A= 3 sur8x(4sur9+2sur3)

b=3sur8x4sur9+2sur3

c=7sur5x(3sur10+1sur5)

d=7sur5x3sur10+1sur5
éeme ex:
l'air est constituéde 39sur50 de diazote,de 1sur5de dioxygène et de gaz rares.
Quelle est la proportion de gaz rares contenue dans l'air?
L'argon est l'un de ces gaz,il reprèsente 9sur10des gaz rares contenue dans l'air quel est le volume en cl d'argon dans 2l d'air?

2

Réponses

2014-05-29T14:48:40+02:00
Bonjour

A= 3 sur8x(4sur9+2sur3)
A = 3/8 x (4/9 + 2/3)
A = 3/8 x  (4/9 + 6/9)
A = 3/8 x 10/9 = 30/ 72 = 5/12

b=3sur8x4sur9+2sur3
b = 3/8 x 4/9 + 2/3
b = 12/72 + 2/3
b = 12/72 + 48/72
b = 60/72 = 5/6

c=7sur5x(3sur10+1sur5)
c = 7/5 x (3/10 +1/5)
c = 7/5 x ( 3/10 + 2/10)
c = 7/5 x 5/10
c = 35/50 = 7/10

d=7sur5x3sur10+1sur5

d = 7/5 x 3/10 + 1/5
d = 21/50 + 1/5
d =21/50 + 10/50
d = 31/50

dioxygène + diazote = 1/5 + 39/50 = 10/50 + 39/50 = 49/50
gaz rares = 50/50 -49/50 = 1/50
argon = 9/10 x 1/50 = 9/500
2 x 9/500 =18/500 = 9/250 = 0.036 L = 3.6 cl

2014-05-29T14:49:52+02:00
A) 3/8 * (4/9 +2/3)
= 3/8 * (4/9 + 6/9)
= 3/8 * 10/9
= 3*5*2/ 2*4*3*3
= 5/12

b) 3/8 * 4/9 + 2/3
= 3*4/4*2*3*3 +2/3
= 1/6 +2/3
= 1/6 + 2*2/3*2
= 1/6 +4/6
= 5/6

c) 7/5 * (3/10 +1/5)
= 7/5 * (3/10 + 1*2/5*2)
= 7/5 * (3/10 + 2/10)
= 7/5 * 5/10
= 7/10

d) 7/5 * 3/10 +1/5
= 21/50 + 1/5
= 21/50 + 1*10/5*10
= 21/50 +10/50
= 31/50
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