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 : 
Answer:
98.6 g/mol.
Explanation:
Equation of the reaction
HX + NaOH--> NaX + H2O
Number of moles = molar concentration × volume
= 0.095 × 0.03
= 0.00285 moles
By stoichiometry, 1 mole of HX reacted with 1 mole of NaOH. Therefore, number of moles of HX = 0.00285 moles.
Molar mass = mass ÷ number of moles
= 0.281 ÷ 0.00285
= 98.6 g/mol.
The limiting reactant is chlorine (Cl2).
<u>Explanation</u>:
Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.
2 Al + 3 Cl2 ==> 2 AlCl3 represents the balanced equation.
Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g
= 1.260 g moles of Al
Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g
= 1.10 g moles of Cl2
Dividing each reactant by it's coefficient in the balanced equation obtains:
1.260 moles Al / 2 = 0.63 g moles of Al
1.11 moles Cl2 / 3 = 0.36 g moles of Cl2
The reactant which produces a lesser amount of product is called as limiting reactant.
Here the Limiting reactant is Cl2.
Answer:
The molar solubility of YF₃ is 4.23 × 10⁻⁶ M.
Explanation:
In order to calculate the molar solubility of YF₃ we will use an ICE chart. We identify 3 stages: Initial, Change and Equilibrium and we complete each row with the concentration of change of concentration. Let's consider the solubilization of YF₃.
YF₃(s) ⇄ Y³⁺(aq) + 3 F⁻(aq)
I 0 0
C +S +3S
E S 3S
The solubility product (Ksp) is:
Ksp = [Y³⁺].[F⁻]³= S . (3S)³ = 27 S⁴
![S=\sqrt[4]{Ksp/27} =\sqrt[4]{8.62 \times 10^{-21} /27}=4.23 \times 10^{-6}M](https://tex.z-dn.net/?f=S%3D%5Csqrt%5B4%5D%7BKsp%2F27%7D%20%3D%5Csqrt%5B4%5D%7B8.62%20%5Ctimes%2010%5E%7B-21%7D%20%20%2F27%7D%3D4.23%20%5Ctimes%2010%5E%7B-6%7DM)