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Answer:
2KClO3 》》2KCl +3O2
C+ O2》》CO2
number of C moles
Required O2 moles (According to the mole ratio )
Relevant to the first equation, find the moles the KClO3, which is used to produce that amount of O2 moles
Now you can find the mass of KClO3
I mentioned the useful steps which can guide you to get the answer.
Explanation:
Answer:
Density is a measure of mass per unit of volume. Density is a measure of mass per volume. The average density of an object equals its total mass divided by its total volume.
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Answer:
Explanation:
a ) Total mixture = 4.656 g
Sand recovered = 2.775 g
percent composition of sand in the mixture
= (2.775 g / 4.656 g ) x 100
= 59.6 % .
b )
Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .
Total mixture = 4.656 g
percent recovery = (3.627 / 4.656 ) x 100
= 77.9 % .
Answer:
Limiting reactant = B2O3
Amount of BCl3 formed = 468 g
Explanation:
The given reaction is:
In order to identify the limiting reagent calculate the moles of B2O3, C and Cl2. The reagent with the lowest moles is the limiting reactant
Since the moles of B2O3 < C < Cl2, the limiting reactant is B2O3
Based on the reaction stoichiometry:
1 mole of B2O3 produces 2 moles of BCl3
Hence, the number of moles of BCl3 produced under the experimental conditions = 2*1.997=3.994 moles
