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Featured 5 months 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 5 months 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 months ago

Several techniques can be used. They include - filtration, centrifugation, chromatography, crystallization...

Mixtures can be separated using various separation methods such filtration,separating funnel,sublimation,simple distillation and paper chromatography.

The methods stated above are all physical methods. There are also chemical methods, which are used by rearranging the particles so a certain substance no longer exists (chemical reaction). However here is my explanation on the four MAIN physical methods (the ones that show up on tests) of separation:

**1. Distillation.**

If two substances have different boiling points and are mixed together, you can boil them and the one with the lower boiling point will evaporate out.

Here is a video of an experiment which uses distillation to purify water from a solution of salt water.

Video from: Noel Pauller

**2. Chromatography** .

If you've ever done the experiments where you draw colored dots on a paper towel and dip it in water, this is that. Chromatography is when a substance is carried away (through the towel in this case) or spread around by the absorption of water. You can also think of a spill on a rug.

Here is a video which shows a paper chromatography experiment which was conducted to separate the pigments found in a black overhead marker.

Video from: Noel Pauller

**3. Crystallization.**

For a supersaturated solution, you can choose to let the solute crystallize out. Example: Rock candy (sugar solution)

**4. Filtration/decanting.**

Sort them out by particle size using a selective membrane such as filter paper.

The video below shows how filtration can be used to separate calcium carbonate (chalk) from water.

Video from: Noel Pauller

Featured 3 months ago

This is a kind of complicated question.

- A
**degenerate state**is a state in which the energy is the same as other states. - A
**degeneracy**is the number of states that have that same energy, and is described as#2l+1# .

*So, the difference is that degeneracy describes how many states, and a degenerate state is specifically which ones count.*

- Any
#ns# orbital is the same energy for the same#n# . - No,
#p# and#d# orbitals of the same#n# don't necessarily have the same energy (not even in hydrogen atom). - I cannot tell you the energy of every orbital, because their energies change throughout the periodic table, and sometimes the actual ordering is different.

**DEGENERATE STATES**

** Atomic** orbitals that share the

(if they share the same *same* orbital!)

On the other hand, if their *differ*, they might be *degenerate by coincidence*, but they are *not* necessarily degenerate.

For example, the

- They do not have the same shape (different
#l# ). - They do not have the same total number of nodes (different
#n - l - 1# ). - Possibly, they do not have the same number of each type of node (radial vs. planar).
- Maybe some combination of all three.

However, the

One might call them degenerate once the energies get close enough. When comparing their *counteracting* (lower *degenerate by coincidence*".

**DEGENERACY**

Like I said, degeneracy is just the number of orbitals of the same energy. Typically we say, for example:

- The
#np# atomic orbitals are, because there are*triply degenerate***three**of them in a free atom, and they are all equivalent orbitals with equivalent energies. - The
#np# atomic orbitals have aof*degeneracy*#2l + 1 = 3# , since their#l# is equal to#1# .

For example, a

#ul(uarr darr) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr))#

#underbrace(" "" "" "" "" "" "" "" ")#

#2p_x" "" "2p_y" "" "2p_z#

Featured 1 month ago

Well, referring to the **Heisenberg Uncertainty Principle**, there is one formulation of it that is fairly easy to use in calculations:

#bb(DeltaxDeltap_x >= ℏ)# ,

or

#bb(DeltaxDeltap_x >= h/(2pi))# ,

where

The **de Broglie wavelength** is:

#lambda = h/(mv)#

If the uncertainty in the position becomes numerically equal to

#cancel(h)/(mv_x)Deltap_x >= cancel(h)/(2pi)#

Since

#1/(cancel(m)v_x)cancel(m)Deltav_x >= 1/(2pi)#

#(Deltav_x)/v_x >= 1/(2pi)#

Flipping both sides, we get:

#color(blue)(v_x/(Deltav_x) <= 2pi)#

Since *small* (on the order of **the uncertainty in the velocity is very large** so that the inequality

That should make sense because if

**EXAMPLE**

For instance, the

#(2.188xx10^6)/(Deltav_x) <= 2pi#

#color(red)(Deltav_x >= 3.483 xx 10^5)# #color(red)("m/s")#

i.e. the uncertainty in the velocity of a ** least**) when you are sure of the position of the electron to within

**It physically means that if we were to try to predict its velocity, we are extremely unsure of which way it's going and at what actual velocity.**

In real life, if this were to be the case, then if you shined a laser through a slit of a few

Of course, the de Broglie relation is for electrons, as photons have no mass, but both behave as waves, and so, the slit experiment applies to both.

Featured 1 month ago

*A conceptual approach is to simply count electrons in a bond and treat each bonding valence electron as half a bond order.*

This works for many cases, except for when the highest-energy electron is in an antibonding molecular orbital.

**SIMPLE CASE**

For example, the **bond order** of *bonding* valence electron contributes half a bond order.

So:

#"BO"_"triple" = "BO"_sigma + 2"BO"_pi = 1/2 xx ("2 electrons") + 2(1/2 xx ("2 electrons")) = 3# for the bond order, as we should expect, since bond order tells you the "degree" of bonding.

**MULTI-ATOM CASE**

Or, in a more complicated example, like

So, in

That means its **bond order** is simply

#"BO" = "BO"_sigma + "BO"_pi = 1 + 0.333 = 1.333# .

Therefore, **three** "**bonds** overall (instead of one double bond and two single bonds), meaning it is one third of the way between a single bond and a double bond.

**EXCEPTION EXAMPLE: O2**

If we were to calculate its bond order, we would get

But what if we wanted the bond order for *What is it actually?*

You may realize that we would have removed one electron from an ** antibonding** molecular orbital. That means we've removed half a bond order corresponding to

So, by *removing* an *antibonding* electron, we've done the equivalent of *adding* a *bonding* electron.

**In other words, we've decreased a bond-weakening factor, thereby increasing the bonding ability of the molecule.**

Therefore, the actual bond order of

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