The color of light emitted during a flame test depends on
2 answers:
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Given parameters:
Mass of KClO₃ = 125g
Unknown:
Total mass of the products = ?
When KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;
2KClO₃ → 2KCl + 3O₂
All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.
The total mass of the products must give us 125g according to this law of conservation of matter.
Now to find the masses of each product,
- Find the number of moles of the given reactant:
Number of moles =
molar mass of KClO₃ = 39 + 35.5 + 3(16) = 122.5g/mol
So number of moles of KClO₃ = = 1.02moles
2. Now, using this number of moles, find the number of moles of the products using this value;
2 moles of KClO₃ produced 2 moles of KCl
1.02 moles of KClO₃ will also produce 1.02moles of KCl
2 moles of KClO₃ produced 3 moles of O₂
1.02 moles of KClO₃ will produce mole = 1.53 moles of O₂
3. Now find the masses of each product;
Mass = number of moles x molar mass
molar mass of KCl = 39 + 35.5 = 74.5g/mol
molar mass of O₂ = 16 x 2 = 32g/mol
Mass of KCl = 74.5 x 1.02 = 75.99g
Mass of O₂ = 32 x 1.53 = 48.96g
Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g
= 124.95g
This value is approximately the same as that of mass of KClO₃
Answer: Option (b) is the correct answer.
Explanation:
When there are more number of hydroxide ions in a solution then there will be high concentration of or hydroxide ions. As a result, more will be the strength of base in that particular solution.
A base is strong when it readily dissociate into its ions in the solution. When a base is strong, then it does not matter at what concentration it is dissolved in the solution because despite of its low concentration it will remain a strong base.
Thus, we can conclude that out of the given options, the statement even at low concentrations, a strong base is strong best relates the strength and concentration of a base.