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## What are the set of four quantum numbers that represent the electron gained to form the #Br# ion from #Br# atom?

Stefan V.
Featured 3 months ago

$n = 4 , l = 1 , {m}_{l} = 0 , {m}_{s} = - \frac{1}{2}$

#### Explanation:

As you know, there are four quantum numbers used to describe the position and spin of an electron in an atom.

Your goal here will be to use the information provided by the electron configuration of a neutral bromine atom, $\text{Br}$, to determine the quantum numbers associated with the electron needed to form the bromide anion, ${\text{Br}}^{-}$.

So, bromine is located in period 4, group 17 of the periodic table and has an atomic number equal to $35$. This tells you that a neutral bromine atom has a total of $35$ electrons surrounding its nucleus.

The electron configuration of a neutral bromine atom looks like this

$\text{Br: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} \textcolor{red}{4} {s}^{2} \textcolor{red}{4} {p}^{5}$

Notice that bromine's outermost electrons, i.e. its valence electrons, are located on the fourth energy level, $n = \textcolor{red}{4}$.

The incoming electron will be added to this energy level, so right from the start you know that it must have $n = 4$.

The angular momentum quantum number, $l$, tells you the subshell in which the electron is located. In this case, the incoming electron will be added to the 4p-subshell, which is characterized, much like any p-subshell, by $l = 1$.

The magnetic quantum number, ${m}_{l}$, tells you the exact orbital in which the electron is located. The 4p-subshell contains a total of $3$ p-orbitals

• $4 {p}_{x} \to {m}_{l} = - 1$
• $4 {p}_{y} \to {m}_{l} = + 1$
• $4 {p}_{z} \to {m}_{l} = \textcolor{w h i t e}{-} 0$

This is used by convention because the wave function associated with ${m}_{l} = 0$ is said to be symmetric about its axis, I.e. it has no component of angular momentum about its axis, which is usually chosen as the $z$ axis.

Since the neutral bromine atom already has 5 electrons in its 4p-subshell, you can say that its $4 {p}_{x}$ and $4 {p}_{y}$ orbitals are completely filled and the $4 {p}_{z}$ contains one electron.

The incoming electron will thus be added to the half-empty $4 {p}_{z}$ orbital, and so it will have ${m}_{l} = 0$.

Finally, the spin quantum number, ${m}_{s}$, tells you the spin of the electron. Because the $4 {p}_{z}$ already contained one electron that has spin-up, or ${m}_{s} = + \frac{1}{2}$, the incoming electron must have an opposite spin, and so ${m}_{s} = - \frac{1}{2}$.

The quantum number set that describes the incoming electron will thus be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{n = \textcolor{red}{4} , l = 1 , {m}_{l} = 0 , {m}_{s} = - \frac{1}{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

SIDE NOTE You'll sometimes see this notation used for the magnetic quantum number

• $4 {p}_{x} \to {m}_{l} = - 1$
• $4 {p}_{y} \to {m}_{l} = \textcolor{w h i t e}{-} 0$
• $4 {p}_{z} \to {m}_{l} = + 1$

You can use this if you want, but make sure that you are consistent.

In this notation, the wave function associated with ${m}_{l} = 0$ has symmetry about the $y$ axis, so make sure that you specify and keep track of this for other orbitals such as the d-orbitals of f-orbitals.

## How many liters are in 16 grams of #H_2# at STP?

Sam
Featured 3 months ago

$V \cong 1.8 \times {10}^{2} \setminus L$

#### Explanation:

Use the ideal gas equation

$P \cdot V = n \cdot R \cdot T$

Where:

$P \to \text{ is the pressure expressed in } a t m$

$V \to \text{ is the volume occupied by the gas expressed in } L$

$n \to \text{ is the number of moles of the gas}$

$R \to \text{ is the universal gas constant} = 0.0821 \setminus L \cdot a t m \cdot m o {l}^{-} 1 \cdot {K}^{-} 1$

$T \to \text{ is the kelvin temperature }$

$- - - - - - - - - - - - - - - - -$

$S . T . P \text{conditions} \implies T = 273 K \mathmr{and} P = 1.00 \setminus a t m$

Rearrange the formula and solve for V.

$V = \frac{n \cdot R \cdot T}{P}$

Find the number of moles of the hydrogen gas present in the 16 grams.

${n}_{{H}_{2}} = 16 \setminus g \setminus {H}_{2} \times \frac{1 \setminus m o l . {H}_{2}}{2.016 \setminus g \setminus {H}_{2}}$

${n}_{{H}_{2}} \cong 7.9 \setminus m o l .$

$V = \frac{7.9 \setminus m o l . \times 0.0821 \setminus L \cdot a t m \cdot m o {l}^{-} 1 \cdot {K}^{-} 1 \times 273 \setminus K}{1.00 \setminus a t m}$

$V = \frac{7.9 \setminus \cancel{m o l .} \times 0.0821 \setminus L \cdot \cancel{a t m} \cdot \cancel{m o {l}^{-} 1} \cdot \cancel{{K}^{-} 1} \times 273 \setminus \cancel{K}}{1.00 \setminus \cancel{a t m}}$

$V \cong 1.8 \times {10}^{2} \setminus L$

$- - - - - - - - - - - - - - - - - - - -$

A quick approach

At S.T.P you can use the following formula:

${n}_{{H}_{2}} = \frac{V}{V} _ M$

${n}_{{H}_{2}} \text{ is the number of moles of the gas.}$

$V \text{ is Volume of the gas under S.T.P conditions}$

${V}_{M} \text{ is the molar volume i.e the volume occupied by 1 mole of }$
$\text{any gas under S.T.P conditions, equal to 22.4 L/mol.}$

${n}_{{H}_{2}} = 16 \setminus g \setminus {H}_{2} \times \frac{1 \setminus m o l . {H}_{2}}{2.016 \setminus g \setminus {H}_{2}}$

${n}_{{H}_{2}} \cong 7.9 \setminus m o l .$

$V = {n}_{{H}_{2}} \times {V}_{M}$

$V = 7.9 \setminus m o l \times 22.4 L \cdot m o {l}^{-} 1$

$V \cong 1.8 \times {10}^{2} \setminus L$

## Given the following information and assuming the final solution will be diluted to 1.00L, how much more HCl should you add to achieve the desired pH?

Stefan V.
Featured 1 month ago

$\text{42.9 mL}$

#### Explanation:

The idea here is that adding sodium hydroxide, $\text{NaOH}$, to your hydrochloric acid solution will neutralize some, if not all, depending on how much you've added, of the acid.

$\text{100.0 mL "-> 6.00 * 10^(-2)"M HCl}$

$\text{100.0 mL " -> 5.00 * 10^(-2)"M NaOH}$

Now, you know that after you realize your error, you're left with $\text{81.0 mL}$ of hydrochloric acid and $\text{89.0 mL}$ of sodium hydroxide. This means that you've added to the resulting solution

$\text{100.0 mL " - " 81.0 mL" = "19.0 mL HCl}$

$\text{100.0 mL " - " 89.0 mL" = "11.0 mL NaOH}$

When you mix hydrochloric acid, a strong acid, and sodium hydroxide, a strong base, a neutralization reaction takes place

${\text{HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

Because hydrochloric acid and sodium hydroxide produce hydrogen cations, ${\text{H}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$, respectively, in a $1 : 1$ mole ratio, you will have

${\text{H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O}}_{\left(l\right)}$

Notice that this reaction consumes hydrogen cations and hydroxide anions in a $1 : 1$ mole ratio, which means that for every mole of hydrochloric acid present it takes one mole of sodium hydroxide to neutralize it.

Use the molarities and volumes of the solutions you've mixed to calculate how many moles of each were added

#19.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace((6.00 * 10^(-2)"moles HCl")/(1color(red)(cancel(color(black)("L")))))^(color(blue)(=6.00 * 10^(-2)"M")) = "0.00114 moles H"^(+)#

#11.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace((5.00 * 10^(-2)"moles NaOH")/(1color(red)(cancel(color(black)("L")))))^(color(blue)(=6.00 * 10^(-2)"M")) = "0.000550 moles OH"^(-)#

So, you know that you've accidentally added $0.00114$ moles of hydrogen cations and $0.000550$ moles of hydroxide anions to your solution.

Since you have fewer moles of hydroxide anions present, you can say that the hydroxide anions will be completely consumed by the neutralization reaction, i.e. they will act as a limiting raegent.

Your resulting solution will thus contain

$0.000550 - 0.000550 = {\text{0 moles OH}}^{-} \to$ completely consumed

$0.00114 - 0.000550 = {\text{0.000590 moles H}}^{+}$

Now, the volume of this solution will be equal to the volume of hydrochloric acid and the volume of sodium hydroxide solutions you've mixed

${V}_{\text{sol" = "19.0 mL" + "11.0 mL" = "30.0 mL}}$

This means that the concentration of hydrogen cations in the resulting solution will be

#["H"^(+)] = "0.000590 moles"/(30.0 * 10^(-3)"L") = "0.01967 M"#

The pH of the solution is given by

#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"^(+)])color(white)(a/a)|)))#

$\text{pH} = - \log \left(0.01967\right) = 1.71$

So, you know that your target solution must have a volume of $\text{1.00 L}$ and a pH of $2.50$. Use the above equation to find the concentration of hydrogen cations needed to have this target solution

#"pH" = - log(["H"^(+)]) implies ["H"^(+)] = 10^(-"pH")#

#["H"^(+)] = 10^(-2.50) = "0.003162 M"#

For a volume equal to $\text{1.00 L}$, this gives you $0.003162$ moles of hydrogen cations -- remember that when dealing with a liter of solution, molarity and number of moles of solute are interchangeable.

Your solution contains $0.000590$ moles of hydrogen cations and needs $0.003162$ moles, which means that you must add

#n_("H"^(+)"needed") = 0.003162 - 0.000590 = "0.002572 moles H"^(+)#

Use the molarity of the stock hydrochloric acid solution to see what volume would contain this many moles of acid

#0.002572 color(red)(cancel(color(black)("moles H"^(+)))) * "1.0 L"/(6.00 * 10^(-2)color(red)(cancel(color(black)("moles H"^(+))))) = "0.0429 L"#

This is equivalent to $\text{42.9 mL}$, which means that in order to prepare your target solution, you must add another $\text{42.9 mL}$ of stock hydrochloric acid solution to the existing $\text{30.0 mL}$ solution.

This will give you a volume of $\text{72.9 mL}$ of solution. To get your target solution, you must add enough distilled water to get the total volume to $\text{1000 mL}$.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{volume of HCl needed " = " 42.9 mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.

## Estimate the value of the equilibrium constant at 610 K for each of the following reactions. ΔH∘f and S∘ for BrCl(g) is 14.6 kJ/mol and 240.0 Jmol⋅K, respectively. 2NO2(g)⇌N2O4(g)?

Ernest Z.
Featured 1 month ago

I get #K = 1.93 × 10^11# for $\text{BrCl}$ and #2.03 × 10^4# for ${\text{N"_2"O}}_{4}$.

#### Explanation:

For $\text{BrCl}$

$\text{Br"_2"(g)" + "Cl"_2"(g)" ⇌ "2BrCl(g)}$
#Δ_fH^° = "14.6 kJ·mol"^"-1"#
#ΔS^° = "240.0 J·K"^"-1""mol"^"-1"#

#Δ_fG^° = Δ_fH^° - TΔS^° = "14.6 kJ·mol"^"-1" - 610 color(red)(cancel(color(black)("K"))) × "0.2400 kJ"·color(red)(cancel(color(black)("K"^"-1")))"mol"^"-1"= "14.6 kJ·mol"^"-1" - "146.4 kJ·mol"^"-1" = "-131.8 kJ·mol"^"-1"#

#ΔG = "-"RTlnK#

#lnK = ("-"ΔG)/(RT) = ("131 800" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 610 color(red)(cancel(color(black)("K")))) = 25.99#

#K = e^25.99 = 1.93 × 10^11#

For $\text{N"_2"O"_4}$

$\textcolor{w h i t e}{m m m m m m m m m} 2 {\text{NO"_2 ⇌ "N"_2"O}}_{4}$
#Δ_fH^°"/kJ·mol"^"-1":color(white)(ll)33.10color(white)(mmm)9.08#
#S^°"/J·K"^"-1""mol"^"-1": color(white)(ll)240.04color(white)(mm)304.38#

The formula for enthalpy of reaction is

#color(blue)(|bar(ul(color(white)(a/a) Δ_rH = sumΔ_fH_text(products) - sumΔ_fH_text(reactants)color(white)(a/a)|)))" "#

# Δ_rH = "[1(9.08) - 2(33.01)] kJ" = "-56.94 kJ"#

The formula for the entropy of reaction is

#color(blue)(|bar(ul(color(white)(a/a)Δ_rS = sumS_text(products) - sumS_text(reactants)color(white)(a/a)|)))" "#

#Δ_rS = "(1×304.38 - 2×240.04) J/K" = "-175.80 J/K"#

#color(blue)(|bar(ul(color(white)(a/a)ΔG = ΔH - TΔScolor(white)(a/a)|)))" "#

#Δ_rG = "-56.94 kJ" - 610 color(red)(cancel(color(black)("K"))) × ("-0.175 80 kJ"·color(red)(cancel(color(black)("K"^"-1")))) = "-56.94 kJ" + "107.238 kJ" = "-50.30 kJ"#

The formula for $K$ is

#color(blue)(|bar(ul(color(white)(a/a)ΔG = -RTlnKcolor(white)(a/a)|)))" "#

#lnK = ("-"ΔG)/(RT) = ("50 300" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 610 color(red)(cancel(color(black)("K")))) = 9.918#

#K = e^"9.918" = 2.03 × 10^4#

## How would you calculate the surface tension of a 2% (w/v) solution of a wetting agent that has a density of 1.008 g/mL and that rises 6.6 cm in a capillary tube having an inside radius of 0.2 mm?

Ernest Z.
Featured 4 weeks ago

The surface tension is $\text{73 mJ·m"^"-2}$.

#### Explanation:

One method to measure the surface tension of a liquid is to measure the height the liquid rises in a capillary tube.

The formula is

#color(blue)(|bar(ul(color(white)(a/a) γ = "rhρg"/"2cosθ" color(white)(a/a)|)))" "#

where

#γ# = the surface tension
$r$ = the radius of the capillary
$h$ = the height the liquid rises in the capillary
#ρ# = the density of the liquid
$g$ = the acceleration due to gravity
#θ# = the angle of contact with the surface

For pure water and clean glass, the contact angle is nearly zero.

If #θ ≈ 0#, then #cosθ ≈ 1#, and the equation reduces to

#color(blue)(|bar(ul(color(white)(a/a) γ = "rhρg"/2 color(white)(a/a)|)))" "#

$r = \text{0.2 mm" = 2 × 10^"-4"color(white)(l) "m}$
$h = \text{6.6 cm" = "0.066 m}$
#ρ = "1.008 g/mL" = "1.008 kg/L" = "1008 kg·m"^"-3"#
$g = \text{9.81 m·s"^"-2}$

#γ = (2 × 10^"-4" color(red)(cancel(color(black)("m"))) × 0.066 color(red)(cancel(color(black)("m"))) × 1008 color(red)(cancel(color(black)("kg·m"^"-3"))) × 9.81 color(red)(cancel(color(black)("m·s"^"-2"))))/2 × ("1 J")/(1 color(red)(cancel(color(black)("kg")))·"m"^2color(red)(cancel(color(black)("s"^"-2"))))= "0.065 J·m"^"-2" = "65 mJ·m"^"-2" #

For comparison, the surface tension of pure water at 20 °C is $\text{73 mJ·m"^"-2}$.

## What is the pH of a a 10 mL, 0.10 M NH3 solution after the a) addition of 0.10 ml 1.00 M HCl b) of another 0.10 ml 1.00M HCl c) addition of 0.10 ml 0.10 M NaOH and d) addition of another 0.10 ml 0.10 M NaOH?

Ernest Z.
Featured 7 hours ago

#### Explanation:

a) After adding 0.10 mL $\text{HCl}$

${\text{Initial moles NH"_3 = 0.010 color(red)(cancel(color(black)("L NH"_3))) × "0.10 mol NH"_3/(1 color(red)(cancel(color(black)("L NH"_3)))) = "0.0010 mol NH}}_{3}$

$\text{Moles HCl added" = "moles NH"_3color(white)(l) "reacted = moles NH"_4^+ "formed}$
$= \text{0.000 10" color(red)(cancel(color(black)("L"))) × "1.00 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.000 10 mol}$

${\text{Moles NH"_3color(white)(l) "remaining = (0.0010 - 0.000 10) mol NH"_3 = "0.0009 mol NH}}_{3}$

$\text{NH"_3 + "H"_2"O" ⇌ "NH"_4^+ "+ OH"^"-"; "pK"_text(b)" = 4.75}$

$\text{B + H"_2"O" ⇌ "BH"^+ "+ OH"^"-}$

#"pOH" = "pK"_"b" + log((["BH"^+])/(["B"]))#

#"pOH" = 4.75 + log(("0.000 10")/0.0009) = 4.75 - 0.95 = 3.80#

$\text{pH = 14.00 - pH = 14.00 - 3.80 = 10.20}$

b) After adding another 0.10 mol $\text{HCl}$

$\text{Total moles HCl added" = "moles NH"_3color(white)(l) "reacted = moles NH"_4^+ "formed" = "0.000 20 mol}$

${\text{Moles NH"_3 "remaining = (0.0010 - 0.000 20) mol NH"_3 = "0.0008 mol NH}}_{3}$

#"pOH" = 4.75 + log((["BH"^+])/(["B"])) = 4.75 + log(0.0002/0.0008) = 4.75 - 0.60 = 4.15#

$\text{pH = 14.00 - pH = 14.00 - 4.15 = 9.85}$

The $\text{pH}$ of this solution should be close to that in Part a), because the solution in Part a) is a buffer.

c) After adding 0.10 mL NaOH

The strong base will dissociate completely.

$\text{Moles of NaOH" = "0.000 10" color(red)(cancel(color(black)("L NaOH"))) × "0.10 mol NaOH"/(1 color(red)(cancel(color(black)("L NaOH")))) = 1.0 × 10^"-5"color(white)(l) "mol}$

$V \text{= (10 + 0.10) mL = 10.1 mL = 0.0101 L}$

#["B"]_0 = "0.0010 mol"/"0.0101 L" = "0.099 mol/L"#

#["OH"^"-"]_0 =(1.0 × 10^"-5" "mol")/("0.0101 L") = 9.9 × 10^"-4"color(white)(l) "mol/L"#

$\textcolor{w h i t e}{m m m m m m m m} \text{B" + "H"_2"O" ⇌ "BH"^+ +color(white)(m) "OH"^"-}$
$\text{I/mol·L"^"-1":color(white)(mm) 0.099color(white)(mmmmml)0 color(white)(mmll)9.9 × 10^"-4}$
$\text{C/mol·L"^"-1":color(white)(mmll) "-"xcolor(white)(mmmmml)"+"xcolor(white)(mmmm)"+} x$
$\text{E/mol·L"^"-1":color(white)(ml)"0.099 -" x color(white)(mmmmll)xcolor(white)(ml)9.9 × 10^"-4} + x$

${K}_{\text{b" = (x(1.0 × 10^"-5" + x))/("0.099 -" x) = 10^"-4.75" = 1.78 × 10^"-5}}$

$x$ is negligible with respect to $0.099$, but not with respect to #9.9 × 10^"-4"#.

#(x(1.0 × 10^"-5" + x))/0.099 = 1.78 × 10^"-5"#

#1.0 × 10^"-5"x + x^2 = 1.78 × 10^"-5" × 0.099 = 1.76 × 10^"-8"#

#x^2 + 1.0 × 10^"-5"x - 1.76 × 10^"-8" = 0#

#x = 1.25 × 10^"-4"#

#["OH"^"-"] = 1.25 × 10^"-4" color(white)(l)"mol/L"#

#"pOH = -log"(1.25 × 10^"-4") = 3.90#

$\text{pH = 14.00 - 3.90 = 10.10}$

d) After adding another 0.10 mL $\text{NaOH}$

#"Moles of NaOH added" = 0.000 20 color(red)(cancel(color(black)("L NaOH"))) × "0.10 mol NaOH"/(1 color(red)(cancel(color(black)("L NaOH"))))#
#= 2.0 × 10^"-5"color(white)(l) "mol"#

$V \text{= (10 + 0.20) mL = 10.2 mL = 0.0102 L}$

#["B"]_0 = "0.0010 mol"/"0.0102 L" = "0.098 mol/L"#

#["OH"^"-"]_0 =(1.0 × 10^"-5" "mol")/("0.0102 L") = 9.8 × 10^"-4"color(white)(l) "mol/L"#

$\textcolor{w h i t e}{m m m m m m m m} \text{B" + "H"_2"O" ⇌ "BH"^+ +color(white)(m) "OH"^"-}$
$\text{I/mol·L"^"-1":color(white)(mm) 0.098color(white)(mmmmml)0 color(white)(mmll)9.8 × 10^"-4}$
$\text{C/mol·L"^"-1":color(white)(mmll) "-"xcolor(white)(mmmmml)"+"xcolor(white)(mmmm)"+} x$
$\text{E/mol·L"^"-1":color(white)(ml)"0.098 -" x color(white)(mmmmll)xcolor(white)(ml)9.8 × 10^"-4} + x$

${K}_{\text{b" = (x(1.0 × 10^"-5" + x))/0.098 = 10^"-4.75" = 1.78 × 10^"-5}}$

$x$ is negligible with respect to 0.098, but not with respect to #9.8 × 10^"-4"#.

#(x(1.0 × 10^"-5" + x))/0.098 = 1.78 × 10^"-5"#

#1.0 × 10^"-5"x + x^2 = 1.78 × 10^"-5" × 0.098 = 1.74 × 10^"-8"#

#x^2 + 1.0 × 10^"-5"x - 1.74 × 10^"-8" = 0#

#x = 1.25 × 10^"-4"#

#["OH"^"-"] = 1.25 × 10^"-4" color(white)(l)"mol/L"#

#"pOH = -log"(1.25 × 10^"-4") = 3.90#

$\text{pH = 14.00 - 3.90 = 10.10}$

In this case, adding the extra $\text{NaOH}$ has no effect because its concentration is negligible compared with that provided by the ${\text{NH}}_{3}$.

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