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kumpel [21]
3 years ago
6

I need a help from this question!!

Chemistry
1 answer:
Alik [6]3 years ago
7 0

Answer:

B

Explanation:

Please take a look at the picture attached for the drawings and structures.

C2H4 is a alkene (C-C double bond). When steam (water) is added, it turns into an alcohol, where the double bond breaks and a (-OH) functional group is attached to one of the Carbons. In this case, C2H4 ethene is turned into ethanol.

When an alcohol undergoes oxidation, primary alcohols turn into aldehyde (-CHO) or carboxylic acids (-COOH). Secondary alcohol turns into ketone. Ethanol is a primary alcohol. And since it later reacts with propanol, it can only form carboxylic acid when it oxidizes. The product in this reaction is ethanoic acid.

Carboxylic acid reacts with alcohol to form an ester (-COO-). the -COOH group from acid combines with the -OH group from alcohol to form an ester bond -COO- while eliminating H2O. Therefore, when propanol undergoes esterification with ethanoic acid, propyl ethanoate is produced. It is the answer of B.

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aalyn [17]

Answer:

The pH of the solution is 4.60.

Explanation:

The pH gives us an idea of the acidity or basicity of a solution. More precisely, it indicates the concentration of H30 + ions present in said solution. The pH scale ranges from 0 to 14: from 0 to 7 corresponds to acid solutions, 7 neutral solutions and between 7 and 14 basic solutions. It is calculated as:

pH = -log (H30 +)

pH= -log (2,5 x 10-5)

<em>pH=4.60</em>

7 0
3 years ago
An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
Tju [1.3M]

Answer:

a) \Delta S

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

Explanation:

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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Help!
babunello [35]
D.mno4- is reuced it loses h atom

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H₂CO₃

This compound is known as carbonic acid.
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3 0
3 years ago
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