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3 years ago

Which of the following are likely to form a covalent bond?

Physics

2 answers:

sesenic [268]3 years ago

8 0

The answer is
"C"
<span>Magnesium and chlorine</span>

Vera_Pavlovna [14]3 years ago

4 0

its c- sulfur and oxygen trust me

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What is cosmic ray spallation

love history [14]

The Cosmic Ray is a natural way for nuclear fission and nucleosynthesis to occur. It refers to the formation of chemical elements from the impact of cosmic rays on an object.

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3 years ago

Read 2 more answers

When an object experiences uniform circular motion the direction of the net force is?

guajiro [1.7K]

In an uniform circular motion, the direction of the net force on the object is radially inward, passing through the center of the circle.

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3 years ago

A bike, a truck, and a train—all without passengers, motors, or engines—roll down the same hill. Put the vehicles in order from

Pepsi [2]

Answer: WAIT WHATTTT i have that same test due today and the answer is in explanation

Explanation:

Bike, truck train. we are in the same school i think. Its imma say the incisal JMES Im Lusi i used to help in the library

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3 years ago

What is the acceleration of a car initially traveling at -5m/s and<br> reaching -22m/s in 3s.

lozanna [386]

Answer:

$a=-5.67\ m/s^2$

Explanation:

Given that,

Initial velocity, u = -5 m/s

Final velocity, v = -22 m/s

Time, t = 3s

We need to find the acceleration of the car. The formula of it is given by :

Acceleration,

$a=\dfrac{v-u}{t}\\\\a=\dfrac{(-22)-(-5)}{3}\\\\a=-5.67\ m/s^2$

So, the acceleration of the car is $-5.67\ m/s^2$ .

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3 years ago

The de broglie wavelength of an electron with a velocity of 6.00 × 106 m/s is ________ m. The mass of the electron is 9.11 × 10-

WINSTONCH [101]

Answer: $1.212(10)^{-10} m$

Explanation:

The de Broglie wavelength $\lambda$ is given by the following formula:

$\lambda=\frac{h}{p}$ (1)

Where:

$h=6.626(10)^{-34}\frac{m^{2}kg}{s}$ is the Planck constant

$p$ is the momentum of the atom, which is given by:

$p=m_{e}v$ (2)

Where:

$m_{e}=9.11(10)^{-28}g=9.11(10)^{-31}kg$ is the mass of the electron

$v=6(10)^{6}m/s$ is the velocity of the electron

This means equation (2) can be written as:

$p=(9.11(10)^{-31}kg)(6(10)^{6}m/s)$ (3)

Substituting (3) in (1):

$\lambda=\frac{6.626(10)^{-34}\frac{m^{2}kg}{s}}{(9.11(10)^{-31}kg)(6(10)^{6}m/s)}$ (4)

Now, we only have to find $\lambda$ :

$\lambda=1.2122(10)^{-10} m$ >>> This is the de Broglie wavelength of the electron

8 0

3 years ago