Astudent increases the temperature of a 200 cm3 balloon from 60°c to 180°c. what will the new volume of the balloon be? a) 100 cm3 b) 272 cm3 c) 600 cm3 d) 236 cm3

I dont remember off the top of my head what the formula is but I do know that 
volume and temperature are directly proportional therefore, since there is an increase in temperature there is going to be an increase in volume. I solved it by doing the following steps:

200          x
 =  
60          180

60x = 36,000
    x = 36,000/60
    x = 600 cm^3

hope this helps and have a great day :)

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