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UNO [17]
2 years ago
8

Which is the correct statement regarding the relative Rf values of the starting methyl benzoate vs the product, methyl m-nitrobe

nzoate on a silica gel TLC plate. The product has a higher Rf value on a silica gel TLC plate because it is less polar than the starting methyl benzoate. The product has a lower Rf value on a silica gel TLC plate because it is less polar than the starting methyl benzoate. The product has a higher Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate. The product has a lower Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate. 1.5 points QUESTION 2 Methyl benzoate is more reactive than benzene in Electrophilic Aromatic Substition [EAS] reactions. True False 1 points QUESTION 3 Methyl m-nitrobenzoate is less reactive than methyl benzoate in Electrophilic Aromatic Substition [EAS] reactions. True False
Chemistry
1 answer:
Marianna [84]2 years ago
5 0

Answer:

1. The product has a higher Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate.

2. False

3. True

Explanation:

In chromatography, there is a stationary phase and a mobile phase. The ratio of the distance moved by a component and the distance moved by the solvent gives the retention factor (Rf).

Since silica gel is a polar solvent, it will retain the more polar product methyl m-nitrobenzoate compared to the methyl benzoate starting material.

In comparing the electrophillic aromatic substitution of m-nitrobenzoate  and methyl benzoate, we must remember that the presence of electron withdrawing groups (such as -NO2 and -CHO) on the aromatic compound deactivates the compound towards electrophillic aromatic substitution hence, methyl m-nitrobenzoate is less reactive than methyl benzoate in Electrophilic Aromatic Substition and Methyl benzoate is less reactive than benzene in Electrophilic Aromatic Substition

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which intensive physical property is observed when droppings of a person seated inside a closed room has able to reach a person
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The right response will be "Speed".

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2 years ago
Give the major organic product of the reaction of o-methylaniline with benzenediazonium chloride [(phn≡n)+ cl-].
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7 0
2 years ago
How many grams of C5H12 must be burned to heat 1.39 kg of water from 21.2 °C to 97.0 °C? Assume that all the heat released durin
faust18 [17]

Answer:

m = 8.9856 g

Explanation:

In order to do this, we need to write the expressions that are to be used. First, to calculate heat:

Q = m*C*ΔT (1)

Where C would be heat capacity of the substance.

The heat can also be relationed with the moles and enthalpy of a compound using the following expression:

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Finally for the mass of any compound, we use the following expression:

m = n*MM (3)

So, in order to calculate the grams of pentane (C5H12), we need to calculate the moles of the compound, and to do that, we need the heat exerted.

So, as we are using water, let's calculate the heat that is been exerted with the water. The C of the water is 4.186 J/g °C so:

Q = (1.39 * 1000) * 4.186 * (21.2 - 97)

Q = -441,045.33 J

This is the heat neccesary to burn pentane and heat water. Now, with this value, let's calculate the moles used of pentane with expression (2). The ΔH of the pentane is -3,535 045.kJ/mol or -3.535x10⁶ J/mol. Solving for n we have:

n = -441,045.3 / -3.535x10⁶

n = 0.1248 moles

Finally, we can calculate the grams needed with expression (3). The molar mass of pentane is 72 g/mol

m = 0.1248 * 72

m = 8.9856 g

This is the mass needed to heat 1.39 kg of water

6 0
2 years ago
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