<h3><u>Answer;</u></h3>
0.002512 moles of H2O
<h3><u>Explanation</u>;</h3>
The reaction between acetic acid ( CH3COOH) and NaOH is given by the equation;
CH3COOH + NaOH ------> CH3COONa + H2O
Number of moles of CH3COOH = molarity × volume in litres
= 0.08 × 31.4/1000
= 2.512 × 10^-3
Similarly number of moles of NaOH = 1 × 24.3/1000
= 0.0243
From the reaction the mole ratio of CH3COOH : NaOH
Therefore; 0.0243 moles of NaOH will react with 0.0243 moles of CH3COOH but no.of moles of CH3COOH given in the question are 0.002512 moles, which is less than what is required.
Thus; CH3COOH is the limiting reagent and amount of products produced will depend on amount of CH3COOH only.
Since; 1 mole of CH3COOH gives 1 mole of water.
Then; 0.002512 moles of CH3COOH will give 0.002512 moles of H2O
Answer:
There are 0.1 moles of solute in 250 mL of 0.4 M solution
Explanation:
because it is
Answer:
Mole Fraction of O2 --> 0.42
Mole Fraction of Ar --> 0.037
Explanation:
Question:
What type of product forms in the intramolecular reaction between the aldehyde portion of the glucose molecule below and its C-5 hydroxyl group?
a. disaccharide
b. carboxylic acid
c. hemiacetal
d. ester
e. stereoisomer
Answer:
hemiacetal forms in the intramolecular reaction between the aldehyde portion of the glucose molecule and its C-5 hydroxyl group
Explanation:
It is an alcohol also an ether that has been attached to the carbon molecule. Here the hydrogen has occupied the fourth bonding position. This hemiacetal has been derived from the aldehyde. Hence, hemiketal being an alcohol as well as ether has been attached to the same carbon and also to the two other carbon.
