Oblicz ile moli stanowi
a) 50g H20
b) 25g FeCl3
c) 100g KMn04
d) 30g NaOH
M H2O = 16u + 2 *1u= 18g/mol
n H2O = 50g : 18g/mol = 2,(7) mola
M FeCl3 = 56u + 3 * 35,5 u = 162,5 g/mol
n FeCl3 = 25g:162,5 g/mol = około 0,15 mola
M KMnO4 = 39u + 55 u + 4 * 16 u = 158 g/mol
n KMnO4 = 100g:158g/mol = około 0,63 mola
M NaOH = 23 u + 16u + 1u = 40g/mol
n NaOH = 30g:40g/mol = 0,75 mola
Mam nadzieję, że pomogłem ;)
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M H2O = 16u + 2 *1u= 18g/mol
n H2O = 50g : 18g/mol = 2,(7) mola
M FeCl3 = 56u + 3 * 35,5 u = 162,5 g/mol
n FeCl3 = 25g:162,5 g/mol = około 0,15 mola
M KMnO4 = 39u + 55 u + 4 * 16 u = 158 g/mol
n KMnO4 = 100g:158g/mol = około 0,63 mola
M NaOH = 23 u + 16u + 1u = 40g/mol
n NaOH = 30g:40g/mol = 0,75 mola
Mam nadzieję, że pomogłem ;)