<u>Answer:</u> The vapor pressure of the liquid is 0.293 atm
<u>Explanation:</u>
To calculate the vapor pressure of the liquid, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm
= pressure of the liquid = ?
= Heat of vaporization = 28.9 kJ/mol = 28900 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = 341.88 K
= final temperature = 305.03 K
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{28900J/mol}{8.314J/mol.K}[\frac{1}{341.88}-\frac{1}{305.03}]\\\\\ln P_2=-1.228atm\\\\P_2=e^{-1.228}=0.293atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B28900J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B341.88%7D-%5Cfrac%7B1%7D%7B305.03%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D-1.228atm%5C%5C%5C%5CP_2%3De%5E%7B-1.228%7D%3D0.293atm)
Hence, the vapor pressure of the liquid is 0.293 atm
The amount of energy in kilocalories released from 49 g of glucose given the data is -4.4 Kcal
How to determine the mole of glucose
Mass of glucose = 49 g
Molar mass of glucose = 180.2 g/mol
Mole of glucose = ?
Mole = mass / molar mass
Mole of glucose = 49 / 180.2
Mole of glucose = 0.272 mole
How to determine the energy released
C₆H₁₂O₆ →2C₂H₆O + 2CO₂ ΔH = -16 kcal/mol
From the balanced equation above,
1 mole of glucose released -16 kcal of energy
Therefore,
0.272 mole of glucose will release = 0.272 × -16 = -4.4 Kcal
Thus, -4.4 Kcal were released from the reaction
Learn more about stoichiometry:
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Answer:
yes the smoke from it scares them it also if, i remeber correctly, the smoke puts them to sleep
Explanation:
B. When a substance freezes, molecules come close to each other and leads to strong intermolecular bonding.
