Answer:
I need more information. What is the question?
Answer:
Aromatic Hydrocarbons
Explanation:
Aromatic (Pleasant Odour) Hydrocarbons are those having pleasant odours.
An element is always trying to have 8 valence electrons. Each element must obtain a different number to come to a total of 8
CaCO3 + 2KCL ⇒ CaCl2 + K2CO3
It is balanced as so based on the charges given on the periodic table and polyatomic ions.
Calcium has the charge of 2 but CO3 also shares the same charge, thus cancelling that out.
Potassium has a charge of 1 while Chlorine also shares a charge of 1, also cancelling it out.
Thus, if it performs a double replacement reaction, they would take these charges to the new elements that do not cancel out their charges.
Therefore, we need the coefficient of 2 in front of Potassium Chloride in order to balance the equation as on the products side of the equation, Potassium and Chlorine both have a subscript of 2.
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Answer:
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Explanation:
tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=
mass compound
mass H
×100%
\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=
mass compound
mass C
×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=
10.0g compound
2.5g H
×100%=25%
\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=
10.0g compound
7.5g C
×100%=75%