The reaction between butanoic acid and ethanol is:
CH3CH2CH2COOH + CH3CH2OH → CH3CH2CH2COOCH3CH2 + H2O
Based on the reaction stoichiometry:
1 mole of butanoic acid forms 1 mole of ethyl butyrate
Now,
Molar mass of Butanoic acid = 88.0 g/mol
Given mass of butanoic acid = 7.20 g
Therefore, # moles of butanoic acid reacted = 7.20/88.0 = 0.0818 moles
# moles of ethyl butyrate formed = 0.0818 moles
Molar mass of ethyl butyrate = 116 g/mol
Mass of ethyl butyrate synthesized = 0.0818 * 116 = 9.49 g
Answer:
La respuesta está en la explicación
Explanation:
Los átomos que no poseen ___<em>CARGA</em>____ son conocidos como ____<em>ÁTOMOS</em>____ ____<em>NEUTROS</em>_____. Los átomos que poseen carga se denominan ___<em>IONES</em>____, aquellos con carga positiva de llaman ____<em>PROTONES</em>____ y son aquellos que __<em>PIERDEN</em>__ electrones. Los ______<em>ANIONES</em>____ son aquellos con carga negativa y son los que _____<em>GANAN</em>___ electrones. Para todos los casos el valor de _<em>MASA</em>__ y _<em>NÚMERO ATÓMICO</em>_ permanecen iguales y son los que sacamos de la tabla periódica.
Answer:
Diluted concentration is 0.5M
Explanation:
Let's solve this with rules of three, although there is a formula to see it easier
In 1000 mL (1L), we have 2 moles of NaOH
In 250 mL we must have (250 . 2) / 1000 = 0.5 moles of NaOH
These moles will be also in 1 L of the final volume of the diluted solution
More easy:
1 L of solution has 0.5 moles of NaOH
Then, molarity is 0.5 M
The formula is: Concentrated M . Conc. volume = Diluted M . Diluted volume
2 M . 0.250L = 1L . Diluted M
0.5M = Diluted M
Explanation:
Since, some of the given sample is stuck inside and behind the pipet. Hence, there will occur a decrease in the percent of acetic acid.
This is because a decrease in concentration of the acid will also lead to a decrease in the amount of sample taken for the estimation. Since. lesser is the amount or concentration present lesser will be its analyte concentration.
For example, we took 10 mg of a pickel sample but 3 mg of the sample remain stuck in the pipet. This means we actually titrating a sample less than 10 mg.
Therefore, the analyte concentration in the pickel will also be less.
The chemical formula for table sugar is C12H22O11. From the formula, you can say that it is an organic molecule where in every molecule contains 12 atoms of carbon, 22 atoms of hydrogen and 11 atoms of oxygen. There are 45 atoms in total in one molecule of sugar.