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Natali5045456 [20]
3 years ago
6

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

Physics
2 answers:
Snowcat [4.5K]3 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]3 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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Una cuerda es puesta a vibrar 400 veces en 4 segundos Cual es la frecuencia del sonido emitido?
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n = number of vibrations set in the string = 400 vibrations

t = total time taken for "n" vibrations set in the string = 4 second

f = frequency of the sound emitted due to vibrations set in the string


frequency of the sound emitted due to vibrations set in the string is given as

f = n/t

inserting the above values in the above formula

f = 400/4

f = 100 Hz

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Read 2 more answers
A parallel plate capacitor with air between the plates has a potential difference of 71.0 V. Determine the potential difference
Gwar [14]

Answer:

15.8 V

Explanation:

The relationship between capacitance and potential difference across a capacitor is:

q=CV

where

q is the charge stored on the capacitor

C is the capacitance

V is the potential difference

Here we call C and V the initial capacitance and potential difference across the capacitor, so that the initial charge stored is q.

Later, a dielectric material is inserted between the two plates, so the capacitance changes according to

C'=kC

where k is the dielectric constant of the material. As a result, the potential difference will change (V'). Since the charge stored by the capacitor remains constant,

q=C'V'

So we can combine the two equations:

CV=CV'\\CV=(kC)V'\\V'=\frac{V}{k}

and since we have

V = 71.0 V

k = 4.50

We find the new potential difference:

V'=\frac{71.0}{4.50}=15.8 V

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