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3 years ago

How many valence electrons does each Cholrine atom have?

Physics

1 answer:

Nat2105 [25]3 years ago

6 0

Answer:

7 valence electrons is right answer

Explanation:

2 elections in first shell and 5 electrons in second shell which total makes seven electrons.

hope it helped you:)

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Water in the oceans may become fresh water available to humans through the processes of

Darya [45]

<span>Water in the oceans may become fresh water available to humans through the processes of evaporation, condensation and precipitation.

In these processes, water is heated to a very high temperature until it evaporates in order to kill the germs and remove the salts which remains after water evaporation. The next step in condensing the water vapor (which is now fresh) and precipitating this vapor to be used by humans.</span>

4 0

3 years ago

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ

pantera1 [17]

Answer:

The period is $T = 0.700 \ s$

Explanation:

From the question we are told that

The mass is $m = 0.350 \ kg$

The extension of the spring is $x = 12.0 \ cm = 0.12 \ m$

The spring constant for this is mathematically represented as

$k = \frac{F}{x}$

Where F is the force on the spring which is mathematically evaluated as

$F = mg = 0.350 * 9.8$

$F =3.43 \ N$

$k = \frac{3.43 }{ 0.12}$

$k = 28.583 \ N/m$

The period of oscillation is mathematically evaluated as

$T = 2 \pi \sqrt{\frac{m}{k} }$

substituting values

$T = 2 * 3.142* \sqrt{\frac{0.35 }{28.583} }$

$T = 0.700 \ s$

7 0

3 years ago

A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a

Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C = 50 - x cm

We know that ,electric field is given as

$E=K\dfrac{q}{r^2}$

$K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\$