RESULTS AND CALCULATIONS: RUN 2 RUN 1 PART A (density of H20 = 1.00 g/mL) 29tHoomL= LOng 1029 %3D 6,222 Tfinal - Tinitial = 3.256 27.49-23.74=3.s°C q-meAT (H22n-23) (1) qsoln 1569J2 1602J %3D Moles of NaOH = 29 NAOH 2-0.05 0.08 Moles NaDIH AHa = Average AHa (For Parts A 1 in the left hand and give
RESULTS AND CALCULATIONS: RUN 2 RUN 1 PART A (density of H20 = 1.00 g/mL) 29tHoomL= LOng 1029 %3D 6,222 Tfinal - Tinitial = 3.256 27.49-23.74=3.s°C q-meAT (H22n-23) (1) qsoln 1569J2 1602J %3D Moles of NaOH = 29 NAOH 2-0.05 0.08 Moles NaDIH AHa = Average AHa (For Parts A 1 in the left hand and give
Introduction to General, Organic and Biochemistry
11th Edition
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter5: Gases, Liquids, And Solids
Section: Chapter Questions
Problem 5.74P
Related questions
Question
How do I answer this.
First picture is my data.
How do I find delta H(a)
Transcribed Image Text:RESULTS AND CALCULATIONS:
RUN 2
RUN 1
PART A
1029
1029
(density of H20 = 1.00 g/mL) 29HO0ML= Long
6.222
Tfinal - Tinitial =
27.49-23.74=3.5°C
3.75C
Goan = = meAT ( 22.n-23) (m)
qsoln
1569J/2,602J
%3D
29 NADH
= 40
Moles of NaOH =
Na DIH
2-0.05
0.05 mda
0.08 Moles
ΔΗ -
Average AHa
%3D
(For Parts A – C for RUN 1 in the left hand and give
Transcribed Image Text:PURPOSE:
DATA:
RUN 1
RUN 2
PART A
2.00g
2.00g
100 mL
Mass of NaOH
100 miL
Volume of water
23,74°c
27.89°C
Initial temperature of water (Tinitial)
27.49C
30.06c
Highest temperature reached (Tfinal)
PART B
RUN 1
RUN 2
Z107g
Mass of NaOH
Volume of HCI
[HCIJ= 0.S00M
100mL
Initial temperature of HCI (Tinitial)
24.14°c
Highest temperature reached (Trinal)
35.69°C
36.14c
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