Recycling one ton of paper is the equivalent of approximately how much of the following? a.) all of the below b.) 460 gallons of
1 answer:
You might be interested in
Answer:
See explanation below
Explanation:
In this reaction we have the ethyl acetoacetate which is reacting with 2 eq of sodium etoxide. The sodium etoxide is a base and it usually behaves as a nucleophyle of many reactions. Therefore, it will atract all the acidics protons in a molecule.
In the case of the ethyl acetoacetate, the protons that are in the methylene group (CH3 - CO - CH2 - COOCH2CH3) are the more acidic protons, therefore the etoxide will substract these protons instead of the protons of the methyl groups. This is because those hydrogens (in the methylene group) are between two carbonile groups, which make them more available and acidic for any reaction. As we have 2 equivalents of etoxide, means that it will substract both of the hydrogen atoms there, and then, reacts with the Br - CH2CH2 - Br and form a product of an aldolic condensation.
The mechanism of this reaction to reach X is shown in the attached picture.
Answer:
Explanation:
To solve this problem, we must understand the relationship between mass of a substance and the number of atoms.
Atoms are the smallest indivisible particles of any matter. A substance can be made up of several number of atoms in their space.
The mass of any substance is a function of the amount of atoms its contains.
The mass of a substance is related in chemistry to the amount of atoms its contains using the parameter called the number of moles.
A mole is the amount of substance that contains the Avogadro's number of particles. This number is 6.02 x 10²³ particles. The particles here can be protons, neutrons, electrons, atoms e.t.c.
Now,
Number of moles =
Molar mass of copper = 63.6g/mole
Number of moles = = 0.03mole
Since 1 mole of a substance contains 6.02 x 10²³atoms
0.03 mole of copper will contain 0.03 x 6.02 x 10²³atoms
= 1.89 x 10²² atoms
He needs to add 1.89 x 10²² atoms to make 2g of the sample.