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Featured 4 months ago

Here's my 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

Where

Where

and

The other equation

Where E is electric field at the given location

J is current density

σ (sigma) is a material-dependent parameter called the conductivity

R is the electrical resistance of a uniform specimen of the material

A is the cross-sectional area of the specimen

Simplifying

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

If the

ρ depends on the conductor and is a constant

Here's a table of ρ of the conductor at

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

Where

ρ is resistivity

There are many version of the Ohm's law

Now lets come to the main problem.

Where Q is charge

PE is energy

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

As

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

The reaction

=

Net ionic reaction

Each mole of

Thus

The whole reaction

As 2C/s is applied for 2s

=

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

1 mole of

Thus moles of

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

=

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

x = mol of electrons used in the process

Featured 3 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 3 months ago

**DISCLAIMER:** *LONG ANSWER!*

#"NAD"^(+)(aq) + "H"^(+)(aq) + 2e^(-) -> "NADH"(aq)#

The **general Nernst equation** relates the non-standard

#E_"cell" = E_"cell"^@ - (RT)/(nF)lnQ#

We now know how to write

#Q = (["NADH"])/(["NAD"^(+)]["H"^(+)])#

Hence, the Nernst equation for this problem, which is for a half-cell, is:

#bb(E_"red" = E_"red"^@ - (RT)/(nF)ln((["NADH"])/(["NAD"^(+)]["H"^(+)]))#

- that the platinum electrode is inert and does not participate in the reaction.
- that the
**concentration of**#bb("H"^(+))# is#["H"^(+)] = 10^(-"pH") = 10^(-7) "M"# . - that from the half-cell reaction, the
**number of electrons transferred**is TWO.

We are also given

So, we just use the equation we wrote down in part

#color(blue)(E_"red") = -"0.11 V" - ("8.314472 J/mol"cdot"K"cdot"298 K")/("2 mol e"^(-)cdot"96485 C/mol e"^(-))ln(("1 M")/("1 M"cdot 10^(-7) "M"))#

#= color(blue)(-"0.32 V")#

This indicates that the reduction of **nonspontaneous** using these concentrations in these conditions.

*reduced by*

Since the important reaction is the reduction of the iron center, let's focus on that half-reaction. However, don't forget our other half-reaction. Since *oxidized*.

Thus, we flip our first half reaction, which was a reduction to begin with, and turn it into an oxidation half-reaction.

#2("Fe"^(3+)(aq) + cancel(e^(-)) -> "Fe"^(2+)(aq))#

#"NADH"(aq) -> "NAD"^(+)(aq) + "H"^(+)(aq) + cancel(2e^(-))#

#"-------------------------------------------------------------"#

#color(blue)(2"Fe"^(3+)(aq) + "NADH"(aq) -> "NAD"^(+)(aq) + "H"^(+)(aq) + 2"Fe"^(2+)(aq))#

I'll leave you to check that the mass and charge are balanced.

#"Fe"^(3+)(aq) + e^(-) -> "Fe"^(2+)(aq)# ,#E_"red"^@ = "0.25 V"#

We now essentially repeat what we did in parts

This is a pretty bulky reaction quotient expression...

#Q = (["NAD"^(+)]["H"^(+)]["Fe"^(2+)]^2)/(["Fe"^(3+)]^2["NADH"])#

Thus, the Nernst equation for this problem is:

#E_"cell" = E_"cell"^@ - (RT)/(nF)ln((["NAD"^(+)]["H"^(+)]["Fe"^(2+)]^2)/(["Fe"^(3+)]^2["NADH"]))#

We first have to figure out

Since we know from part

#E_"cell"^@ = E_"red"^@ + E_"ox"^@#

#= "0.25 V" + [-stackrel(E_"red"^@)overbrace((-"0.11 V"))]#

#=# #+"0.36 V"#

Now, we can calculate the **non-standard cell potential** at

#color(blue)(E_"cell") = "0.36 V" - ("8.314472 J/mol"cdot"K"cdot"298 K")/("2 mol e"^(-)cdot"96485 C/mol e"^(-))ln(("1 M"cdot10^(-7)"M"cdot("0.004 M")^2)/(("0.01 M")^2cdot"1 M"))#

#= color(blue)(+"0.59 V")#

Thus, the reduction of the iron center in cytochrome c by **spontaneous**.

Featured 3 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 2 months ago

The basic formula is

To answer this, we can take a look at the **ionic charges** of the elements involved.

Since oxygen is in group 16, it will have

Recall that an atom will want to gain or lose electrons to fulfill the *octet rule*; it wants to do whatever is easiest to obtain

For an oxygen atom, the easiest way to get

For the alkali metals, which are located in group 1 of the periodic table, they each have

In order to obtain an octet for each of the alkali metals, the easiest way to do it is by removing its one valence electron. This will cause each of the atoms (symbol

**The formula**:

The compound the alkali metal *electrically neutral*. That is to say, the charges must balance out to

The easiest way to write a formula of any ionic compound if you know the ionic charges of each species is to **put the charge of the anion (the negative ion, in this case is #"O"^(2-)#), #2-# as the ***. Here's what I mean, using sodium (

Featured 1 month ago

The **main postulates** or assumptions are (for ideal gases):

- Gases are constantly in motion, colliding
**elastically**in their container. - They
**do not interact**(i.e. no intermolecular forces are considered), and are assumed to be point masses with**negligible volume**compared to the size of their container. - Ensembles of gases produce a
**distribution of speeds**, and the average kinetic energy of this ensemble is*proportional*to the translational kinetic energy of the sample.

As for "one kinetic gas equation", there is not just one. But I can derive the Maxwell-Boltzmann distribution, and from there one could derive many other equations... some of them being:

- collision frequency
- average speed
- RMS speed
- most probable speed

The latter three are derived in more detail here.

**The following derivation is adapted from Statistical Mechanics by Norman Davidson (1969). Fairly old, but I kinda like it.**

Consider the **classical Hamiltonian** for the free particle in three dimensions:

#H = (p_x^2)/(2m) + (p_y^2)/(2m) + (p_z^2)/(2m)# ,where

#p = mv# is the momentum and#m# is the mass of the particle.

The **distribution function** will be denoted as

From *Statistical Mechanics* by Norman Davidson (pg. 150), we have that

#(dN)/N = (e^(-betaH(p_1, . . . , p_n; q_1, . . . , q_n)) dp_1 cdots dp_n dq_1 cdots dq_n)/zeta# ,where:

#beta = 1//k_BT# is a constant, with#k_B# as the Boltzmann constant and#T# as the temperature in#"K"# .#H(p_1, . . . , p_n; q_1, . . . , q_n)# is the Hamiltonian for a system with#3N# momentum coordinates and#3N# position coordinates, containing#N# particles in 3 dimensions.#zeta = int cdots int e^(-betaH(p_1, . . . , p_n; q_1, . . . , q_n)) dp_1 cdots dp_n dq_1 cdots dq_n# is the classical phase integral.

This seems like a lot, but fortunately we can look at three coordinates for simplicity.

A **phase integral** is an integral over *phase space*, a coordinate system where two conjugate variables are orthogonal to each other. In this case, position

With

#(dN(p_x,p_y,p_z; x,y,z))/(N) = ("exp"(-(p_x^2 + p_x^2 + p_z^2)/(2mk_BT))dp_xdp_ydp_z)/(int_(-oo)^(oo) int_(-oo)^(oo) int_(-oo)^(oo) "exp"(-(p_x^2 + p_x^2 + p_z^2)/(2mk_BT))dp_xdp_ydp_z)#

Now, let's take the bottom integral and separate it out:

#zeta = int_(-oo)^(oo) e^(-p_x^2//2mk_BT)dp_x cdot int_(-oo)^(oo) e^(-p_y^2//2mk_BT)dp_y cdot int_(-oo)^(oo) e^(-p_z^2//2mk_BT)dp_z#

In phase space, each of these integrals are identical, except for the particular notation. These all are of this form, which is tabulated:

#2int_(0)^(oo) e^(-alphax^2)dx = cancel(2 cdot 1/2) (pi/alpha)^(1//2)#

Notice how the answer has no

#zeta = (2pimk_BT)^(3//2)#

So far, we then have:

#(dN(p_x,p_y,p_z; x,y,z))/(N) = ("exp"(-(p_x^2 + p_x^2 + p_z^2)/(2mk_BT))dp_xdp_ydp_z)/(2pimk_BT)^(3//2)#

Next, it is convenient to transform into velocity coordinates, so let

#(dN(v_x,v_y,v_z; x,y,z))/(N) = (e^(-m^2(v_x^2 + v_y^2 + v_z^2)//2mk_BT)m^3dv_xdv_ydv_z)/(2pimk_BT)^(3//2)#

#= (m/(2pik_BT))^(3//2)e^(-m(v_x^2 + v_x^2 + v_z^2)//2k_BT)dv_xdv_ydv_z#

*Now, we assume that the gases are isotropic, meaning that no direction is any different from any other.*

To enforce that, suppose we enter **velocity space**, a coordinate system analogous to spherical coordinates, but with components of

#v_x = vsintheta_vcosphi_v#

#v_y = vsintheta_vsinphi_v#

#v_z = vcostheta_v#

With the differential volume element being

#(dN(v, theta_v, phi_v))/(N)#

#= (m/(2pik_BT))^(3//2) cdot e^(-m(v^2sin^2theta_v cos^2phi_v + v^2 sin^2theta_v sin^2phi_v + v^2 cos^2theta_v)//2k_BT)v^2 dv sintheta_v d theta_v d phi_v#

Fortunately, using

#= (m/(2pik_BT))^(3//2) cdot e^(-m(v^2sin^2theta_v(cos^2phi_v + sin^2phi_v) + v^2 cos^2theta_v)//2k_BT)v^2 dv sintheta_v d theta_v d phi_v#

#= (m/(2pik_BT))^(3//2) cdot e^(-m(v^2sin^2theta_v + v^2 cos^2theta_v)//2k_BT)v^2 dv sintheta_v d theta_v d phi_v#

#=> ul((dN(v, theta_v, phi_v))/(N) = (m/(2pik_BT))^(3//2)e^(-mv^2//2k_BT)v^2 dv sintheta_v d theta_v d phi_v)#

We're almost there! Now, the integral over allspace for

#int_(0)^(pi) sintheta_vd theta_v = 2#

And the integral over allspace for

#int_(0)^(2pi) dphi_v = 2pi#

So, we integrate both sides to get:

#color(blue)(barul(|stackrel(" ")(" "(dN(v))/(N) -= f(v)dv = 4pi (m/(2pik_BT))^(3//2)e^(-mv^2//2k_BT)v^2 dv" ")|))#

And this is the **Maxwell-Boltzmann distribution**, the one you see here:

**DERIVATION OF SOME OTHER EQUATIONS**

This is only a summary.

**Gas Speed Equations**

#<< v >> = int_(0)^(oo) vf(v)dv = sqrt((8k_BT)/(pim)#

#v_(RMS) = << v^2 >>^(1//2) = (int_(0)^(oo) v^2f(v)dv)^(1//2) = sqrt((3k_BT)/m)#

#v_(mp)# : Take the derivative of#f(v)# , set the result equal to#0# , and solve for#v# to get

#v_(mp) = sqrt((2k_BT)/m)#

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