The reaction of p-aminophenol with one mole of actyl chloride in presence of pyridine gives:

Answers

Answer 1
Answer: The reaction of p-aminophenol with one mole of acetyl chloride in the presence of pyridine results in the formation of an amide compound called N-acetyl-p-aminophenol.
Here is a step-by-step explanation of the reaction:

1. Acetyl chloride (CH3COCI) is an acylating agent that reacts with the amino group (-NH2) of p-
aminophenol (C6H4NH2OH). The
acetyl chloride is a source of the acety group (CH3CO-), which will replace the amino group in p-aminophenol.

2. Pyridine (C5H5N) is used as a catalyst in this reaction. It helps in the formation of the acety group by removing the hydrogen ion (Ht) produced during the reaction, making the reaction proceed more efficiently.

3. Electrons on the amino group of p-aminophenol acts as the nucleophile and attacks the carbony carbon of the acetyl chloride.

4. The carbonyl carbon is electron-deficient, so it readily accepts the lone pair of electrons from the amino group, forming a new bond. At the same time, the chlorine atom of acetyl chloride is displaced as a leaving group.

5. The resulting product is N-acety-p-aminophenol (C6H4NHCOCH3). In
this compound, the acetyl group is attached to the nitrogen atom of p-aminophenol, replacing the amino group. The pyridine catalyst remains unchanged and is not part of the final product.

Overall, the reaction between p-aminophenol and acetyl chloride in the presence of pyridine leads to the formation of N-acety -p-aminophenol, an amide compound.
Answer 2
Answer:

Final answer:

The reaction of p-aminophenol with one mole of acetyl chloride in the presence of pyridine results in the formation of an amide compound.

Explanation:

The reaction of p-aminophenol with one mole of acetyl chloride in the presence of pyridine leads to the formation of an amide compound. This reaction is known as acylation, where the acetyl chloride acts as the acylating agent and reacts with the amino group of p-aminophenol to form an amide bond.

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Atoms gain or lose electrons to increase:a. reactivity
c. stability
b. nucleus size
d. atomic mass

Answers

Atoms gain or lose electrons toincrease their stability. The answer is letter C. Remember noble gases? Theyfollow the octet rule in which their shells are electrons are filled with eightelectrons. So the atoms that do not belong to the noble gas tend to berestless. They want their electrons to be filled to follow the octet rule. Thatis why they tend to share or gain electrons. Suppose you have a compound ofCCl4. You know that chlorine can only share 1 electron because 7 of itselectrons are filled. Also, in carbon, it can only share 4 electrons because 4of it are already filled. That is why carbon needs four chlorine to form CCl4.

List the following in smallest to largest:-Glucose
-Carbonic Acid
-Hydrogen Gas
-Atom
-Proton
-Sodium Chloride
-Triglyceride
-Electron
-Protein
-Amino Acid

Answers

Here you are! The correct sequence is being shown below : 

- Electron
 - Proton
 - Atom
 - Hydrogen Gas
- Amino Acid
 - Glucose
 -Triglyceride
 - Sodium Chloride
 - Carbonic Acid
 - Protein

Using the atomic masses and relative abundance of the isotopes of nitrogen given below, determine the average atomic mass of nitrogen. N-14: 14.003 amu; 99.63% N-15: 15.000 amu; 0.37%

Answers

The formula for average atomic mass is :

mass of isotope A * % of isotope A + mass of isotope B * % of isotope B + ....

Now,
Here,
Average atomic mass of nitrogen = (14.003 * 99.63%) + (15.000 * 0.37%)
                                                 = (14.003 * 0.9963) + ( 15.000 * 0.0037)
                                                 = 13.951 + 0.056
                                                 = 14.007 a.m.u.                                                 

What information does a chemical formula of a compound provide?a. The type and amount of elements
b. The different kinds of compounds included within the chemical formula
c. The types of molecules present in a formula
d. The correct amounts of substances to mix together to make a formula

Answers

I think the best option is A. A chemical formula of a compound provides information on the type and amounts of elements in that compound. For instance, we have NaOH. From the given formula, we see that it contains the elements sodium, oxygen and hydrogen. Also, their corresponding number of molecules.

Convert the following numbers into Scientific Notation1) 3.400
2) 0.000023
3) 101,000
4) 0.010
5) 45.01
6) 1,000,000
7) 0.00671
8) 4.50

Answers

3400 x 10^-3
2.3 x 10^-5
1.01 x 10^5
1.0 x10^-2
4.5 x 10^1
1.0 x 10^6
6.71 x10^-3
450 x 10^-2

Final answer:

The question asked about converting normal numbers into scientific notation. The answer provided the eight scientific notation equivalents for each of the numbers provided in the question, explaining the process of calculating them.

Explanation:

To write numbers in scientific notation, we have to express them in the form of a x 10^n, where 1 <= a < 10 and n represents the number of places the decimal place has been moved. Let's proceed with each given number:

  1. 3.400 is already in the form. Hence, it is written as 3.4 x 10^0.
  2. To convert 0.000023 into scientific notation, move the decimal place 5 places to the right. So, this is 2.3 x 10^-5.
  3. 101,000 converts to 1.01 x 10^5.
  4. For 0.010, we move the decimal place 2 places to the right, giving 1.0 x 10^-2.
  5. 45.01 in scientific notation is 4.501 x 10^1.
  6. 1,000,000 becomes 1.0 x 10^6.
  7. To translate 0.00671 into scientific notation, you move the decimal place 3 places to the right, leaving 6.71 x 10^-3.
  8. Lastly, 4.50 remains as 4.5 x 10^0.

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How many moles of oxygen are produced when 26.5 mol of aluminum oxide are decomposed?

Answers

The balanced chemical reaction is:

 

2Al2O3 = 4Al +3 O2

 

We are given the amount of the reactant. This will be the starting point of our calculations.

 

26.5 mol Al2O3 ( 3 mol O2 / 2 mol Al2O3 ) = 39.75 mol O2 needed

 

Thus, 0.014 mol of I2 is needed to form the given amount of NI3.

the balanced chemical equation for the decomposition of Al₂O₃ is as follows

2Al₂O₃ --> 4Al + 3O₂

stoichiometry of Al₂O₃ to O₂ is 2:3

when 2 mol of Al₂O₃ decomposes - 3 mol of O₂ are formed

therefore when 26.5 mol of Al₂O₃ decomposes - 3/2 x 26.5 mol = 39.75 mol

the number of moles of O₂ formed are 39.75 mol