Answer : The vapor pressure of propane at 
is 17.73 atm.
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of propane at = ?
= vapor pressure of propane at normal boiling point = 1 atm
= temperature of propane = 
= normal boiling point of propane = 
= heat of vaporization = 24.54 kJ/mole = 24540 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the vapor pressure of propane at is 17.73 atm.
Answer:
34.8 moles of CO₂ are produced
Explanation:
This is the reaction:
1C₅H₁₂ + 8O₂ → 6H₂O + 5CO₂
Ratio is 1:5. We make a rule of three
1 mol of pentane can produce 5 moles of CO₂
Then, 6.96 moles of pentane may produce (6.96 .5) / 1 = 34.8 moles
Potassium carbonate, K 2CO 3, sodium iodide, NaI, potassium bromide, KBr, methanol, CH 3OH, and ammonium chloride, NH 4Cl, are s
slava [35]
Answer:
Potassium carbonate (K₂CO₃)
Explanation:
The compounds dissociate into ions in water, as follows:
K₂CO₃ → 2 K⁺ + CO₃⁻ ⇒ 3 dissolved particles per mole
NaI → Na⁺ + I⁻ ⇒ 2 dissolved particles per mole
KBr → K⁺ + Br⁻ ⇒ 2 dissolved particles per mole
CH₃OH → CH₃O⁻ + H⁺ ⇒ 2 dissolved particles per mole
NH₄Cl → NH₄⁺ + Cl⁻ ⇒ 2 dissolved particles per mole
Therefore, the largest number of dissolved particles per mole of dissolved solute is produced by potassium carbonate (K₂CO₃).
Answer:
450. g of 0.173 % KCN solution contains 779 mg of KCN.
Explanation:
Mass of the solution = m
Mass of the KCN in solution = 779 mg
Mass by mass percentage of KCN solution = 0.173%
1 mg = 0.001 g
m = 450,289 mg × 0.001 g = 450.289 mg ≈ 450. g
450. g of 0.173 % KCN solution contains 779 mg of KCN.
