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Chemistry: Post your doubts here!

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Please help! O/N 2005 p1 question 12

One mole of magnesium, aluminium, and sulphur are each completely burnt in an excess of oxygen gas. Which graph shows the moles of oxygen used in each case?
Answer is D but when i worked the equations out, I got
Mg + 1/2 O2 -> MgO
Al + 3/4 O2 -> 1/2 Al2O3
S + O2 -> SO2

And i noticed that there is a difference of 0.25 between each so i assumed it's a direct proportion hence my answer was A ( a lineat graph). However the answer is D. Please help!
 
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HNO3= Oxidising agent ( O.A)
Nh3= Reducing agent ( R.A)
NO2 =Oxidising agent
NO & SO2 =R.A
H2S= R.A

^ are these correct?

& Cl2 is a better oxidising agent than H2so4 , right?
 
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Please help! O/N 2005 p1 question 12

One mole of magnesium, aluminium, and sulphur are each completely burnt in an excess of oxygen gas. Which graph shows the moles of oxygen used in each case?
Answer is D but when i worked the equations out, I got
Mg + 1/2 O2 -> MgO
Al + 3/4 O2 -> 1/2 Al2O3
S + O2 -> SO2

And i noticed that there is a difference of 0.25 between each so i assumed it's a direct proportion hence my answer was A ( a lineat graph). However the answer is D. Please help!
I thought the same! But apparently you're meant to consider SO3 formed and not SO2
 
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Amount of heat produced = 80 * 1000 * 4.18 = 334000J
Based on energy per mole information given, you can find the number of moles of each fuel.
For methane, it's 334000/890000 = 0.375, and for methanol it comes to 0.468
Obviously, since there is lesser moles of methane, it will produce less carbon dioxide.
Thank you so much :D
 
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Can you explain these:
40 Fats and grease that build up on pans used in cooking are esters. Pans which are dirty from fats
or grease may be cleaned by heating them with a reagent that will react with the ester group.
What may be used to clean such pans by this reaction?
1 vinegar - aqueous ethanoic acid, CH3CO2H
2 alcohol - ethanol, C2H5OH
3 baking powder - sodium hydrogencarbonate, NaHCO3


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U still want explanation? :p I slept just 10 mins before u've asked ...
 
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