C because
Explanation Plato
The first word in the name of an ester is derived from the alcohol used in the esterification.
<h3>What is esterification?</h3>
Esterification is a chemical process where an organic acid with the formula is combined with an alcohol molecule having the chemical formula (ROH).
The process of esterification is known to produce an ester molecule and during this phenomenon is released water (H2O).
An example of an esterification reaction occurs when ethanoic acid (i.e., the active ingredient of vinegar) can react with C2H5OH (i.e., ethanol) in order to form the ethyl ethanoate molecule, which is a well-known ester molecule.
In conclusion, the first word in the name of an ester is derived from the Alcohol used in the esterification.
Learn more about esterification here:
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Answer: 7.98 grams of
are produced if 10.7 grams of
are reacted.
Explanation:
To calculate the number of moles, we use the equation:
.....(1)
Putting values in equation 1, we get:
The chemical equation for the reaction is
By Stoichiometry of the reaction:
2 moles of
produce = 1 mole of
So, 0.100 moles of
produce=
of
Mass of
=
Hence 7.98 grams of
are produced if 10.7 grams of
are reacted.
Answer:
6.68 X 10^-11
Explanation:
From the second Ka, you can calculate pKa = -log (Ka2) = 6.187
The pH at the second equivalence point (8.181) will be the average of pKa2 and pKa3. So,
8.181 = (6.187 + pKa3) / 2
Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11
Answer:
The correct answer is 1.21 L.
Explanation:
Based on the given information, the reaction will be,
CS2 (l) + 3Cl2 (g) ⇒ CCl4 (l) + S2Cl2 (l)
By using the standard values of the substances, the standard enthalpy of the reaction is,
ΔH° = [(-139.5) + (-58.5) – 0 – (87.3)] kJ/mol
= -285.3 kJ/mol
The amount of heat evolved for 3 moles of chlorine reacted us 285.3 kJ.
Now the number of moles of chlorine needed to react to produce 5.00 kJ is,
= 5.00 kJ × 3 mol Cl2/285.3 kJ
= 0.0526 mol Cl2
Now the volume of chlorine gas at 27degree C and 812 mmHg will be,
Volume = 0.0526 mol Cl2 × 0.0821 Latm/mol K × 300 K/ 1.07 atm
= 1.21 L