second compound
Let molar mass of x is = X
Let molar mass of y is = Y
Moles of x in second compound = Mass / molar mass = 7 / X
Moles of y in second compound = Mass / molar mass = 4.5 / Y
For second compound
7 / X : 4.5/ Y = 1:1
Therefore
X / Y = 7/4.5
Y / X = 4.5/ 7
The mass of x in first compound = 14g
moles of x in first compound = 14/X
Mass of y in first compound = 3
moles of y in first compound = 3 / Y
14 / X : 3/ Y = 14Y / 3X = 14 X 4.5 / 3 X 7 = 3 :1
Thus molar ratio in first compound = moles of x / Moles of y = 3:2
Formula = x3y
The organism would no longer grow.
Answer is: the absolute pressure of the air in the balloon is 1.015 atm (102.84 kPa).
n = 0.250 mol; amount of substance.
V = 6.23 L; volume of the balloon.
T = 35°C = 308.15 K; temperature.
R = 0.08206 L·atm/mol·K, universal gas constant.
Ideal gas law: p·V = n·R·T.
p = n·R·T / V.
p = 0.250 mol · 0.08206 L·atm/mol·K · 308.15 K / 6.23 L.
p = 1.015 atm; presure of the air.
