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jarptica [38.1K]
3 years ago
12

Which statement is true regarding a titration? It measures the amount of heat given off in combustion reactions. It determines h

ow much water to add in a dilution. It determines the concentration of an unknown substance in neutralization reactions. It determines if a reaction will take place in single replacement reactions. It is used to make a solution of designated molarity.
Chemistry
2 answers:
exis [7]3 years ago
3 0
C) It determines the concentration of an unknown substance in neutralization reactions.
pishuonlain [190]3 years ago
3 0

Answer:  It determines the concentration of an unknown substance in neutralization reactions.

Explanation: Titration is defined as the process where a solution of known concentration is used to determine the concentration of an unknown solution.

Volumetric titrations are usually used for neutralization reactions.

When an acid and base reacts to give salt and water, the reaction is called as  neutralization reaction.

The normality equation used during the titration is:

N_1V_1=V_2V_2

N_1= Normality of acid solution

V_1= volume of acid solution

N_2= Normality of basic solution

V_2= Volume of basic solution

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The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
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Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

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