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Nataly_w [17]
2 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]2 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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The work function for metallic magnesium is 3.68 eV. Calculate the velocity in km/s for electrons ejected from a metallic magnes
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Answer:

v = 301.62 km/s

Explanation:

Given that:

The work function of the magnesium = 3.68 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function = 3.68\times 1.6022\times 10^{-19}\ J=5.8961\times 10^{-19}\ J

Using the equation for photoelectric effect as:

E=\psi _0+\frac {1}{2}\times m\times v^2

Also, E=\frac {h\times c}{\lambda}

Applying the equation as:

\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

m is the mass of electron having value 9.11\times 10^{-31}\ kg

\lambda is the wavelength of the light being bombarded

\psi _0=Work\ function

v is the velocity of electron

Given, \lambda=315\ nm=315\times 10^{-9}\ m

Thus, applying values as:

\frac{6.626\times 10^{-34}\times 3\times 10^8}{315\times 10^{-9}}=5.8961\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2

5.8961\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{315\times \:10^{-9}}

\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=6.31047619\times 10^{-19}-5.8961\times 10^{-19}

\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=0.41437619\times 10^{-19}

v^2=0.0909717212\times 10^{12}

v = 3.0162 × 10⁵ m/s

Also, 1 m = 0.001 km

So, v = 301.62 km/s

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