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## What are examples of colloids?

Ernest Z.
Featured 5 months ago

Here are some common examples.

#### Explanation:

Colloids

A colloid is a mixture whose particles range in size from 1 nm to 1000 nm and do not settle out on standing.

The colloidal particles are distributed in a dispersing medium, which can be a solid, liquid, or a gas.

Examples of colloids

(a) Gas dispersed in liquid (foams)

Whipped cream consists of tiny air bubbles dispersed in the liquid cream.

(b) Gas dispersed in a solid (solid foams)

Marshmallows consist of tiny air bubbles dispersed in a solid consisting of sugar and gelatin.

(c) Liquid dispersed in gas (fogs or aerosols)

Common fog is a dispersion of water droplets in the air.

(d) Liquid dispersed in liquid (emulsions)

Milk is an emulsion of liquid butterfat globules dispersed in a water-based solution.

(e) Liquid dispersed in solid (solid emulsions)

Butter is an emulsion of water droplets in solid butterfat.

(f) Solid dispersed in gas (smokes)

Smoke is a dispersion of finely divided soot particles in the atmosphere.

(g) Solid dispersed in liquid (sols and gels)

Gelatin is a dispersion of the protein molecules of gelatin in water.

## What are three colligative properties of solutions?

Truong-Son N.
Featured 1 month ago

1) The lowering of the solvent's vapor pressure.
2) The decrease in the solvent freezing point.
3) The increase in the solvent boiling point.

Heck, I could list a fourth:

4) The increase in osmotic pressure.

VAPOR PRESSURE REDUCTION

This follows from Raoult's Law for ideal solutions:

${P}_{A} = {\chi}_{A \left(v\right)} P = {\chi}_{A \left(l\right)} {P}_{A}^{\text{*}}$

where:

• ${\chi}_{A \left(l\right)}$ is the mol fraction of the solvent $A$ in the liquid phase.
• ${\chi}_{A \left(v\right)}$ is the mol fraction of the solvent $A$ in the vapor phase.
• ${P}_{A}$ is the partial pressure of the solvent $A$ above the solution.
• ${P}_{A}^{\text{*}}$ is the partial pressure of the pure solvent $A$ at the surface of the liquid phase.

Raoult's law relates the change in vapor pressure (the partial pressure above the liquid phase) to the mol fraction of the dissolved solute (${\chi}_{B \left(l\right)}$) and the mol fraction of the vaporized solvent (${\chi}_{A \left(v\right)}$).

Given that ${\chi}_{A \left(l\right)}$ is always between $0$ and $1$, it follows that ${P}_{A} \le {P}_{A}^{\text{*}}$. For a two-component solution, ${\chi}_{A \left(l\right)} = 1 - {\chi}_{B \left(l\right)}$, where $B$ is the other component (generally the solute).

Thus, for any quantity of solute added, the vapor pressure of the solvent above the solution decreases.

FREEZING POINT DEPRESSION AND BOILING POINT ELEVATION

The chemical potential ${\mu}_{A}$ for some solvent $A$ in the solution as it relates to the mol fraction of $A$ in an ideal solution is:

${\mu}_{A} = {\mu}_{A}^{\text{*}} + R T \ln {\chi}_{A}$

where:

• ${\mu}_{A}$ is the chemical potential of the solvent in solution.
• ${\mu}_{A}^{\text{*}}$ is the chemical potential of the pure solvent.
• ${\chi}_{A}$ is the mol fraction of $A$ in the solution. ${\chi}_{A} = 1$ if no solute is dissolved.

Chemical potential is just the molar Gibbs' free energy; it tends downwards, i.e. $\Delta {\mu}_{A} < 0$ indicates a spontaneous process, or we say that ${\mu}_{A}$ runs downhill.

Given that any typical phase transition occurs at constant pressure, consider the following graph of ${\mu}_{A}$ vs. $T$:

Now, note that in the above equation, we know that ${\chi}_{A} \le 1$. It follows that $\ln {\chi}_{A} \le 0$, and that ${\mu}_{A} < {\mu}_{A}^{\text{*}}$. Therefore, $\mu$ for the solvent must decrease when solute is dissolved into the solvent.

When $\mu$ decreases for a liquid, the liquid line in the above diagram moves downwards, which shifts ${T}_{f}$ leftwards and ${T}_{b}$ rightwards.

This simultaneously explains why the solvent freezing point decreases when a solute is dissolved into a solvent, and why the solvent boiling point also increases (by less).

OSMOTIC PRESSURE ELEVATION

Imagine a U-shaped tube separated in the middle by a semi-permeable membrane.

Osmotic pressure is the pressure that needs to be applied to prevent a pure solvent on one side from flowing to the other side due to a concentration gradient.

Osmotic pressure is given by:

$\Pi = i M R T$

where $i$ is the van't Hoff factor (the effective number of particles per dissolved solute particle), $M$ is the molarity of the solution, and $R$ and $T$ are the universal gas constant and absolute temperature, respectively.

One can readily see that if $M \uparrow$, $\Pi \uparrow$ as well. Thus, higher concentration leads to higher osmotic pressure. The van't Hoff factor accentuates that, as it increases the effective concentration of the solute ($i \ge 1$).

Conceptually, when a higher concentration of solute is on one side of the semi-permeable membrane, a higher concentration gradient exists with respect to the other side; the solvent wishes to flow into the side where the concentration is higher (the side with higher concentration "thirsts" for more solvent).

Therefore, higher concentration of solute increases the osmotic pressure, since one needs to apply a greater amount of pressure to prevent the solvent from flowing to the side with higher concentration.

## Standard conditions to produce #"K"_2"MnO"_4# ?

Michael
Featured 1 month ago

You can make Mn(VI) from Mn(VII) and Mn(IV) in alkaline conditions.

#### Explanation:

We can use standard electrode potentials ( $\textsf{{E}^{\circ}}$) to set this up. In alkaline conditions we have:

$\textsf{\text{ "E^@" } \left(V\right)}$

$\textsf{\text{ "MnO_4^(-)+erightleftharpoonsMnO_4^(2-)" } + 0.56}$

$\textsf{2 {H}_{2} O + M n {O}_{4}^{2 -} + 2 e r i g h t \le f t h a r p \infty n s M n {O}_{2} + 4 O {H}^{-} \text{ } + 0.59}$

When listed -ve to +ve like this a general rule to predict the outcome of a redox reaction is to say "bottom left will oxidise top right."

On this basis we can say that $\textsf{M n {O}_{4}^{-}}$ will not be able to oxidise $\textsf{M n {O}_{2}}$ to #sf(MnO_4^(2-)# because the $\textsf{{E}^{\circ}}$ value is not +ve enough.

However, these values refer to standard conditions. Chemists are able to alter reaction conditions to drive the reaction in the direction they want.

Although $\textsf{+ 0.59 > + 0.56}$ the values are very close. If we raise the concentration of $\textsf{O {H}^{-}}$ Le Chatelier's Principle tells us that the 2nd 1/2 reaction should be driven to the left.

This will push out more electrons which will reduce the E value which makes the reaction thermodynamically feasible.

The two 1/2 reactions now become:

$\textsf{M n {O}_{2} + 4 O {H}^{-} \rightarrow 2 {H}_{2} O + M n {O}_{4}^{2 -} + 2 e \text{ } \textcolor{red}{\left(1\right)}}$

$\textsf{\text{ "MnO_4^(-)+erarrMnO_4^(2-)" } \textcolor{red}{\left(2\right)}}$

To get the electrons to balance we multiply #sf(color(red)((2))# by 2 then add to #sf(color(red)((1))rArr#

$\textsf{M n {O}_{2} + 4 O {H}^{-} + 2 M n {O}_{4}^{-} + \cancel{2 e} \rightarrow 2 {H}_{2} O + 3 M n {O}_{4}^{2 -} + \cancel{2 e}}$

To do this take 10 ml of $\textsf{0.01 \textcolor{w h i t e}{x} M}$ $\textsf{K M n {O}_{4}^{-}}$ solution and 5 ml of 1 M NaOH solution.

Add a little manganese(IV) oxide (#sf(MnO_2))# and shake for a few minutes.

Filter the solution. You should see green $\textsf{M n {O}_{4}^{2 -}}$ that looks like this:

This ion is only stable at high pH conditions. If you try adding a solution of dilute sulfuric acid then the process is reversed.

You see the purple color of #sf(MnO_4^(-)# return.

The net effect is Mn(VI) $\rightarrow$ Mn(VII) + Mn(IV)

When a species is simultaneously oxidised and reduced like this it is termed "disproportionation".

This all illustrates why many redox reactions are pH sensitive.

## Balance this reaction using ion electron method in shortest way possible . but must be clear .. #Fe^(2+)+MnO_4^(-)+H^(+) -> Fe^(3+)+Mn^(2+)+H_2O# ?

Truong-Son N.
Featured 1 month ago

$\boldsymbol{5 \stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) + 8stackrel(color(blue)(+1))("H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 5stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right)}$

Here are the primary steps for balancing in ACIDIC solution:

1. Pick out the elements being oxidized and reduced and write out their unbalanced half-reactions (this should require almost zero effort as this is the easy part).
2. Balance the non-$\boldsymbol{\text{O}}$ and non-$\boldsymbol{\text{H}}$ atoms.
3. Balance $\boldsymbol{\text{O}}$ using $\text{H"_2"O}$ molecules, since the solution is aqueous.
4. Balance $\boldsymbol{\text{H}}$ using ${\text{H}}^{+}$ ions, since the solution is acidic.
5. Balance the remaining charge using electrons on the side with more positive charge, since electrons must cancel out in the overall reaction.
6. Scale one or both reactions to cancel out the electrons.

In BASIC solution, add the step of adding ${\text{OH}}^{-}$ to both sides of the final reaction, combining ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ to make water, then canceling out common waters on each side.

Usually you don't have to do all of these steps, and the nice thing is that if you have to balance another reaction that involves the same oxidation or reduction half-reaction, you can recycle the same half-reaction.

OXIDATION HALF-REACTION

Iron gets oxidized from a $+ 2$ to $+ 3$ oxidation state. For pure ions, the charge is the oxidation state.

$\stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) -> stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right)$ (step 1)

Everything is balanced except for the charge, so we can skip straight to step 5.

$\stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) -> stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right) + {e}^{-}$ (step 5)

The total charge on each side is now:

$\left(+ 2\right) = \left(+ 3\right) + \left(- 1\right)$ #color(blue)(sqrt"")#

Done with the oxidation half-reaction.

REDUCTION HALF-REACTION

In some respects, this is the easier one to identify. Reduction can sometimes be thought of as the loss of oxygen atoms, and oxidation the gain of oxygen atoms.

Manganese can be readily seen to have lost oxygen atoms going from ${\text{MnO}}_{4}^{-}$ to ${\text{Mn}}^{2 +}$, which is why I suggested this easier way to spot reduction if applicable.

$\stackrel{\textcolor{b l u e}{+ 7}}{{\text{Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn}}^{2 +}} \left(a q\right)$ (step 1)

The non-oxygen and non-hydrogen atoms (i.e. $\text{Mn}$) are balanced, so we skip to step 3. Add water to balance the oxygens.

$\stackrel{\textcolor{b l u e}{+ 7}}{\text{Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)$ (step 3)

Add ${\text{H}}^{+}$ to balance the hydrogens.

$8 \stackrel{\textcolor{b l u e}{+ 1}}{\text{H"^(+))(aq) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)$ (step 4)

Balance the charge using electrons on the more positive side. Each side has:

#(??) + (+8) + (-1) = (+2)#

Hence, add $5 {e}^{-}$ to the left side.

$5 {e}^{-} + 8 \stackrel{\textcolor{b l u e}{+ 1}}{\text{H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)$ (step 5)

#(-5) + (+8) + (-1) = (+2) color(blue)(sqrt"")#

Done with the reduction half-reaction.

OVERALL REACTION

Now make the electrons cancel out. (Step 6)

$5 \left(\stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) -> stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right) + \cancel{{e}^{-}}\right)$
$\cancel{5 {e}^{-}} + 8 \stackrel{\textcolor{b l u e}{+ 1}}{\text{H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)$

Add the reactions to get: (Step 7)

$\boldsymbol{5 \stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) + 8stackrel(color(blue)(+1))("H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 5stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right)}$

## How does voltage effect the mass plated during electrolysis?

Professor Sam
Featured 4 weeks ago

Here's my explanation

#### Explanation:

According to Ohm's law the current passed through a material is directly proportional to the voltage.

You can see the mathematical equation that describes this proportionality

$I = \frac{V}{R}$

$V = I R$

Where $I$ is the current in the units of "amperes"
Where $V$ is voltage
and $R$ is the resistance of the material in units of "ohms"

The other equation

$E = J \sigma$

Where E is electric field at the given location
J is current density
σ (sigma) is a material-dependent parameter called the conductivity

$R = \setminus \rho \setminus \frac{l}{A}$

R is the electrical resistance of a uniform specimen of the material
$l$ is the length of the piece of material
A is the cross-sectional area of the specimen

Simplifying

$\setminus \rho = \setminus R \setminus \frac{A}{l}$

The resistance of a given material increases with length, but decreases with increasing cross-sectional area

If the $A = 1 {m}^{2}$ and $l = 1$ the resistance of the conductor is equal to the the ρ

ρ depends on the conductor and is a constant

Here's a table of ρ of the conductor at [email protected]/* */#

http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/rstiv.html

Note that ρ is different for conductors at different temperature

ρ increases as the temperature increases
ρ decreases as the temperature decreases

#sigma = 1/ρ#

Where $\sigma$ is conductivity
ρ is resistivity

There are many version of the Ohm's law

#P = (ΔV^2) / R#

$\Delta V$ is the potential difference between the two points in the conductor

#P = I^2 • R#

Now lets come to the main problem.

#ΔV =( ΔPE )/ Q#

Where Q is charge
PE is energy

By this you can understand that Voltage is proportional to charge which is proportional to coulombs

As

$C = V \cdot Q$
There is a relationship between V and C

As the V increases and the Q is thought to be constant the C increases. Means more voltage more coulombs

Now let me show you an example

Suppose that $2 C$/s is applied to a solution of $C u S O 4$ solution for 2 seconds.

The reaction

= $2 {H}_{2} O + 2 C u S {O}_{4} \rightarrow {O}_{2} + 2 {H}_{2} S O 4 + 2 C u$

Net ionic reaction

$2 {H}_{2} O + C {u}^{2 +} \rightarrow 2 {H}_{2} + 2 C u$

Each mole of $C {u}^{2 +}$ produces one mole of $C u$

$2 {H}_{2} O$ is reduced to ${H}_{2}$

${H}_{2} O + \rightarrow {H}_{2} + 2 {e}^{-}$
#Cu^(2+) + 2e^-) rarr Cu #

Thus
$2 {H}_{2} O \rightarrow 2 {H}_{2} + 4 {e}^{-}$
#2Cu^(2+) + 4e^-) rarr Cu #

The whole reaction

#2H_2O + 2Cu^(2+) + 4e^-) rarr 2H_2 + Cu + 4e^-)#
$2 {H}_{2} O + 2 C {u}^{2 +} \rightarrow 2 {H}_{2} + C u$

As 2C/s is applied for 2s

$\frac{2 C}{\cancel{s}} \times 2 \cancel{s}$

= $4 C$

Use the Faraday's constant to calculate the moles of electrons lost and gained in total.

#( 4 cancel"Coulombs")xx ("1 mole of electrons") / (96500 cancel"Coulombs") = "0.00004145077 mole of electrons"#

1 mole of $C {u}^{2 +}$ is reduced per 2 mole electrons

Thus moles of $C u$ formed

$\text{1 mol of Cu "/(2 molcancel( e^-)) xx "0.00004145077 mol of} \cancel{{e}^{-}}$

$\frac{1}{2} \times 0.00004145077 = \text{0.00002072538 mol of Cu}$

Now see what happens if we apply more charge

Consider the charge as 3C/s for 2s

Do the same calculations

As 3C/s is applied for 2s

$\frac{3 C}{\cancel{s}} \times 2 \cancel{s}$

= $6 C$

#( 6 cancel"Coulombs")xx ("1 mole of electrons") / (96500 cancel"Coulombs") = "0.00006217616 mole of electrons"#

$\text{1 mol of Cu "/(2 molcancel( e^-)) xx "0.00004145077 mol of" cancel(e^-)=" 0.00003108808 mol of Cu}$

$\text{0.00003108808 mol of Cu" > "0.00002072538 mol of Cu}$

Thus more the current,voltage and coulomb the more the mol of copper formed thus more the mass of copper formmed

Plotting a graph

If my handwriting is too bad

y = mol of Cu formed from reduction of $C {u}^{2 +}$
x = mol of electrons used in the process

## What is meant by a stress on a reaction at equilibrium?

Doc048
Featured 3 weeks ago

There are three 'Stress Factors' that affect the stability of an equilibrium:
1. Concentration Effects
2. Temperature Effects
3. Pressure-Volume Effects

#### Explanation:

Kinda long, but I had JPEGs in my files => Don't mind sharing.

A system (Reaction) at equilibrium is producing products as fast as it is producing reactants. This is known as the Law of Mass Action. That is ...

Rate of forward reaction = Rate of reverse reaction

When this condition exists, the reaction system is said to be in a 'stable' Dynamic Equilibrium'. While the concentrations of reactants and products appear to be constant, the chemical or physical process is still taking place.

To visualize this, think of the equilibrium process as being balanced on a See-Saw. Such as ...

If a stress is applied, the balance will be disturbed and the See-Saw will tilt toward the side of the equation to which the stress factor is applied.

The reaction will then shift away from the applied stress and establish a 'new' equilibrium with different concentration values for the reactants and products. => LeChatelier's Principle

Consider Concentration Effects:

Concept Check:
Consider the Rxn:
$F e O \left(s\right)$ + $C O \left(g\right)$ $\to$ $F {e}^{o} \left(s\right)$+$C {O}_{2} \left(g\right)$
Stress Factor => Removing $C O \left(g\right)$ (lighter reactant side) => Right tilt => Rxn Shifts Left to to establish new equilibrium.

Consider the Rxn:
$C O C {l}_{2} \left(g\right) \to C O \left(g\right) + C {l}_{2} \left(g\right)$
Stress Factor => Adding $C {l}_{2} \left(g\right)$ into the equilibrium mixture => (heavier product side) => Rxn balance tilts right => Rxn Shifts Left to to establish new equilibrium.

Stress Factor => removing $C {l}_{2} \left(g\right)$ from the from the equilibrium mixture => (Lighter Product Side) => Rxn balance would tilt Left => Rxn Shifts Right to establish new equilibrium.

Consider Temperature Effects:

Concept Check:
Consider the reaction:
$C O \left(g\right)$ + $2 {H}_{2} \left(g\right)$ -> $C {H}_{3} O H \left(G\right)$ + 21.7Kj
Stress Factor => Cooling rxn environment => product side is lighter => Rxn balance would tilt left => Rxn Shifts Right to establish new equilibrium.

Consider the reaction:
$2 {H}_{2} O$ + 484Kj -> $2 {H}_{2} \left(g\right)$+${O}_{2} \left(g\right)$
To increase decomposition => Stress Factor => Increase temperature around reaction environment => Rxn balance would tilt left => Rxn Shifts Right to establish new equilibrium.

Consider Pressure-Volume Effects:

Concept Check: Increasing pressure will ...
$2 {H}_{2} \left(g\right)$ + ${O}_{2} \left(g\right)$ $\to$ $2 {H}_{2} O \left(g\right)$ => Shifts Right

$2 {H}_{2} O \left(g\right)$ -> $2 {H}_{2} O \left(l\right)$ => Shifts Right [${V}_{\text{m" H_"2}} O \left(l\right) = 0$]

$C {H}_{4} \left(g\right)$+$2 {O}_{2} \left(g\right)$ -> $C {O}_{2} \left(g\right)$+$2 {H}_{2} O \left(l\right)$ => Shifts Right (P-V Effects apply to gas phase components only)

Same Rxn but note => all gas phase components.
$C {H}_{4} \left(g\right)$+$2 {O}_{2} \left(g\right)$ -> $C {O}_{2} \left(g\right)$+$2 {H}_{2} O \left(g\right)$ => No Shift Occurs
$\sum \left({V}_{m}\right) \left(R e a c \tan t s\right)$ = $\sum \left({V}_{m}\right) \left(P r o \mathrm{du} c t s\right)$

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