12. P x V = n x R x T
R = 0.0821 ( L x atm/mol K)
T = 25 + 273 = K
P = 1 atm
Masa = 40 x 1000 = 40000 g
Moles = masa/Mm
Mm metano = 16 g/mol
Moles = 40000 g / 16 g/mol = 2500 mol
V = n x R x T / P
V = 2500moles x 0.0821 ( L x atm/mol K) x 298 K / 1 atm
V = 61164.5 Litros
La Ley de Graham: v1 / v2 = (M2 / M1)-1/2
V = Velocidad y MM = MASA MOLECULAR.
1 min --- 60 seg
X ---- 90 seg
X = 1.5 min
(t2 / t1)2 = MM2/ MM1. Si T2 = 5 minutos; t1=1,5 minutos, PM2 = 160g /mol.
Despejar MM1 = MM2( t1 / t2)2 , MM1 = Masa de 1mol-g del compuesto orgánico desconocido: PM1= 160g /mol ( 1,5 / 5)2 PM1 = 16,0 g/mol
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12. P x V = n x R x T
R = 0.0821 ( L x atm/mol K)
T = 25 + 273 = K
P = 1 atm
Masa = 40 x 1000 = 40000 g
Moles = masa/Mm
Mm metano = 16 g/mol
Moles = 40000 g / 16 g/mol = 2500 mol
V = n x R x T / P
V = 2500moles x 0.0821 ( L x atm/mol K) x 298 K / 1 atm
V = 61164.5 Litros
13.La Ley de Graham: v1 / v2 = (M2 / M1)-1/2
V = Velocidad y MM = MASA MOLECULAR.
1 min --- 60 seg
X ---- 90 seg
X = 1.5 min
(t2 / t1)2 = MM2/ MM1. Si T2 = 5 minutos; t1=1,5 minutos, PM2 = 160g /mol.
Despejar MM1 = MM2( t1 / t2)2 , MM1 = Masa de 1mol-g del compuesto orgánico desconocido:
14.PM1= 160g /mol ( 1,5 / 5)2
PM1 = 16,0 g/mol
a.
moles H2 = 5 mol NH3 x 3 mol H2
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2 mol NH3
mol H2 = 7.5 mol