Chemistry: Post your doubts here!

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Hi everyone, AsSalamoAlaikum Wr Wb..

To get things organized in a better way, I am making this thread. As othewise, some queries remain unanswered!

So post your CHEMISTRY doubts in this thread. InshaAllah members around will help you.

Any Chemistry related notes and links will be added here in this post. Feel free to provide the links to your notes around the forum, or any other websites!


Chemistry Notes:


http://www.chemguide.co.uk This is the website, which contains almost everything classified according to the syllabus.

Tips for solving chemistry MCQs 9701/01


Chemistry P5 Tips and Notes

Some links & Notes - by 'destined007'

Chemistry worksheets Link shared by hassam

Chemistry Application Booklet: Mistakes and Corrections!

Calculations for A level Chemistry, author E.N. Ramsden third edition ebook download.



Regards,

XPC Staff.

I want to help my type of confusion.
 
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The full Cambridge AS Chemistry course is given in an organized way.
The outline of Cambridge O level Chemistry from the following links:

 
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The outline of Cambridge O level Chemistry from the following links:

Cambridge International A2 Level Chemistry (9701) preparation guideline:

 
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Hi can you explain why the enthalpy change of neutralisation of one mole of sulfuric acid, H2SO4, is not the standard enthalpy change of neutralisation.
 
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Hi can you explain why the enthalpy change of neutralisation of one mole of sulfuric acid, H2SO4, is not the standard enthalpy change of neutralisation.
Because of the term 'standard' which tells us the conditions which the neutralisation occurs in . The enthalpy change of any reaction is dependent on the conditions which the reaction occurs in . The standard enthalpy change refers to when the concentration of the sulfuric acid and the base are at 1 moldm^-3 and the temperature which the reaction occurs in at 25 degrees Celsius and the sulfuric acid in its most standard state which is when it is dissolved in water. When such things are altered they change the enthalpy change of neutralisation to give a slightly different value.
I don't know if I have answered you but I hope I have
 
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CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ∆H = -890 kJmol-1

what’s the entalphy of formation of methane
 
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Because of the term 'standard' which tells us the conditions which the neutralisation occurs in . The enthalpy change of any reaction is dependent on the conditions which the reaction occurs in . The standard enthalpy change refers to when the concentration of the sulfuric acid and the base are at 1 moldm^-3 and the temperature which the reaction occurs in at 25 degrees Celsius and the sulfuric acid in its most standard state which is when it is dissolved in water. When such things are altered they change the enthalpy change of neutralisation to give a slightly different value.
I don't know if I have answered you but I hope I have
thank you
 
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CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ∆H = -890 kJmol-1

what’s the entalphy of formation of methane
Hello .We would need the enthalpy change of formation of carbon dioxide and water so we may draw a hess cyclee diagram since we are not told that the reaction is performed at standard conditions and cannot just blindly assume
 
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Hello .We would need the enthalpy change of formation of carbon dioxide and water so we may draw a hess cyclee diagram since we are not told that the reaction is performed at standard conditions and cannot just blindly assume
exactly😂 I thought the same way because this question was given by my teacher and I thought I’m missing out something
 
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Hello .We would need the enthalpy change of formation of carbon dioxide and water so we may draw a hess cyclee diagram since we are not told that the reaction is performed at standard conditions and cannot just blindly assume
have a question , what’s the purpose of drawing the Hess’a cycle if we can directly do the subtraction method
 
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have a question , what’s the purpose of drawing the Hess’a cycle if we can directly do the subtraction method
That's an alternative method .... A faster one for that matter 🔥🔥 Both of them work accurately well but I always prefer drawing the cycle then using the formula to confirm my answer . Unless if it's multiple choice !!!
 
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That's an alternative method .... A faster one for that matter 🔥🔥 Both of them work accurately well but I always prefer drawing the cycle then using the formula to confirm my answer . Unless if it's multiple choice !!!
What if we get this question CO(g) + 1/2O2(g) CO2(g) ∆H = -284 kJmol-1 , isn’t it oxygen gas is 0 , CO results to -284
 
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What if we get this question CO(g) + 1/2O2(g) CO2(g) ∆H = -284 kJmol-1 , isn’t it oxygen gas is 0 , CO results to -284
.
Yes the oxygen is 0 . The backward reaction gives +284 i.e the enthalpy of formation of Carbon monoxide from carbon dioxide is +284 but care is to be taken that this isn't the standard enthalpy of formation of neither CO nor CO2 ... I hope you understand that there's a difference as this are formed from non standard Conditions
 
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