Additional information
Relative atomic mass(Ar) : A=7, G=16
The empirical formula : A₂G
<h3>Further explanation</h3>
Given
3.5g of element A
4.0g of element G
Required
the empirical formula for this compound
Solution
The empirical formula is the smallest comparison of atoms of compound forming elements.
The empirical formula also shows the simplest mole ratio of the constituent elements of the compound
mol of element A :
mol of element G :
mol ratio A : G = 0.5 : 0.25 = 2 : 1
Answer: measure the mass (48.425g) of KCl
Explanation:
To prepare the solution 0.65M KCl we must measure the mass of KCl that would be dissolved in 1L of the solution. This can be achieved by:
Molar Mass of KCl = 39 + 35.5 = 74.5g/mol
Number of mole (n) = 0.65
Mass conc of KCl = n x molar Mass
Mass conc of KCl = 0.65 x 74.5 = 48.425g
Therefore, to make 0.65M KCl, we must measure 48.425g
Three groups
gases, metals, metalliods/nonmetals
Answer:
1.
Explanation:
1. They are made up of two or more Pure substances that are not chemically bonded together and appear non-uniform
A heterogeneous mixture is not chemically combined and its components are visible and can be seen.
Answer: 16.32 g of 
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of 
require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of 
Thus 16.32 g of as excess reagent are left.
