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Hydrochloric acid + copper ll?

Michael
Featured 4 months ago

The green complex $\textsf{{\left[C u C {l}_{4}\right]}^{2 -}}$ is formed.

Explanation:

I am assuming we are in aqueous conditions.

The aqueous copper(II) ion consists of a central $\textsf{C {u}^{2 +}}$ ion surrounded by 6 $\textsf{{H}_{2} O}$ ligands:

It has the formula $\textsf{{\left[C u {\left({H}_{2} O\right)}_{6}\right]}^{2 +}}$ and is blue in colour. In solution it looks like this:

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

$\textsf{{\left[C u {\left({H}_{2} O\right)}_{6}\right]}^{2 +} + 4 C {l}^{-} r i g h t \le f t h a r p \infty n s C u C {l}_{4}^{2 -} + 6 {H}_{2} O}$

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(rarr)#

$\textsf{\text{ "color(blue)(blue)" } \textcolor{g r e e n}{g r e e n}}$

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 $\textsf{C u C {l}_{4}^{2 -}}$ ions.

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.

$\textsf{{\left[C u {\left({H}_{2} O\right)}_{6}\right]}^{2 +} + 4 C {l}^{-} r i g h t \le f t h a r p \infty n s C u C {l}_{4}^{2 -} + 6 {H}_{2} O}$

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#

$\textsf{\text{ "color(blue)(blue)" } \textcolor{g r e e n}{g r e e n}}$

Can someone explain how do I set up a molecular orbital diagram?

Truong-Son N.
Featured 4 months ago

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

1. Choose the set of atomic valence orbitals that each atom comes in with. Assume core orbitals don't interact. Determine your coordinate axes.
2. Have an approximate idea of the relative orbital energies if working with a heteronuclear diatomic molecule.
3. By conservation of orbitals, each of the two corresponding atomic orbital interacts to produce one bonding and one antibonding molecular orbital in the middle.
4. 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.
5. 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 $\text{CO}$.

CHOOSE YOUR ATOMIC ORBITALS & AXES

• Carbon has $2 s$ and $2 p$ atomic valence orbitals.
• Oxygen has $2 s$ and $2 p$ 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 $n s$ interaction:

For lithium through nitrogen (including nitrogen), orbital mixing occurs, so that the ${\sigma}_{n {p}_{z}}$ orbital is higher in energy than the ${\pi}_{n {p}_{x}}$ and ${\pi}_{n {p}_{y}}$ orbitals.

Therefore, carbon has its energy ordering switched compared to oxygen, and so, the $n p$ interactions look like this:

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.

Please can someone help with this electrochemistry problem involving the Nernst equation?

Michael
Featured 4 months ago

See below:

Explanation:

$\textsf{N A {D}^{+} + {H}^{+} + 2 e r i g h t \le f t h a r p \infty n s N a D H \text{ } {E}^{\circ} = - 0.11 \textcolor{w h i t e}{x} V}$

$\textsf{\left(a\right)}$

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

#sf(E=E^@-(RT)/(zF)ln([["reduced form"]]/(["oxidised form"]))#

At 298K this can be simplified to:

$\textsf{E = {E}^{\circ} - \frac{0.0591}{z} \log \left(\frac{\left[N a D H\right]}{\left[N a {D}^{+}\right] \left[{H}^{+}\right]}\right)}$

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

$\textsf{\left(b\right)}$

Putting in the numbers:

#sf(E=-0.11-0.591/2log((1)/((1)xx(10^(-7))))#

$\textsf{E = - 0.11 - 0.2068 = \textcolor{red}{- 0.32 \textcolor{w h i t e}{x} V}}$

$\textsf{\left(c\right)}$

It helps to consider the $\textsf{{E}^{\circ}}$ values:

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#

$\textsf{N A {D}^{+} + {H}^{+} + 2 e r i g h t \le f t h a r p \infty n s N a D H \text{ } {E}^{\circ} = - 0.11 \textcolor{w h i t e}{x} V}$

$\textsf{c F {e}^{3 +} + e r i g h t \le f t h a r p \infty n s c F {e}^{2 +} \text{ } {E}^{\circ} = + 0.25 \textcolor{w h i t e}{x} \textcolor{w h i t e}{x} V}$

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)#

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:

$\textsf{N a D H \rightarrow N a {D}^{+} + {H}^{+} + 2 e}$

$\textsf{c F {e}^{3 +} + e \rightarrow c F {e}^{2 +}}$

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

$\textsf{N a D H + 2 c F {e}^{3 +} + \cancel{2 e} \rightarrow N a {D}^{+} + {H}^{+} + \cancel{2 e} + 2 c F {e}^{2 +}}$

$\textsf{\left(d\right)}$

Now we have to use the full Nernst Equation:

$\textsf{{E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{z F} \ln Q}$

This can be simplified at 298K to:

$\textsf{{E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{0.0591}{z} \log Q}$

Where z = 2.

To find $\textsf{{E}_{c e l l}^{\circ}}$ there are different conventions around but they all rely on finding the arithmetic difference between the two $\textsf{{E}^{\circ}}$ values for the 1/2 cells.

The clearest way is to subtract the least +ve $\textsf{{E}^{\circ}}$ value from the most +ve.

Using the values in (c):

$\textsf{{E}_{c e l l}^{\circ} = + 0.25 - \left(- 0.11\right) = + 0.36 \textcolor{w h i t e}{x} V}$

From the complete equation the reaction quotient Q is written:

$\textsf{Q = \frac{\left[N a D\right] \left[{H}^{+}\right] {\left[c F {e}^{2 +}\right]}^{2}}{{\left[c F {e}^{3 +}\right]}^{2} \left[N a D H\right]}}$

pH = 7 $\therefore$$\textsf{\left[{H}^{+}\right] = {10}^{- 7} \textcolor{w h i t e}{x} \text{mol/l}}$

Putting the numbers into The Nernst Equation:

$\textsf{{E}_{c e l l} = + 0.36 - \frac{0.0591}{2} \log \left(\frac{1 \times {10}^{- 7} \times {0.004}^{2}}{{0.01}^{2} \times 1}\right)}$

$\textsf{{E}_{c e l l} = + 0.36 + 0.22885 \textcolor{w h i t e}{x} V}$

#sf(E_(cell)=color(red)(+0.59color(white)(x)V)#

What is the maximum number of grams of #PH_3# that can be formed when 6.2 g of phosphorus reacts with 4.0 g of hydrogen to form #PH_3#?

Meave60
Featured 4 months ago

The maximum mass of phosphane $\left({\text{PH}}_{3}\right)$ that can be produced under the conditions stated in the question is $\text{6.8 g}$.

Explanation:

$\text{P"_4 + "6H"_2}$$\rightarrow$$\text{4PH"_3}$

This is a limiting reactant problem. The reactant that produces the least $\text{PH"_3}$ determines the maximum number of grams of $\text{PH"_3}$

The process will go as follows:

#color(blue)("given mass P"_4"#$\rightarrow$#color(blue)("mole P"_4"#$\rightarrow$#color(red)("mol PH"_3"#$\rightarrow$#color(purple)("mass PH"_3"#

and

#color(blue)("given mass H"_2"#$\rightarrow$#color(blue)("mole H"_2"#$\rightarrow$#color(red)("mol PH"_3"#$\rightarrow$#color(purple)("mass PH"_3"#

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.

#color(green)("Molar Masses"#
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.

${\text{P}}_{4} :$#(4xx30.974"g/mol P")="123.896 g/mol P"_4"#

${\text{H}}_{2} :$#(2xx1.008"g/mol H")="2.016 g/mol H"_2"#

${\text{PH}}_{3} :$#(1xx30.974"g/mol P")+(3xx1.008"g/mol H")="33.998 g/mol PH"_3"#

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

Now you need to determine the mass of phosphane, $\text{PH"_3}$, produced individually by ${\text{6.2 g P}}_{4}$ and by $\text{4.0 g H"_2}$. I'm going to start with $\text{P"_4}$ since it is first in the equation.

#color(blue)("Moles of Phosphorus"#
Multiply the given mass of $\text{P"_4}$ by the inverse of its molar mass.

#6.2color(red)cancel(color(black)("g P"_4))xx(1"mol P"_4)/(123.896color(red)cancel(color(black)("g P"_4)))="0.05004 mol P"_4#

#color(red)("Moles of Phosphane"#
Multiply mol ${\text{P}}_{4}$ by the mole ratio,in the balanced equation, between $\text{P"_4}$ and ${\text{PH}}_{3}$ that will cancel $\text{P"_4}$.

#0.05004color(red)cancel(color(black)("mol P"_4))xx(4"mol PH"_3)/(1color(red)cancel(color(black)("mol P"_4)))="0.20016 mole PH"_3"#

#color(purple)("Mass of Phosphane"#
Multiply mol ${\text{PH}}_{3}$ by its molar mass.

#0.20016color(red)cancel(color(black)("mol PH"_3))xx(33.998"g PH"_3)/(1color(red)cancel(color(black)("mol PH"_3)))="6.8 g PH"_3"# (rounded to two sig figs due to 6.2 g and 4.0 g)

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

Now you need to determine the mass of phosphane that ${\text{4.0 g H}}_{2}$ can produce. I'm going to put the steps together into one equation, but it will contain all of the steps required.

#color(black)cancel(color(blue)(4.0"g H"_2))xxcolor(blue)((1color(black)cancel(color(blue)("mol H"_2)))/(2.016color(black)cancel(color(blue)("g H"_2)))xxcolor(red)((4color(black)cancel(color(red)("mol PH"_3)))/(6color(black)cancel(color(red)("mol H"_2)))xxcolor(purple)((33.998"g PH"_3)/(1color(black)cancel(color(purple)("mol PH"_3)))=color(purple)("45 g PH"_3#

Phosphorus is the limiting reactant, which means the maximum amount of phosphane that can be produced is $\text{6.8 g}$.

Which transition occurs when light with a wavelength of 434 nm is emitted by a hydrogen atom?

Stefan V.
Featured 4 months ago

$n = 5 \to n = 2$

Explanation:

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

$\frac{1}{l a m {\mathrm{da}}_{\text{e}}} = R \cdot \left(\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2}\right)$

Here

• $l a m {\mathrm{da}}_{\text{e}}$ is the wavelength of the emitted photon (in a vacuum)
• $R$ is the Rydberg constant, equal to $1.097 \cdot {10}^{7}$ ${\text{m}}^{- 1}$
• ${n}_{1}$ represents the principal quantum number of the orbital that is lower in energy
• ${n}_{2}$ represents the principal quantum number of the orbital that is higher in energy

Notice that you need to have ${n}_{1} < {n}_{2}$ in order to avoid getting a negative value for the wavelength of the emitted photon.

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)# $\text{m}$

Rearrange the Rydberg formula to isolate the unknown variables

$\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2} = \frac{1}{l} a m {\mathrm{da}}_{\text{e}} \cdot \frac{1}{R}$

Plug in your values to find

$\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2} = \frac{1}{4.34 \cdot \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{{10}^{- 7}}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{m"))) * 1.097 * color(blue)(cancel(color(black)(10^7)))color(red)(cancel(color(black)("m}}^{- 1}}}}}$

$\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2} = 0.21$

This will be equivalent to

$\frac{{n}_{2}^{2} - {n}_{1}^{2}}{{n}_{1} \cdot {n}_{2}} ^ 2 = \frac{21}{100}$

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

$\left\{\begin{matrix}{n}_{2}^{2} - {n}_{1}^{2} = 21 \\ {n}_{1} \cdot {n}_{2} = \sqrt{100}\end{matrix}\right.$

Use the second equation to write

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

Plug this into the first equation to get

${n}_{2}^{2} - {\left(\frac{10}{n} _ 2\right)}^{2} = 21$

${n}_{2}^{2} - \frac{100}{n} _ {2}^{2} = 21$

This is equivalent to

${n}_{2}^{4} - 100 = 21 \cdot {n}_{2}^{2}$

${n}_{2}^{4} - 21 {n}_{2}^{2} - 100 = 0$

If you take ${n}_{2}^{2} = x$, you will have

${x}^{2} - 21 x - 100 = 0$

Use the quadratic equation to get

${x}_{1 , 2} = \frac{- \left(- 21\right) \pm \sqrt{{\left(- 21\right)}^{2} - 4 \cdot 1 \cdot \left(- 100\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{21 \pm \sqrt{841}}{2}$

${x}_{1 , 2} = \frac{21 \pm 29}{2} \implies \left\{\begin{matrix}{x}_{1} = \frac{21 - 29}{2} = - 4 \\ {x}_{2} = \frac{21 + 29}{2} = 25\end{matrix}\right.$

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 $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$, you will have

${n}_{1} = \frac{10}{5} = 2$

This means that your transition is taking place from

$n = 5 \to n = 2$

which is part of the Balmer series.

Calculate the change in melting point of ice at 0°c when pressure is increased by 2 ATM. L(fusion)=796*4.186*10^7 erg/gm. Specific volume of water =1.0001cm^3 Specific volume of ice=1.0908 cm^3 ?

Truong-Son N.
Featured 4 days ago

$\Delta {T}_{f} \approx - {0.015}^{\circ} \text{C}$

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 ${10}^{10} \text{ergs}$ in $\text{1 kJ}$, so you should have had $\Delta {H}_{\text{fus" = 796 * 4.186 * 10^6 "erg/g}}$, or $\text{6.02 kJ/mol}$.
• 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 ${\text{1 g/cm}}^{3}$... and they are:

${\rho}_{w} = \frac{1}{1.0001} {\text{g/cm"^3 = "0.9999 g/cm}}^{3}$

${\rho}_{\text{ice" = 1/1.0908 "g/cm"^3 = "0.9168 g/cm}}^{3}$

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

$\frac{\mathrm{dP}}{\mathrm{dT}} = \left(\Delta {\overline{H}}_{\text{fus")/(T_fDeltabarV_"fus}}\right)$

where:

• $\frac{\mathrm{dP}}{\mathrm{dT}}$ is the slope of the two-phase coexistence curve on a phase diagram.
• $\Delta {\overline{H}}_{\text{fus}}$ is the change in molar enthalpy of fusion in $\text{kJ/mol}$.
• ${T}_{f}$ is the freezing point of the substance in $\text{K}$.
• $\Delta {\overline{V}}_{\text{fus" = barV_w - barV_"ice}}$ is the change in molar volume in $\text{L/mol}$ due to melting ice.

We already converted $\Delta {\overline{H}}_{\text{fus}}$ to the proper units. Now for the densities.

${\overline{V}}_{w} = {\left[\left(0.9999 \cancel{\text{g H"_2"O"))/cancel("cm"^3) xx cancel("1 cm"^3)/cancel"mL" xx (1000 cancel"mL")/"L" xx ("1 mol")/(18.015 cancel("g H"_2"O}}\right)\right]}^{- 1}$

#= 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 $\frac{\mathrm{dP}}{\mathrm{dT}}$.

In some small interval where the temperature and pressure change, we consider the reference temperature and pressure to be ${0}^{\circ} \text{C}$ at $\text{1 atm}$, the normal melting point:

$\frac{\mathrm{dP}}{{\mathrm{dT}}_{f}} \approx \frac{\Delta P}{\Delta {T}_{f}} = \left(\text{3 atm" - "1 atm")/(T_f' - "273.15 K}\right)$

The right-hand side of the equation becomes:

$\frac{\Delta P}{\Delta {T}_{f}} \approx \left(6.02 \cancel{\text{kJ""/"cancel"mol")/("273.15 K" cdot (0.01802 cancel"L""/"cancel"mol" - 0.01965 cancel"L""/"cancel"mol")) xx (0.082057 cancel"L"cdot"atm")/(8.314472 xx 10^(-3) cancel"kJ}}\right)$

$= - \text{133.44 atm/K}$

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. $\Delta {\overline{H}}_{\text{fus}} > 0$, while $\Delta {\overline{V}}_{\text{fus}} < 0$ and $T > 0$. This would be experimentally showing that 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:

${T}_{f} ' = \text{2 atm"/(-"133.44 atm/K") + "273.15 K}$

$=$ $\underline{\text{273.14 K}}$

Or $\textcolor{b l u e}{\Delta {T}_{f} \approx - {0.015}^{\circ} \text{C}}$. It should make sense that the increase in pressure makes ice easier to melt; the added pressure is like you squeezing hard on the ice.

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