Answer:
V = 240.79 L
Explanation:
Given data:
Volume of butane = ?
Temperature = 293°C
Pressure = 10.934 Kpa
Mass of butane = 33.25 g
Solution:
Number of moles of butane:
Number of moles = mass/ molar mass
Number of moles = 33.25 g/ 58.12 g/mol
Number of mole s= 0.57 mol
Now we will convert the temperature and pressure units.
293 +273 = 566 K
Pressure = 10.934/101 = 0.11 atm
Volume of butane:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
V = nRT/P
V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm
V = 26.49 L/0.11
V = 240.79 L
You did not include the list but F is fluorine. The first halogen.
So, you can expect that the other members of the same group (halogens, column 17 of the periodic table) exhibit similar chemical behavior (reactivity).
So, I am sure your list contains one or more of theses elements: Cl (chlorine), Br (bromine), and I (iodine).
All of them you can expect to also be reactive non metal.
K2MnO4+(2)H2O —>KMnO4+(2)KOH+(2)H2
Coefficients are in parentheses and subscripts are just normal letters