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:
17.5moles
Explanation:
The number of moles in a substance can be calculated by using the formula;
Number of moles (n) = mass (m) ÷ molar mass (MM)
According to this question, mass of ammonia (NH3) = 297g
Molar Mass of NH3 = 14 + 1(3)
= 17g/mol
n = 297/17
n = 17.47
Number of moles of NH3 = 17.5moles
Answer:
Explanation: Nuclear fusion
Answer:
11.0 L
Explanation:
The equation for this reaction is given as;
2H2 + O2 --> 2H2O
2 mol of H2 reacts with 1 mol of O2 to form 2 mol of H2O
At STP;
1 mol = 22.4 L
This means;
44.8 L of H2 reacts with 22.4 L of O2 to form 44.8 L of H2O
In this reaction, the limiting reactant is H2 as O2 is in excess.
The relationship between H2 and H2O;
44.8 L = 44.8 L
11.0 L would produce x
Solving for x;
x = 11 * 44.8 / 44.8
x = 11.0 L
Answer:
3.09kg
Explanation:
First, let us write a balanced equation for the reaction. This is illustrated below:
2C8H18 + 25O2 —> 16CO2 + 18H2O
Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol
Mass of C8H18 from the balanced equation = 2 x 114 = 228g
Converting 228g of C8H18 to kg, we obtained:
228/1000 = 0.228kg
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 16 x 44 = 704g
Converting 704g of CO2 to kg, we obtained:
704/1000 = 0.704kg
From the equation,
0.228kg of C8H18 produced 0.704kg of CO2.
Therefore, 1kg of C8H18 will produce = 0.704/0.228 = 3.09kg of CO2