1) Balanced equation
C3H8 + 5O2 -> 3 CO2 + 4 H2O
2) 0.700 L C3H8
Given the pressure and temperature do not change, the molar ratio is equivalent to volume ratio
1molC3H8 / 5 mol O2 => 1 L C3H8 / 5 L O2
0.700 L C3H8 / x L O2 = 1 L C3H8 / 5 L O2 => x = 0.700 L C3H8 * 5 L O2 / 1 L C3H8
x = 3.500 L O2
3) CO2 produced
1 L C3H8 / 3 L CO2 = 0.700 L C3H8 / x L CO2 =>
x = 0.700 L C3H8 * 3 L CO2 / 1 L C3H8 = 2.100 L CO2
4) Water vapor produced
1) 1 L C3H8 / 4 L H2O = 0.700 LC3H8 / x L H2O =>
x = 0.700 L C3H8 * 4 L H20 / 1 L C3H8 = 2.800 L H2O
Answer:
dipole-dipole interactions.
Explanation:
hope this helps
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Answer:
Cccccccccccccccccccccccccccccccccccccccccccc
