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Lecture

CHEM 1000 Lecture Notes - Silver Chloride, Joule


Department
Chemistry
Course Code
CHEM 1000
Professor
Hameed A.Mirza

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1. A 100g coffee sample (specific heat = 0.385 J/gC) at 100C is added to 50.0g of water at
26.5C what is the final T of the mixture? (T in Fahrenheit could be asked)
qcoffee + qwater = 0
qcoffee = - qwater
100g x 0.385J/gC x (Xc 100C) = -50g x 4.184T/gC x (XC 26.5C)
38.5Xc -3850J = -209Xc +5539J
Xc =37.9C
2. 10g ice is heated to 60C… what is the total ene involved in this sample? Specific heat of
water 4.184 J/gC and enthalpy of fusion = 6.02 KJ/mol
Initial temp = -10C. desity of water is 1.00 g/L
-10C q1 0C q2water q3 60C
3. The heat of combustion of benzoic acid in -26.42 KJ/g. The combustion of 1.176g sample
of benzoic acid causes a temp increase of 4.96C in bomb calorimeter. What is the heat
capacity of the of the calorimeter assembly?
qrxn = 1.176g x -26.42 KJ/g = -31.07 KJ = - qcal
qcal = c x T
so, c = 6.26 KJ/C
Aqueous silver ion reacts with aqueous chloride ion to yield a white precipitate of
silver chloride.
Ag+ + cl- agcl
When 10.0 ml of 1.00 M AgNo3 is added to 10.0 ml of 1.00 M Nacl solution at 25C in a
calorimeter a white precipitate of Agcl forms and the temperature of the aqueous mixture
increases to 32.6C Assuming the specific heat of the aqueous mixture is 4.18 J/gC the density of
the mixture is 1.00 g/ml, calorimeter absorbs negligible heat. Calculate H in kJ in the reaction.
Ans.
* Balance the equation first! In such cases
m = 20.0 ml x 1 g/ml = 20 g
T = 32.6C 25.0C = 7.6C
q = 4.18 J/gC x 20.0 g x 7.6C = 6.4 x 102 J
moles of Ag+ = 10.0 ml X (1.00 mol Ag+ / 1000 ml ) = 1.00 X 102 mol Ag+
Therefore, moles of AgCl
Heat evolved / mole AgCl = 6.4 X102 J / (1.00 X 10-2 mol AgCl) = -64 KJ/ mol Agcl
*Either mention heat is evolved or put the negative sign
4) Oxyac welding torches burn acetylene gas C2H2O. calculate H in KJ for combustion reaction
of acetylene to yield Co2 and H2O
Hf , C2H2 = 227.4 KJ/mol
H2O = -241.8 KJ/mol
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