Question

Chemistry 1516R Problem Set QUESTIONS For water, the molar heat of fusion is 6.01 kJ/mol and the molar heat of vaporization is 40.7 kJ/mol. al Calculate the amount of energy needed to melt 52.4g of ice at O'C b Calculate the amount of energy needed to evaporate 52.4 g of water at 100°C c If the specific heats of ice, water, and steam are 2.09.4.18 and 1.84J ' respectively, calculate the energy required to convert 100.0 g of iscat - 20.0C to steam at 140.0°C and I am. Which of the following gases can be liquified by applying pressure at 27'C? For those that cannot describe the conditions under which they can be liquified. 4 xenon, Xe: critical point: 17°C.58 atm bradon, Rn: critical point: 104°C, 62 atm nitric oxide, NO: critical point:-93C, 64 atm How many atoms are there in a body-centered unit cell of an atomic crystal with atoms at all lattice points? What type of crystal (molecular, ionic, covalent network, or metallic) do the following represent? al An element melts at temperatures below 100.0°C: neither solid nor melt conduct electricity bAn clement conducts heat and electricity well. 10 c An element melts at high temperatures, >1000.0°C; neither solid nor melt conduct electricity Classify crystals of each of the following substances by type: als, d/ Xe b Ni Sio | Mgo El Camond 11 For each pair, decide which substance has the lower melting point and explain your answer: a CH,CH,CH,OH or HOCH,CH,OH | CH or SiH c MgO or CH P-2 CHEMISTRY 1516L

Answer 1

Q6

(a)

heat required to melt ice = n.∆H_f = (52.4g/18g/mol )×6.01kJ/mol = 17.496kJ = 17.5 kJ. (Answer)

(b)

heat required to evaporate = n.∆H_v =(52.4g/18g/mol)× 40.7kJ/mol = 118.48kJ = 118.5 kJ. (Answer)

(c)

Energy needed = 100.0g×2.09J/g.℃ × 20℃ + (100.0g/18g/mol)×6.01×1000J/mol + 100.0g×4.18J/g.℃×100℃ + (100.0g/18g/mol)×40.7×1000J/mol + 100.0g×1.84J/g.℃×40.0℃ = 312840 J

= 312.84 kJ. (Answer)

Q7:

Only radon can be liquified at temperature 27℃ because Rn has critical temperature greater than 27℃ .

Answer (b)

Q8:

Number of atoms in BCC unit cell = 8 × 1/8 + 1×1 = 2