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how was chemistry p21?????!!!

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wonderland said:
I WROTE THAT WHITE SOLID FORMS FOR CA(OH)2 FOR BOTH OF THEM ...
AND FOR THE SULPHURIC ACID ONE I WROTE H2SO3(SULPHUROUS ACID)
bt i think they said other than acid,, not sure
 
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libra94 said:
there was water cuz they told us that oxide of different elements were formed!!
and then they asked that ONLY 2 OF THE OXIDES ARE DANGEROUS TO THE ENVIRONMENT which was SO2 and CO2


THTS WRONG!! how can co2 be dangerous!!

we breathe iut ut aftr all

and so2 can cause acid rain so its dangerous
 
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WellWIshER said:
libra94 said:
there was water cuz they told us that oxide of different elements were formed!!
and then they asked that ONLY 2 OF THE OXIDES ARE DANGEROUS TO THE ENVIRONMENT which was SO2 and CO2


THTS WRONG!! how can co2 be dangerous!!

we breathe iut ut aftr all

and so2 can cause acid rain so its dangerous
CO2 causes global warming and greenhouse effect!!
 
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I DONT SEEN TO REMEMBER THAT IT WAS THIS WAY ALTHOUGH I THINK THEY ASK WRITE CAUSE FOR EACH OXIDE SOMETHING
SO HOW CAN BE H2O DANGEROUS
 
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Here were the few answers that i can remember:

The compounds of Calcium were:

CaO
CaCl2
CaCO3
Ca(OH)2
Ca(NO3)2

And it asked how to make CaO from CaCO3 so I said you should heat it.

The reagent used for reaction has to be a sulfate salt of a metal less reactive than Ca so i wrote magnesium sulfate.

And I thought Ca(OH)2 was insoluble so i wrote a white precipitate forms as my observation for both.

The compound with an Mr of 44 was CH3CHO and the compound with an Mr of 60 was CH3COOH.

Reaction one was hydration because a water molecule was added and reaction 2 was oxidation.

For the addition of propanone to those 2 molecules a water molecule was removes and a C=N bond formed.

The reagents to make thel C2H2 from the halogenoalkane i wrote alcoholic concentrated NaOH and the conditions were heat under reflux because I thought it was the same as getting an alkene from a halogenoalkane.

For the enthalpy change of formation (last question):

Enthalpy change of formation = 2(enthalpy change of formation of CO2) + (enthalpy change of formation of H2O) - (enthalpy change of combustion of C2H2) = +227 kJ mol-1

That's all i remember write now :p Feel free to tell me where I'm wrong.
 
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DIS question is out of context but if we have da equation y<x then how should we do da shading???
 
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@exampasser i wrote reduction, not hydration!!
and i dont remember the reagents to make alcohol from C2H2!?
 
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no the first one wasnt oxidation cuz there were GAIN OF HYDROGEN which is reduction!! and yes hydrolysis can also be correct i think
 
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i did a mistake, i told u, in balancing..there must be 1 mole of C2H2 formed, and i had 2 moles of it :(
anyways, it was reactants minus products
u did products minus reactancts thts why u got it wrong :(
 
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@libra94 sorry it was to make C2H2 from the halogenoalkane. I've edited my post. Thanks.

@wonderland I Don't remember my value but I know the answer was = 2 * enthalpy of formation of CO2 + enthalpy of formation of H2O - enthalpy of combustion of C2H2.

It was a +ve value. I think about +27 or something like that.
 
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