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Someone Ask me AP Chem Questions
Ahmed- NH Admin
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Justice- NH Member
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Ahmed there is a mistake with your units -3 for units!!
Justice- NH Member
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You basically have (g)(g^-1c^-1)(c)-> kj/mol
Ahmed- NH Admin
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The sig figs are correct. WHat are you talking about??
Also I see now. The units for specific heat is J/(g-k)
So all those numbers are Joules. But the percentage still shouldn't change.
Also I see now. The units for specific heat is J/(g-k)
So all those numbers are Joules. But the percentage still shouldn't change.
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i was talking about the units not sig figs lol
Justice- NH Member
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Mailman took of -3 in the beginning of the year cause i did that too. You have to a careful, diligent, studious student
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what is the pH of 3M of HCl?
you do -log[3] beacuse the HCl is a strong acid. But u get a negative number. Explain it?
you do -log[3] beacuse the HCl is a strong acid. But u get a negative number. Explain it?
Ahmed- NH Admin
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Ok my diligient studious comrade. 
As for your Q:
-log([H+]) = -log(3) = -0.477
However, the experimentally derived data should be a bit higher, because at such high concentrations not all of the HCl ionizes.
As for your Q:
-log([H+]) = -log(3) = -0.477
However, the experimentally derived data should be a bit higher, because at such high concentrations not all of the HCl ionizes.
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But how can the pH level be negative? Isn't the scale from 1-14?
Ahmed- NH Admin
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The molarity is so large so it won't dissociate completely. However, if you do the experiment in real life the number will be greater by a little.
Ask me a good question please
Ask me a good question please
The Banker- NH Member
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the answer to the Q i asked was +/- 167 J 
Ahmed- NH Admin
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That can't be right. It has to be 83.6 J...and I thought you wanted a percentage btw.
The Banker- NH Member
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temperature rise= (24.5-17.4)+/-0.4 = 7.1+/-0.4 degrees Celsius
Percentage error caused by the thermometer = +/-0.4+ (100/7.1)= +/-5.63%
Heat produced= 100g*4.18jg^-1C^-1= 2968J
Error in value of heat produced= +/-5.63% of 2968J= +/-167J
Percentage error caused by the thermometer = +/-0.4+ (100/7.1)= +/-5.63%
Heat produced= 100g*4.18jg^-1C^-1= 2968J
Error in value of heat produced= +/-5.63% of 2968J= +/-167J
Ahmed- NH Admin
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That's not the heat produced.
The forumla is Q=mC(delta)T
You didn't use Delta T at all.
The forumla is Q=mC(delta)T
You didn't use Delta T at all.
The Banker- NH Member
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Marwa wrote:temperature rise= (24.5-17.4)+/-0.4 = 7.1+/-0.4 degrees Celsius
Percentage error caused by the thermometer = +/-0.4+ (100/7.1)= +/-5.63%
Heat produced= 100g*4.18jg^-1C^-1= 2968J
Error in value of heat produced= +/-5.63% of 2968J= +/-167J
Ahmed- NH Admin
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^^^Where did you use it here?Heat produced= 100g*4.18jg^-1C^-1= 2968J
The Banker- NH Member
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my bad..
Heat produced= 100g*4.18jg^-1C^-1*7.1= 2968J
Heat produced= 100g*4.18jg^-1C^-1*7.1= 2968J
Ahmed- NH Admin
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In the 2nd step why did you do add the percentage error to (100/7.1)?
The Banker- NH Member
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my bad again.. multiply em..
+/-0.4*(100/7.1)= +/-5.63%
+/-0.4*(100/7.1)= +/-5.63%
Ahmed- NH Admin
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But why did you do this fraction: (100/7.1)?
The Banker- NH Member
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ahmed, i honestly have no idea.. it would make more sense if it was 7.1/100
but thats what the answer says, and i'de trust the answers more
but thats what the answer says, and i'de trust the answers more
Ahmed- NH Admin
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Ok...another Q?
The Banker- NH Member
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calculate the mass of sodium carbonate decahydrate, Na2CO3.10H2O, needed to make up 250 cm^3 of 0.0500 mol dm^-3 solution.
this one is nice n easy..
this one is nice n easy..
Ahmed- NH Admin
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Remind me again...1cm^3=1 mL right?
The Banker- NH Member
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yup.. but why would u need that in this Q?
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