Calorimetry calculations

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Sarwat.
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#1
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#1
How would I answer this question this is all I’ve done so far.
The answer for q2 is -57kjmol-1
And the answer for q3 is -525kjmol-1.
I don’t understand how they’ve gotten this can someone please explain what I’ve done wrong/need to do.Name:  image.jpg
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BlueChicken
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#2
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#2
(Original post by Sarwat.)
How would I answer this question this is all I’ve done so far.
The answer for q2 is -57kjmol-1
And the answer for q3 is -525kjmol-1.
I don’t understand how they’ve gotten this can someone please explain what I’ve done wrong/need to do.Name:  image.jpg
Views: 21
Size:  152.7 KB
So, for the first question, your answer is 0.057. Note, this is the actual answer of 57 divided by 1000 - so where do you think the extra factor of 1000 is coming from in your calculation?
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Sarwat.
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(Original post by BlueChicken)
So, for the first question, your answer is 0.057. Note, this is the actual answer of 57 divided by 1000 - so where do you think the extra factor of 1000 is coming from in your calculation?
Converting it from j to Kj right because enthalpy change is measured in kj
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BlueChicken
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(Original post by Sarwat.)
Converting it from j to Kj right because enthalpy change is measured in kj
That is a true statement and does need to be done. However, are you sure your units are correct in q = mcT?
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Turning_A_Corner
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Make sure you know which is the limiting reagent as well. Write out the equation and do a molar calculation before you do anything else.
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Sarwat.
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(Original post by BlueChicken)
That is a true statement and does need to be done. However, are you sure your units are correct in q = mcT?
yes? I think so..?
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Sarwat.
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(Original post by Turning_A_Corner)
Make sure you know which is the limiting reagent as well. Write out the equation and do a molar calculation before you do anything else.
ohhh ok I get it now so you divide the enthalpy change by the limiting reagent?
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Turning_A_Corner
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(Original post by Sarwat.)
ohhh ok I get it now so you divide the enthalpy change by the limiting reagent?
The maximum number of moles that can react is determined by the limiting reagent. So the joules/kilojoules calculated by m=c(delta)T will be divided by the moles of the limiting reagent, not the excess reagent.
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BlueChicken
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(Original post by Sarwat.)
yes? I think so..?
Look at the units - your mass should be in grams, as the specific heat capacity is given as 4.18 J g-1 K-1 (based on what you have written, you have mass in kg)
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