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2 years ago

Which types of chemical reactions are considered opposites of one another?

Chemistry

2 answers:

gladu [14]2 years ago

6 0

Explanation:

Decomposition chemical reactions

Decomposition reactions are really the opposite of combination reactions. In decomposition reactions, a single compound breaks down into two or more simpler substances (elements and/or compounds).

The decomposition of water into hydrogen and oxygen gases

and the decomposition of hydrogen peroxide t form oxygen gas and water

are examples of decomposition reactions.

mestny [16]2 years ago

4 0

Answer:

Explanation:

Decomposition reactions are really the opposite of combination reactions. In decomposition reactions, a single compound breaks down into two or more simpler substances (elements and/or compounds).

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Scrat [10]

Explanation:

c2H5

cH3 _ c _ c H2 _ c H3

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2 years ago

For the following electrochemical reaction: Al3+(aq) + 3e -> Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -> 2F (aq) Calculate

makvit [3.9K]

<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.

<u>Explanation:</u>

We are given:

$E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

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

$E^o_{cell}=2.87-(-1.66)=4.53V$

Hence, the standard electrode potential of the cell is 4.53 V.

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3 years ago

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Sati [7]

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3 years ago

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For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80

Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

$A\rightleftharpoons B$

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

$K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%} =0.25$