On the outermost layer of the rings (where the electrons are)
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!
68 kg. There are 58 kg salt in 500 gal seawater.
<em>Step 1.</em> Convert gallons to litres
1 US gal = 3.79 L (1 Imp gal = 4.55 L)
<em>Step 2</em>. Find the volume of the seawater
Volume = 500 gal × (3.79 L/1 gal) = 1895 L
<em>Step 3</em>. Find the mass of the seawater
Mass = 1895 L × (1.025 kg/1 L) = 1942 kg
<em>Step 4</em>. Find the mass of the salt
Mass of salt = 1942 kg seawater × (3.5 kg salt/100 kg seawater) = 68 kg salt
volume is mass divide by density
100/0.8= 125mL
