Answer:
MECHANISM:
1) The lone pair on oxygen attacks the H-Br molecule forming a hydronium ion.
2) Formation of carbocation.
3) Attack of Nucleophile Br − .
Explanation:
The percent composition of this compound :
Mg = 72.182%
N = 27.818%
<h3>Further explanation</h3>
Given
9.03 g Mg
3.48 g N
Required
The percent composition
Solution
Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements
Total mass of the compound :
= 9.03 g + 3.48 g
= 12.51 g
The percent composition :
Mg : 9.03/ 12.51 g x 100% = 72.182%
N : 3.48 / 12.51 g x 100% = 27.818%
Hello!
Understanding how much of a product is produced in a reaction is referred to as Stoichiometrically understanding the reaction.
Stoichiometry is the calculation of the quantitative relationships between reactants and products in a chemical reaction. The first to talk about stoichiometry was Jeremias Benjamin Ritcher, who said that "Stoichiometry is the science that measures the quantitative proportions or mass ratios of chemical elements that are involved in a chemical reaction".
To calculate how much of a product is produced in a reaction, Stoichiometry is used, applying the law of conservation of mass. That means that the amount of product can be calculated from the amounts of reactants if they are known.
Answer: 7.07 grams
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require 1 mole of 
Thus 0.052 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 1 mole of give = 1 mole of 
Thus 0.052 moles of give = of
Mass of 
Thus 7.07 g of will be produced from the given masses of both reactants.
Answer:
d. 8 moles of H2O on the product side
Explanation:
Hello,
In this case, we need to balance the given redox reaction in acidic media as shown below:
![MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\](https://tex.z-dn.net/?f=MnO_4%5E%7B1-%7D%20%28aq%29%20%2B%20Cl%5E%7B1-%7D%20%28aq%29%20%5Crightarrow%20%20Mn%5E%7B2%2B%7D%20%28aq%29%20%2B%20Cl_2%20%28g%29%5C%5C%5C%5C%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%20%2B%20Cl%5E%7B1-%7D%20%28aq%29%20%5Crightarrow%20%20Mn%5E%7B2%2B%7D%20%28aq%29%20%2B%20Cl_2%20%28g%29%5C%5C%5C%5C%5C%5C%5C%5C%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%2B8H%5E%2B%2B5e%5E-%20%5Crightarrow%20Mn%5E%7B2%2B%7D%2B4H_2O%5C%5C%5C%5C2Cl%5E%7B1-%7D%5Crightarrow%20Cl_2%5E0%2B2e%5E-%5C%5C%5C%5C2%2A%5B%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%2B8H%5E%2B%2B5e%5E-%20%5Crightarrow%20Mn%5E%7B2%2B%7D%2B4H_2O%5D%5C%5C%5C%5C5%2A%5B2Cl%5E%7B1-%7D%5Crightarrow%20Cl_2%5E0%2B2e%5E-%5D%5C%5C%5C%5C%5C%5C%5C%5C2%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%2B16H%5E%2B%2B10e%5E-%20%5Crightarrow%202Mn%5E%7B2%2B%7D%2B8H_2O%5C%5C%5C%5C10Cl%5E%7B1-%7D%5Crightarrow%205Cl_2%5E0%2B10e%5E-%5C%5C)
Then, we add the half reactions:
Thereby, we can see d. 8 moles of H2O on the product side.
Best regards.
