This question is describing two chemical equations whereby the concentration of ammonia has to be determined. The first reaction is between 25.00 mL of ammonia and 50.00 mL of 0.100-M HCl whose excess was neutralized with 21.50 mL of 0.050-M Na₂CO₃ and thus, the concentration ammonia in the cloudy solution was determined as 0.114 M.
First of all we need to go over the titration of the excess HCl with Na₂CO₃ by writing the chemical equation it takes place when they react:

Whereas the mole ratio of HCl to Na₂CO₃ is 2:1 and the volume of the HCl leftover is determined as follows:

Next, we infer that the consumed volume of HCl by the ammonia solution was:

Then, we write the chemical equation that takes place between ammonia and HCl:

Whereas the mole ratio is now 1:1, which means that the concentration of ammonia was:

Learn more:
More water is not always better. For example; cacti only need water about every two weeks, the same with succulent plants. Not every plant needs to be watered every day, or the plant may be overwatered and die.
Answer:
Carbon - 14
Oxygen - 16
Nitrogen - 15
Sulphur - 16
Explanation:
The question above is related to the "Periodic Table of Elements" which shows the proper arrangement of elements in a table. Every element falls on a<em> group/family</em> within the table. Each group has its own number, and the table has a total of<u> 18 groups</u><em> (from left to right). </em>They are classified according to <em>similarities in their characteristics</em>. For example, group 1 is composed of <em>alkali metals</em> while group 2 is composed of<em> alkali earth metals</em>.
Answer:
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Explanation:
Answer:
Explanation:
Concepts and reason
This problem is based on the concept of hydrolysis of esters.
An ester is hydrolyzed to a carboxylic acid and an alcohol when treated with aqueous acid or aqueous base. Under alkaline conditions, the carboxylic acid is obtained in the form as its salt.
Fundamentals
Alkaline hydrolysis of ester is done with strong base {\\rm{NaOH}}NaOHand {\\rm{O}}{{\\rm{H}}^ - }OH\u2212acts as nucleophilic reagent. This reaction is reversible, since carboxylate anion has tendency to react with an alcohol and gives back ester.
Step-by-step
Step 1 of 2
Attack of [{ m{O}}{{ m{H}}^ - }]on carbonyl take place as follows resulting formation of tetrahedral intermediate:
\u043e\u043d\u043e\u043d
Explanation | Hint for next step
The {\\rm{O}}{{\\rm{H}}^ - }OH\u2212nucleophile attacks on the electrophilic carbon of an ester {\\rm{C}} = {\\rm{O}}C=O and forms tetrahedral intermediate after breaking the \\pi\u03c0-bond.
Step 2 of 2
Hydrolysis of product formed from step 1 followed by reaction with {\\rm{NaOH}}NaOHis as follows:
The products of the reaction are:
Explanation | Common mistakes
On the reaction of octyl acetate with aqueous sodium hydroxide, the products octyl alcohol and acetate ion are formed after omitting the {\\rm{N}}{{\\rm{a}}^ + }Na+ions.