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melomori [17]
3 years ago
9

How do lone pairs of electrons affect the bond angle differently than electrons shared in a bond?

Physics
1 answer:
lubasha [3.4K]3 years ago
3 0

Answer:

Lone pairs cause bond angles to deviate away from the ideal bond angles

Explanation:

Bonded electrons are stabilized and clustered between the bonding electrons meaning they are much closer together. Non-bonding electrons however are not being shared between any atoms which allows them to roam a little further spreading the charge density over a larger space and therefore interfering with what would be an expected bond angle

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When is the next total solar eclipse in illinois
atroni [7]

April 8, 2024

Eclipse will be total over the southern part of the state.  Roughly everything south of Decatur and Springfield.

7 0
3 years ago
When are ionic bonds formed
allochka39001 [22]

Answer:

ionic bonds formed from the electrostatic attraction between oppositely charged ions in a chemical compound.

Explanation:

5 0
3 years ago
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on
g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

6 0
3 years ago
Set the radius to 2.0 m and the velocity to 1.0 m/s. Keeping the radius the same, record the magnitude of centripetal accelerati
jek_recluse [69]

Answer:

a=4\ m/s^2

Explanation:

Given that,

Radius, r = 2 m

Velocity, v = 1 m/s

We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{1}\\\\=4\ m/s^2

So, the magnitude of centripetal acceleration is 4\ m/s^2.

5 0
3 years ago
the amount of surface area of the block contact with the surface is 2.03*10^-2*m2 what is the average pressure exerted on the su
CaHeK987 [17]

Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ =  1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

6 0
3 years ago
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