mohon d bantu dgn cara

17) A. Gay Lussac

18) V O2 = n O2/n N2 × V N2 = 0,25 mol/0,5 mol × 14 L = 7 L (A)

19) Mr CaSO4 = Ar Ca + Ar S + 4 Ar O = 40 + 32 + 4(16) = 136 (E)

20) Mr = 6 Ar C + 12 Ar H + 11 Ar O = 6(12) + 12(1) + 11(16) = 260
M = m/Mr × 1000/V = 13/260 × 1000/500 = 0,1 M

21) yang mengandung jmlh molekul terdikit => Mr terbesar
Mr H2 = 2 ; Mr O2 = 32 ; Mr N2 = 28 ; Mr H2S = 34 ; Mr CO2 = 44
maka jawabannya : CO2 (E)

22) Mr CO(NH2)2 = 12+16+2(14)+2(1) = 60
% N = 2 Ar N/Mr × 100% = 2 × 14 / 60 × 100% = 46,6 % (C)

23) Mr CaCO3 = 40 + 12 + 3(16) = 100
n = m / Mr = 20 gram / 100 = 0,2 mol
jmlh molekul = n × NA = 0,2 mol × 6,02 × 10^23 = 1,204 × 10^23 molekul (B)

24) MV = MV
100 mL × 0,5 M = 0,1 M × V
V = 500 mL
jmlh air yg ditambah = 500 - 100 = 400 mL (D)

25) V O2 = 2/1 × 2,5 L = 5,0 L (C)