there you go, hope that helps
A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.
<h3>What is a decomposition reaction?</h3>
A decomposition reaction can be defined as a chemical reaction in which one reactant breaks down into two or more products.
- Step 1: Write the balanced equation for the decomposition of KHCO₃.
2 KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l)
The mass loss of 0.275 g is due to the gaseous CO₂ that escapes the sample.
- Step 2: Calculate the mass of KHCO₃ that formed 0.275 g of CO₂.
In the balanced equation, the mass ratio of KHCO₃ to CO₂ is 200.24:44.01.
0.275 g CO₂ × 200.24 g KHCO₃/44.01 g CO₂ = 1.25 g KHCO₃
- Step 3: Calculate the mass percentage of KHCO₃ in the sample.
There are 1.25 g of KHCO₃ in the 1.775 g sample.
%KHCO₃ = 1.25 g/1.775 g × 100% = 70.4%
A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.
Learn more about decomposition reactions here: brainly.com/question/14219426
The percent yield of carbon dioxide will be 49.0 %.
<h3>Percent yield</h3>
First, let's look at the equation of the reaction:

The mole ratio of octane to oxygen is 2:25.
Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol
Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol
Thus, octane is limiting.
Mole ratio of octane to carbon dioxide = 2:16.
Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol
Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams
Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %
More on percent yield can be found here: brainly.com/question/17042787
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Answer:
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It can lead to skin can in case of extreme exposures.
I hope the answer is helpful.
Thanks for answering.
Answer:
1.66 × 10⁻¹⁸ Moles
Explanation:
As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.
The relation between Moles, Number of Atoms and Avogadro's Number is given as,
Number of Moles = Number of Atoms ÷ 6.022 × 10²³ Atoms/mol
Putting values,
Number of Moles = 1.0 × 10⁶ Atoms ÷ 6.022 × 10²³ Atoms/mol
Number of Moles = 1.66 × 10⁻¹⁸ Moles