About 3% of the water on Earth is freshwater. Only about 40% of that freshwater is available for human use. Why is so much fresh
2 answers:
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Answer:
The structures are shown below.
Explanation:
When the acid reacts with water, it loses one proton (H⁺) and forms a base, which is the conjugate base of its acid.
The formal charge of an atom can be calculated by:
FC = X - (Y + Z/2)
Where X is the valence electrons of the neutral atom, Y is the unshared electrons, and Z is the shared electrons in the molecule.
a) When HCl deprotonates, it forms Cl⁻ as the conjugate base. The neutral atom Cl has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1 The structure is shown below in figure a.
b) When Hbr deprotonates it forms Br- as the conjugate base. The neutral atom has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1. The structure is shown below in figure b.
c) When CH3COOH loses a proton, it forms the conjugate base CH3COO⁻. The carbon as 4 valence electrons, hydrogen has 1 valence electron and oxygen has 6 valence electrons. The first carbon make simple bonds with each hydrogen and with the second carbon, and so, all the electrons are shared, and it has FC = 4 - (0 + 8/2) = 0, as so the hydrogens have FC = 1 - (0 + 2/2) = 0.
The second carbon does 1 simple bond with the first carbon, a double bond with one oxygen, and a simple bond with the other oxygen, and so doesn't have unshared electrons, and FC = 4 - (0 + 8/2) = 0.
The first oxygen does a double bond with the carbon, and so it has 4 unshared electrons, so FC = 6 - (4 + 4/2) = 0. The second oxygen does a simple bond with the carbon, and so has 5 unshared electrons, so FC = 6 - (5 + 2/2) = 0.
The structure is shown in figure c.
Answer:
(1) addition of HBr to 2-methyl-2-pentene
Explanation:
In this case, we will have the formation of a <u>carbocation</u> for each molecule. For molecule 1 we will have a <u>tertiary carbocation</u> and for molecule 2 we will have a <u>secondary carbocation</u>.
Therefore the <u>most stable carbocation</u> is the one produced by the 2-methyl-2-pentene. So, this molecule would react faster than 4-methyl-1-pentene. (See figure)
