Vanadium can lose 1 to 5 electrons to give possible positive charges 1, 2, 3, 4, or 5. A previous worker in your lab gave you a vial labelled "Vanadium fluoride", but did not include the chemical formula on the label and did not indicate the charge of the vanadium ion. {Always be complete when you label things!!} The label does indicate that the formula mass is 126.94g/mol. What is the charge of the vanadium in this "vanadium fluoride"

The answer is conduction

First up write the aim
then write the materials reqd.
then the procedure which you follow
then observation
then conclusion
then you can write lab safety rules or precautions(you can avoid if u want).
lemme' give an example
aim: to observe the double displacement reaction
materials reqd: bacl2 and na2so4, test tubes and a dropper
procedure:
1. take bacl2 in a test tube and na2so4 in the other.
2. then pour bacl2 into na2so4 solution.
3. observe
observation:
you will see that a white precipitate is formed called baso4. along with that common salt or nacl is formed.
the equation is
bacl2+na2so4=baso4+2nacl
conclusion:
a double displacement reaction is observed.
hope it