You can solve this by dividing the mass by the molar mass. The molar mass of CuF2 is about 101.5 g/mol. Therefore there are 100.0/101.5 = 0.985 mol.
Hey there!:
Given % of Mn=59.1% means 59.1 g of Mn present in 100 g of manganese fluoride.
Molar mass of Mn= 54.938 g/mol
Moles of Mn = mass / molar mass
59.1 /54.938 => 1.07 ≈ 1 mol.
and % of F=40.9% means 40.9 g of of F present in 100 g of manganese fluoride.
Molar mass of F=18.998 g/mol
Moles of F :
40.9 / 18.999 => 2.15 mol ≈ 2 mol.
The mole ratio between Mn:F= 1 : 2
Therefore the empirical formula of manganese fluoride:
=> MnF2=Mn1F2
Hope that helps!
The best way to balance an equation is to balance one atom at a time.
You start with two Au atoms on the left, so you know the coefficient of Au on the right has to be 2. So at first we get,
Au2S3 + H2 --> 2Au + H2S
Then, notice you have 3 sulfur atoms on the left, so you need three on the right.
Our equation becomes
Au2S3 + H2 --> 2Au + 3H2S
Lastly, we now have six hydrogen atoms on the right, and only two on the left, so we assign a three to the H2 on the left
Au2S3 + 3H2 --> 2Au + 3H2S Is the balanced final equation.
It's not the protons neutrons or electrons I know for sure but not the energy levels but it's the only one left so it has to be
