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Nataly_w [17]
3 years ago
10

A student is studying the rate of the following reaction: C2H4O NaOH --> H20 NaC2H3O Knowing that this is an exothermic react

ion, he is measuring the rate of the reaction by timing how quickly the reaction vessel heats up. He notices that if he adds HCl to this reaction, the rate increases dramatically. He also determines that the HCl is being used up during the reaction. Is the HCl a catalyst for this reaction.
Chemistry
1 answer:
emmasim [6.3K]3 years ago
8 0

Answer:

HCl is not a catalyst because these are not used up during the chemical reactions.

Explanation:

Hello there!

In this case, according to the performed experiments, it is possible for us to realize that HCl cannot be a catalyst for this reaction because it is used up during the reaction. This is explained by the fact that catalyst are able to return to the original form once the reaction has gone to completion; this is the example of palladium in the hydrogenation or dehydrogenation of hydrocarbons depending on the case. Moreover, we know that the catalysts increase the reaction rate because they decrease the activation energy of the reaction and therefore the student observed such increase.

Best regards!

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During evaporation, the volume of the liquid decreases and the liquid becomes what??? ​
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Read 2 more answers
at atmoshperic pressure, a balloon contains 2.00L of nitrogen of gas. How would the volume change if the Kelvin temperature were
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<u>Answer:</u> The percent change in volume will be 25 %

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=2L\\T_1=T_1\\V_2=?\\T_2=75\% \text{ of }T_1=0.75\times T_1

Putting values in above equation, we get:

\frac{2L}{T_1}=\frac{V_2}{0.75\times T_1}\\\\V_2=\frac{2\times 0.75\times T_1}{T_1}=1.5L

Percent change of volume = \frac{\text{Change in volume}}{\text{Initial volume}}\times 100

Percent change of volume = \frac{(2-1.5)}{2}\times 100=25\%

Hence, the percent change in volume will be 25 %

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