- 407.4 kJ of heat is released.
<u>Explanation:</u>
We have to write the balanced equation as,
2 C₂H₆(g) + 7O₂ → 4CO₂ + 6H₂O
Here 2 moles of ethane reacts in this reaction.
Now we have to find out the amount of ethane reacted using its given mass and molar mass as,
2 mol C₂H₆ × 30.07 g of C₂H₆ / 1 mol C₂H₆ = 60.14 g of C₂H₆
Heat released = ΔH × given mass / 60.14
= - 1560. 7 kj ×15.7 g / 60. 14 g = -407. 4 kJ
Answer:
the fuel efficiency in kilometers per liter is 16.561 kilometer per liter
Explanation:
The computation of the full efficiency in kilometers per liter is shown below:
39.0 miles ÷ gallon = (39.0 miles ÷ gallon) × (1.6094 km ÷ 1 miles) × (1 gallon ÷ 3.79 L)
Now cut the opposite miles and gallons
So, the fuel efficiency would be
= 16.561 kilometers per liter
Hence, the fuel efficiency in kilometers per liter is 16.561 kilometer per liter
Evaporation and transpiration
