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MariettaO [177]
3 years ago
11

PLEASE HELP <3

Chemistry
1 answer:
frutty [35]3 years ago
8 0
In order for carbon to be stable and have 8 electrons, it must make 4 total covalent bonds.

In prefer for oxygen to be stable and have 8 electrons, it must make 2 covalent bonds.

So, we can deduce that CO2 looks like this:

O=C=O

This molecule has two double bonds.

Pssst...Can I get a brainliest?
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What is the Electron Center Geometry of a CH3 carbon.
wel

Answer:

tetrahedral geometry

<h3>CHCH2O- CH2CH3</h3>

Explanation:

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An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
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Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

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a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

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So the human died about 13235.5 years ago

3 0
3 years ago
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