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Grace [21]
4 years ago
10

How many electrons are being shared between the atoms in this molecule

Physics
1 answer:
NemiM [27]4 years ago
5 0
. The formula for methane is CH4. There are 8 total valence electrons - 1 each for the hydrogen atoms, and 4 for the carbon atom. There are four single bonds, and no lone electron pairs in the methane molecule. With 8 valence electrons and 4 single bonds, we get 8/4=2 electrons per bond. 

<span>2. The methane molecule contains four single, or sigma, bonds. Each sigma bond is formed between the 1s orbital of a hydrogen atom, which contains one electron, and one of the four hybrid sp3 orbitals of the carbon atom, each of which contains one electron. That gives us 2 electrons per bond.

</span>
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La aceleración del camión que parte del reposo y alcanza la velocidad de 40 km/h en 5 segundos es de 2.22 m/s².

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Still don’t know what to do with this stuff
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What are things that we can do to protect our climate for future generations?
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6 0
3 years ago
Eva is in a closed, dark room. She uses her arm muscles to turn on a lamp. When she moves her hand closer to the lamp, the light
12345 [234]

Well, we know that the total energy in a closed system remains constant.

The problem with the story of Eva is that she is not in a closed system. 
If the dark room were really a closed system, then she could press the
button or turn the switch all day, and the lamp could not light.  It needs
electrical energy coming in from somewhere in order to turn on.

Let's say that Eva used her arm muscles to strike a match and light the
candle on the table.  Then we would have have food energy, muscle
energy, chemical energy in the match, chemical energy in the candle,
heat and light energy coming out of the candle, heat energy soaking into
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4 0
4 years ago
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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
3 years ago
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