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vladimir1956 [14]
3 years ago
13

The compound ethylene (C2H4) is represented by this diagram. What statement best describes the arrangement of the atoms in an et

hylene molecule? Each carbon atom shares two electrons with one hydrogen atom and also shares two electrons with the other carbon atom. Each carbon atom shares four electrons with one hydrogen atom and also shares four electrons with the other carbon atom. One electron is shared between each hydrogen atom and the carbon atom bonded to it, and two electrons are shared between the carbon atoms. Two electrons are shared between each hydrogen atom and the carbon atom bonded to it, and four electrons are shared between the carbon atoms.

Physics
2 answers:
Effectus [21]3 years ago
6 0

<u>Answer:</u> The correct answer is two electrons are shared between each hydrogen atom and the carbon atom bonded to it, and four electrons are shared between the carbon atoms.

<u>Explanation:</u>

Ethylene is a compound given by the chemical formula H_2C=CH_2.

The bond present between hydrogen and carbon atoms or carbon and carbon atoms are covalent bonds. A covalent bond is formed by the sharing of electrons between the atoms combining.

A double bond is present between carbon and carbon atoms. So 2 pairs of electrons are shared which means in total of 4 electrons are shared.

Bond present between hydrogen and carbon atoms are single bonds. So, a pair of electrons is shared which means that in total of 2 electrons are shared.

Hence, the correct answer is two electrons are shared between each hydrogen atom and the carbon atom bonded to it, and four electrons are shared between the carbon atoms.

tatuchka [14]3 years ago
6 0

Answer:

Two electrons are shared between each hydrogen atom and the carbon atom bonded to it, and four electrons are shared between the carbon atoms.

Explanation:

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Savatey [412]

Answer:

a). E_{kA}=1.2635 J

b). V_{B}=4.535\frac{m}{s}

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Explanation:

ΔE=kinetic energy

a).

E_{kA}=\frac{1}{2}*m*v_{A} ^{2} \\ v_{A}=1.9 \frac{m}{s}\\ m=0.70kg\\E_{kA}=\frac{1}{2}*0.70kg*(1.9 \frac{m}{s})^{2} \\E_{kA}=1.2635 J

b).

E_{kB}=\frac{1}{2}*m*v_{B} ^{2}

V_{B}^{2}=\frac{E_{kB}*2}{m} \\V_{B}=\sqrt{\frac{E_{kB}*2}{m}} \\V_{B}=\sqrt{\frac{7.2J*2}{0.70kg}} \\V_{B}=4.53 \frac{m}{s}

c).

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3 0
3 years ago
An object with a mass of 1.5 kg changes its velocity +15 m/s during a time interval of 3.5 seconds what impulse was delivered to
Arada [10]
The answer is 10.5 kg m/s

Impulse (I) is the multiplication of force (F) and time interval (Δt): I = F · Δt

Force (F) is the multiplication of mass (m) and acceleration (a): F = m · a

Acceleration (a) can be expressed as change in velocity (v) divided by time interval (Δt): a = Δv/Δt

So: 
a = Δv/Δt         ⇒ F = m · a = m · Δv/Δt
F = m · Δv/Δt   ⇒ I = m · Δv/Δt · Δt
Since Δt can be cancelled out, impulse can be expressed as:
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It is given:
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I = 1.5 · (22 - 15) = 1.5 · 7 = 10.5 kgm/s.
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Answer:

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