A. n = 6/60 = 0,1 mol b. jumlah molekul = 0,1 x 6,02.10^23 = 6,02.10^22 molekul c. V (STP) = 0,1 x 22,4 = 2,24 L d. T = 25 + 273 = 298 K P V = n R T 2 x V = 0,1 x 0,082 x 298 V = 1,2218 L
Mr CO(NH₂)₂ = Ar C + Ar O + 2 . Ar N + 4 . Ar H = 12 + 16 + 2 . 14 + 4 . 1 = 60 g/mol a. n = 6 / 60 = 0.1 mol b. JM = n . L = 0.1 . 6.02 . 10²³ = 6.02 . 10²² molekul c. v (STP) = n . 22.4 = 0.1 . 22.4 = 2.24 L d. p . V = n . R . T 2 . V = 0.1 . 0.082 . (25 + 273) V = 1.2218 L ⇒ Jawab
Verified answer
A. n = 6/60 = 0,1 molb. jumlah molekul = 0,1 x 6,02.10^23 = 6,02.10^22 molekul
c. V (STP) = 0,1 x 22,4 = 2,24 L
d. T = 25 + 273 = 298 K
P V = n R T
2 x V = 0,1 x 0,082 x 298
V = 1,2218 L
Verified answer
Mr CO(NH₂)₂ = Ar C + Ar O + 2 . Ar N + 4 . Ar H = 12 + 16 + 2 . 14 + 4 . 1 = 60 g/mola. n = 6 / 60 = 0.1 mol
b. JM = n . L = 0.1 . 6.02 . 10²³ = 6.02 . 10²² molekul
c. v (STP) = n . 22.4 = 0.1 . 22.4 = 2.24 L
d. p . V = n . R . T
2 . V = 0.1 . 0.082 . (25 + 273)
V = 1.2218 L ⇒ Jawab