Answer:
a. 123.9°C
b.
c.
Explanation:
Hello, I'm attaching a picture with the numerical development of this exercise.
a. Since the steam is overheated vapour, the specific volume is gotten from the corresponding table. Then, as it became a saturated vapour, we look for the interval in which the same volume of state 1 is, then we interpolate and get the temperature.
b. Now, at 80°C, since it is about a rigid tank (constant volume for every thermodynamic process), the specific volume of the mixture is 0.79645 m^3/kg as well, so the specific volume for the liquid and the vapour are taken into account to get the quality of 0.234.
c. Now,since this is an isocoric process, the heat transfer per kg of steam is computed as the difference in the internal energy, considering the initial condition (showed in a. part) and the final one computed here.
** The thermodynamic data were obtained from Cengel's thermodynamics book 7th edition.
Best regards.
Answer:
B.Lone pair in pyrrolidine ring is localized and, therefore, is expected to be more reactive.
Explanation:
There are two nitrogen atoms bearing lone pairs of electrons in the structure of nicotine as shown in the image attached.
One nitrogen atom is found in the pyrrolidine ring. The lone pair on this nitrogen atom is localized hence it is more reactive than the lone pair of electrons found on the nitrogen atom in the pyridine ring which is delocalized a shown in the image attached to this answer.
Answer:
molar mass M(s) = 65.326 g/mol
Explanation:
- M(s) + H2SO4(aq) → MSO4(aq) + H2(g)
∴ VH2(g) = 231 mL = 0.231 L
∴ P atm = 1.0079 bar
∴ PvH2O(25°C) = 0.03167 bar
Graham´s law:
⇒ PH2(g) = P atm - PvH2O(25°C)
⇒ PH2(g) = 1.0079 bar - 0.03167 bar = 0.97623 bar = 0.9635 atm
∴ nH2(g) = PV/RT
⇒ nH2(g) = ((0.9635 atm)(0.231 L))/((0.082 atmL/Kmol)(298 K))
⇒ nH2(g) = 9.1082 E-3 mol
⇒ n M(s) = ( 9.1082 E-3 mol H2(g) )(mol M(s)/mol H2(g))
⇒ n M(s) = 9.1082 E-3 mol
∴ molar mass M(s) [=] g/mol
⇒ molar mass M(s) = (0.595 g) / (9.1082 E-3 mol)
⇒ molar mass M(s) = 65.326 g/mol
