You start out with 800 moles of a radioactive substance. After 24 hours, 100 moles remain. What is the half-life of the substanc
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The question is incomplete, here is the complete question:
Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide , nitrogen dioxide
, and no other substances. A small sample gives 2.389 g CaO, 1.876 g
, and 3.921 g
Determine the empirical formula of the compound.
<u>Answer:</u> The empirical formula for the given compound is
<u>Explanation:</u>
The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:
where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.
We are given:
Mass of CaO = 2.389 g
Mass of
Mass of
We know that:
Molar mass of calcium oxide = 56 g/mol
Molar mass of carbon dioxide = 44 g/mol
Molar mass of nitrogen dioxide = 46 g/mol
<u>For calculating the mass of carbon:</u>
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 1.876 g of carbon dioxide, of carbon will be contained.
<u>For calculating the mass of nitrogen:</u>
In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.
So, in 3.921 g of nitrogen dioxide, of nitrogen will be contained.
<u>For calculating the mass of calcium:</u>
In 56 g of calcium oxide, 40 g of calcium is contained.
So, in 2.389 g of calcium oxide, of calcium will be contained.
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Calcium =
Moles of Carbon =
Moles of Nitrogen =
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.
For Calcium =
For Carbon =
For Nitrogen =
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of Ca : C : N = 1 : 1 : 2
Hence, the empirical formula for the given compound is