Answer:
207g
Explanation:
Given compound:
Ca(ClO₃)₂
The equivalent mass can be derived by summing the molar masses of each atom
Molar mass of Ca = 40
Cl = 35.5
O = 16
Now solve;
Molar mass = 40 + 2(35.5 + 3(16)) = 207g
Order.
Hope this helps! :)
Answer: Option (3) is the correct answer.
Explanation:
When there is a negative charge on an atom then we add the charge with the number of electrons. Whereas when there is a positive charge on an atom then we subtract the charge from the number of electrons.
Atomic number of chlorine is 17. So, number of electrons present in 
is 17 + 1 = 18 electrons.
Atomic number of cobalt is 27. So, number of electrons present in 
is 27 - 4 = 23 electrons.
Atomic number of iron is 26. So, number of electrons present in 
is 26 - 2 = 24 electrons.
Atomic number of vanadium is 23. So, number of electrons present in V is 23 electrons.
Atomic number of scandium is 21. So, number of electrons present in 
is 21 + 2 = 23 electrons.
Thus, we can conclude that out of the given species, has the greatest number of electrons.
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%
