Answer:
5 × 10^-4 L
Explanation:
The equation of the reaction is;
2KClO3 = 2KCl + 3O2
Number of moles of KClO3 = 13.5g/122.5 g / mol = 0.11 moles
From the stoichiometry of the reaction;
2 moles of KClO3 yields 3 moles of O2
0.11 moles of KClO3 yields 0.11 × 3/2 = 0.165 moles of oxygen gas
From the ideal gas equation;
PV= nRT
P= 85.4 × 10^4 KPa
V=?
n= 0.165
R= 8.314 J K-1 mol-1
T= 40+273 = 313K
V= 0.165 ×8.134 × 313/85.4 × 10^4
V=429.4/85.4 × 10^4
V= 5 × 10^-4 L
Answer:
Normalidad = 4N
%p/V = 27.6%
Explanation:
La solución 2M de carbonato de potasio contiene 2moles de carbonato por litro de solución. La normalidad son los equivalente de carbonato de potasio (2eq/mol) por litro de solución:
2moles * (2eq/mol) = 4eq / 1L = 4N
El porcentaje peso volumen es el peso de carbonato en gramos dividido en el volumen en mL por 100:
%p/V:
Masa K2CO3 -Masa molar: 138.205g/mol-
2moles * (138.205g/mol) = 276g K2CO3
Volumen:
1L * (1000mL/1L) = 1000mL
%p/V:
276g K2CO3 / 1000mL * 100
<h3>%p/V = 27.6%</h3>
Answer:
Product: ethyl L-valinate
Explanation:
If we want to understand what it is the molecule produced we have to an<u>alyze the reagents</u>. We have valine an <u>amino acid</u>, in this kind of compounds we have an <em>amine group</em> (
) and a <em>carboxylic acid</em> group (
). Additionally, we have an <u>alcohol </u>(
) in the presence of HCl (a <u>strong acid</u>) in the first step, and a base (
).
When we have an acid and an alcohol in a vessel we will have an <u>esterification reaction</u>. In other words, an ester is produced. As the <em>first step,</em> the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the <em>second step</em>, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In <em>step 3</em>, a proton is transferred to produce a better leaving group (
). In <em>step 4</em>, a water molecule leaves the main structure to produce again the double bond C=O. <em>Finally</em>, a base () removes the hydrogen from the C=O bond to produce ethyl L-valinate
See figure 1
I hope it helps!
Answer:
25.89 × 10²³ molecules
Explanation:
Given data:
Mass of CoCl₂ = 560 g
Number of molecules present = ?
Solution:
Number of moles of CoCl₂:
Number of moles = mass/molar mass
Number of moles = 560 g/ 129.84 g/mol
Number of moles = 4.3 mol
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
4.3 mol × 6.022 × 10²³ molecules /1 mol
25.89 × 10²³ molecules
