Ionic compounds are formed between oppositely charged ions.
A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).
To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.
First empirical formula of binary ionic compound is written between
First Formula would be 
Second empirical formula is between 
Second Formula would be 
Note : When the subscript are same they get cancel out, so
would be written as 
Third empirical formula is between 
Third Formula would be :
Forth empirical formula is between 
Forth Formula would be :
or 
Note- The subscript will be simplified and the formula will be written as
.
The empirical formula of four binary ionic compounds are : 
6 Atoms!
Mg = 1 atom.
O = 4 atoms.
A = 1 atom.
Answer:
Boron. The answer is boron.
Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.
Since its volume you do what it says 55 cm x 100 cm x 80cm and do 100 x 50 which is 5,000. then 5,000 x 80 which is 40,000. sorry i cant show work on here.