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:
1.67g H2CO3 are produced
Explanation:
Based on the reaction:
2NaHCO3 → Na2CO3 + H2CO3
<em>2 moles of NaHCO3 produce 1 mole of Na2CO3 and 1 mole of H2CO3</em>
To solve this question we need to find the moles of Na2CO3 = Moles of H2CO3. With their moles we can find the mass of H2CO3 as follows:
<em>Moles Na2CO3 -Molar mass: 105.99g/mol-</em>
2.86g Na2CO3 * (1mol/105.99g) = 0.02698 moles Na2CO3 = Moles H2CO3
<em>Mass H2CO3 -Molar mass: 62.03g/mol-</em>
0.02698 moles * (62.03g/mol) =
<h3>1.67g H2CO3 are produced</h3>
Answer:
you didnt give the question
Explanation:
Answer:
False
Explanation:
False. The molecules of liquid are hold in the liquid state due to intermolecular forces or Van de Waals forces , without affecting the molecule itself and its atomic bonds (covalent bonds). When the temperature increases the kinetic energy of the molecules is higher , therefore they have more possibilities to escape from the attractive intermolecular forces and go to the gas state.
Note however that this is caused because the intermolecular forces are really weak compared to covalent bonds, therefore is easier to break the first one first and go to the gas state before any covalent bond breaks ( if it happens).
A temperature increase can increase vaporisation rate if any reaction is triggered that decomposes the liquid into more volatile compounds , but nevertheless, this effect is generally insignificant compared with the effect that temperature has in vaporisation due to Van der Waals forces.
