The volume of CO2 at STP =124.298 L
<h3>Further explanation</h3>
Given
Reaction
4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O
701,52 g of KMnO4
Required
volume of CO2 at STP
Solution
mol KMnO4 (MW=158,034 g/mol) :
mol = mass : MW
mol = 701.52 : 158.034
mol = 4.439
mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549
At STP 1 mol = 22.4 L, so for 5.549 moles :
=5.549 x 22.4
=124.298 L
Answer:
vague symptoms are characteristic of an acute toxin, because of the of the lack of well defined consistency that these symptoms have in relation to the course of the disease progress.
The longest hydrocarbon chain in the given compound is hexane, therefore it is the parent chain to be considered with one methyl group attached to the 3rd carbon and one chloro attached on the 2nd carbon, therefore the name of the compound is 2-chloro-3-methylhexane
Answer:
893 moles
Explanation:
An ideal gas at STP occupies 22.4 liters. Calculating Oxygen as if it were an ideal gas there are . 893 moles of Oxygen in 20.0 liters.
