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kow [346]
4 years ago
9

For this assignment, the target compound that you should synthesize is 1-methyl-4-nitro-benzene. This is an electrophilic aromat

ic substitution reaction. Examine the product carefully and determine the substitution pattern. Which group will already be present in the substrate? Keep in mind the mechanism and how that will control the selectivity of the process. Remember, you can easily separate ortho and para isomers.

Chemistry
1 answer:
Liono4ka [1.6K]4 years ago
5 0

Answer:

The methyl group will already be present  

Explanation:

You want to make 1-methyl-4-nitrobenzene.

The question is, "Do I start with the methyl group on the ring and then nitrate, or do I start with the nitro group on the ring and then add the methyl group?"

The methyl group is activating and ortho, para directing.

The nitro group is deactivating and meta directing.

You want a para-substituted product, so you choose to nitrate toluene, as in the reaction scheme below.

 

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How many formula units are present in 473.7 g lead(ii) chloride?
pantera1 [17]
First, divide the given mass of lead (ii) chloride by the molar mass of the compound. The molar mass is equal to 278.1 g/mol.

    n = m/M

where n is the number of moles, m is the given mass, and M is the molar mass. Substituting the known values,

   n = 473.7 g / 278.1 g/mol
   n = 1.7 mols

Then, it is to be remember that as per Avogadro, every mol of a substance contains 6.022 x 10^23 formula units. 

     N = (1.7 mols)(6.022 x 10^23 formula units/mol)
     N = 1.02 x 10^24 formula units

Answer: 1.02 x 10^24 formula units
5 0
4 years ago
Plzz help meee!
Juliette [100K]
Answer: B, 8 kilometers per hour
5 0
3 years ago
Read 2 more answers
How is common salt formed
tia_tia [17]

Answer:

Hello

Explanation:

Common salt is obtained from sea-water by the process of evaporation. Sea water is trapped in large, shallow pools and allowed to stand there. The sun's heat evaporates the water slowly and common salt is left behind.

This impure common salt obtained has impurities and is hence purified to obtain pure common salt by recrystallization.

7 0
3 years ago
Read 2 more answers
When nitroglycerine (227.1 g/mol) explodes, N2, CO2, H2O, and O2 gases are released initially. Assume that the gases from the ex
denis23 [38]

Answer:

a. 561L of N₂

b. 14,80L of CO₂

c. 715,0L of gases

Explanation:

The explosion of nitroglycerine gives the reaction:

4C₃H₅N₃O₉(s) → 6N₂(g) + 12CO₂(g) + 10H₂O(g) + O₂(g)

a. 4 moles of nitroglycerine produce 6 moles of N₂. Thus, 16,7 moles of nitrofglycerine produce:

16,7 moles C₃H₅N₃O₉ × (6 moles N₂ / 4 moles C₃H₅N₃O₉) = <em><u>25,05 moles N₂</u></em>

At STP, 1 mole of gas occupies 22,4 L. 25,05 moles are:

25,05 moles N₂ × (22,4L / 1mole) = <em>561 L</em>

b. 100,0g of C₃H₅N₃O₉ are:

100,0g C₃H₅N₃O₉ × (1mole / 227,1g) = 0,4403 moles of C₃H₅N₃O₉.

As 4 moles of C₃H₅N₃O₉ produce 6 mole of CO₂. Moles of CO₂ are:

0,4403 moles of C₃H₅N₃O₉ × ( 6mol CO₂ / 4mol C₃H₅N₃O₉) = 0,6605 moles CO₂ Again, as 1 mol of gas occupies 22,4L:

0,6605 mol CO₂ × (22,4L / 1mol) = <em>14,80 L</em>

c.<em> </em>1,000 kg ≈ 1000g of nytroglicerine are:

1000g C₃H₅N₃O₉ × (1mole / 227,1g) = 4,403 moles of C₃H₅N₃O₉.

As 4 moles of C₃H₅N₃O₉ produce 29 moles of gases. Moles of CO₂ are:

4,403 moles of C₃H₅N₃O₉ × ( 29mol gases / 4mol C₃H₅N₃O₉) = 31,92 moles of gases. Again, as 1 mol of gas occupies 22,4L:

31,92 mol CO₂ × (22,4L / 1mol) = <em>715,0 L </em>

<em />

I hope it helps!

3 0
3 years ago
The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen
tino4ka555 [31]

Answer: The mass percentage of Tl_2SO_4 is 5.86%

Explanation:

To calculate the mass percentage of Tl_2SO_4 in the sample it is necessary to know the mass of the solute (Tl_2SO_4 in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the TlI precipitate.  We can establish a relation between the mass of TlI and Tl_2SO_4 using the stoichiometry of the compounds:

moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.

Since for every mole of Tl in TlI there are two moles of Tl in Tl_2SO_4, we have:

moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol

Using the molar mass of Tl_2SO_4 we have:

mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g

Finally, we can use the mass percentage formula:

mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%

6 0
3 years ago
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