Answer:
-2.3 ºC
Explanation:
Kf (benzene) = 5.12 ° C kg mol – 1
1st - We calculate the moles of condensed gas using the ideal gas equation:
n = PV / (RT)
P = 748/760 = 0.984 atm
T = 270 + 273.15 = 543.15 K
V = 4 L
R = 0.082 atm.L / mol.K
n = (0.984atm * 4L) / (0.082atm.L / K.mol * 543.15K) = 0.088 mol
Then, you calculate the molality of the solution:
m = n / kg solvent
m = 0.088 mol / 0.058 kg = 1.52mol / kg
Then you calculate the decrease in freezing point (DT)
DT = m * Kf
DT = 1.52 * 5.12 = 7.8 ° C
Knowing that the freezing point of pure benzene is 5.5 ºC, we calculate the freezing point of the solution:
T = 5.5 - 7.8 = -2.3 ºC
Answer: The new pressure is 7.1 atm
Explanation:
To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:
Putting values in above equation, we get:
Hence, the new pressure is 7.1 atm
Answer:
The smallest unit of a compound is a molecule, which is made up of atoms held together by bonds.
Answer:
2 and 4 if its more than one answerd and if not 2
Explanation:
brainleist would be nice
Answer:
pH = 4.543
Explanation:
- CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
- pKa = - Log Ka
∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
∴ pKa = 4.87
⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
added 300 mL 0f 0.02 M NaOH:
⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)
⇒ <em>C</em> CH3CH2COOH = 0.048 M
⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M
mass balance:
⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)
charge balance:
⇒ [H3O+] + [Na+] = [CH3CH2COO-]
∴ [Na+] = 0.02 M
⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)
(2) in (1):
⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]
replacing in Ka:
⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])
⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]
⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]
⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0
⇒ [H3O+ ] = 2.867 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = 4.543
