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    Hi
    Can anyone help me on how to do part (ii) of the question

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    (Original post by Marccs)
    Hi
    Can anyone help me on how to do part (ii) of the question

    Thanks
    You need to use m/Mr = n to find the mass of [Cu(H20)6]2+
    Then convert that into mg and then divide that by the volume of water you have.

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    (Original post by pineneedles)
    You need to use m/Mr = n to find the mass of [Cu(H20)6]2+
    Then convert that into mg and then divide that by the volume of water you have.

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    Why do you divide by the volume of water after working out the mg , is there an equation for this like m= mr * n
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    (Original post by Marccs)
    Why do you divide by the volume of water after working out the mg , is there an equation for this like m= mr * n
    It's similar to the equation we use for concentration:
    C = n/v
    It's just that instead of finding the concentration in terms of moles, we're finding the concentration in terms of mass.
    It also helps if you look at the units they want the answer in: mgdm-3 which means mg/dm³ (mg divided by volume)

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    (Original post by pineneedles)
    It's similar to the equation we use for concentration:
    C = n/v
    It's just that instead of finding the concentration in terms of moles, we're finding the concentration in terms of mass.
    It also helps if you look at the units they want the answer in: mgdm-3 which means mg/dm³ (mg divided by volume)

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    Thank You , This makes a lot of sense now
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    (Original post by pineneedles)
    It's similar to the equation we use for concentration:
    C = n/v
    It's just that instead of finding the concentration in terms of moles, we're finding the concentration in terms of mass.
    It also helps if you look at the units they want the answer in: mgdm-3 which means mg/dm³ (mg divided by volume)

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    Hi
    Can you help me with this question
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    (Original post by Marccs)
    Hi
    Can you help me with this question
    Sure thing. You're told that 8 moles of solute are needed to lower the boiling point of water to -15 °C, so all you need to do is compare the mass of 8 moles of methanol to 8 moles of 1,2-ethan-di-ol. Remember that m/Mr = n

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    (Original post by pineneedles)
    Sure thing. You're told that 8 moles of solute are needed to lower the boiling point of water to -15 °C, so all you need to do is compare the mass of 8 moles of methanol to 8 moles of 1,2-ethan-di-ol. Remember that m/Mr = n

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    Ohh yes I get it now
    Thank you again
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    (Original post by pineneedles)
    Sure thing. You're told that 8 moles of solute are needed to lower the boiling point of water to -15 °C, so all you need to do is compare the mass of 8 moles of methanol to 8 moles of 1,2-ethan-di-ol. Remember that m/Mr = n

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    Hi
    Can you help me with this question, its only part (V) I'm asking about
    Thank You
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    (Original post by Marccs)
    Hi
    Can you help me with this question, its only part (V) I'm asking about
    Thank You
    So the equation shows you that one mole of sulphuric acid releases two moles of protons into solution (it is dibasic, so it requires two moles of a base to neutralise it)
    [H2SO4] = 2[H+]
    Does this help?
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    (Original post by pineneedles)
    So the equation shows you that one mole of sulphuric acid releases two moles of protons into solution (it is dibasic, so it requires two moles of a base to neutralise it)
    [H2SO4] = 2[H+]
    Does this help?
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    I understand that it is 2 moles of H+ so is that [H+] ^2 ?
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    (Original post by Marccs)
    I understand that it is 2 moles of H+ so is that [H+] ^2 ?
    Not quite, I think you may be getting confused with Ka equations.
    Just think about it like this:
    Every molecule of sulphuric acid releases two H+ into solution. So if you have, for instance, a 0.300 moldm-3 solution of sulphuric acid, this means that the concentration of H+ ions is 0.600 moldm-3. Now, can you apply that to your question? 😊

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    (Original post by pineneedles)
    Not quite, I think you may be getting confused with Ka equations.
    Just think about it like this:
    Every molecule of sulphuric acid releases two H+ into solution. So if you have, for instance, a 0.300 moldm-3 solution of sulphuric acid, this means that the concentration of H+ ions is 0.600 moldm-3. Now, can you apply that to your question? 😊

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    Yes that makes sense now , but why do we ignore So42- ?
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    (Original post by Marccs)
    Yes that makes sense now , but why do we ignore So42- ?
    Good, and we ignore that because pH is a measure of hydrogen ion concentration, SO4(2-) is irrelevant. It's just like how when you find the pH of HCl :
    HCl > H+ + Cl-
    We don't consider the Cl-, do we? And with a weak acid,
    HCOOH <-> H+ + HCOO- we don't consider the HCOO- ion either.

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    (Original post by pineneedles)
    Good, and we ignore that because pH is a measure of hydrogen ion concentration, SO4(2-) is irrelevant. It's just like how when you find the pH of HCl :
    HCl > H+ + Cl-
    We don't consider the Cl-, do we? And with a weak acid,
    HCOOH <-> H+ + HCOO- we don't consider the HCOO- ion either.

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    Ohh yeah thanks
 
 
 
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