only reactants are present
If equllibrium constant is greater than 10^3 then product
predominates
If equllibrium constant is greater than 10^ -3 then reactant predominates
in between 10^3 and 10^-3 the reaction is in equllibrium
and we see that the above value is 10^-31, it is very small so the answer is reactants are presents
Answer:
cyclohexilic acid
Explanation:
The malonic ester synthesis is often used to form different types of carboxilic acid. In this case, we have excess of base, so we can expect a ring formed in the molecule.
In the first step, the base substract the more acidic hydrogen of the ester. In the next step we have the ester reacting with the dibromopentane. Then, in the next step, the excess of base will substract again the acidic hydrogen remaining of the carbon, and then, will promove a Sn2 reaction with the bromine in the pentane (That it was previously attached to the molecule). In this step, a ring id formed. Then, the hydrolisis with acid, will form carboxilic acid in both sides of the molecule, and finally the decarboxilation in heat will separate the molecule and formed the cyclohexilic acid.
The picture below shows the mechanism
Answer:
Hydrogen is a very reactive element. It doesn't exist as a single atom in nature. Neither do any of the other binary nonmetals -- nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. They're too prone to react with something. They react with each other and form binary molecules because the binary molecules are more stable than single atoms (by a lot!). Compounds with other atoms are even more stable, so hydrogen reacts with oxygen to form water, and chlorine reacts with sodium to form table salt.
Explanation:
Answer:
Sodium was first discovered in Britain in 1807, where a chemist named Sir Humphry Davy found it by electrolysis of caustic soda.
Explanation:
Hope this helped!
Answer is: hydrogen (cathode), iodine (anode).
The balanced
oxidation half reaction: 2I⁻(aq) →
I₂(s) + 2e⁻.<span>
Iodine is oxidized (lost electrons) from -1
to neutral charge (0).
The balanced reduction
half-reaction: 2H</span>₂O(l) + 2e⁻ → H₂(g) + 2OH⁻.<span>
<span>Hydrogen is reduced (gain electrons) from
+1 to neutral charge.</span></span>
