This set up of a conversion table should show you that if you multiply the grams of BeI2 times .02 moles, it equals <span>5.256 g (your answer) </span>
Use atomic properties to explain the reduction of a less active metal by a more active one:(b) in the molten state. Give a speci
1 answer:
The reduction of a less active metal by a more active one is called metal displacement reactions. For example:
Fe + CuSO4 → FeSO4 + Cu
<h3>What is metal displacement reaction? </h3>Displacement reactions is a reaction which includes a metal and the compound of a other metal. A more reactive metal will push or displace out a less reactive metal from its compound in this displacement reaction. The metal which is less reactive left uncombined after the reaction.
As we know that, electrons are the basis of the chemical reactions. If chemical compound or element A is more easily oxidized than B, then according to the terms of the activity series, the elements which are more easily oxidized can react with more chemicals, since they are able to act as a reducing agents for more chemicals.
Since, Metal ions are positively charged ions as they lose electrons. Some metals give up their electrons more readily than others and become more reactive.
Thus, we concluded that the reduction of a less active metal by a more active one is called metal displacement reactions. For example:
Fe + CuSO4 → FeSO4 + Cu
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Answer:
See explanation below
Explanation:
The question is incomplete. However in picture 1, you have the starting materials and the structure of the product, which you miss in this part.
Now, in picture 2, you have the starting reactant and the product, and the mechanism that is taking place here.
First, all what we have here is an acid base reaction. In the first step, we are using the acid medium to convert the reactant into an alcohol. The bromine there, is not leaving the molecule yet, because it's neccesary for the next step. The starting reactant is an alkene, in that way, we can convert the reactant in the first step into a secondary alcohol. In other words, the first reaction is a alkene hydration.
In the second step, we use a strong base. You may say this is a strong nucleophile and will do a Sn2 reaction to form another alcohol there, but it's not the case, because, before any kind of reaction happens, the priority here is always the acid base, so the base will react with the acidic hydrogen. In this case, it will substract an hydrogen from the OH. When this happens, the lone pair will do an auto condensation here, and attacks the bromine in the molecule. In this way, the molecule will become a cyclomolecule, and that way it form the final product.
See picture 2, for mechanism
