For Hydrocarbon combustion:
CₓHₐ + O₂ → xCO₂ + a/2 H₂O
Moles of CO₂ = 16.2 / 44 = 0.37
Moles of Carbon = 0.37
Moles of H₂O = 4.976 / 18 = 0.28
Moles of Hydrogen = 0.28 x 2 = 0.56
Molar ratio = C : H = 1 : 1.5
= 2 : 3
C₂H₃
Mass of empirical unit = 12 x 2 + 1 x 3
= 27
Mr = 54.09
Number of empirical units repeated: 54.09 / 27
= 2
Molecular formula = C₄H₆
Answer:
17 ppm
Explanation:
Paso 1: Información brindada
Masa de soluto: 0.025 g
Volumen de solución: 1.5 L
Paso 2: Convertir la masa de soluto a miligramos
Usaremos el factor de conversión 1 g = 10³ mg.
0.025 g × (10³ mg/1 g) = 25 mg
Paso 3: Calcular las partes por millon (ppm) del soluto
Usaremos la siguiente expresión.
ppm = mg de soluto / litros de solución
ppm = 25 mg / 1.5 L
ppm = 17 mg/L
Answer:
Heterogenous
Explanation:
Mud is a mixture of water and soil
The London dispersion force is "the weakest intermolecular force"
Answer:
To consume the 2.8 moles of CH4 we need 5.6 moles of O2 since the molar ratio is 1:2. We have only 3 moles of O2 ; therefore, O2 is the limiting reactant.
Explanation: