n / V = P / RT = (4.77 atm) / ((0.08205746 L atm/K mol) x (118 K)) = 0.4926 mol/L
(2.016 g/mol) x (0.4926 mol/L) = 0.993 g/L
Answer:
A) 4P + 5O₂ → 2P₂O₅
Explanation:
A) 4P + 5O₂ → 2P₂O₅
This equation is balanced. There are four phosphorus and ten oxygen atoms are on both side of equation.
B) 5P + 4O₂ → 2P₄O₅
This equation is not balanced. There are five phosphorus and eight oxygen atoms on left, eight phosphorus ten oxygen on right side of equation.
C) 2P + O₂ → P₂O₅
This equation is not balanced. There are two phosphorus, two oxygen atoms on left and two phosphorus five oxygen on right side of equation.
D) 4P + 2O₂ → 2P₄O₅
This equation is not balanced. There are four phosphorus, four oxygen atoms on left and eight phosphorus ten oxygen on right side of equation.
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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:
223.6539
Explanation:
1 mole is equal to 1 moles KCl, or 74.5513 grams