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Schach [20]
3 years ago
7

To what class of organic compounds does ethyl acetate belong

Chemistry
2 answers:
earnstyle [38]3 years ago
8 0

Your Answer Is Ester

Ivahew [28]3 years ago
5 0

Question: To which class of organic compounds does ethyl acetates belong?

Answer: Ester

Hope this helps. c;

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Does not react with aqueous solution of bromine <br> (a)C2H6<br> (b)C2H4<br> (c)C2H2<br> (d)C3H6
Damm [24]

Answer:

Hi, there!

which does not react with aqueous solution of bromine?

<em>c)C2H2</em>

<em />

<em>xXxAnimexXx</em>

<em>Happy Labor day! :)</em>

3 0
3 years ago
An automobile gasoline tank holds 23 kg of gasoline. When the gasoline burns, 86 kg of oxygen is consumed, and carbon dioxide an
ira [324]

Question:

What is the total combined mass of carbon dioxide and water that is produced?

Answer:

<u><em>109 kg</em></u>

Explanation:

When 23 kg of gasoline burns by consuming 86 kg oxygen, they produce carbon dioxide and water. To find the total combined mass of carbon dioxide and water, we will use mass conversation law.

According to mass conversation law, the mass of the product is equal to the mass of reagent.

Mass of reagent = Mass of product

In this reaction,

Gasoline + O2 → CO2 + H2O

23 kg + 86 kg → ?

23 kg + 86 kg =  109 kg

Combined mass of carbon dioxide and water will be 109 kg.

4 0
3 years ago
Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

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4 years ago
A chemistry student needs 75.0ml of pentane for an experiment. by consulting the crc handbook of chemistry and physics, the stud
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3 years ago
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