Answer:
10.6 moles of CO₂ are produced in this combustion
Explanation:
The combustion reaction is:
2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (g)
We assume the ethane as the limiting reactant because the excersise states that the O₂ is in excess.
We make a rule of three:
2 moles of ethane can produce 4 moles of CO₂
Therefore 5.30 moles of ethane will produce (5.3 . 4) /2 = 10.6 moles
It is B I think but I’m not completely sure
Answer:
V = 0.5 L
Explanation:
Given data:
Moles of nitrogen = 2.23×10⁻² mol (0.0223 mol)
Temperature = 273 K
Pressure = 1 atm
Volume = ?
Solution:
PV = nRT
V = nRT / P
V = 0.0223 mol × 0.0821 atm. mol⁻¹. L . k⁻¹ × 273 K / 1 atm
V = 0.5 L
Answer:
Diazanium sulfate is the correct answer.
Explanation:
Hope this helped Mark BRAINLIEST!!
Answer:
The voltage delivered by a primary battery is unrelated to its size.
