oblicz(warunki normalne) mase 200cm3 wodoru objętość 170g amoniaku NH3
a)
M(H₂)=2g/mol
22,4dm³ --- 2g
0,2dm³ --- x
x=2g·0,2dm³/22,4dm³≈0,018g=18mg
b)
M(NH₃)=17g/mol
17g --- 22,4dm³
170g --- y
y=170g·22,4dm³/17g=224dm³
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
a)
M(H₂)=2g/mol
22,4dm³ --- 2g
0,2dm³ --- x
x=2g·0,2dm³/22,4dm³≈0,018g=18mg
b)
M(NH₃)=17g/mol
17g --- 22,4dm³
170g --- y
y=170g·22,4dm³/17g=224dm³