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    I can't remember the paper, but you know the question with 'Snottites' and it was a moles question, I didn't get how the last part was found, could somone help aha

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    (Original post by nukethemaly)
    really wish i'd gone to bed earlier, ugh
    Awwh, you'll be fine, still got till one have a power nap?

    (Original post by Branny101)
    I can't remember the paper, but you know the question with 'Snottites' and it was a moles question, I didn't get how the last part was found, could somone help aha

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    If I can get the question i'd be happy to help I don't know which question this is
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    (Original post by Branny101)
    I can't remember the paper, but you know the question with 'Snottites' and it was a moles question, I didn't get how the last part was found, could somone help aha

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    Found it Its Jan 12 one sec I'll post it

    Moleh of NaOH=  \frac{26.4}{1000}*0.05=0.00132 Moles= concentration*volume, make sure you convert to dm3 by dividing by 1000

    for part 2 divide by 2 as the ratio

    finally divide your moles by the volume for the concentration
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    17) N2O is present in air at a concentration of 310 ppbv.

    (a) Given a cubic metre contains 10 6 cm3, calculate how many cm3 of N2O would be in a cubic metre of air

    how do we do this?
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    (Original post by CoolRunner)
    17) N2O is present in air at a concentration of 310 ppbv.

    (a) Given a cubic metre contains 10 6 cm3, calculate how many cm3 of N2O would be in a cubic metre of air

    how do we do this?
    310/ 10^9*10^6
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    How is HNO3 removed from the atmosphere via wet deposition ?

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    (Original post by tigerz)
    Found it Its Jan 12 one sec I'll post it

    Moleh of NaOH=  \frac{26.4}{1000}*0.05=0.00132 Moles= concentration*volume, make sure you convert to dm3 by dividing by 1000

    for part 2 divide by 2 as the ratio

    finally divide your moles by the volume for the concentration
    And aaaah thanks tigerz, I'll check this out now

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    (Original post by SyedaK)
    310/ 10^9*10^6
    Can you please explain it?
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    (Original post by Branny101)
    How is HNO3 removed from the atmosphere via wet deposition ?

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    Erm acid rain, snow, fog?
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    (Original post by Whostolemycookie)
    Erm acid rain, snow, fog?
    Ahh duuuh, HNO3 is Nitric Acid, right ?

    Thanks BTW

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    Enothermic has positive sign.

    Describe and explain the effect of
    i) an increase in pressure on the rate of reaction.
    ii) an increase in pressure on the equilibrium yield of the reaction.
    iii.)an increase in temperature on the equilibrium yield of the reaction.


    Can someone explain to me I'm so confused
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    (Original post by michmic)
    Enothermic has positive sign.

    Describe and explain the effect of
    i) an increase in pressure on the rate of reaction.
    ii) an increase in pressure on the equilibrium yield of the reaction.
    iii.)an increase in temperature on the equilibrium yield of the reaction.


    Can someone explain to me I'm so confused
    ROR = Rate of Reaction

    i) Pressure increases the ROR as it increases the concentration
    Hence, More frequent collisions

    ii) As the pressure increases, the equilibrium will shift to the side with more gas molecules

    iii) As the temperature increases, the equilibrium will shift to the endothermic side.
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    (Original post by tigerz)
    Found it Its Jan 12 one sec I'll post it

    Moleh of NaOH=  \frac{26.4}{1000}*0.05=0.00132 Moles= concentration*volume, make sure you convert to dm3 by dividing by 1000

    for part 2 divide by 2 as the ratio

    finally divide your moles by the volume for the concentration
    Ahhh no I was talking about part v), I don't get the calculation

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    So what is the difference between increase on rate and on yield if it is endothermic reaction?
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    (Original post by michmic)
    So what is the difference between increase on rate and on yield if it is endothermic reaction?
    Increasing pressure increases the rate of reaction, regardless whether it's an endo/exothermic reaction.

    Increasing the pressure will shift the position of equilibrium to the side with fewer gas molecules. If there's less on the left hand side, then the yield of reactants increase whilst the yield of products decrease. If there's less on the right hand side, the yield of products increases, whilst the yield of reactants decrease.

    Pressure isn't related to endo/exo, only temperature is

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    (Original post by michmic)
    So what is the difference between increase on rate and on yield if it is endothermic reaction?
    The endothermic reaction doesn't affect it, but the rate is how fast the equilibrium is achieved, and the yield is how much of the final product you get
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    Does anyone have Jan13 pastpaper??
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    (Original post by CoolRunner)
    ROR = Rate of Reaction

    i) Pressure increases the ROR as it increases the concentration
    Hence, More frequent collisions

    ii) As the pressure increases, the equilibrium will shift to the side with more gas molecules

    iii) As the temperature increases, the equilibrium will shift to the endothermic side.
    Your wrong, in ii), it shifts to the side with less gas molecules
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    (Original post by shahida07)
    Your wrong, in ii), it shifts to the side with less gas molecules
    Thanks for correcting me. I'm really tired atm :L
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    (Original post by CoolRunner)
    Thanks for correcting me. I'm really tired atm :L
    Haha, no worries
    think we all are
 
 
 
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