All categories

3 years ago

What electron configurations do atoms usually achieve by sharing electrons to form covalent bonds?

Physics

1 answer:

Pani-rosa [81]3 years ago

7 0

Explanation: The atoms attain the configuration of inert gas by sharing electrons.

Lets take an example of HCl,

Atomic number of H is 1 and it has only 1 valance electron. whereas atomic number of chlorine is 17 and it has 7 valance electron.

When hydrogen atom combine with chlorine atom then they share electrons through covalent bonding in order to complete their octet.

According to figure,

Electronic configuration of H = 1

Electronic configuration of Cl = 2,8,7

Now, after share of electron

Electronic configuration of H = 2

Electronic configuration of Cl = 2,8,8

Hence, we can conclude that when atoms share electrons then their octet gets complete and atoms achieve noble gas electronic configuration.

You might be interested in

UDAY WAS TOLD TO PUT SOME CONTAINERS IN ONE OF THE COLD STORES AT WORK. THE LABLES ON THE CONTAINERS READ STORE BELOW -5 C.THERE

Alja [10]

He should choose the room that’s 15 F.

-5 C = 23 F
Meaning that 15 F is below -5 C and 25 F is not.

3 0

3 years ago

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a dire

svp [43]

Answer:

(A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

Explanation:

Given that,

Distance = 20.0 m

Frictional force = 35.0 N

Angle = 25.0°

(A). We need to calculate the work done on the cart by the friction

Using formula of work done

$W_{fr} = -F\cdot d$

Where, F = force

d = distance

Put the value into the formula

$W_{fr}=-35.0\times20$

$W_{fr}=−700\ J$

(B). The work done by the gravity is perpendicular to the direction of the motion

We need to calculate the work done on the cart by the gravitational force

Using formula of work done

$W=fd\cos\theta$

Put the value into the formula

$W=35.0\times20\cos90$

$W=0\ J$

(C). We need to calculate the work done on the cart by the shopper

Using formula of work done

$W_{sh}=W_{net}-W_{fr}$

Put the value into the formula

$W_{sh}=0-(-700)$

$W_{sh}=700\ J$

(D). We need to calculate the force the shopper exerts

Using formula of force

$F_{sh}=\dfrac{W_{fr}}{d\cos\theta}$

Put the value into the formula

$F_{sh}=\dfrac{700}{20\cos25}$

$F_{sh}=38.61\ N$

(E). We need to calculate the total work done on the cart

Using formula of work done

$W_{cart}=W_{fr}+W_{sh}$

Put the value into the formula

$W_{cart}=700-(-700)$

$W_{cart}=0\ J$

Hence, (A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

6 0

3 years ago

A basketball player shoots toward a basket 4.9 m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor a

Zolol [24]

Answer:

v₀ = 6.64 m / s

Explanation:

This is a projectile throwing exercise

x = v₀ₓ t

y = y₀ + v_{oy} t - ½ g t²

In this case they indicate that y₀ = 1.8 m and the point of the basket is x=4.9m y = 3.0 m

the time to reach the basket is

t = x / v₀ₓ

we substitute

y- y₀ = $\frac{ v_o \ x \ sin \theta }{ v_o \ cos \theta} - \frac{1}{2} g \ \frac{x^2 }{v_o^2 \ cos^2 \theta }$

y - y₀ = x tan θ - $\frac{ g \ x^2 }{ 2 \ cos^2 \theta } \ \frac{1}{v_o^2 }$

we substitute the values

3 -1.8 = 3.0 tan 60 - $\frac{ 9.8 \ 3^2 }{2 \ cos^2 60 } \ \frac{1}{v_o^2}$

1.2 = 5.196 - 176.4 1 / v₀²

176.4 1 / v₀² = 3.996

v₀ = $\sqrt{ \frac{ 176.4}{3.996} }$

v₀ = 6.64 m / s

6 0

3 years ago

11. If a man is standing more than one focal length away from the focal point of a concave mirror, how will his image form in th

Vladimir79 [104]

Upside down if its further than 1 focal point you have seen this with a spoon and enlarged right side up if closer than 1 focal point

5 0

3 years ago

Lori wants to send a box of oranges to a friend by mail. The box of oranges cannot exceed a mass of 10.222 Kg. If each orange ha

Sergeu [11.5K]

Explanation:

Given that,

The box of oranges cannot exceed a mass of 10.222 Kg if we are sending to a friend by mail.

The mass of each orange is 198 g

We know that,

1 kg = 1000 g

10.222 kg = 10.222×1000 g

Let there are n number of oranges. So,

$n=\dfrac{10.222\times 1000\ g}{198\ g}\\\\n=51.92\approx 52\ \text{oranges}$

It means she can send 52 oranges and it is maximum quantity.

4 0

3 years ago