Chemistry  permalink

2009 q19

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Messages: 1 - 5 of 5
  • Message 1. 

    Posted by Honeybee (U15216393) on Friday, 24th May 2013

    I'm wondering if you could take the time to help me with this question?

    I've always been taught that when you do reacting volumes calculations, you write out the balanced equation, write down the volume required for each beneath it and then place the smallest volume under each substance (according to the number of moles.) and up until now, that has always worked- so I'm confused as to why I am getting this wrong. Could you please help me?

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  • Message 2

    , in reply to message 1.

    Posted by MissChemTeacher_Bitesize (U15172316) on Saturday, 25th May 2013

    Going to run out of time to give you a full response. I'll be back on tomorrow and will explain it to you. The clue to the question is the complete combustion of the hydrogen and carbon monoxide mixture.

    So 3l of hydrogen needs to be burned.

    I'll explain more tomorrow.

    Night

    Miss_Chem smiley - cool

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  • Message 3

    , in reply to message 2.

    Posted by MissChemTeacher_Bitesize (U15172316) on Monday, 27th May 2013

    Sorry I'm just getting back to you now. I was running the Edinburgh Half Marathon yesterday and it was a bit of a scorcher!

    So you need to start off by writing a simple combustion equation:

    CO + H₂ + ?O₂ ----------> CO₂ + H₂O

    I put the values for each gas as the as the number of moles in the balanced equation. This effects the products too:

    CO + 3H₂ + ?O₂ ----------> CO₂ + 3H₂O

    You can then balance this equation to work out how much oxygen would be needed.

    CO + 3H₂ + 2O₂ ----------> CO₂ + 3H₂O

    So that would give us 2litres of oxygen.

    Normally you'd have a balanced equation in these types of questions.

    Hope this helps and you understand what I've done. It's a bit of a thinker! I normally do the same thing as you, putting the values (volumes) of each gas under the balanced equation.

    Cheers Miss_Chem smiley - cool

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  • Message 4

    , in reply to message 3.

    Posted by Honeybee (U15216393) on Thursday, 30th May 2013

    Thank you so much! If you have the time, could you also please help me with q34 in the 2012 paper?

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  • Message 5

    , in reply to message 4.

    Posted by MissChemTeacher_Bitesize (U15172316) on Thursday, 30th May 2013

    2012 Q34

    Hey There!

    A solution of hydrochloric acid with a pH of 6 and another of sodium hydroxide with a pH of 8 are each diluted by a factor of 100.

    After dilution, when tested using pH indicator paper,

    A the pH of the acid drops by 2
    B the pH of the alkali rises by 2
    C the pH of the acid equals that of the alkali
    D the pH of the acid is 2 below that of the alkali.


    Diluting hydrochloric acid moves the pH towards 7.

    Diluting sodium hydroxide moves the pH towards 7

    Both the acid and the base will be at pH 7 as they've both been diluted so the pH of the acid will equal that of the alkali.

    Hope that makes sense to you.

    All the best for tomorrow!
    Miss_Chem smiley - cool

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