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3 years ago

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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Two resistors, R1 and R2, are

dlinn [17]

The reciprocal of the total resistance is equal to the sum of the reciprocals of the component resistances:

1/(120.7 Ω) = 1/R₁ + 1/(221.0 Ω)

1/R₁ = 1/(120.7 Ω) - 1/(221.0 Ω)

R₁ = 1 / (1/(120.7 Ω) - 1/(221.0 Ω)) ≈ 265.9 Ω

3 0

3 years ago

Rank these temperatures from hottest to coldest: 32° F,32° C, and 32 K 320 F> 32° C>32 K 32°C 32° F 32 K 32° F 32 K 32° c

Leviafan [203]

Answer:

32 C > 32 F > 32 K

Explanation:

32 F, 32 C, 32 K

Let T1 = 32 F

T2 = 32 C

T3 = 32 K

Convert all the temperatures in degree C

The relation between F and C is given by

(F - 32) / 9 = C / 100

so, (32 - 32) / 9 = C / 100

C = 0

So, T1 = 32 F = 0 C

The relation between c and K is given by

C = K - 273 = 32 - 273 = - 241

So, T3 = 32 K = - 241 C

So, T 1 = 0 C, T2 = 32 c, T3 = - 241 C

Thus, T2 > T1 > T3

32C > 32 F > 32 K

3 0

3 years ago

A 5 N force pushes on the right side of a box. At the same time, a 10 N force pushes on the left side of the box. What happens t

olya-2409 [2.1K]

The answer is C as there is more force on the left side ( excess of 5 N) which therefore pushed it to the right with a force of 5 N!

8 0

3 years ago

Read 2 more answers

A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit

Ratling [72]

Answer:

The magnitude of angular acceleration is $232.38\ rad/s^2$ .

Explanation:

Given that,

Initial angular velocity, $\omega_i=3500\ rev/min=366.5\ rad/s$

When it switched off, it comes o rest, $\omega_f=0$

Number of revolution, $\theta=46=289.02\ rad$

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2$

So, the magnitude of angular acceleration is $232.38\ rad/s^2$ . Hence, this is the required solution.

6 0

3 years ago

A sled starts from rest,

Marrrta [24]

Answer:

25.6 m/s

Explanation:

Draw a free body diagram of the sled. There are two forces acting on the sled:

Normal force pushing perpendicular to the hill

Weight force pulling straight down

Take sum of the forces parallel to the hill:

∑F = ma

mg sin θ = ma

a = g sin θ

a = (9.8 m/s²) (sin 38.0°)

a = 6.03 m/s²

Given:

v₀ = 0 m/s

a = 6.03 m/s²

t = 4.24 s

Find: v

v = at + v₀

v = (6.03 m/s²) (4.24 s) + (0 m/s)

v = 25.6 m/s

5 0

3 years ago