Explanation:
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Match: Substance ....... Definition Lewis acid A. Provides H+ in water Bronsted-Lowry base B. Provides OH - in water Arrhenius b
1 answer:
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<u>Answer:</u> The empirical formula for the given compound is
<u>Explanation:</u>
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Conversion factor: 1 g = 1000 mg
Mass of
Mass of
Mass of compound = 2.78 mg = 0.00278 g
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
- <u>For calculating the mass of carbon:</u>
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.00632 g of carbon dioxide, of carbon will be contained.
- <u>For calculating the mass of hydrogen:</u>
In 18g of water, 2 g of hydrogen is contained.
So, in 0.00258 g of water, of hydrogen will be contained.
- Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Oxygen =
- <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
For Carbon =
For Hydrogen =
For Oxygen =
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : H : O = 3 : 6 : 1
Hence, the empirical formula for the given compound is
Answer:
1) 1,1,1-trichloropropane
2) 1,1,2-trichloropropane
3) 1,2,2-trichloropropane
4) 1,2,3-trichloropropane
Explanation:
For this question, we must remember that isomers are molecules that have the <em>same formula but different structure</em>s. For the formula we can draw a <u>linear chain of three carbons</u> and change the position of the chlorine atoms in the carbon chain.
With this in mind, if we put all the chlorine atoms on the same carbon we will get <u>1,1,1-trichloropropane</u>. If we change an atom from chlorine to carbon 2 we will obtain <u>1,1,2-trichloropropane</u>. If we move another chlorine atom to carbon two we will get <u>1,2,2-trichloropropane</u>. Finally, if we put a chlorine atom in each carbon we will obtain <u>1,2,3-trichloropropane</u>.
See figure one for further explanations
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