Jon is running for city council and wants to campaign on human development goals for his community. Which issue should he use? A
2 answers:
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Answer:
- 10.19 g CO₂
- 4.69 g H₂O
Explanation:
The combustion reaction of Octane is:
- C₈H₁₈ → 8CO₂ + 9H₂O
To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.
We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:
- 1 gallon = 3.785 L
- 1 L = 1000 mL
Now we<u> convert 1.24 gallons to mL</u>:
- 1.24 gallon *
4693.4 mL
We <u>calculate the mass of Octane</u>:
- 4693.4 mL * 0.703 g/mL = 3.30 g Octane
Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:
- CO₂ ⇒ 3.30 g Octane ÷ 114g/mol *
* 44 g/mol = 10.19 g CO₂
- H₂O ⇒ 3.30 g Octane ÷ 114g/mol *
* 18 g/mol = 4.69 g H₂O
Answer:
1120 L.
Explanation:
Hello!
In this case, as no conditions of pressure of temperature are given for this problem, we can assume that the scuba diver dives at STP (1 atm and 273.15 K), which means that 1 mole of air would occupy a volume of 22.4 L.
In such a way, since she needs 50.0 moles of air, the following ratio is useful to compute the size (volume) of the tank she needs:
Thereby, we plug in to obtain:
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