Question

Exercitii de clasa a 9a , chimie , vă rog mult , dau coroană

Answer 1

I.

1-procentuala

2-2

4/2 = 2 mol/L

3-20 g

md = mdxc%/100 = 200x10/100 = 20 g

4-180g

m.apa = ms-md = 200-20 = 180 g

5-6,022x10la24

1 mol H2O ...... 18g H2O ........ 6,022x10la23 molecule

                     180 g  ................ n = 6,022x10la24

II.

Cm = roxc%c10/miu, miu.HNO3 = 63 g/mol

= 1,53x63x10/63 = 15,3 mol/L

ro = ms/Vs => ms = roxVs = 1,53x4000

= 6120 g sol HNO3

md =msxc%/100 = 6120x63/100 = 3855,6 g HNO3 dizolvat

m.apa = ms-md = 6120-3855,2 = 2264,4 g apa

n moli = masa/miu = 3855,6/63 = 61,2 moli HNO3

1-b, 2-d, 3-d, 4-a, 5-b

III.

a)

ro = ms/Vs => ms1 = roxVs = 782,4x1,503 = 1176 g

md1 = ms1xc1%/100 = 1176x60/100 = 705,6 g H2SO4

md2 = ms2xc2%/100 = 2450x88/100 = 2156 g H2SO4

md.tot = md1+md2 = 2861,6 g

ms.tot = ms1+ms2 = 3626 g

c%.finala = 2861,6x100/3626 = 79% (aproximat)

Vs2 = ms2/ro2 = 2450/1,808 = 1355 cm3

Vs.tot = Vs1+Vs2 = 782,4+1355 = 2137,4 cm3 = 2,1374 L

Cm.final = md.tot/miuxVs.tot

= 2861,6/98x2,1374 = 13,7 M

b)

  md g      2861,6 g

2NaOH + H2SO4 --> Na2SO4 + 2H2O

2x40           98

=> md.NaOH = 2x40x2861,6/98 = 2336 g

=> ms = mdx100/c% = 2336x100/40 = 5840 g sol. NaOH

=> m.apa = ms-md = 5840-2336 = 3504 g apa