All categories

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

You might be interested in

Of the eight most abundant elements in earths crust, what percentage does the six next most abundant elements of earths crust ma

Mrac [35]

Answer:

the sixth most abundant element in the earth's crust is sodium and it occupies 2.8%

5 0

3 years ago

A human services professional has worked with a particular client for a couple of years. The client

zhuklara [117]

Answer:

Respect the client’s decision

Explanation:

just took the test

4 0

3 years ago

A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl

Sergeu [11.5K]

Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon $m_1$ and the mass of the basket $m_2$ , then according to newton's second law:

$(1). \:F = (m_1+m_2)a$ ,

where $a =0.265m/s^2$ is the upward acceleration, and $F = 688N$ is the net propelling force (counts the gravitational force).

Also, the tension $T$ in the rope is 79.8 N more than the basket's weight; therefore,

$(2). \:T = m_2g+79.8$

and this tension must equal

$T -m_2g =m_2a$

$(3). \:T = m_2g +m_2a$

Combining equations (2) and (3) we get:

$m_2a = 79.8$

since $a =0.265m/s^2$ , we have

$\boxed{m_2 = 301.13kg}$

Putting this into equation (1) and substituting the numerical values of $F$ and $a$ , we get:

$688N = (m_1+301.13kg)(0.265m/s^2)$

$\boxed{m_1 = 2295 kg}$

Thus, the mass of the balloon and the basket is 2295 kg and 301 kg respectively.

8 0

3 years ago

PLEASE HELP! I'LL GIVE BRAINLEST

melisa1 [442]

Answer:

Weight = 8.162 Newton.

Explanation:

Given the following data;

Mass = 2.2 kg

Acceleration due to gravity = 3.71 N/kg

To find the weight of the textbook;

Weight = mass * acceleration due to gravity

Weight = 2.2 * 3.71

Weight = 8.162 N

Therefore, the weight of the science textbook in mars is 8.162 Newton.

3 0

3 years ago

In 8.4 s a fisherman winds 2.9 m of fishing line onto a reel whose radius is 3.0 cm (assumed to be constant as an approximation)

alukav5142 [94]

Answer:

The angular speed of the reel is 11.33 rad/s

Explanation:

Given

The fisherman takes t = 8.4 s to wind distance x = 2.9 m into a circle radius of r = 3 cm = 0.03 m

Than the tangencial speed equals the change in the distance to the time

v = $\frac{x}{t}$ = $\frac{2.9 m}{8.4 s} = 0.34 \frac{m}{s}$

Knowing the tangencial velocity is proportional to the radius r and the angular velocity

v = r*w

w = $\frac{0.34 m/s}{0.03 m} = 11.33 \frac{rad}{s}$

3 0

3 years ago