• Asignatura: Química
  • Autor: ricardo202
  • hace 8 años

alguien que me explique o resuelva? favor es temario para examen

Adjuntos:

Respuestas

Respuesta dada por: snorye
1
1.

n CaCO3 = 0.75  mol Ca(OH)2  x  1 mol CaCO3
                                                      ````````````````````
                                                       1 mol Ca(OH)2

n CaCO3 = 0.75 mol

2.

n CO2 = 5.3 mol C2H2  x  4 mol CO2
                                         ````````````````
                                           2 mol C2H2

n CO2 = 10.6 moles

3. Mm NaOH = 40 g/mol    ;  H2SO4 = 98 g/mol

g NaOH = 75 g H2SO4 x 1 mol H2SO4 x 2 mol NaOH x 40 g NaOH
                                        ```````````````````   ``````````````````  ````````````````
                                          98 g H2SO4     1 mol H2SO4  1 mol NaOH

g NaOH = 61.22

4.  %m/m = masa soluto  x  100
                   ``````````````````
                     masa solc.

% m/m =  8.6 g   x   100
               ``````````
                 95 g

% = 9.05 %m/m

5. 

a) prepara la solución

m solc. = m souto + m solvente

m solc. = 25 g + 225 g = 250 g

b) calcular %m/m

% m/m =  25 g   x   100
              `````````
                250 g

%m/m = 10 %


ricardo202: gracias
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