 \times {10}^{15} = 2.18 \times {10}^{- 18} \text{J}$

This is the energy needed to remove 1 electron from 1 hydrogen atom. To find the energy required to ionise 1 mole of H atoms we multiply by the Avogadro Constant:

$E = 2.18 \times {10}^{- 18} \times 6.02 \times {10}^{23} = 13.123 \times {10}^{5} \text{J""/""mol}$

$\textcolor{red}{E = 1312 \text{ ""kJ/mol}}$

$\textcolor{b l u e}{\left(2\right)}$

Use the convergence limit of the u.v spectrum in an experimental method.

You can see from the diagram that the energy levels converge and coalesce to a continuum. This means that the emission lines will converge also.

The frequency at which this happens can give us the ionisation energy.

The 1st column shows the frequency of the line.

The second column gives the difference in frequency which is getting less and less.

If you plot $\nu$ against $\Delta \nu$ you get a graph like this:

You can actually get 2 lines using upper and lower values. The important point is that they converge to the same point where $\Delta \nu$ tends to zero.

You can the read the convergence limit off the x axis.

This gives $\nu = 3.28 \times {10}^{15} {\text{s}}^{- 1}$ which is very close to the value from method $\textcolor{b l u e}{\left(1\right)}$.

As in method $\textcolor{b l u e}{\left(1\right)}$ we can convert to Joules using the Planck expression:

$E = h \nu = 6.626 \times {10}^{- 34} \times 3.28 \times {10}^{15} = 2.173 \times {10}^{- 18} \text{ ""J}$

So for 1 mole:

$E = 2.173 \times {10}^{- 18} \times 6.02 \times {10}^{23} = 13.08 \times {10}^{5} \text{J/mol}$

$\textcolor{red}{E = 1308 \text{ ""kJ/mol}}$

## What is beta decay in terms of quarks?

Ernest Z.
Featured 1 month ago

In β-decay, a quark decays into another type of quark, releasing a β particle and a neutrino.

#### Explanation:

Protons and neutrons consist of quarks.

(from quantumpulse.com)

A proton consists of two up quarks and a down quark ($\text{p = uud}$).

In positron (#β^+#) decay, an up quark changes to a down quark, with the release of a positron and a neutrino.

#"u → d" color(white)(l)+ β^+ + ν#

The overall result us that a proton becomes a neutron:

#"p → n"color(white)(l) + β^+ + ν#

A neutron consists of two down quarks and a up quark ($\text{n = ddu}$).

In beta (#β^"-"#) decay, a down quark changes to an up quark, with the release of an electron (#β^"-"#) and an antineutrino.

#"d → u"color(white)(l) + β^"-" + bar(ν)#

The overall result us that a neutron becomes a proton:

#"n → p"color(white)(l) + β^"-" + bar(ν)#

In pictures, β-decay looks like this:

(from schoolphysics.co.uk)

## What determines the probability pattern of an orbital?

Truong-Son N.
Featured 1 month ago

It just turned out that way. We can't define equations to influence nature, only describe it.

You can imagine anything you want for what "determines" the probability distribution, but ultimately, nature turned out that way, and we described these orbitals after-the-fact.

DISCLAIMER: This is going to be a long answer.

WHAT IS THE WAVE FUNCTION FOR AN ORBITAL?

Erwin Schrodinger published the wave function $\psi$, which describes the state of a quantum mechanical system.

For each orbital, its radial density distribution describes the regions with particular probabilities for finding an electron in that particular orbital.

In general, the wave function for spherical harmonics coordinates can be written as:

$\setminus m a t h b f \left({\psi}_{n l {m}_{l}} \left(r , \theta , \phi\right) = {R}_{n l} \left(r\right) {Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)\right)$

where:

• $r$, $\theta$, and $\phi$ are spherical coordinates:

• ${\psi}_{n l {m}_{l}}$ is a wave function that can be constructed to describe what the orbital's electron distribution looks like.

It depends on the quantum numbers $\setminus m a t h b f \left(n\right)$, $\setminus m a t h b f \left(l\right)$, and $\setminus m a t h b f \left({m}_{l}\right)$.

• ${R}_{n l}$ is the radial component of the wave function, describing the variation in the distance from the center of the orbital (the radius!).

It depends on the quantum numbers $\setminus m a t h b f \left(n\right)$ and $\setminus m a t h b f \left(l\right)$.

• ${Y}_{l}^{{m}_{l}}$ is the angular component of the wave function, describing the aspects of the orbital electron distribution that can possibly give it a non-spherical shape.

It depends on the quantum numbers $\setminus m a t h b f \left(l\right)$ and $\setminus m a t h b f \left({m}_{l}\right)$.

OKAY, WHAT DOES IT HAVE TO DO WITH PROBABILITY PATTERNS??

Focus on the ${R}_{n l} \left(r\right)$ portion of $\psi$.

When we plot the radial density distribution (or "probability pattern") of an orbital, we plot $4 \pi {r}^{2} {\left({R}_{n l} \left(r\right)\right)}^{2}$ vs. $r$ in units of ${a}_{0} = 5.29177 \times {10}^{- 11} \text{m}$ (Bohr radii) like so:

The wave functions themselves, as we defined them, generate these "probability patterns".

HOW DO I KNOW YOU'RE NOT MAKING THIS UP?

Well, with an example, I guess. This will give a result that should prove familiar.

The simplest example of an orbital wave function (used by $\text{Real Chemists"^"TM}$, folks!) is the one for the $\setminus m a t h b f \left(1 s\right)$ orbital, which happens to be:

#color(blue)(psi_(1s) = R_(10)(r)Y_(0)^(0)(theta,phi)#

$= \stackrel{{R}_{10} \left(r\right)}{\overbrace{\left(2 {Z}^{\text{3/2")/(a_0^"3/2")e^(-"Zr/}} {a}_{0}\right)}} \cdot \stackrel{{Y}_{0}^{0} \left(\theta , \phi\right)}{\overbrace{\frac{1}{\sqrt{4 \pi}}}}$

#= color(blue)(1/sqrt(pi) (Z/(a_0))^"3/2" e^(-"Zr/"a_0))#

where:

• $Z$ is the atomic number of the atom.
• ${a}_{0}$ is the same Bohr radius we just defined as $5.29177 \times {10}^{- 11} \text{m}$.

In order to plot the "probability pattern", we have to grab what we need and turn it into $4 \pi {r}^{2} {\left({R}_{n l} \left(r\right)\right)}^{2}$.

$4 \pi {r}^{2} {\left({R}_{n l} \left(r\right)\right)}^{2}$

$= 4 \pi {r}^{2} {\left(\stackrel{{R}_{n l} \left(r\right)}{\overbrace{\left(2 {Z}^{\text{3/2")/(a_0^"3/2")e^(-"Zr/}} {a}_{0}\right)}}\right)}^{2}$

$= 4 \pi {r}^{2} \frac{4 {Z}^{3}}{{a}_{0}^{3}} {e}^{- \text{2Zr/} {a}_{0}}$

$\implies \textcolor{red}{4 \pi {r}^{2} {\left({R}_{n l} \left(r\right)\right)}^{2} = 16 \pi {\left(\frac{Z}{{a}_{0}}\right)}^{3} {r}^{2} {e}^{- \text{2Zr/} {a}_{0}}}$

This is the function that corresponds to the red curve in the diagram above.

Now, since we have the plot for $4 \pi {r}^{2} {R}_{n l}^{2} \left(r\right)$ vs. $r$, one thing we can do is find the region of highest electron density, which is just the highest point in the $1 s$ orbital graph above.

So...

1. Take the derivative with respect to $r$.
2. Set it equal to $0$, since the peak has a slope of $0$.

$\implies 16 \pi {\left(\frac{Z}{{a}_{0}}\right)}^{3} \frac{d}{\mathrm{dr}} \left[{r}^{2} {e}^{- \text{2Zr/} {a}_{0}}\right]$

Utilize the product rule:

$\implies 16 \pi {\left(\frac{Z}{{a}_{0}}\right)}^{3} \left[\frac{- 2 Z}{{a}_{0}} {r}^{2} {e}^{- \text{2Zr/"a_0) + 2re^(-"2Zr/} {a}_{0}}\right]$

Now setting it equal to $0$ and crossing out nonzero terms:

$0 = \cancel{16 \pi {\left(\frac{Z}{{a}_{0}}\right)}^{3}} \left[\frac{- 2 Z}{{a}_{0}} {r}^{2} {e}^{- \text{2Zr/"a_0) + 2re^(-"2Zr/} {a}_{0}}\right]$

$0 = \frac{- 2 Z}{{a}_{0}} {r}^{2} {e}^{- \text{2Zr/"a_0) + 2re^(-"2Zr/} {a}_{0}}$

Although ${e}^{- x}$ reaches $0$ at very large $r$, it is nonzero for small $r$, and we are at very small $r$. So...

$0 = \left(\frac{- 2 Z}{{a}_{0}} {r}^{2} + 2 r\right) \cancel{{e}^{- \text{2Zr/} {a}_{0}}}$

$0 = \frac{- 2 Z}{{a}_{0}} {r}^{2} + 2 r$

$\cancel{2} r = \frac{\cancel{2} Z}{{a}_{0}} {r}^{2}$

$r = \frac{Z}{{a}_{0}} {r}^{2} \implies \textcolor{g r e e n}{r = \frac{{a}_{0}}{Z}}$

...for a Hydrogen-like atom (e.g. ${\text{He}}^{+}$, ${\text{Li}}^{2 +}$, etc).

But for the Hydrogen atom, $Z = 1$, so

$\setminus m a t h b f \left(\textcolor{b l u e}{r = {a}_{0} = 5.29177 \times {10}^{- 11} \text{m}}\right)$.

That tells us that the electron in the $1 s$ orbital of Hydrogen atom is most often one bohr radius distance away from the center of the nucleus, just as we would expect. Great!

## Can someone check my work please and please tell me if I made a mistake where I error?

Stefan V.
Featured 1 month ago

Here's my take on this.

#### Explanation:

The problem wants you to predict the precipitate produced by mixing two solutions, one that contains aluminium nitrate, #"Al"("NO"_3)_3#, a soluble ionic compound, and one that contains sodium hydroxide, $\text{NaOH}$, another soluble ionic compound.

The reaction takes place in aqueous solution, so right from the start you should be aware that you're dealing with ions.

Soluble ionic compounds dissociate completely in aqueous solution to form cations, which are positively charged ions, and anions, which are negatively charged ions.

The two solutions can thus be written as

${\text{Al"("NO"_ 3)_ (3(aq)) -> "Al"_ ((aq))^(3+) + 3"NO}}_{3 \left(a q\right)}^{-}$

It's absolutely crucial to make sure that you add the charges of the ions to the chemical equation.

${\text{NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

Now, you should be familiar with the solubility rules for aqueous solutions.

The aluminium cations, ${\text{Al}}^{3 +}$, will combine with the hydroxide anions, ${\text{OH}}^{-}$, to form the insoluble aluminium hydroxide, which precipitates out of solution.

Notice that the aluminium cations have a $3 +$ charge and the hydroxide anions have a $1 -$ charge. This means that you're going to need $3$ hydroxide anions in order to balance the positive charge of the cation.

Therefore, the chemical formula for aluminium hydroxide is #"Al"("OH")_3#, .

The other product of the reaction will be aqueous sodium nitrate. The sodium cations have a $1 +$ charge and the nitrate anions have a $1 -$ charge, so the chemical formula for sodium nitrate will be ${\text{NaNO}}_{3}$.

The balanced chemical equation will thus looks like this

${\text{Al"("NO"_ 3)_ (3(aq)) + 3"NaOH"_ ((aq)) -> "Al"("OH")_ (3(s)) darr + 3"NaNO}}_{3 \left(a q\right)}$

The complete ionic equation for this reaction looks like this

${\text{Al"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-) + 3"Na"_ ((aq))^(+) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr + 3"Na"_ ((aq))^(+) + 3"NO}}_{3 \left(a q\right)}^{-}$

To get the net ionic equation, eliminate spectator ions, which are those ions that are found on both sides of the equation

#"Al"_ ((aq))^(3+) + color(red)(cancel(color(black)(3"NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)(3"Na"_ ((aq))^(+)))) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr + color(red)(cancel(color(black)(3"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)(3"NO"_ (3(aq))^(-))))#

This will get you

#color(green)(|bar(ul(color(white)(a/a)color(black)("Al"_ ((aq))^(3+) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr)color(white)(a/a)|)))#

Now, I'm not really sure if the math formatting is to blame for some of the equations you wrote, but you're missing subscripts and charges in all of them.

Take the first equation, for example. #"Al"("NO"_3)# is actually #"Al"("NO"_3)_3#.

I'm assuming that $\text{Al"^3"OH}$ is actually ${\text{Al"^(3+)"OH}}^{-}$. If that is the case, then the correct version would be

${\text{Al"^(3+)("OH"^(-))_3 implies "Al"^(3+) + 3"OH}}^{-}$

There's no such thing as ${\text{Na}}_{3}$. Sodium cannot form molecules, it can exist either as a solid, $\text{Na}$, or a cation, ${\text{Na}}^{+}$.

In your case, the sodium cation was present in solution, so ${\text{Na}}_{3}$ is actually $3 {\text{Na}}^{+}$.

Once again, remember to always add charges when dealing with ions. An ion without an added charge is not actually an ion.

For example, there's no such thing as ${\text{NO}}_{3}$. Instead, add the $1 -$ charge that belongs to the ions to get ${\text{NO}}_{3}^{-}$ anion.

Also, keep in mind that a coefficient add to a soluble ionic compound gets distributed to all the ions that are produced by said compound in solution.

For example,

$\textcolor{red}{3} {\text{NaOH" = color(red)(3)("Na"^(+) + "OH"^(-)) = color(red)(3)"Na"^(+) + color(red)(3)"OH}}^{-}$

$\textcolor{b l u e}{2} {\text{Al"("NO"_3)_3 = color(blue)(2)("Al"^(3+) + 3"NO"_3^(-)) = color(blue)(2)"Al"^(3+) + 6"NO}}_{3}^{-}$

Notice that the charges must remain balanced at all times.

All in all, you should review ionic compounds before diving into complete and net ionic equations.

## Consider the following reaction: #Xe(g) + 2F_2(g) -> XeF_4(g)#. A reaction mixture initially contains 2.24 atm #Xe# and 4.27 atm #F_2#. lf the equilibrium pressure of #Xe# is 0.34 atm, how do you find the equilibrium constant (#K_p#) for the reaction?

Stefan V.
Featured 2 days ago

${K}_{p} = 25.3$

#### Explanation:

The equilibrium reaction given to you looks like this

${\text{Xe"_ ((g)) + color(red)(2)"F"_ (2(g)) rightleftharpoons "XeF}}_{4 \left(g\right)}$

You know that the mixture initially contains $\text{2.24 atm}$ of xenon, $\text{Xe}$, and $\text{4.27 atm}$ of fluorine gas, ${\text{F}}_{2}$.

It's worth mentioning that the problem provides you with pressures instead of moles because when the volume of the container remains unchanged, and the temperature of the reaction is kept constant, the pressure of a gas is directly proportional to the number of moles of gas present.

Now, notice that the pressure of xenon is decreasing to an equilibrium value of $\text{0.34 atm}$. This is a significant decrease, which can only mean that the equilibrium constant, ${K}_{p}$, is greater than $1$.

In other words, the forward reaction is favored at the temperature at which the reaction takes place.

Consequently, you can expect the equilibrium pressure of fluorine gas to be significantly lower than its initial value.

So, use an ICE table to find the equilibrium pressures of fluorine gas and xenon tetrafluoride

${\text{ ""Xe"_ ((g)) " "+" " color(red)(2)"F"_ (2(g)) " "rightleftharpoons" " "XeF}}_{4 \left(g\right)}$

#color(purple)("I")color(white)(aaaaacolor(black)(2.24)aaaaaaaacolor(black)(4.27)aaaaaaaaaaacolor(black)(0)#
#color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaacolor(black)((-color(red)(2)x))aaaaaaaaacolor(black)((+x))#
#color(purple)("E")color(white)(aaacolor(black)(2.24-x)aaaacolor(black)(4.27-color(red)(2)x)aaaaaaaaacolor(black)(x)#

Now, you know that the equilibrium concentration of xenon is $\text{0.34 atm}$. This means that you have

$2.24 - x = 0.34 \implies x = 1.90$

The equilibrium pressures of fluorine gas and xenon tetrafluoride will thus be

#("F"_2) = 4.27 - color(red)(2) * 1.90 = "0.47 atm"#

#("XeF"_4) = "1.90 atm"#

By definition, ${K}_{p}$ will be equal to

${K}_{p} = \left({\left({\text{XeF"_4))/(("Xe") * ("F}}_{2}\right)}^{\textcolor{red}{2}}\right)$

Plug in your values to find

#K_p = (1.90 color(red)(cancel(color(black)("atm"))))/(0.34color(red)(cancel(color(black)("atm"))) * ("0.47 atm")^color(red)(2)) = "25.3 atm"^(-2)#

${K}_{p} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{25.3} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ rounded to three sig figs

As predicted, the equilibrium constant turned out to be grater than $1$.

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