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Answer:

Because polarity is a continuum, and solubility lies somewhere on that continuum.......

Explanation:

Ethanol is miscible in water in all concentrations; and so is methanol. On the other hand petroleum ether, hexanes, is completely miscible in ethanol, whereas, believe it or not, methanol is IMMISCIBLE in hexanes. By the same token, should we have an aqueous solution of brine; addition of ethanol would precipitate the brine.

Why? Well, clearly ethanol has polar and non-polar functionality. The hydroxyl group promotes water solubility, whereas the hydrocarbyl tail allows some solubility in a non-polar solvent such as hexanes. The effect of the hydroxyl group is quite startling with respect to volatility. Ethanol has a normal boiling point of #78.4# #""^@C#; but ethane has a normal boiling point of #-89.0# #""^@C#; and propane has a normal boiling point of #-42.0# #""^@C#.

Throw in some polarity (but not hydrogen bonding), and we find that #"dimethyl ether"# has a normal boiling point of #-21.0# #""^@C#, and #"diethyl ether"# has a normal boiling point of #34.6# #""^@C#.

As regards the use of ethanol as a solvent for non-polar solutes, it is one of the most useful solvents in the laboratory, and it is probably the first solvent we would turn to for recrystallization of organic solutes. Organic solutes tend to have some solubility in HOT ethanol, whereas, upon cooling, the solutes tend to crystallize out - which is arguably the effect of the hydroxyl group (and thus ethanol is a preferred solvent to recrystallize such solutes).

Ethanol is also none too flammable; certainly less so than hexanes; I would happily use a heat gun on an ethanolic solution, whereas I would think twice about using it on a hexanes solution. As an important bonus, ethanol smells nice, and is not too hard on your hands.

Just to add for clarity. The ethanol we use in a lab is often known supplied as industrial methylated spirit. Why? Because when it is supplied it is adulterated with so-called denaturants such as phenol, or methanol (hence #"methylated spirit"#). And if you drink this stuff you go blind. But why should anyone want to drink ethanol?

It would be approximately half.

This physically says that at higher temperatures, the particle-wave moves faster. A faster particle-wave has a larger linear momentum, which means it has a larger average kinetic energy.

Therefore, since temperature is directly proportional to average kinetic energy of linear motion, the higher the temperature, the smaller the wavelength of the particle-wave.

This makes sense because as #lambda -> 0#, we reach the classical limit, where the particle-wave behaves less like a wave and more like a particle. This is also known as the high-temperature limit.

This agrees with the correspondence principle, that as #lambda->0#, discrete energies converge to become a continuum of energy:

http://hyperphysics.phy-astr.gsu.edu/


We first consider that the de Broglie wavelength is given by:

#lambda = h/(mv) = h/p#

where:

  • #m# is the mass of the particle in #"kg"#.
  • #v# is its speed in #"m/s"#.
  • #h = 6.626 xx 10^(-34) "J"cdot"s"# is Planck's constant.
  • #p# is the momentum of the particle in #"kg"cdot"m/s"#.
  • #lambda# is the wavelength in #"m"#.

Now, it is important to recognize that the kinetic energy of a particle with mass (so not photons) is given by:

#K = 1/2 mv^2 = p^2/(2m)#

And so, #p = sqrt(2mK)# is the forward momentum. Therefore, we can rewrite #lambda# in terms of #K#:

#lambda = h/sqrt(2mK)#

Next, consider that the translational average kinetic energy is given by the equipartition theorem in the classical limit:

#<< K >>_(tr) = K/N = 3/2 nRT#

where #N# is the number of particles in an ensemble, and #n#, #R#, and #T# are known from the ideal gas law. The #3# indicates the three dimensions in Cartesian coordinates, #x,y,z#.

Assuming that

  • A neutron has only translational motion,
  • We are at a suitably high temperature that the "classical limit" applies at both #27^@ "C"# and #927^@ "C"#,

then for a single neutron, #N = 1#, so that

#<< K >>_(tr) = K/1 = 3/2 nRT#.

We then have

#n = "1 particle" xx "1 mol"/(6.022 xx 10^23 "particles")#

#= 1.6605 xx 10^(-24) "mols"# of particles.

This can then be related back to the wavelength. Consider the wavelength at #27^@ "C" = "300.15 K"#:

#lambda_"300.15 K" = h/sqrt(3mnR cdot "300.15 K")#

And now, compare it to the wavelength at #"1200.15 K"#:

#lambda_"1200.15 K" = h/sqrt(3mnR cdot "1200.15 K")#

If we then represent the room-temperature wavelength as #lambda# and the high-temperature wavelength as #lambda'#, then the high-temperature wavelength is about half of the room-temperature wavelength:

#color(blue)(lambda') ~~ h/sqrt(3mnR cdot "300.15 K" cdot 4)#

#~~ 1/2 h/sqrt(3mnR cdot "300.15 K")#

#~~ color(blue)(lambda/2)#

Answer:

Here's what I find.

Explanation:

What it does

Lunar caustic is an effective oxidizing agent for many organic compounds

It kills microorganisms and reacts with the compounds in human flesh.

Thus, it is used for removing warts, destroying damaged or diseased tissues, and for stopping superficial bleeding.

Origin of "caustic"

Caustic means "able to corrode organic tissue by chemical action".

This is just what lunar caustic does.

Improper use of lunar caustic can cause chemical burns.

Thus the origin of "caustic" in the name is understandable.

Origin of "lunar"

The alchemists believed that silver was associated with the moon.

They called silver luna (from Latin Luna = "the Moon"), so their name for silver nitrate was "lunar caustic".

Yes, that is true, but only guaranteed for the system, not the surroundings. It is also important to note that since the system is isolated, #DeltaS_(surr) = 0#, and thus #DeltaS_(univ) > 0#.

[If the process was instead reversible (and consequently not spontaneous), then it's possible that #DeltaS_(sys) = 0#. When the reversible process is in an isolated system, it follows that #DeltaS_(univ) = 0#.]


A spontaneous process can only be reversed if a different path is taken than what was used going forward (since the microscopic reverse is nonspontaneous). Hence, this process must be irreversible.

Take this spontaneous reaction for example:

http://www.chem.ucla.edu/

The reverse must be nonspontaneous, since the role of products and reactants are switched.

An irreversible process (any real process) is inefficient. (A truly reversible process keeps the entropy of the universe constant.)

https://chem.libretexts.org/

Thus, some energy disperses as a result that was not used for the process itself.

https://www.slideshare.net/erletshaqe1/lecture-19-entropy

Since the system is isolated, no heat can escape it (the process is thus adiabatic), so when this flow of energy disperses inside the system, the entropy of the system increases, i.e. #DeltaS_(sys) > 0#.

Hence, the entropy of the system must increase for a spontaneous process in this isolated system.

And this occurs until the system has reached equilibrium, which is when the system entropy has maxed out, subject to the constraints placed upon the system.

In other words, the distribution of states is as spread out as it can get, and there is no macroscopic indication of an ongoing process.

They are quite complicated, and can often do combinations of #sigma#, #pi#, #delta#, and even #phi# bonding.

For an introduction into these kinds of bonds:

Inorganic Chemistry, Miessler et al.

#sigma# bonds are in every chemical bond. #pi# bonds start showing up in double and triple bonds (e.g. #"O"_2#, #"N"_2#, etc), #delta# bonds start showing up in quadruple bonds (see link), and #phi# bonds aren't seen until a sextuple bond is made (e.g. in #"Mo"_2# or #"W"_2#).


The #4f# orbitals can be separated into three types (here, we use the convention that outer atoms point their #y# axes inwards and #z# axes upwards):

#1)# Two lobes - #sigma# bonding only (#m_l = 0#)

  • The #f_(z^3)# (#m_l = 0#) is the only one that only #sigma# bonds. It can bond head-on along the #z# axis.

#2)# Six lobes - #sigma# and #pi# bonding, OR #phi# bonding only (#m_l = -3, +3, -1, +1#)

  • The #f_(y(3x^2 - y^2))# (#m_l = -3#) can #sigma# bond along the #x# axes (for example, with a #p_y# orbital) AND #pi# bond along the #y# axes (for example, with a #p_x# orbital, or a #d_(xy)# orbital).

It can alternatively form a #phi# bond (a six-lobed side-on overlap) along the #xy# plane (with another #f_(y(3x^2 - y^2))# orbital in a bimetallic complex).

  • The #f_(x(x^2 - 3y^2))# (#m_l = +3#) can #sigma# bond along the #y# axes (for example, with a #p_y# orbital) AND #pi# bond along the #x# axes (for example, with a #p_x# orbital, or a #d_(xy)# orbital).

It can alternatively form a #phi# bond (a six-lobed side-on overlap) along the #xy# plane (with another #f_(x(x^2 - 3y^2))# orbital in a bimetallic complex).

  • The #f_(yz^2)# (#m_l = -1#) can form decent #sigma# bonds along the #y# axes, AND/OR #pi# bonds along the #y# AND #z# axes.

It can alternatively form a #phi# bond (a six-lobed side-on overlap) along the #yz# plane (with another #f_(yz^2)# orbital in a bimetallic complex).

  • The #f_(xz^2)# (#m_l = +1#) can form decent #sigma# bonds along the #x# axes, AND/OR #pi# bonds along the #x# AND #z# axes.

It can alternatively form a #phi# bond (a six-lobed side-on overlap) along the #xz# plane (with another #f_(xz^2)# orbital in a bimetallic complex).

#3)# Eight lobes - #pi# bonding OR #delta# bonding (#m_l = -2, +2#)

  • The #f_(z(x^2 - y^2))# (#m_l = -2#) is for #pi# bonding along ANY of the axes, #x,y#, or #z#. The lobes lie above and below each of the axes, but also along them.

It can alternatively form a #delta# bond with another #f_(z(x^2 - y^2))# orbital in a bimetallic complex.

  • The #f_(xyz)# (#m_l = +2#) is for #delta# bonding along ANY of the planes (#xz, yz, xy#) (for example, with #d_(xy)#, #d_(xz)#, or #d_(yz)# orbitals).

It can alternatively form a #pi# bond with another #f_(xyz)# orbital in a bimetallic complex.

Answer:

Explanation:

From the reaction formula:
Copper is reduced as its oxidation state decreases from #+2# in #Cu^(2+)(aq)# to #0# in #Cu(s)#.
Cobalt is oxidized; its oxidation state increases from #0# in #Co(s)# to #+2# in #Co^(2+)(aq)#

Direction of electron flow
An element gains electrons as it undergoes reduction and loses electron when it undergoes oxidation. Therefore there's going to be a flow of electron from cobalt to copper through the external circuit.

Anode or cathode
"The cathode is where the reduction take place and oxidation takes place at the anode". (Chemistry Libretexts) [2]
Cobalt is being oxidized to form cobalt (II) ions so the cobalt electrode would be the anode. Copper (II) ions are reduced to elementary copper at the copper electrode, so that would be the cathode.

The way I memorize this is by considering where the two names for the voltaic electrodes came from. The #color(blue)("An")"ode"# of a cell, voltaic or electrochemical, attracts anions (ions with negative charges) and thus must carries some positive charges. So it is losing electrons and thus undergoing oxidation. Similarly, the #color(red)("Cat")"hode"# attracts cations (positively-charged ions), possesses excessive negative charges and therefore undergoes reduction. It's apparently a rule of thumb but it works for me.

Positive or negative terminal
Electrons always flow from the negative terminal to the positive terminal- exactly the opposite as the direction of the current flow. Therefore the cathode would be the positive terminal and the anode the negative terminal.

References
[1] "Galvanic Cells", CK-12 Editor, https://www.ck12.org/section/Galvanic-Cells/

[2] "Voltaic Cells", LibreText Chemistry, https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Voltaic_Cells

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