`3.`
`n_{NaOH\ bd}=1,25.0,25=0,3125(mol)`
`V_{dd\ NaOH\ sau}={0,3125}/{0,5}=0,625(l)=625(ml)`
`V_{H_2O\ them}=625-250=375(ml)`
`4.`
`n_{HCl\ bd}={5,6}/{22,4}=0,25(mol)`
`m_{dd\ HCl\ sau}=0,25.36,5+0,1.1000=109,125(g)`
`C\%_{HCl\ sau}={0,25.36,5}/{109,125}.100≈8,36\%`
`C_{M\ HCl\ sau}={0,25}/{0,1}=2,5M`
`5.`
`C\%_{NaOH\ sau}={500.3\%+300.10\%}/{500+300}.100=5,625\%`
`6.`
`m_{dd\ HNO_3\ bd}=500.1,2=600(g)`
`C\%_{HNO_3\ sau}={600.20\%}/{300}.100=40\%`