step by step solution please

1 13 14 15 16 13 2 H 1.00 3 2 He 4,003 10 Ne 20,10 > F 10.00 2 O 16.00 C 6,94 11 Na 22.99 4 Be 9.012 12 Mg 24,31 3 16 S 32,05 17 CI 35.45 3 4 5 6 2 9 10 11 12 16 Ar 39,95 36 Kr 33,80 35 4 21 SC 44,95 22 TI 20 Ca 40,00 24 Cr 52.00 30 Zn 65,38 Br 34 Se 70,97 1 K 39,10 37 Rb 3547 23 V 50,94 41 Nb 92.91 25 Mn 54, 43 To 26 Fe 55.95 44 Ru 101.1 27 58.93 45 Rh 102.9 s 6 7 B N 10,01 12,01 14,01 13 10 15 AI Si P 26,90 28,00 30,97 31 32 33 Ga Ge As 69,2272.6374.92 50 51 In Sn Sb 110, 121,8 32 83 TI Pb BI 2014 2072 209,0 114 Nh FI Mc 28 NI 58,62 46 Pd 1064 38 Sr 5 30 Y 8,91 40 zr 91.22 29 Cu 63.55 49 Ag 107, 19 AU 192,0 42 95,95 29,90 53 1 126,9 52 Te 127.6 cd 112.4 54 Xe 131,3 5. 27 56 Ba 72 HE 120,5 6 74 w 183 75 Re 106,2 16 Os 100, 70 PL 1951 34 Po 36 Rn AL 73 Ta L10,9 105 Db Hg 2006 55 Cs 132,9 07 Fr 192,2 -100 106 107 103 115 00 Ra 304 RI 100 HS 100 ME 110 Ds 112 Cn 116 LV 117 TS Bh 118 Og Sg RS 57 La 18.0 50 Pr 1400 60 Nd 1442 S1 Pm 62 Sm 66 Dy 1625 63 Eu 1920 Us Am 07 Ho 164.9 70 Yb 123.0 1504 58 Ce 140.1 0 Th 22,0 65 Tb 150 2 BK 71 Lu 175.0 Gd 192 36 Cm 60 Er 16,3 100 Fm 69 Tm 16,9 101 Md 103 80 Ac 1 231,0 92 U 210.0 09 Np Pu CF 99 Es 102 No ur C = 2.998 x 10 ms 1 atm = 760 torr 8 = 9,8 ms 1 atm = 760 mmlig R = 8.314 ) mol-'K! = 10-12 m Khwater) = 0.512 Km R = 0.08206 L atm mol-K-1 1 nm = 10m Cwater = 4.18g"K-1 h = 6.626 X 100) 1 atm = 101325 Pa 11 = 1 kg m s- 1 g = 108 Rewater = 1.86 Km-1 R = 2.179 x 10-18 1 atm = 1.01325 bar 1 cal = 4.181 NA = 6.02 x 1023 1 pm I 3. The vapor pressure of water at 25C is 23.8 mmHg and normal boiling point is 100 C. As every 1 km increase in altitude causes a 10 torr decrease in atmospheric pressure, calculate the altitude as km where the boiling point of water is 98.5 C