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
The molarity is count by dividing the mole of the solute within 1 liter of solvent. In this case, the KNO3 is 16.8g with 101.11 g/mol molar mass. Then we need to find the mol first. The calculation would be: 16.8g / (101.11g/mol)= 0.0166 mol.
Then the molarity would be: 0.0166mol/ 0.3l= 0.0498= 0.0553 M
I think it’s oxygen and carbon
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