Question

calculați masa oxidului de carbon(IV) si a azotului luate cu volumul 7,84l

Answer 1

niu = masa/M.molec.

niu = V/Vm (Vm = volumul molar = 22,4 L in c.n.)

=> m/M.molec. = V/Vm

=> m (CO2) = M.molec. x V/Vm = 44x7,84/22,4 = 15,4 g

=> m (N2) = M.molec. x V/Vm = 28x7,84/22,4 = 9,8 g

Answer 2

$raspuns \\ \\ explicatie \: pas \: cu \: pas$
niu = masa/M.molec
niu=V/Vm (Vm = volumul molar =22,4 L in c.n)
=>m/M.molec =V/Vm
=> m (CO2) =M.molec x V/Vm = 44×7,84/22,4=15,4 g
=>m (N2) =M.molec x V/Vm =28×7,84/22,4=9,8 g