6 Flashcards
(21 cards)
calculate max number go benzoic acid molecules that can dissolve in 50cm3 of cold water If the solubility is 1.7g per 1000cm3
mass = 1.7 x 50/1000 = 0.085g moles = mass/rfm = 0.085/122 = 6.967x10-4 moles N= moles x avogadros constant = 4.2x1020
melting temp was between 116-121 compare this result with value
wide range
it is lower doe to impurities that are present
explain one procedural error that resulted in student 2 obtaining a great molar mass than student 1
the bung was not replaced quickly enough
CO2 lost to surroundings
why using higher vol of acid would have not affected students 2s results
the acid was already in excess
how using powered carbonate would have affected rate of reaction
Faster ROR. Powder has greater SA
no effect on final volume of gas
moles of carbonate are unchanged
equation for the thermal decomposition of G2 carbonated MCO3
MCO3(s) = MO(s) + CO2(g)
why mass of carbonate measured by student 3 has a greater % uncertainty that student 1s
student 3 used smaller masses
how do you ensure all sodium hydroxide was transferred to volumetric flask
rinse beaker
rinse funnel
transfer washings to volumetric flask
students added water above mark why does procedure have to be started rather than just pipetting out excess water
removal of water will remove some of dissolved NaOH
give two steps student needs to take before the student takes the initial burette reading
ensure tip of burette is filled with solution
read from eye level (in line with meniscus)
remove funnel
student 2 cleaned the burette by rinsing it with deionised water before filling with NaOH - state effect
titre will be larger because NaOH will be diluted
asses suitability of methyl red as an indicator for this titration
vertical part of a graph is at pH7-10
mid point of colour change of methyl red is 5.1
pH range of methyl red does not lie within vertical range of pH curve
colour change will occur before equivalence point
calculate total % uncertainty each burette reading is 0.05cm3 12.4
%= (0.05x4)/12.4 x100= 1.6%
butan-1-ol + HBr= 1-bromobutane
why cool mixture
why antibumping granules
why heat under reflux
cool mixture because reaction is very exothermic
antidumping granule to promote formation of small bubbles
heat under reflux to prevent loss of any volatile products
identify errors of reflux
water is flowing wrong way through condenser so doesn’t fill up with water
gap between condenser and flask so gas will escape
stopper on condenser so there will be a build up of pressure if gap between condenser and flask was closed
why brown vapour forms
bromine
bromide ions oxidised by conc sulphuric acids
state position of aqueous layer in separating funnel
aq layer is on top because H20 has a lower density than 1-bromobutane
why step 8 = add aq sodium hydrogen carbonate
why step 9 = shake and invert and open tap
why step 10 = add anhydrous sodium sulphate and swirl until clear
step 8 = react with any H+ ions in mixture
9= tap opened to release CO2
10= to remove any water
give sutibel temp range to collect 1-bromobutane in redistilation in step 11
100-105
why bromine has a higher boiling temperature than chlorine
bromine had more electrons than chlorine
so bromine had stronger LDF
more energy needed to overcome the LDF between bromine molecules
carry out experiments to determine order of reactivity of 3 halogens
1) mix Br2 with KCl
2) mix Br2 with KI
colours of halogen in cyclohexane
1) bromine is orange
2) iodine is purple
Br2 + 2I- = 2Br- + I2
