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Sindy
Questions (11)
The average score for a business examination comprising 300 students is 54%. If the marks are normally distributed with a
1 answer
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The coordinates of triangle ABC areA(0,0), B(2,6), and C(4,2). Using coordinates geometry; prove that, if the midpoints of sides
1 answer
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The coordinates of triangle ABC areA(0,0), B(2,6), and C(4,2). Using coordinates geometry; prove that, if the midpoints of sides
1 answer
678 views
Any help with the following would be greatly appreciated. I don't understand how to use the information given to create the
1 answer
695 views
I'm really struggling with word problems right now. Got through 20 just three left. Can someone please give me some pointers or
1 answer
653 views
Two bike riders X and Y both start at 2 pm riding towards each other from 40 km apart.
X rides at 30 km/h, Y at 20 km/h. If they
2 answers
1,076 views
The combustion of benzoic acid releases 26.38kJ/g and is often used to calibrate a bomb calorimete. The combustion of 1.045g of
9 answers
1,237 views
What is the mass of La2(C2O4)3*9H2O, can be obtained from 650mL of 0.0170 M aq solution of LaCl3 by adding a stoichiometric
1 answer
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If a student needs to prepare 50g of bromobenzene and expects no more than a 75% yieldl, how much benzene should the student
1 answer
589 views
animals cause ___________________ weathering by burrowing in and moving underground soil around.
0 answers
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I got 2 questions and I've been stuck on them for days and its due in 2 days.
1. For a quick way of doing multiplication, you can
3 answers
1,181 views
Answers (8)
Thanks!
I did is again: Benzoic acid: q= m*Cs*T Cs = 4.217kJ/g°C q1 + q2 = 0 [m*Cs*T] + [26.38kJ] = 0 [.876g*Cs*8.518°C]+ [26.38kJ] = 0 7.462g°C * Cs = -26.38kJ Cs= -3.535kJ/g°C to get the molar enthalpy: -3.535kJ/g°C * 114g/mol= 403.03kJ/mol* °C
Benzoic is 26.38 and the temp rises 8.518
this is so confusing =(
36913.6 J/g * 114g/mol= -4208150.4J/mol and to convert that to kJ= -4208150400 kJ/mol. the correct answer is -5116 kJ/mol
Okie I get 36913.6J/g* degree C but how do I change that into kJ/mol?
Okie i get Cp= 42178.8J/g delta C and for the octanol im using q= m*Cs*T q= 0.876*Cs*8.158 q= 7.1395g degreeC * Cs In your work i don't understand thisequation: [mass octanol x specific heat octanol x delta T] + [Cp x delta T] = 0
Ok thanks. The problem is, I'm from Australia and we don't use those measurements. But I MIGHT be able to use other Australian measurements.
