The combustion reaction of propane would be expressed as:
C3H8 + 5O2 = 3CO2 + 4H2O
To determine the mass of water that is produced from the given amount of propane, we use the mass of propane and the relation of the substances from the balanced reaction. We do as follows:
moles propane = 22 g C3H8 ( 1 mol / 44.1 g ) = 0.50 mol C3H8
moles H2O = 0.50 mol C3H8 ( 4 mol H2O / 1 mol C3H8) = 2 mol H2O
mass H2O = 2 mol H2O ( 18.02 g / 1 mol ) = 36.04 g H2O
Therefore, the mass of water that is produced from 22 grams of propane would be 36.04 g.
Answer:
–2.23 L
Explanation:
We'll begin by calculating the final volume. This can be obtained as follow:
Initial pressure (P₁) = 1.03 atm
Initial volume (V₁) = 3.62 L
Final pressure (P₂) = 2.68 atm
Final volume (V₂) =?
P₁V₁ = P₂V₂
1.03 × 3.62 = 2.68 × V₂
3.7286 = 2.68 × V₂
Divide both side by 2.68
V₂ = 3.7286 / 2.68
V₂ = 1.39 L
Finally, we shall determine the change in volume. This can be obtained as follow:
Initial volume (V₁) = 3.62 L
Final volume (V₂) = 1.39 L
Change in volume (ΔV) =?
ΔV = V₂ – V₁
ΔV = 1.39 – 3.62
ΔV = –2.23 L
Thus, the change in the volume of her lung is –2.23 L.
NOTE: The negative sign indicate that the volume of her lung reduced as she goes below the surface!
Answer:
38
Explanation:
Symbol: Sr
Atomic mass: 87.62 u
Electrons per shell: 2,8,18,8,2
Atomic number: 38
Electron configuration: [Kr] 5s2
Van der Waals radius: 255 pm
Valence electrons: two
Each of the following are descriptions of physical properties except C. Flammability
