O slideshow foi denunciado.
Seu SlideShare está sendo baixado. ×

adding a catalyst d- removing NHs(g) A 2500ml s the concentration of t.docx

Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio

Confira estes a seguir

1 de 2 Anúncio

adding a catalyst d- removing NHs(g) A 2500ml s the concentration of t.docx

Baixar para ler offline

adding a catalyst d. removing NHs(g) A 2500ml s the concentration of the 10. ple ofKOHis titrated with 21.83 mL of 0.2120 M?CI(aq, what is KOH solution? a. 0.0006720 M 0.002574 M c. 0.1851 NM d 0.2428 M
Solution
10.
KOH (aq) + HCl(aq) -------------> KCl(aq) + H2O(l)
1 mole        1 mole
KOH                                                               HCl
M1 =                                                               M2 = 0.212M
V1   = 25ml                                                      V2 = 21.83ml
n1 = 1                                                             n2 =1
M1V1/n1    =    M2V2/n2
M1         =    M2V2n1/V1n2
= 0.212*21.83*1/25*1    = 0.1851M
c. 0.1851M >>>>answer
.

adding a catalyst d. removing NHs(g) A 2500ml s the concentration of the 10. ple ofKOHis titrated with 21.83 mL of 0.2120 M?CI(aq, what is KOH solution? a. 0.0006720 M 0.002574 M c. 0.1851 NM d 0.2428 M
Solution
10.
KOH (aq) + HCl(aq) -------------> KCl(aq) + H2O(l)
1 mole        1 mole
KOH                                                               HCl
M1 =                                                               M2 = 0.212M
V1   = 25ml                                                      V2 = 21.83ml
n1 = 1                                                             n2 =1
M1V1/n1    =    M2V2/n2
M1         =    M2V2n1/V1n2
= 0.212*21.83*1/25*1    = 0.1851M
c. 0.1851M >>>>answer
.

Anúncio
Anúncio

Mais Conteúdo rRelacionado

Mais de wviola (20)

Mais recentes (20)

Anúncio

adding a catalyst d- removing NHs(g) A 2500ml s the concentration of t.docx

  1. 1. adding a catalyst d. removing NHs(g) A 2500ml s the concentration of the 10. ple ofKOHis titrated with 21.83 mL of 0.2120 M?CI(aq, what is KOH solution? a. 0.0006720 M 0.002574 M c. 0.1851 NM d 0.2428 M Solution 10. KOH (aq) + HCl(aq) -------------> KCl(aq) + H2O(l) 1 mole       1 mole KOH                                                              HCl M1 =                                                              M2 = 0.212M V1  = 25ml                                                     V2 = 21.83ml n1 = 1                                                            n2 =1 M1V1/n1   =   M2V2/n2 M1        =   M2V2n1/V1n2 = 0.212*21.83*1/25*1   = 0.1851M c. 0.1851M >>>>answer

×