mr oxygen= 16. 16 x 3= 48
mr Al2O3= 27 + 27 + 48= 102
(48/102) X 100%= 47.05% oxygen
Answer:
A. The partial pressure for CH4 = 0.0925atm
B. The partial pressure for C2H6 = 0.925atm
C. The partial pressure for C3H8 = 0.346atm
D. The partial pressure for C4H10 = 0.115atm
Explanation:
Total pressure = 1.48atm
Total mole = 0.4+4+1.5+0.5=6.4
A. Mole fraction of CH4 = 0.4/6.4 = 0.0625
The partial pressure for CH4 = 0.0625 x 1.48 = 0.0925atm
B. Mole fraction of C2H6 = 4/6.4 = 0.625
The partial pressure for C2H6 = 0.625 x 1.48 = 0.925atm
C. Mole fraction of C3H8 = 1.5/6.4 = 0.234
The partial pressure for C3H8 = 0.234 x 1.48 = 0.346atm
D. Mole fraction of C4H10 = 0.5/6.4 = 0.078
The partial pressure for C4H10 = 0.078 x 1.48 = 0.115atm
Answer:
Explanation:
The combustion reaction of Octane is:
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>:
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:
OPTION B ,COMPOUND CONTAINING AMMONIUM
Answer:
the answer should 126 859.2 m2