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Chemistry: Post your doubts here!

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It's answer is B need explanation
the equation for complete combustion is as follows
1C3H7OH + 4.5O2 -> 3CO2 +4H2O

so if we use 0.1 instead the equation will be like
0.1C3H7OH + 0.45O2 -> 0.3CO2 + 0.4H2O

so 0.3 moles of CO2 are produced which means 0.3 x 24 i.e. 7.2dm3 of CO2 is produced
however since only 0.45 moles of O2 reacted out of 0.5, 0.05 moles of O2 r still left
so adding its volume as well v have (0.05 x 24) + 7.2
this equals 8.4dm3

hoping this helps :)
 
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the equation for complete combustion is as follows
1C3H7OH + 4.5O2 -> 3CO2 +4H2O

so if we use 0.1 instead the equation will be like
0.1C3H7OH + 0.45O2 -> 0.3CO2 + 0.4H2O

so 0.3 moles of CO2 are produced which means 0.3 x 24 i.e. 7.2dm3 of CO2 is produced
however since only 0.45 moles of O2 reacted out of 0.5, 0.05 moles of O2 r still left
so adding its volume as well v have (0.05 x 24) + 7.2
this equals 8.4dm3

hoping this helps :)
Thanks a lot! ☺
 
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Cambridge International AS/A Level – May/June 2016 9701 35
QP:
ii) A student suggested that the accuracy of the experiment in (b)(i) would be improved by weighing FA 5 using a balance measuring to two decimal places. State and explain whether or not the student is correct.
MS:
(ii) The student is wrong because MgO is in excess or The student is wrong because H2SO4 is the limiting reagent

Can someone explain?
 
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Cambridge International AS/A Level – May/June 2016 9701 35
QP:
ii) A student suggested that the accuracy of the experiment in (b)(i) would be improved by weighing FA 5 using a balance measuring to two decimal places. State and explain whether or not the student is correct.
MS:
(ii) The student is wrong because MgO is in excess or The student is wrong because H2SO4 is the limiting reagent

Can someone explain?
it will not be more accurate becuz here v have to find the enthalpy change using the mass of MgO that reacted with H2SO4
but not all of it reacted some of it remained
this shows it is in excess

so the mass of all the MgO doesnt help rather the mass of the MgO which reacted will help obtain an accurate result
hope this helps
 
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--> Place a small spatula measure of FB 5 in a boiling tube, add a 1cm depth of dilute hydrochloric acid and warm the contents of the tube gently.
Observation : FB 5 + HCl: effervescence/ fizzing/ bubbling gas pops with lighted splint

-->Place small spatula measures of FB 5 and FB 6 in a single boiling tube. Use a test-tube holder to hold the tube. Add a 2cm depth of aqueous sodium hydroxide. CARE
Observation: FB 5 + FB 6 + NaOH: vigorous / violent/ exothermic / great/ extreme/ lots of and effervescence/ fizzing/ bubbling gas/NH3 turns (damp) red litmus (paper) blue.

How do we identify FB 5 is???

Fb5 is Al or Zn (s)
Is there a logical way to deduce this???

--> "Al and Zn reduce NO3- to ammonia in basic solutions"?

I searched, found out that this reaction is called "a nitrate test using Devarda's alloy", but there is no specific description of the process in our syllabus so how do the examiners expect us to know this!!
 
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Guys do we have to wash the conical flask before performing the next titration or nah?
 
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Please can anybody help me with these question . I cant able to understand them. Question 20 in first screenshot and question 10 in the second one .thanks
 

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