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-BARSIC- [3]

Gregor Mendel

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6 0

2 years ago

As noted in Assignment 1.4.1., alkylation of benzene with 1-chlorobutane in the presence of AlCl3 gives both butylbenzene and (1

Sunny_sXe [5.5K]

Answer:

Use the Bromotriflouride catalyst, BF₃

Explanation:

The BF₃ is most likely to yield less desired side products. The effect lies in the reaction mechanism.

BF₃ is a Lewis acid. Its role is to promote the ionization of the HF. This is achieved through the electrophilic mechanism. The reaction mechanism is as follows:

2 - methylpropene + H-F-BF₃ → H-F + H₃C + benzene

butylbenzene + F-BF₃ → tert-butylbenzene + H-F + BF₃ (regenerated catalyst)

6 0

3 years ago

How are all cells similar?

Reika [66]

Answer:

All cells have structural and functional similarities. Structures shared by all cells include a cell membrane, an aqueous cytosol, ribosomes, and genetic material (DNA). All cells are composed of the same four types of organic molecules: carbohydrates, lipids, nucleic acids, and proteins.

Explanation:

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8 0

2 years ago

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$

Hence, hydrogen would produce 2.31 moles of ammonia.

Therefore, hydrogen is the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

I hope it helps you!

6 0

2 years ago

Explain how the processes that occur at mid-ocean ridges and ocean trenches illustrate the cycling of matter.

Nimfa-mama [501]

Explanation:

At one end, new crusts are being produced, at other end the crust is being destroyed and this strikes a unique balance.

At the mid-ocean ridges, the lithospheric plates are diverging. This is implies that the earth is pulling apart here. When the earth is pulling apart, new materials from the asthenosphere comes to the surface thereby creating new lithospheric plate.

As new plates are formed, they push back against the old one. New plates are found very close to the margin and it begins to age away from the margin.

On the other end, old plates are taken away from this center to ocean trenches. At oceanic trenches subduction is occurring.

In a subduction, the lithospheric plate plunges deep into the asthenosphere where they are being melted back.

This is a covergent margin.

This process continues in a dynamic manner to cycle matter on earth.

learn more:

Sea floor spreading brainly.com/question/9912731

#learnwithBrainly

3 0

2 years ago