Class Notes (1,100,000)
CA (620,000)
UTM (20,000)
Chemistry (700)
CHM110H5 (200)
Lecture

CHM110H5 Lecture Notes - Sodium Hydroxide


Department
Chemistry
Course Code
CHM110H5
Professor
Thottackad Radhakrishnan

This preview shows half of the first page. to view the full 2 pages of the document.
Name: Cassy Latchman Student No.: 999839769
Lab Section No.: 0101
CHM110 Pre-lab Questions #4
Question 1
Assume that no heat was lost the enviornment.
Water at 46.90C released heat=Calorimeter absorbrd heat+Water at 25.10C absorbed heat
Let the total heat capacity of calorimeter be C.
50 ×1×4.18 ×
(
46.930.1
)
=C ×(30.125.1)+50×1×4.18 ×
(
30.125.1
)
C ×5K=3511.2J1200J
C=462.24J K1
Question 2
Assume that no heat was lost the enviornment.
Assume that the calorimeter isinitally at 250C.
Assume that ρH2O=ρNaOH =4.18J g1K1
Number of NaOH moles 5g of NaOH=5
40 =0.125moles
Molar heat of NaOH
(
aq
)
=11003
0.125 =44.012kJ
You're Reading a Preview

Unlock to view full version