Possible products in this reaction are the products 2 and 3.
<h3>What is the Friedel Crafts reaction?</h3>
The Friedel Crafts reaction is one in which the electrophile is created by a Lewis acid reaction between the AlCl3 and the alkylhalide reactant .
Now we know that there is the possibility of two products in this reaction due to a resonance shift as such possible products in this reaction are the products 2 and 3.
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Answer:
(A) first order reaction
Explanation:
A first order reaction is a type of reaction in which the rate of the reaction depends only on the concentration on one of the reactants. Since A is the only reactant we have, it is right to deduce that this reaction is a first order reaction.
Note: while the order of this reaction is 1, its molecularity is 2. The molecularity of a reaction is the number of moles of reactants that is actually reacting.
(B) is wrong
This is because a zero order reaction is one in which the rate of reaction is not influenced by the concentrations of the reactants and hence remains constant irrespective. Since we were not furnished with this idea in the question, it is only right that we reject this answer.
(C) is wrong.
C is outrightly wrong as we have only one reactant.
(D) is wrong
We have only one reactant.
A pure crystalline substance is a substance with an almost perfect regular and periodic pattern in a solid state. This makes this type of substance a hard one compared to an amorphous substance which is soft because of the irregular pattern within.
<h3>Answer:</h3>
Strontium (Sr)
<h3>Explanation:</h3>
The condition given in statement is the presence of two valence electron. Hence, first we found the electronic configuration of given atoms as follow;
Rubidium [Kr] 5s¹
Strontium [Kr] 5s²
Zirconium [Kr] 4d² 5s²
Silver [Kr] 4d¹⁰ 5s¹
From above configurations it is cleared that only Strontium and Zirconium has two electrons in its valence shell.
We also know that s-block elements are more reactive than transition elements due to less shielding effect in transition elements hence, making it difficult for transition metals to loose electrons as compared to s-block elements. Therefore, we can conclude that Strontium present in s-block with two valence electrons is the correct answer.
