Alicia

2

QUÍMICA ANALÍTICA. FORMULARIO Y TABLAS PARA EL EXAMEN

ν=

 ( x − µ )2 
1
exp  −

2σ 2 
σ 2π


∑ (x -µ)

∑ (x - x )

σ=

b=

2

2

i

s=

i

N

2

i

∑(x i − x )

2

i

texp erimental =

i

n −1

r ⋅ n−2
1− r 2

a = y − b⋅ x

 ∂y  2  ∂y  2  ∂y  2
 σ a +   σ b +   σ c + ......
 ∂a 
 ∂b 
 ∂c 
2

∑ (x i − x )⋅(y i − y )

i

2

σ y2 = 

∑( y

sy/ x =

i

ˆ
− yi ) 2

sb =

i

n−2

sy/x

∑(x

i

− x)2

i

tex =

(x

)

− xB
1
1
s⋅
+
n A nB

donde

A

s

2(x

tex =

(n − 1)s A 2 + (n B − 1)sB 2
=A
(n A + n B − 2 )

A

− xB

2

)
2

sA
s
+B
nA
nB

ν=

ν = (n A + nB -2 )

 2
sdentro _ de _ las _ muestras ⋅   ⋅ t h⋅(n −1)n

σF2 =

W = 1.7 W1/2 = 4 s ;

CM entre − CM dentro
n

uA = u ⋅

1
1+ K ⋅

As =

V FE
V FM

H = 2·λ ·dp +

Rs =

∆t R
W promedio

s 2 injector =

Rs =

∆tinj

2

3 ⋅  Pi  − 1
 P



j=
3



2 ⋅  Pi  − 1
P



k2 '
= 10
k1 '

( P2 ' − P1 ' )
2

;

α=

t' R 2
t' R 1

N=

tR

2

σt2

distancia entre elmáximo y la rama descendent e b
=
distancia entre la rama ascendente y el máximo
a

2·γ ·DM
df 2 · f E ( k' ) dp 2 · f M ( k' )
+(
+
)·u
u
DE
DM

N  α − 1   k' B 

⋅
 ⋅
4  α  1 + k' B 



2

12

tR'
tM

s 2 detector =

Vg =

∆t det
12

K

ρ FE

16·Rs 2 ·H  α  (1 + k' B )
·
·
2
u
k' B
 α −1 
2

t RB =

2

⋅

273
Tc

11( yo − y )2
++
m n b2 ∑( xi − x )2

m e ⋅ B2 ⋅ r2
=
z
2 ⋅V

3

∑(x

1

1i

C ac o

k' =

⋅

s xE =

sy / x
b

2

⋅

1
y
+
n b2 ⋅ ∑( xi − x )2
i

(b1 − b2 )
sp ⋅i

W1/2 = 2.35 s

b

texp =

2

2

3



Vac

C ac i = 
V + V D
org
 ac


sy / x

i

 sA
sB 


n + n 
B
A
4
4


sA
sB


 n 2 (n − 1)...