Chapter 5) Molecular weight of gas Using ideal gas equation, PV = nRT = (mass of

Question

Chapter 5)

Molecular weight of gas

Using ideal gas equation,

PV = nRT = (mass of gas/molecular weight of gas)RT

with,

P = 732 mmHg/760 = 0.963 atm [1 atm = 760 mmHg]

V = 180 ml = 0.180 L [1 L = 1000 ml]

R = gas constant = 0.08205 L.atm/mol.K

T = 100 oC + 273 = 373 K [1 K = 273 oC]

mass of gas = 0.504 g

Feeding values,

0.963 x 0.180 = (0.504/molecular weight of gas)0.08205 x 373

molecular weight of gas = 89.1 g/mol

Answer : d) 89.1 g/mol

Chapter 6)

If reacting ammonia with oxygen releases energy of 1170 kJ.

The reaction is exothermic in nature (heat evolved).

In endothermic reaction, heat is absorbed.

a) Rocket propellant

reacting hydrazine (N2H4) with dinitrogen tetroxide (N2O4)

Thermochemical equation,

N2H4(g) + 2N2O4(g) ---> 3N2(g) + 4H2O(g)     dH = 2 x 1049 = 2098 kJ [2 moles of N2O4 is reacting here]

Explanation / Answer

AB Chapter 5 eses the flask, the differenee in a) 24.23hmel I.121o-mat)89.1 mal in the presence calalyst ta ge espathermie react heat at a camatant dinitragem reacti wride the thermachemica epecist andl this was

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.