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Samenvatting - Farmaceutische Bachelorproef (J000512A)

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Deze uitgebreide samenvatting voor het vak "Farmaceutische Bachelorproef" bevat alle belangrijke theorie uit de oefeningen, op basis van kernwoorden en georganiseerd volgens de volgorde van de oefeningen. Het is de ultieme samenvatting om snel de essentiële concepten te begrijpen.

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1


KERNWOORDEN CURIOS : SESSIE 1

Concentratie
ℎ𝑣𝑙ℎ 𝑠𝑡𝑜𝑓
» C = 𝑡𝑜𝑡𝑎𝑙𝑒 ℎ𝑣𝑙ℎ (𝑠𝑜𝑙𝑣𝑒𝑛𝑡)

Molariteit : aantal mol (mmol) per liter (mL) oplossing
𝑔
o M=
𝑀𝐺 𝑥 𝑉

densiteit (ρ)
𝑚
- ρ= 𝑉
o Eenheid: g/mL of kg/L of mg/µL



Vaste (m/m) Vloeistoffen (m/V) Vloeistoffen (V/V) Gassen (V/V)

ppu g/g g/mL mL/mL mL/mL

% g/100g g/100mL mL/100mL mL/100mL

ppm µg/g µg/mL nL/mL nL/mL

ppb ng/g ng/mL pL/mL pL/mL



Formules voor de bereidingen van oplossingen :

Verdunnen en vermengen

C x V = C’ x V’

» Verdunnen → aanlengen v V ml v C in totaal volume Vt
𝐶𝑥𝑉
▪ C x V = Cx x Vt  Cx = 𝑉𝑡

» Toevoegen → toevoegen v V0 ml H2O aan V ml van concentratie C
𝐶𝑥𝑉
▪ C x V = Cx x (V + V0)  Cx =
𝑉+ 𝑉0

» Mengen van oplossingen met hetzelfde product
o Toevoegen v V2 ml met concentratie C2 aan V1 met concentratie C1



o Mengen v V1 ml met conc. C1 met V2 ml met conc. C2 + aanlengen met V0 ml H2O:




o Mengen v V1 ml met conc. C1, V2 ml met conc. C2 aangelengd tot totaal volume Vt:




Massapercentage : ([massagevraagde stof]/[totale massa])*100

, 2


Titratieformules

- Titratieschema: aA + bB → cC + dD
- Aan SP geldt: a mmol A equivalent met b mmol B
- Toepassing regel van 3:


- Aangezien we VA ml van A getitreed hebben geldt:




Omzettingen : 1 mL = 1 cm³ = 1000 mm³ & 1L = 1 dm³




Gasdruk = onderlinge botsingen / met de wand

p.V=nRT

o p ↑ = KE ↑ = V↑
o V↓ = gasdruk dezelfde aantal molec ↑
o Gas mol. ↑ = V↑ = p↑
o T↑ = p↑ = V↑

- Partieeldruk
o P = xA dP + xB dP + xc dP + …
𝑛
▪ XA = 𝐴
𝑛𝑡𝑜𝑡

- De absolute druk : P = P0 + Pmanometer

- 1 atm = 1,013 x 105 N/m2 --> bijna gelijk aan 1 bar
o 1 bar = 1,00 x 10 N/m². = 100 pKa = 105 Pa = 105 J/m³
5




Absorptiewetten, Lambert beer

→ Transmissie (T of T%): T = I/I0; T% = 100.I/I0
» Extinctie (E) of absorbantie (A): E(A) = -log T = -log I/I0 = k’ . c . d


gemeten absorptie/emissie is rechtevenredig met concentratie
▪ ε x conversiefactoren MG & 1/10
→ Hoe groter ε bij een bep λ --> hoe groter E bij die λ

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Uploaded on
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Samenvattingen Farmaceutische Wetenschappen

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