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2 years ago

8

Now imagine you have several of such dipoles, and place them regularly between the plates. For this part of the pre-lab, you can

ignore any interactions between the dipoles themselves, and think of each dipole as interacting with the electric field of the plates only. What will be different now

1 answer:

Dmitry_Shevchenko [17]2 years ago

6 0

Answer:

Dipole is two equal but opposite charges in one. As like charges repels and unlike charges attracts each other. If dipole are in electric field region, it positive side will be attracted to the negative field and vis-versa. This will result to different movement or stationary based on the orientation of the field. If the interaction between other dipole is considered this effect may be obstructed

Explanation:

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When 0.610 g of titanium is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 °C to 50.50

PtichkaEL [24]

Answer:

(ΔHrxn)= 0.15kJ/mol

Explanation:

(ΔHrxn) = mc∆t

=0.00061×9.84×(50.5-25)

= 0.153kJ/mol

8 0

2 years ago

One mole of an ideal gas in a closed system, initially at 25°C and 10 bar, is first expanded adiabatically, then heated isochori

Igoryamba

Answer:

$P_2=0.398bar=39800Pa$

$T_2=118.7K\\$

$Q=-3729.9J$

$W=-61753.24J$

Δ $U_T=0J$

Δ $H_T=0J$

Explanation:

Hello,

At the first state, the molar volume is:

$v_1=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^6Pa}=2.48x10^{-3}m^3$

The volume in both the second and third state:

$v_2=v_3=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^5Pa}=2.48x10^{-2}m^3$

Now, as it is about an adiabatic process, one remembers the following relationships:

$PV^\alpha =K\\TV^{\alpha-1}\\\alpha=\frac{Cp}{Cv}=\frac{7/2R}{5/2R}=1.4$

- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:

$P_2=\frac{P_1V_1^\alpha }{V_2^\alpha} =\frac{10bar*(2.48x10^{-3}m^3)^{1.4}}{(2.48x10^{-2}m^3)^{1.4}} =0.398bar=39800Pa$

- And the temperature:

$T_2=\frac{T_1V_1^{\alpha-1}}{V_2^{\alpha-1}} =\frac{298.15K*(2.48x10^{-3}m^3)^{1.4-1}}{(2.48x10^{-2}m^3)^{1.4-1}} =118.7K\\$

- Q:

It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:

$Q_2=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J$

Then the total heat:

$Q=Q_1+Q_2=0-3729.9J=-3729.9J$

- The work for the first process is:

$W_1=\frac{P_2V_2-P_1V_1}{1-\alpha }=\frac{39800Pa*2.48x10^{-3}m^3-1x10^6Pa*2.48x10^{-2}m^3}{0.4} \\W_1=-61753.24J$

It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:

$W=W_1+W_2=-61753.24J+0J=-61753.24J$

- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:

$U_1=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J\\U_2=nCv(T_3-T_2)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=3729.9J\\$

- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:

$H_1=nCp(T_2-T_1)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-5221.86J\\H_2=nCv(T_3-T_2)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=5221.86J\\$

Best regards.

8 0

2 years ago

A radioactive nuclide that is used for geological dating has an atomic number of 19 and mass number 40. Which is the symbol of t

Alina [70]

The mass number goes on top, and the atomic number goes on bottom.
Therefore, the answer is C:
40
K
19

7 0

2 years ago

Read 2 more answers

Joe is lifting boxes to load a trailer. To follow proper lifting procedures, Joe should:

masha68 [24]

<span>oe is lifting boxes to load a trailer. To follow proper lifting procedures, Joe should:</span>

3 0

2 years ago

Rusting of iron is a very common chemical reaction. It results in one form from Fe reacting with oxygen gas to produce iron (III

Vlada [557]

<u>Answer:</u> The given amount of iron reacts with 9.0 moles of $O_2$ and produce 6.0 moles of $Fe_2O_3$

<u>Explanation:</u>

We are given:

Moles of iron = 12.0 moles

The chemical equation for the rusting of iron follows:

$4Fe+3O_2\rightarrow 2Fe_2O_3$

<u>For oxygen gas:</u>

By Stoichiometry of the reaction:

4 moles of iron reacts with 3 moles of oxygen gas

So, 12.0 moles of iron will react with = $\frac{3}{4}\times 12.0=9.0mol$ of oxygen gas

<u>For iron (III) oxide:</u>

By Stoichiometry of the reaction:

4 moles of iron produces 2 moles of iron (III) oxide

So, 12.0 moles of iron will produce = $\frac{2}{4}\times 12.0=6.0mol$ of iron (III) oxide

Hence, the given amount of iron reacts with 9.0 moles of $O_2$ and produce 6.0 moles of $Fe_2O_3$

5 0

2 years ago