Answer:
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<h3>Saponification is a process that involves conversion of fat, oil or lipid into soap and alcohol by the action of heat in the presence of aqueous alkali. Soaps are salts of fatty acids and fatty acids are monocarboxylic acids that have long carbon chains e.g. sodium palmitate.</h3>
Answer:
0.55 mol Au₂S₃
Explanation:
Normally, we would need a balanced equation with masses, moles, and molar masses, but we can get by with a partial equation, if the S atoms are balanced.
1. Gather all the information in one place:
M_r: 34.08
Au₂S₃ + … ⟶ 3H₂S + …
m/g: 56
2. Calculate the moles of H₂S
Moles of H₂S = 56 g H₂S × (34.08 g H₂S/1 mol H₂S)
= 1.64 mol H₂S
3. Calculate the moles of Au₂S₃
The molar ratio is 1 mol Au₂S₃/3 mol H₂S.
Moles of Au₂S₃ = 1.64 mol H₂S × (1 mol Au₂S₃/3 mol H₂S)
= 0.55 mol Au₂S₃
True. For example, electron domain geometry and molecular geometry of water and ammonia are different.
The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of stomach acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

Hence, the volume of HCl neutralized is 1.25 mL