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complete the word doc entitled "LAB7REPORTFORM" using the data that is attached. no plagiarism is allowed and no work should be posted online after the work is completed.

1 answer below »
·    Name                                                    
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·    Group                
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PLEASE ENABLE EDITING BEFORE FILLING OUT THIS REPORT!
No report will be accepted for credit after that date/time.
Grade:            
Results: (60 pts)
Room Temperature
Reactant
Mg
Mg
MgO
MgO
Trial Numbe
1
2
3
4
Mass of reactant (from data)
Moles of reactant (calculate)
Initial time (ti) (from graph)
Initial Temperature (Ti)
(from the graph)
Linear Curve fit slope (m)
(from the graph)
Linear Curve fit intercept (b)
(from the graph)
Final Temperature (Tf)
(calculate from linear fit data)
∆T = Tf – Ti
qwate
qcalorimete
qrxn
∆Hrxn
Use the following to determine the enthalpy of formation of magnesium oxide:
Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)
ΔH = (calculated in the experiment)
MgO (s) + 2HCl (aq) → MgCl2 (aq)+ H2O (l)
ΔH = (calculated in the experiment)
H2 (g) + ½O2 (g) → H2O (l)
ΔH = XXXXXXXXXXkJ/mol
Mg (s) + ½ O2 (g)     MgO (s)
ΔH = (determine using Hess’s Law)
Calculations: (40 pts)
·    Using the enthalpies of reaction computed in the lab and the value given for the enthalpy of formation of liquid water, show the Hess’s Law determination of the enthalpy of formation of solid magnesium oxide. Show your work. If you include a scan of your handwritten work, it must be legible. (30 pts)
·    Compare the value you calculated with the actual value XXXXXXXXXXkJ/mol). Show your determination of percent e
or. Use the standard equation for this: (10 pts)
Room Temperature (use a thermometer to 1 decimal place) A NOC
If you record the data from the graph in the table below, there is no need to print the graph.

Reactant Mg Mg MgO MgO
Trial Number XXXXXXXXXX
0330S [oB3BE A
Mass of reactant
Goad 6


(4 decimal places)

Data from graph
26.20¢|3l-loc
Initial Temp (to) (4 5)

Initial Time (Te) (4 59)
Sos 36.505

Linear Curve fit slope (m)
sh S2.55y XXXXXXXXXX”
Linear Curve fit intercept (b)
as =
Notes:
29.9% ¢| 3465 ¢
Answered 2 days After Dec 11, 2023

Solution

Dr Shweta answered on Dec 14 2023
14 Votes
·    Name                                                    
·    
·    
·    Group                
·    
·    
PLEASE ENABLE EDITING BEFORE FILLING OUT THIS REPORT!
No report will be accepted for credit after that date/time.
Grade:            
Results: (60 pts)
Room Temperature
24.10°C
24.10°C
24.10°C
24.10°C
Reactant
Mg
Mg
MgO
MgO
Trial Numbe
1
2
3
4
Mass of reactant (from data)
0.2345 g
0.2328 g
0.5033 g
0.5091 g
Moles of reactant (calculate)
0.0096
0.0095
0.0124
0.0126
Initial time (ti) (from graph)
33.00 s
35.00 s
57.50s
36.50 s
Initial Temperature (Ti) (in °C)
(from the graph)
23.40
24.00
26.20
31.10
Linear Curve fit slope (m)
(from the graph)
-5.794 *10-3
-5.952 *10-3
-2.553 *10-3
-5.275 *10-3
Linear Curve fit intercept (b)
(from the graph) (in °C)
35.13
35.87
29.73
34.65
Final Temperature (Tf)
(calculate from linear fit data)
32.12
32.54
30.24
33.89
∆T = Tf – Ti
8.72
8.54
7.92
6.79
qwate
4.2
4.2
4.2...
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