Answer:
Final temperature of calorimeter is 25.36^{0}\textrm{C}
Explanation:
Molar mass of anethole = 148.2 g/mol
So, 0.840 g of anethole = 
of anethole = 0.00567 moles of anethole
1 mol of anethole releases 5539 kJ of heat upon combustion
So, 0.00567 moles of anethole release 
of heat or 31.41 kJ of heat
6.60 kJ of heat increases 
temperature of calorimeter.
So, 31.41 kJ of heat increases 
or 
temperature of calorimeter
So, the final temperature of calorimeter = 
Answer:
No
Explanation:
No, but the total mass of reactants must equal the total mass of products to be a balanced equation.
Example: Consider the following reaction ...
3H₂ + N₂ => 2NH₃ and 'amu' is atomic mass units (formula weights from periodic table)
In terms of molecules, there are 4 molecules on the left (3 molecular hydrogens (H₂) and 1 molecular nitrogen (N₂) and 2 molecules of ammonia on the right side of equation arrow. ∑reactant molecules ≠ ∑product molecules.
In terms of mass of reactants & mass of products, the 3H₂ + N₂ => 6amu + 28amu = 34amu & mass of products (2NH₃) => 2(14amu) + 6(1amu) = 34amu for sum of product masses.
∑mass reactants = ∑mass products <=> 34amu = 34amu.
The expression '∑mass reactants = ∑mass products' as applied to chemical equations is generally known as 'The Law of Mass Balance'.
Answer:
dS= 1.79*169.504
j/k = 303.41 j/k
Explanation:
Fe3O4(s) + 4H2(g) --> 3Fe (s)+ 4H2O(g)
dS(Fe3O4) =146.4 j/k
dS(H2) =130.684
dS(Fe) =27.78
dS(H2O) =188.825
dSrxn = dS[product]-dS[reactants]
= 3*dS(Fe)+ 4*dS(H2O)-[1*dS(Fe3O4)+ 4dS(H2)]
= [3*27.78 +4*188.825-146.4 -4*130.684] j/k = 169.504 j/k
This is the dS for 1mole Fe3O4
for 1.79 mols Fe3O4
dS= 1.79*169.504 j/k = 303.41 j/k
