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Uitgewerkte oude tentamens - Chemische Analysemethoden (CAM, 4051CHAN3) - MST $10.26
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Uitgewerkte oude tentamens - Chemische Analysemethoden (CAM, 4051CHAN3) - MST

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  • Course
  • Chemische Analysemethoden (4051CHAN3)
  • Institution
  • Technische Universiteit Delft (TU Delft)
  • Book
  • Quantitative Chemical Analysis

Het vak Chemische Analysemethoden (CAM, 4051CHAN3) omvat de basisbeginselen van chemische analyse waaronder monstervoorbereiding, foutenanalyse, basisprincipes van gas- en vloeistofchromatografie, elektroforese, spectrofotometrie en massaspectrometrie. Het vak wordt gegeven in de derde periode van ...

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  • July 12, 2020
  • 11
  • 2019/2020
  • Exam (elaborations)
  • Unknown

Subjects

  • mst
  • molecular
  • science
  • technology
  • chemische
  • analysemethoden
  • 4051chan3
  • cam

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  • Technische Universiteit Delft (TU Delft)
  • Molecular Science and Technology
  • Chemische Analysemethoden (4051CHAN3)
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By: kcbotma • 8 months ago

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By: lenasanten • 1 year ago

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4051CHAN3 UITWERKINGEN TENTAMENS




Chemische analysemethoden
4051CHAN3

Uitwerkingen oude tentamens




Pagina 1 van 11

, 4051CHAN3 UITWERKINGEN TENTAMENS


Inhoud
Inhoud 2
2020 Oefententamen 3
2014 Tentamen 4
2013 Tentamen 7
2010 Tentamen 8
2006 Tentamen 10




Pagina 2 van 11

, 4051CHAN3 UITWERKINGEN TENTAMENS


2020 Oefententamen
Opdracht 1
A.
156,26 + 156,20 + 156, 40 + 156, 30 + 156,15 + 156,10
x= = 156,235 g / mol
6

s=
(156,26 − 156,235 )2 + (156,20 − 156,235 )2 + (156, 40 − 156,235 )2 + (156, 30 − 156,235 )2 + (156,15 − 156,235 )2 + (156,10 − 156,235 )2 = 0,10839...
6 −1
ttabel = 2,571
2,571• 0,10839...
156,235 + = 156, 349 g / mol
6
2,571• 0,10839...
156,235 − = 156,121 g / mol
6
152,46 g/mol ligt niet binnen het betrouwbaarheidsinterval.

B.
Ja, deze waarden zijn nauwkeurig. Er is namelijk sprake van een kleine standaarddeviatie.

C.
Nee, deze waarden zijn niet accuraat. De waarden liggen ver van de werkelijke waarde af.

D.
De verschillen zijn veroorzaakt door systematische fouten. Er is namelijk sprake van telkens een te
hoge waarde. Dit kan niet komen door een incidentele fout want dan waren er ook lagere waarden
gevonden.


Opdracht 2
A.
mn+1 + 1,008 1022,13 + 1,008
n= = = 8,023
mn − mn+1 1149,65 − 1022,13
m + 1,008 1149,65 + 1,008
n = n+1 = = 7,012
mn − mn+1 1313, 75 − 1149,65
m
De lading van het ion met = 1149,65 is dus 8e.
z
m
De lading van het ion met = 1313, 75 is dus 7e.
z

B.
M = n • ( mn − 1,008 )
M = 7 • (1313, 75 − 1,008 ) = 9205 u
M = 8 • (1149,65 − 1,008 ) = 9189 u
9189 + 9205
= 9197 u
2
De nominale massa van dit eiwit is dus 9197 Da.




Pagina 3 van 11

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