To melt the ice on your driveway, you can use two moles of rock salt (NaCl) or two moles of calcium chloride (CaCl2). Which solu
2 answers:
because it produces more moles of solute particles per mole of compound.
<u>Explanation:</u>
Freezing point depression depend on the "amount of solute particles" . Freezing point is also a colligative property. As we know, calcium chloride will dissociate into 3 particles whereas sodium chloride dissociate into only 2 particles. So, an aqueous solution calcium chloride would be effective than sodium chloride.
Also the water's boiling point with calcium chloride is higher. So is considered as a good road salt since it produces more ions per formula unit of salt that is dissolved than NaCl
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Answer: The molecular of the compound is,
Explanation:
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:
Mass of
Mass of
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in of carbon dioxide,
of carbon will be contained.
For calculating the mass of hydrogen:
In 18 g of water, 2 g of hydrogen is contained.
So, in of water,
of hydrogen will be contained.
For calculating the mass of oxygen:
Mass of oxygen in the compound =
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Oxygen =
Step 2: 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 moles.
For Carbon =
For Hydrogen =
For Oxygen =
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 2 : 3 : 1
Hence, the empirical formula for the given compound is
The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
Molecular formula =
Therefore, the molecular of the compound is,