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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!
Answer:
C Is the correct answer!
Took the Cumulative Exam Review :/
Answer:
A
Explanation:
The total number of each type of atom always stay the same.
The added KI does not have any impact
The reaction invovles Titration of vitaminc ( Ascorbic acid)
ascorbic acid + I₂ → 2 I⁻ + dehydroascorbic acid
the excess iodine is free reacts with the starch indicator, forming the blue-black starch-iodine complex.
This is the endpoint of the titration. since alreay excess KI is added ( the source of Iodine), it does not have an influence.
Answer B
Hope this helps!
