Ethylene Burns in the presence of O₂ to produce CO₂ and H₂O vapors;
C₂H₄ + 3 O₂ → 2 CO₂ + 2 H₂O
According to equation,
22.4 L (1 mole) C₂H₄ reacts completely to produce = 44.8 L (2 moles) of H₂O
So,
1.65 L of C₂H₄ on complete reaction will produce = X L of H₂O
Solving for X,
X = (1.65 L × 44.8 L) ÷ 22.4 L
X = 3.3 L of H₂O
Answer:
The Limiting Reactant is the reactant that when consumed the reaction stops.
Explanation:
Answer:
[ Ga ] = 1.163 E-8 Kg/m³
Explanation:
- %wt = [(mass Ga)/(mass Si)]*100 = 5.0 E-7 %
⇒ 5.0 E-9 = m Ga/m Si
assuming: m Si = 100 g = 0.1 Kg
⇒ m Ga = (5.0 E-9)*(0.1 Kg) = 5 E-10 Kg
∴ density (δ) Si = 2.33 Kg/m³
⇒ Volume Si = (0.1 Kg)*(m³/2.33 Kg) = 0.043 m³
⇒ [ Ga ] = (5 E-10 Kg)/(0.043 m³) = 1.163 E-8 Kg/m³
⇒ [ Ga ] =
Boiling points are related to the type of intermolecular forces between the compounds. if napatha has a lower boiling point hen it must have weaker bonds in it. Stronger bonds equal higher boiling points
