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GarryVolchara [31]
3 years ago
11

What is the difference between Polar and Non-Polar molecules?

Chemistry
1 answer:
Alekssandra [29.7K]3 years ago
7 0
<span>When you have a polar molecule, your bonds will not cancel out. This means that in a polar bond, the electronegativity of the atoms will attend to be different. For non-polar bonds the electro-negativity of the atoms will also be equal. In a polar bond you will have an unequal sharing of electron pairs which causes a molecular dipole. I hope this helps.

</span>
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Electrons in conductors, like copper wire, are free to roam throughout the metal in the presence of the ions that have released
Scrat [10]

Answer:

  • <u><em>No, I would not consider a metal to be a plasma because plasma is just another state of matter, and the copper wire is in solid state.</em></u>

Explanation:

Metal is not a state of matter. Metals can be solid or liquid (molten) depending on their melting point and the temperature at which they are.

Plasma is a state of matter, similar to gas, but it is reached only at very high temperatures like in the Sun. The particles in plasma state are not neutral atoms or molecules but negatively charged  ions and electrons.

The copper wire is yet a solid, thus it cannot be considered a plasma.

Metals can be in plasma state only if the temperature is too high, like the temperatures in the stars. In fact, the metals in the Sun and other hotter stars are in plasma state.

5 0
3 years ago
_________ are rocky materials mixed with very thick oil.
True [87]

Tar pits.

Need more help? Join our growing community here: https://questioncove.com/invite/098ZDB

4 0
3 years ago
Read 2 more answers
A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C・g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('
polet [3.4K]

Answer:

a=5.65cm

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

Q_{Pt}=-Q_{Deu}

We can represent the heats in terms of mass, heat capacities and temperatures:

m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})

Thus, we solve for the mass of platinum:

m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g

Next, by using the density of platinum we compute the volume:

V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3

Which computed in terms of the edge length is:

V=a^3

Therefore, the edge length turns out:

a=\sqrt[3]{180cm^3}\\ \\a=5.65cm

Best regards.

6 0
3 years ago
Which element is found in period 6 group 14<br>​
kicyunya [14]

Answer:

Lead

that can be found in the periodic table of elements

Explanation:

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3 years ago
You have two solutions, both with a concentration of 0.1M. Solution A contains a weak acid with a pKa of 5. The pH of solution A
viktelen [127]

Answer:guguhi

Explanation:

Guggugj

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3 years ago
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