this is a repost.
What is the change in enthalpy for magnesium after the reaction with vinegar?
vinegar is 60ml
magnesium is 0.7g
T1 = ?
T2 = 31.8°C
find the initial temperature if 123.4 KJ energy has been released.
Mg + 2HC2H3O2 ==> Mg(C2H3O2)2 + H2
1 mole of mg is 24.31g
mols Mg = 0.7/24.3 = about 0.0288
q = mass soln x specific heat soln x (Tfinal-Tinitial)
-123400 J = 60 g x 4.184 J/oC*g x (31.8 - Ti)
I really don't know how to solve for Ti. can someone help me!
To solve for the initial temperature (Ti) in the equation -123400 J = 60 g x 4.184 J/oC*g x (31.8 - Ti), we can follow the steps below:
1. Rearrange the equation: -123400 J = 250.44 J/oC x (31.8 - Ti)
2. Divide both sides of the equation by 250.44 J/oC: -123400 J / 250.44 J/oC = 31.8 - Ti
3. Simplify the left side of the equation: -492.8 oC = 31.8 - Ti
4. Subtract 31.8 from both sides of the equation: -492.8 oC - 31.8 oC = -Ti
5. Simplify: -524.6 oC = -Ti
6. Multiply both sides of the equation by -1 to isolate Ti: (-524.6 oC) x (-1) = Ti
7. Calculate Ti: Ti ≈ 524.6 oC
Therefore, the initial temperature (Ti) is approximately 524.6 degrees Celsius.