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Natali5045456 [20]
2 years ago
6

Why would you expect sodium (Na) to react strongly with chlorine (Cl)?

Physics
2 answers:
Snowcat [4.5K]2 years ago
5 0
3. is the answer, <span>Sodium needs to lose one electron, and chlorine needs to gain one electron. This is because Sodium's row always wants to give away an electron, while Chlorine's row wants to gain an electron.</span>
ASHA 777 [7]2 years ago
5 0

Answer: The correct option is (3) " Sodium needs to lose one electron, and chlorine needs to gain one electron ".

Explanation :

The electronic configuration of sodium (Na) is 1s^2\ 2s^2\ 2p^6\ 3s^1. It has only one vacant electron.

While the electronic configuration of chlorine is 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^5. It has seven valance electrons.

When Na reacts with Cl, Na will lose one electron and this lost electron gets transferred to chlorine. This forms a strong ionic bond.

So, the correct option is (3) " Sodium needs to lose one electron, and chlorine needs to gain one electron ".

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Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at
aliina [53]

Answer:

a) 8.99*10³ V  b) 4.5*10⁻² J c) 0 d) 0

Explanation:

a)

  • The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
  • For a point charge, it can be expressed as follows:

        V =\frac{k*q}{d}

  • As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
  • This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
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        V = \frac{2*q*k}{d} = \frac{2*8.99e9N*m2/C2*4e-6C}{8m} =\\ \\ V= 8.99e3 V

  • The potential at point C is 8.99*10³ V

b)

  • The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

        W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J

  • The work needed is 0.045 J.

c)

  • If we replace one of the charges creating the potential at the point  C, by one of the same magnitude, but opposite sign, we will have the following equation:

       V = \frac{8.99e9N*m2/C2*(4e-6C)}{8m}  + (\frac{8.99e9N*m2/C2*(-4e-6C)}{8m}) = 0

  • This means that the potential due to both charges is 0, at point C.

d)

  • If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.
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Answer:

Hey there!

It is false. Theories can be repeatedly tested to give people an explanation of the world.

Let  me know if this helps :)

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C.

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Answer: a. air pollution

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The nuclear waste are radioactive and are non-biodegradable these wastes are disposed off deep in geospheres and in water. They have potential to contaminate both land and water. Radioactive wastes can cause mutations in the genome of the organisms exposed to these wastes which generate deadly diseases and disorders.  Therefore, these wastes are hazardous.

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Answer:

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However, methane can condense on Uranus and Neptune because they are farther from the sun and hence colder.

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