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2 years ago

When four atomic orbitals are mixed to form hybrid orbitals, how many hybrid orbitals are formed?

Physics

2 answers:

lilavasa [31]2 years ago

6 0

When four atomic orbitals are mixed to form hybrid orbitals, 4 x sp3 hybrid orbitals are formed.

Carbon's 2s and all three of its 3p orbitals hybridize to form four sp3 orbitals. These orbitals then bond with four hydrogen atoms through sp3-s orbital overlap, creating methane.

Snezhnost [94]2 years ago

5 0

Answer:

Four

Explanation:

When atomic orbitals are combined, hybrid orbitals are formed. Hybrid orbitals exist only in covalent-bonded atoms and not in isolated atoms. The shapes and orientations of hybrids are different from that of atomic orbitals. The number of hybrid orbitals is same as the number of atomic orbitals that combined to form atomic orbitals. Thus, when 4 atomic orbitals are mixed then 4 hybrid orbitals are formed.

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Diferencie energia cinética de energia potencial gravitacional.

Vikentia [17]

Answer:

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The atoms and molecules in it are in constant motion. The kinetic energy of such a body is the measure of its temperature. Potential energy is classified depending on the applicable restoring force. Gravitational potential energy – potential energy of an object which is associated with gravitational force

4 0

2 years ago

Select the correct location on the image. Identify the magnetic north pole of Earth’s magnet.

OverLord2011 [107]

The magnetic north pole of the earth's magnet is in the geographic south pole.

There are two magnetic and geographic poles each, north and south
The two geographic poles are the locations where the earth's axis of rotation passes through which is imaginary
The magnetic north and south poles are not the same as the geographic north and south poles
In a compass, the needle points to the magnetic north pole
By convention, the magnetic north pole corresponds to the geographic south pole
The magnetic south pole corresponds to the geographic north pole
The magnetic field lines of a magnet start from the magnetic north pole and end at the magnetic south pole

The magnetic north pole of the earth's magnet is the geographic south pole.

Learn more about earth's magnetism here:

brainly.com/question/3928159

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8 0

1 year ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$

where

m = 950 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\Delta h$ is the change in height, which is

$\Delta h = L sin 30^{\circ}$

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

$\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J$

8 0

2 years ago

During an eclipse the sun the earth and the moon act as a light source,an obstacle or screen=

Zigmanuir [339]

Answer:

Un eclipse solar se produce cuando la luna se interpone en el camino de la luz del sol y proyecta su sombra en la Tierra. Eso significa que durante el día, la luna se mueve por delante del sol y se pone oscuro. ... Este eclipse total se produce aproximadamente cada año y medio en algún lugar de la Tierra.

Explanation:

espero que te

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6 0

2 years ago

What would be the escape speed for a craft launched from a space elevator at a height of 54,000 km?

Natasha_Volkova [10]

Answer: 3.63 km/s

Explanation:

The escape velocity equation for a craft launched from the Earth surface is:

$V_{e}=\sqrt{\frac{2GM}{R}}$

Where:

$V_{e}$ is the escape velocity

$G=6.67(10)^{-11} Nm^{2}/kg^{2}$ is the Universal Gravitational constant

$M=5.976(10)^{24}kg$ is the mass of the Earth

$R=6371 km=6371000 m$ is the Earth's radius

However, in this situation the craft would be launched at a height $h=54000 km=54000000 m$ over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:

$V_{e}=\sqrt{\frac{2GM}{R+h}}$

$V_{e}=\sqrt{\frac{2(6.67(10)^{-11} Nm^{2}/kg^{2})(5.976(10)^{24}kg)}{6371000 m+54000000 m}}$

Finally:

$V_{e}=3633.86 m/s \approx 3.63 km/s$

7 0

2 years ago