Answer:
Vapor pressure of Eugenol = 667.04torr
Explanation:
By applying dalton's law of partial pressure;
at 100degree celsius, total pressure = 744.67torr
vapor pressure of water at 100 degree celsius = 760torr
mole fraction of eugenol = 0.1648
mole fraction of water = 1 - 0.1648 = 0.8352
Total pressure = vapor P(water) + vapor P( Eugenol)
for water; vapour pressure = mole fraction x total pressure
for eugenol; vapor pressure = mole fraction x total pressure
substituting into the above gives the vapor pressure of eugenol = 667.04torr
Answer:
787.3 g of CO₂
Explanation:
The combustion of octane is this one:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
2 moles of octane react to 25 moles of oxygen in order to produce 16 moles of CO₂ and 18 moles of water.
The products in combustion reactions are always CO₂ and H₂O. The oxygen is always a reactant.
We convert mass to moles: 255 g . 1mol /114g = 2.24 moles
2 moles of C₈H₁₈ can produce 16 moles of CO₂
Then, 2.24 moles may produce (2.24 . 16)/2 = 17.9 moles of CO₂
We convert moles to mass:
17.9 mol . 44g/mol = 787.3 g of CO₂
Answer:
5.555.
Explanation:
∵ pH = - log[H⁺],
<em>[H⁺] for weak acids = √Ka.C.</em>
Ka for CH₃COOH = 1.74 x 10⁻⁵, C = 0.16 M.
∴ [H⁺] = √Ka.C = √(1.74 x 10⁻⁵)(0.16 M) = 2.784 x 10⁻⁶ M.
∴ pH = - log[H⁺] = - log(2.784 x 10⁻⁶ M) = 5.555.
Answer:
1) The neutralization reaction will mot be completed.
2) pH = 0.6.
Explanation:
<em>1) Will the neutralization reaction be completed?</em>
For the neutralization reaction be completed; The no. of millimoles of the acid must be equal the no. of millimoles of the base.
The no. of millimoles of 125 mL of 2.0 M HCl = MV = (2.0 M)(125.0 mL) = 250.0 mmol.
The no. of millmoles of 175 mL of 1.0 M NaOH = MV = (1.0 M)(175.0 mL) = 175.0 mmol.
∴ HCl will be in excess.
∴ The neutralization reaction will mot be completed.
<em>2) If not what is the pH of the final solution?</em>
[H⁺] =[ (MV)HCl - (MV)NaOH]/V total = (250.0 mmol - 175.0 mmol) / (300.0 mL) = 0.25 M.
∵ pH = - log[H⁺]
<em>∴ pH =</em> - log(0.25) =<em> 0.6.</em>
Answer: oh my gosh, yes I relate with this so much. Don’t worry, you are not alone and you will get through this.
Explanation:
