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Featured 3 months ago

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,

So, bromine is located in period 4, group 17 of the periodic table and has an atomic number equal to **electrons** surrounding its nucleus.

The electron configuration of a neutral bromine atom looks like this

#"Br: " 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 color(red)(4) s^2 color(red)(4)p^5#

Notice that bromine's *outermost electrons*, i.e. its valence electrons, are located on the **fourth energy level**,

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

The *angular momentum quantum number*, **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

The *magnetic quantum number*, **exact orbital** in which the electron is located. The 4p-subshell contains a total of **p-orbitals**

#4p_x -> m_l = -1# #4p_y -> m_l = +1# #4p_z -> m_l = color(white)(-)0#

This is used *by convention* because the wave function associated with **symmetric** about its axis, I.e. it has no component of angular momentum about its axis, which is usually **chosen** as the

Since the neutral bromine atom already has **5** electrons in its 4p-subshell, you can say that its *completely filled* and the

The incoming electron will thus be added to the half-empty

Finally, the *spin quantum number*, *spin-up*, or **must** have an opposite spin, and so

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

#color(green)(|bar(ul(color(white)(a/a)color(black)(n = color(red)(4), l = 1, m_l = 0, m_s = -1/2)color(white)(a/a)|)))#

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

#4p_x -> m_l = -1# #4p_y -> m_l = color(white)(-)0# #4p_z -> 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* *has symmetry about the* *axis, so make sure that you specify and keep track of this for other orbitals such as the d-orbitals of f-orbitals*.

Featured 3 months ago

Use the ideal gas equation

Where:

Rearrange the formula and solve for V.

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

A quick approach

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

Featured 1 month ago

**!! LONG ANSWER !!**

The idea here is that adding *sodium hydroxide*, **neutralize** some, if not all, depending on how much you've added, of the acid.

You know that you start with

#"100.0 mL "-> 6.00 * 10^(-2)"M HCl"#

#"100.0 mL " -> 5.00 * 10^(-2)"M NaOH"#

Now, you know that after you realize your error, you're left with **added** to the resulting solution

#"100.0 mL " - " 81.0 mL" = "19.0 mL HCl"#

#"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

#"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))#

Because hydrochloric acid and sodium hydroxide produce hydrogen cations,

#"H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l))#

Notice that this reaction consumes hydrogen cations and hydroxide anions in a **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 **moles** of hydrogen cations and **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 = "0 moles OH"^(-) -># completely consumed

#0.00114 - 0.000550 = "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_"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)|)))#

In your case, you have

#"pH" = - log(0.01967) = 1.71#

So, you know that your **target solution** must have a volume of

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

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

For a volume equal to **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 **moles** of hydrogen cations and needs **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 **another**

This will give you a volume of

#color(green)(|bar(ul(color(white)(a/a)color(black)("volume of HCl needed " = " 42.9 mL")color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.

Featured 1 month ago

I get

**For #"BrCl"#**

**For #"N"_2"O"_4"#**

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)|)))" "#

∴

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)|)))" "#

∴

The formula for free energy change is

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

∴

The formula for

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

Featured 4 weeks ago

The surface tension is

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

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

If

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

In your problem,

For comparison, the surface tension of pure water at 20 °C is

Featured 7 hours ago

**WARNING! Very long answer!**

**a) After adding 0.10 mL #"HCl"#**

**b) After adding another 0.10 mol #"HCl"#**

The

**c) After adding 0.10 mL NaOH**

The strong base will dissociate completely.

**d) After adding another 0.10 mL #"NaOH"#**

In this case, adding the extra

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