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vivado [14]
2 years ago
12

Which of the following describes the electron sharing between hydrogen and fluorine A.Hydrogen and fluorine share one electron w

ith each other B.
Hydrogen and fluorine share seven electrons with each other C.Hydrogen shares seven electrons, while fluorine shares one electron. D.Hydrogen shares one electron, while fluorine shares seven electrons
Physics
1 answer:
atroni [7]2 years ago
5 0
Hydrogen has one electron in its outermost shell, while fluorine has seven electron in its outermost shell, hence both hydrogen and fluorine needs a single electron to complete its outermost shell. 
That's why there is a single bond between hydrogen and fluorine.
Hence both hydrogen and fluorine share one electron with each other, so option "A" is correct.
You might be interested in
The burj Khalifa in Dubai is the worlds tallest building. It rises to an amazing 828M above the ground and if you were to get to
antoniya [11.8K]
I don't understand what you mean by "depth" of the steps.  The flat part of the step has a front-to-back dimension, and the 'riser' has a height.  I don't care about the horizontal dimension of the step because it doesn't add anything to the climber's potential energy.  And if the riser of each step is 20cm high, then 3,234 of them only take him (3,234 x 0.2) = 646.8 meters up off the ground.  So something is definitely fishy about the steps.

Fortunately, we don't need to worry at all about the steps in order to derive a first approximation to the answer ... one that's certainly good enough for high school Physics.

In order to lift his bulk 828 meters from the street to the top of the Burj, the climber has to provide a force of 800 newtons, and maintain it through a distance of 828 meters.  The work [s]he does is (force) x (distance) = <em>662,400 joules. </em>
6 0
2 years ago
The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 600°C is 1 × 10-6. Calculate the number of vacancies
algol [13]

Answer :

The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 × 10^6cm^3/m^3

= 10.35 × 10^6 g/m^3

Avogadro's number = 6.022 × 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 × 10^-6

Nag = (Na * Da)/Aag

Where,

Nag = Total number of lattice sites in Ag

Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6

= 5.778 × 10^22/m^3.

6 0
2 years ago
A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second
Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

v_x = 15 m/s

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

d= v_x t =(15 m/s)(5 s)=75 m

5 0
2 years ago
An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular accele
leva [86]

To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.

Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.

The average angular acceleration

\alpha = \frac{\omega_f - \omega_0}{t}

Here

\alpha = Angular acceleration

\omega_{f,i} = Initial and final angular velocity

There is not initial angular velocity,then

\alpha = \frac{\omega_f}{t}

We know that the relation between the tangential velocity with the angular velocity is given by,

v = r\omega

Here,

r = Radius

\omega = Angular velocity,

Rearranging to find the angular velocity

\omega = \frac{v}{r}}

\omega = \frac{30}{0.20} \rightarrow Remember that the radius is half te diameter.

Now replacing this expression at the first equation we have,

\alpha = \frac{30}{0.20*6}

\alpha = 25 rad /s^2

Therefore teh average angular acceleration of each wheel is 25rad/s^2

3 0
2 years ago
Here are the positions at three different times for a bee in flight (a bee's top speed is about 7 m/s). Time 6.6 s 6.9 s 7.2 s P
Ber [7]

Answer:

(A.) (- 4.33, 6.33 , 0); (B.) (- 3.66, 7.5, 0); (C.) average at (A) (- 4.33, 6.33 , 0) ; (D.) (- 0.2165, 0.3165, 0)

Explanation:

Given the following :

Time - - - - - - - 6.6s - - - - - - - - - 6.9s - - - - - 7.2s

Position - (1.8,5.0,0) - (0.5,6.9,0) - - (−0.4,9.5,0)

(a) Between 6.6 s and 6.9 s, what was the bee's average velocity?

Vavg = Distance / time

[(0.5,6.9,0) - (1.8,5.0,0)] / 6.9 - 6.6

Vavg = [(0.5 - 1.8), (6.9 - 5.0), (0 - 0)] / 0.3

Vavg = - 1.3 / 0.3, 1.9/0.3, 0/3

Vavg = (- 4.33, 6.33 , 0)

b) Between 6.6 s and 7.2 s, what was the bee's average velocity?

Vavg = [(−0.4,9.5,0) - (1.8,5.0,0)] / 7.2 - 6.6

Vavg = - 2. 2/0.6, 4.5/0.6, 0/0.6

Vavg = (- 3.66, 7.5, 0)

c.) Of the two averages (- 4.3, 6.3 , 0) is closer to the instantaneous Velocity at 6.6s

D.) (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s?

Displacement = Velocity * time

Vavg between 6.6 to 6.9 ; time = (6.65 - 6.6) = 0.05 s

= (- 4.33, 6.33 , 0) * 0.05

= (- 0.2165, 0.3165, 0)

5 0
2 years ago
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