Usloviya: - traditional ponimaem kak vse krome Ni v sootnoshenii (po ob'emu!) PG:VG:AD = 55:35:10 - Polagaem, chto aptechnyj VG soderzhit vody 13% (vmesto 15.4 +- 2.3%). Dolya bazy dolzhna byt' 36/100 =.360 =.036 N + 0.324 PG vse, krome nikotina: 1.036 = 0.964 = 0.530 PG +.337 VG + 0.096 AD Chistogo PG nado dobavit' 0.530 - 0.324 = 0.206 Aptechnogo VG nado dobavit'.337/.87 =.387 Aptechnyj VG budet soderzhat AD.387*.13 = 0.050 Chistoj AD nado dobavit' 0.096-0.050 = 0.046 Itogo: baza 0.360 5.4 10.8 PG 0.206 3.09 6.18 Apt.VG 0.387 5.81 11.62 AD 0.046 0.69 1.38 0.999 15. 29.98 Prigotovlenie Traditional 0mg iz Apt.VG. Prince the hits. Razvedenie 100mg PG (+ pure VG) -> 36mg traditional ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Usloviya: - traditional ponimaem kak vse krome Ni v sootnoshenii (po ob'emu!) PG:VG:AD = 55:35:10 dolya bazy dolzhna byt' 36/100 =.360 =.036 N + 0.324 PG vse, krome nikotina: 1.036 = 0.964 = 0.530 PG +.337 VG + 0.096 AD Chistogo PG nado dobavit' 0.530 - 0.324 = 0.206 Chistogo VG nado dobavit' 0.337 Chistoj AD nado dobavit' 0.096 Itogo: baza 0.360 5.4 10.8 PG 0.206 3.09 6.18 VG 0.337 5.06 10.12 AD 0.096 1.44 2.88 0.999 14.99 29.98 Prigotovlenie 0mg Traditional iz pure VG. Razvedenie 100mg PG (+ pure VG) -> 10mg traditional ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Usloviya: - traditional ponimaem kak vse krome Ni v sootnoshenii (po ob'emu!) PG:VG:AD = 55:35:10 T.e. Hotim imet' 0.01 NI + 0.99 *(.55 PG +.35 VG + 0.1 AD) = = 0.1 (.1 NI +.9 PG) -.09 PG + 0.99 *(.55 PG +.35 VG + 0.1 AD) = = 0.1 (100mg PG) + (.99*.55 -.09) PG +.99*.35 VG +.099 AD = = 0.1 (100mg PG) + 0.4545 PG + 0.3465 VG +.099 AD Itogo: baza 0.1000 50.00 PG 0.4545 227.25 VG 0.3465 173.25 AD 0.0990 49.50 1.

500.00 Razvedenie 100mg PG (+ pure VG) -> 8mg traditional. Razvedenie 100mg PG (+ pure VG) -> 8mg traditional ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Usloviya: - traditional ponimaem kak vse krome Ni v sootnoshenii (po ob'emu!) PG:VG:AD = 55:35:10 T.e. Hotim imet' 0.008 NI + 0.992 *(.55 PG +.35 VG + 0.1 AD) = = 0.08 (.1 NI +.9 PG) -.072 PG + 0.992 *(.55 PG +.35 VG + 0.1 AD) = = 0.08 (100mg PG) + (.992*.55 -.072) PG +.992*.35 VG +.0992 AD = = 0.08 (100mg PG) + 0.4736 PG + 0.3472 VG +.0992 AD Itogo: volume density mass/g m(510mL) m(1020mL) baza 0.0800 40.00 ~1.0363 0.0829 42.28 84.56 PG 0.4736 236.80 1.0363 0.4908 250.31 500.62 VG 0.3472 173.60 1.261 0.4378 223.28 446.56 AD 0.0992 49.60 1.

Oct 14, 2018 - PDF Tranformations in the systems with chemical reaction are discussed. The condition of such situations, as it follows equation (5.12), is the. Balance between the. There is pure oscillations according to the law. (1969) Kurs khimicheskoy kinetiki, Izdanie vtoroe, is.

0.0992 50.59 101.18 1. 500.00 1.1107 5.91 plotnost' smesi (.08+.4736)*1.0363 +0.3472*1.261 +.0992 = 1.1107 Razvedenie 100mg PG (+ pure VG) -> 5mg traditional. Razvedenie 100mg PG (+ pure VG) -> 5mg traditional ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Usloviya: - traditional ponimaem kak vse krome Ni v sootnoshenii (po ob'emu!) PG:VG:AD = 55:35:10 T.e. Hotim imet' 0.005 NI + 0.995 *(.55 PG +.35 VG + 0.1 AD) = = 0.05 (.1 NI +.9 PG) -.045 PG + 0.995 *(.55 PG +.35 VG + 0.1 AD) = = 0.05 (100mg PG) + (.995*.55 -.045) PG +.995*.35 VG +.0995 AD = = 0.05 (100mg PG) + 0.5023 PG + 0.3483 VG +.0995 AD Itogo: volume density mass/g m(510mL) m(1020mL) =m(1mL) baza 0.0500 25.00 ~1.0363 0.0518 26.42 52.84 PG 0.5023 251.15 1.0363 0.5205 265.46 530.91 VG 0.3483 174.15 1.261 0.4392 223.99 447.98 AD 0.0995 49.75 1.

0.0995 50.75 101.49 1.0001 500.00 1.1110 5.22 plotnost' smesi (.05+.5023)*1.0363 +0.3483*1.261 +.0995 = 1.1110.