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valina [46]
4 years ago
5

1) Which action defines a double-displacement reaction?

Chemistry
2 answers:
Ronch [10]4 years ago
5 0

Answer:

The action that defines a double displacement reaction is <u><em>option B) formation of a precipitate.</em></u>

Explanation:

A double displacement reaction occurs when two elements in different compounds exchange positions, forming two new compounds. In general, the exchange occurs between reagents that are ionic compounds in solution, where the anions and cations of the compounds are exchanged.

Double displacement reactions follow the following general chemical equation:

AB + CD ⇒ AD + CB

The initial compounds AB and CD react by exchanging "their partners" and therefore form two new compounds: AD and CB. For this type of reaction to occur, the products formed must be more stable or related than the reagents.

Since the reaction is a simple ion exchange, none of these gains or loses electrons. That is, these chemical reactions do not show changes in the number of oxidation or relative charge of the element. Therefore, if A is a +1 charge cation in compound AB, it will have the same +1 charge in compound AD.

On the other hand, a precipitation reaction occurs when two aqueous ionic compounds form a new ionic compound that is not soluble in water. That is, the reaction can occur between two ionic compounds (salts), if after the ion exchange one of the products is insoluble in water. This insoluble compound is called precipitate, since it will precipitate in the solution. So, it is an example of double displacement reactions.

Then, <u><em>the action that defines a double displacement reaction is option B) formation of a precipitate.</em></u>

jekas [21]4 years ago
3 0

Answer:

formation of a precipitate

Explanation:

When two salt solutions are mixed, an observable reaction is seen when two types of ions attract each other, they clump together and fall to the bottom of the container as a solid precipitate.

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Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

k = Ae^{\frac{-Ea}{RT}}

<u>Where:</u>

k: is the rate constant

A: is the frequency factor    

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

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The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

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