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Featured 4 months ago

The green complex

I am assuming we are in aqueous conditions.

The aqueous copper(II) ion consists of a central

It has the formula

If a large excess of chloride ions is added the water ligands are displaced as the following equilibrium is established:

Concentrated hydrochloric acid contains a large amount of chloride ions so Le Chatelier's Principle tells us that adding this will cause the position of equilibrium to shift to the right producing green

They have a tetrahedral structure:

The solution looks like this:

If excess water is now added the position of equilibrium is driven back to the left and the blue colour returns.

Featured 4 months ago

I will stick to diatomic molecules for simplicity. Some general steps are:

- Choose the set of
**atomic valence orbitals**that each atom comes in with. Assume core orbitals*don't*interact. Determine your coordinate axes. - Have an approximate idea of the
**relative orbital energies**if working with a heteronuclear diatomic molecule. - By conservation of orbitals, each of the
**two**corresponding atomic orbital interacts to produce**one bonding and one antibonding**molecular orbital in the middle. - By conservation of electrons, the valence electrons that are contributed by each atom
**equal the total number**of valence electrons distributed in the MO diagram. **Fill the diagram with electrons**as usual in accordance with the Aufbau principle, Hund's rule, etc.

Some general rules or tips are:

**Antibonding**orbitals*usually*are approximately as**higher**in energy as bonding orbitals are lower in energy than the original atomic orbital energies.**Nonbonding**orbitals have either a**similar**energy to the original atomic orbital energy, or have**both bonding and antibonding contributions**from the interacting atomic orbitals.- Large atomic orbital
*energy offsets*correspond to relative molecular orbital*energy offsets*. This helps with orbital energy ordering.

Let's construct an MO diagram for

**CHOOSE YOUR ATOMIC ORBITALS & AXES**

- Carbon has
#2s# and#2p# atomic valence orbitals. - Oxygen has
#2s# and#2p# atomic valence orbitals.

**ATOMIC ORBITAL ENERGY DIAGRAM**

From the original atomic orbital energies, we then construct the two **atomic orbital energy diagrams**:

(On an exam, you may only be expected to work with homonuclear diatomic molecules, in which case you probably don't have to worry about relative energies.)

**GENERATION OF MOLECULAR ORBITALS**

Then, each pair of atomic orbitals generates molecular orbitals as follows. Here is an

For lithium through nitrogen (including nitrogen), orbital mixing occurs, so that the

Therefore, carbon has its energy ordering switched compared to oxygen, and so, the

**OVERALL MO DIAGRAM**

Now put it together to get:

Note that on an exam, you probably aren't expected or required to know exactly how the interactions go. It's enough to know the relative energies of the molecular orbitals compared to the atomic orbitals, and to know how many valence electrons go in.

Featured 4 months ago

See below:

The potential **E** for a 1/2 cell is given by:

At 298K this can be simplified to:

Where **z** is the no. of moles of electrons transferred which, in this case =2.

Putting in the numbers:

It helps to consider the

The most +ve 1/2 cell is the one that will take in the electrons so you can see that the 1st 1/2 cell will move right to left and the 2nd 1/2 cell will move left to right in accordance with the arrows.

The two 1/2 equations are therefore:

To get the electrons to balance we X the 2nd 1/2 cell by 2 then add:

Now we have to use the full Nernst Equation:

This can be simplified at 298K to:

Where **z** = 2.

To find

The clearest way is to subtract the **least** +ve **most** +ve.

Using the values in (c):

From the complete equation the reaction quotient **Q** is written:

pH = 7

Putting the numbers into The Nernst Equation:

Featured 4 months ago

The maximum mass of phosphane

Start with a balanced equation.

This is a limiting reactant problem. The reactant that produces the least

The process will go as follows:

and

The molar masses of each substance is needed. Molar mass is the mass of one mole of an element, molecule or ionic compound in g/mol.

Multiply the subscript for each element by its atomic weight on the periodic table in g/mol. If there is more than one element, add the molar masses together.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now you need to determine the mass of phosphane,

Multiply the given mass of

Multiply mol

Multiply mol

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now you need to determine the mass of phosphane that

Phosphorus is the limiting reactant, which means the maximum amount of phosphane that can be produced is

Featured 4 months ago

All you have to do here is to use the **Rydberg formula** for a hydrogen atom

#1/(lamda_"e") = R * (1/n_1^2 - 1/n_2^2)#

Here

#lamda_"e"# is thewavelengthof the emitted photon (in a vacuum)#R# is theRydberg constant, equal to#1.097 * 10^(7)# #"m"^(-1)# #n_1# represents theprincipal quantum numberof the orbital that islower in energy#n_2# represents theprincipal quantum numberof the orbital that ishigher in energy

Notice that you need to have

So, convert the wavelength from *nanometers* to *meters*

#434 color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9color(red)(cancel(color(black)("nm")))) = 4.34 * 10^(-7)# #"m"#

Rearrange the Rydberg formula to isolate the unknown variables

#1/n_1^2 - 1/n_2^2 = 1/lamda_"e" * 1/R#

Plug in your values to find

#1/n_1^2 - 1/n_2^2 = 1/(4.34 * color(blue)(cancel(color(black)(10^(-7))))color(red)(cancel(color(black)("m"))) * 1.097 * color(blue)(cancel(color(black)(10^7)))color(red)(cancel(color(black)("m"^(-1)))))#

#1/n_1^2 - 1/n_2^2 = 0.21#

This will be equivalent to

#(n_2^2 - n_1^2)/(n_1 * n_2)^2 = 21/100#

At this point, you have two equations with two unknowns, since

#{(n_2^2 - n_1^2 = 21), (n_1 * n_2 = sqrt(100)) :}#

Use the second equation to write

#n_1 = 10/n_2" "color(darkorange)("(*)")#

Plug this into the first equation to get

#n_2^2 - (10/n_2)^2 = 21#

#n_2^2 - 100/n_2^2 = 21#

This is equivalent to

#n_2^4 - 100 = 21 * n_2^2#

#n_2^4 - 21n_2^2 - 100 = 0#

If you take

#x^2 - 21x - 100 = 0#

Use the **quadratic equation** to get

#x_(1,2) = ( -(-21) +- sqrt( (-21)^2 - 4 * 1 * (-100)))/(2 * 1)#

#x_(1,2) = (21 +- sqrt(841))/2#

#x_(1,2) = (21 +- 29)/2 implies {( x_1 = (21 - 29)/2 = -4), (x_2 = (21 + 29)/2 = 25) :}#

Since we know that

#n_2^2 = x#

you can only use the *positive* solution here, so

#n_2^2 = 25 implies n_2 = 5#

According to equation

#n_1 = 10/5 = 2#

This means that your transition is taking place from

#n=5 -> n=2#

which is part of the **Balmer series**.

Featured 4 days ago

#DeltaT_f ~~ -0.015^@ "C"#

This change was so small since the solid-liquid coexistence curve is often very steep. So the small pressure change typically doesn't alter the melting point that much.

A few things to keep in order:

- Seems like there's a typo. There are
#10^10 "ergs"# in#"1 kJ"# , so you should have had#DeltaH_"fus" = 796 * 4.186 * 10^6 "erg/g"# , or#"6.02 kJ/mol"# . - The specific volume of a substance just an old way to specify a reciprocal density. Not that it matters, because we know that the densities should be close to
#"1 g/cm"^3# ... and they are:

#rho_w = 1/1.0001 "g/cm"^3 = "0.9999 g/cm"^3#

#rho_"ice" = 1/1.0908 "g/cm"^3 = "0.9168 g/cm"^3#

Now, since you have to examine the change in melting point with pressure variance, consider the **Clapeyron equation**:

#(dP)/(dT) = (DeltabarH_"fus")/(T_fDeltabarV_"fus")# where:

#(dP)/(dT)# is theslope of the two-phase coexistence curveon a phase diagram.#DeltabarH_"fus"# is thechange in molar enthalpy of fusionin#"kJ/mol"# .#T_f# is thefreezing pointof the substance in#"K"# .#DeltabarV_"fus" = barV_w - barV_"ice"# is thechange in molar volumein#"L/mol"# due to melting ice.

We already converted

#barV_w = [(0.9999 cancel("g H"_2"O"))/cancel("cm"^3) xx cancel("1 cm"^3)/cancel"mL" xx (1000 cancel"mL")/"L" xx ("1 mol")/(18.015 cancel("g H"_2"O"))]^(-1)#

#= 1/("55.504 mol/L") = "0.01802 L/mol"#

#barV_"ice" = [(0.9168 cancel("g H"_2"O"))/cancel("cm"^3) xx cancel("1 cm"^3)/cancel"mL" xx (1000 cancel"mL")/"L" xx ("1 mol")/(18.015 cancel("g H"_2"O"))]^(-1)#

#= 1/("50.891 mol/L") = "0.01965 L/mol"#

Now, to calculate the change in melting point, we consider the slope given by

In some small interval where the temperature and pressure change, we consider the reference temperature and pressure to be **normal melting point**:

#(dP)/(dT_f) ~~ (DeltaP)/(DeltaT_f) = ("3 atm" - "1 atm")/(T_f' - "273.15 K")#

The right-hand side of the equation becomes:

#(DeltaP)/(DeltaT_f) ~~ (6.02 cancel"kJ""/"cancel"mol")/("273.15 K" cdot (0.01802 cancel"L""/"cancel"mol" - 0.01965 cancel"L""/"cancel"mol")) xx (0.082057 cancel"L"cdot"atm")/(8.314472 xx 10^(-3) cancel"kJ")#

#= -"133.44 atm/K"# where we multiplied by a ratio of universal gas constants to ensure the units worked out.

And if you look on the water phase diagram, this slope SHOULD be negative. This is because it requires **INPUT** of heat to melt (i.e. ** ice contracts when it melts**.

As a result, the left-hand side becomes:

#("2 atm")/(T_f' - "273.15 K") = -"133.44 atm/K"#

Solve for the **new freezing point** to get that:

#T_f' = "2 atm"/(-"133.44 atm/K") + "273.15 K"#

#=# #ul"273.14 K"#

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