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3 years ago

Which of these changes to a modern lifestyle would promote the

Chemistry

1 answer:

lidiya [134]3 years ago

7 0

I think it’s B. At least that’s what makes sense to me

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Write sentences that completely describe each of the chemical reactions shown in these skeleton equations. 11.1

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Please upload a photo of your question!

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3 years ago

When the reaction mixture is worked-up, it is first washed three times with 5% sodium bicarbonate, and then with a saturated nac

Ann [662]

Solution:

After the reaction of mixture is worked-up Washing three times the organic with sodium carbonate helps to decrease the solubility of the organic layer into the aqueous layer. This allows the organic layer to be separated more easily.

And then the reaction washed by saturated NACL we have The bulk of the water can often be removed by shaking or "washing" the organic layer with saturated aqueous sodium chloride (otherwise known as brine). The salt water works to pull the water from the organic layer to the water layer.

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3 years ago

How does the temperature of absolute zero relate to the kinetic energy of a substance?

NeX [460]

At absolute zero the kinetic energy of the substance will be 0

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3 years ago

Granite is which pure substance and mixture

DanielleElmas [232]

Window cleaner is ammonia dissolved in water. This is an example of a _______.

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3 years ago

The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42

denis23 [38]

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

T₂ = 20 °C = 293 K

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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3 years ago