Question

urgent help!!!!

31. Helium atoms do not combine to form Hemolecules, yet He atoms do attract one another weakly through A) dipole-dipole forces. B) ion-dipole forces. C) dispersion forces. D) dipole-induced dipole forces. E) hydrogen bonding. 32. Copper crystallizes in a face-centered cubic unit cell. The density of copper is 8.94 g/cm'. Calculate the length of the edge of the unit cell in pm. 33. The heat capacity of liquid water is 4.18 J/g.°C and the heat of vaporization is 40.7 kJ/mol. How many kilojoules of heat must be provided to convert 1.00 g of liquid water at 67°C into 1.00 g of steam at 100°C? A) 22.7 kJ B) 40.8 kJ C) 2.2 kJ D) 2,400 E) 40.8 J

Answer 1

(31) Helium is non Polar molecule. The forces exist between the non polar molecules are Dispezsion forces ( London forces) These are the weakest forces and type of vanderwaal forces. These type of forces arises when a temporary dipole is induced in a non-polar molecule due to electron movements option c is correct (32) Density of unit cell is given by the expression ZXM d = " I where Z → no of atoms per unit cell - We know that for f.c.c Ż=4 M atomic Mass of copper = 63.5 g No avogado number (6.023 x 1023) a edge length of unit cell. d density of copper & 8.99 g/cm3 Put all these values abore eq? 8.949. = 44 63.59 93 % 6023 x 1023 254 x1623 cm 4X63.58 8.94 8% 3 X 6.01181023 53.84 lhau 23 t = 4.7x102 cm3 a - 2 10 = (47)3 x (1624) cm a-36* 168 cm Convert cm to lm 1pm= 1000 lm or lcm = 1010 pm =a = 3.6x 108 x 100 pm a = 3.6 X 102 pm = 360pm ✓ edge length of unit cell is 360pm

(33) For a process that involves a Temperatux change, q is given by : 19= m.c.ot-0 where m = mass Given Ig c = heat capacity Given 4.189/go at is the change in Temperature AT= T2-T1 = 100°C-610 = 33°c Put all these values in een q= 18 *4.183 x 33'¢ = 137.94 3 Heat sequise to vaporize water Tood For a process that involves phase change, qis given by q = n. AH vap Where n is the no. of males of water n= weight : de OR 137-94 KJ = 0-131 kJ 1 -mol S. Molecular mass of - 1X2+16 Molecular mass =189 ) Given Atrap = 4017kolmor substitute the value of na atrap in eq" q = temer x 40-7 KI [mete = 2.26 K3 Total heat required in KJ = 2:26+ 0.137 K7 = 2.3 kJ (approx) options