Hola aquí va la respuesta!
Se fomentó la inmigración europea porque la economía no era buena y vinieron a Paraguay para mejorar su situación.
We are asked to provide an equation for the transformation of 2-phenylethanoic acid to 2-phenylethanol. This type of a reaction is converting a carboxylic acid to an alcohol, which is classified as a reduction reaction since we are decreasing the number of bonds to oxygen in the molecule. In order to reduce a carbonyl to an alcohol, we need a source of hydride, H⁻. Reducing the carboxylic acid once will convert it to the aldehyde. However, we need to reduce the functional group all the way down to an alcohol, which is another reduction step after aldehyde formation. Therefore, the hydride source of choice is lithium aluminum hydride, LiAlH₄.
A reaction scheme is provided to show the reaction of the reduction of carboxylic acid to alcohol. The first step is addition of lithium aluminum hydride which does the reduction, and the second step is a work-up of acid which protonates the alcohol to get the final product.
<span>vibration of particles decreases as the temperature decreases It also decreases during phase change but temperature does not</span>
I think they test ideas about chemical evolution by doing experiments. Hope this helps
Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻