Tarea pág. 125 no. 4.53, 4.55, 4.56, 4.59

P.125 nr. 4.53
4.53.:
KI= Yoduro de Potasio
# de gramos = 2.80M x (5.00x100mL)x 166
# de gramos = 2.80M x 0.5 lt x 166
# de gramos = 232.4 gr

4.55
MgCl2-> Cloruro de Magnesio (II)

Molaridad = # de moles / 1 lt de sal

0.100M = # moles / 0.6 lt
# de moles = 0.6 lt / 0.100M
# de moles = 6 moles

4.56

# gramos = 5.50M x 0.035 mL x 56
# gramos = 107.8
= 10.78gr

4.59
A. NaCl
# de gramos = M x V x Pm
2.14g = 0.270 x 58
2.14g / (0.270 x 58) = V
136.6 lt = V

B.Etanol

4.30gr / (1.50 x 46) = V
0.0623 ml = V
623 lt = V

C. Ácido acético

0.85gr / (0.30 x 60) = V
0.0472mL = V
472 lt = V