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labwork [276]
3 years ago
8

How can a molecule have a momentary dipole?????

Chemistry
2 answers:
chubhunter [2.5K]3 years ago
8 0
Here is my answer to your question <span> it is by a random </span>momentary<span> clustering of electrons in one particular point of an atom or </span>molecule that is how a molecule have a momentary dipole. AND.... hope u enjoy my answer.
lara [203]3 years ago
4 0
Electrons are constantly in movement. Temporarily dipoles can occur in non-polar molecules when electrons move around and spontaneously come into close proximity with each other and become concentrated on one atom. When this happens one atom becomes temporarily more negatively charged than the other atom. Thus creating a momentary dipole.
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How do scientist prevent biases from affecting their data?
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Answer:

Answer should be D. Scientists ignore their own person feelings and interpret data objectively.

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3 years ago
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An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr
Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa

Rate of flow of ideal gas , n = 4 kmol/hr = \frac{4\times 1000mol}{3600s}=1.11mol/s    (Conversion factors used:  1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW     (Conversion factor:  1 kW = 1000 W)

We know that:

\Delta U=0   (For isothermal process)

So, by applying first law of thermodynamics:

\Delta U=\Delta q-\Delta W

\Delta q=\Delta W      .......(1)

Now, calculating the work done for isothermal process, we use the equation:

\Delta W=nRT\ln (\frac{P_1}{P_2})

where,

\Delta W = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

P_1 = initial pressure = 100 kPa

P_2 = final pressure = 50 kPa

Putting values in above equation, we get:

\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0
3 years ago
A group of students is investigating whether aluminum is a better thermal conductor than steel. The students take an aluminum wi
Mademuasel [1]

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Ans B) Analyze the data

7 0
3 years ago
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- Which of the following is an example of physical change?
dalvyx [7]
B as the glass just changes form as it shattered but the chemical composition is same as it was before
5 0
4 years ago
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Answer:

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