Answer: The enthalpy change for formation of butane is -125 kJ/mol
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f%7Breactants%7D%5D)
Putting the values we get :
![\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B4%5Ctimes%20H_f_%7BCO_2%7D%2B5%5Ctimes%20H_f_%7BH_2O%7D%5D-%5B1%5Ctimes%20H_f_%7BC_4H_%7B10%7D%7D%2B%5Cfrac%7B13%7D%7B2%7D%5Ctimes%20H_f_%7BO_2%7D%5D)
![-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]](https://tex.z-dn.net/?f=-2877%3D%5B%284%5Ctimes%20-393%29%2B%285%5Ctimes%20-286%29%5D-%5B1%5Ctimes%20H_f_%7BC_4H_%7B10%7D%7D%2B%5Cfrac%7B13%7D%7B2%7D%5Ctimes%200%5D)
Thus enthalpy change for formation of butane is -125 kJ/mol
Answer:
Rareness, the others describe what the gold is and its properties.
Answer:
209.68
Explanation:
The only number that is relevant (though the rest are quite interesting) is the last one 1.98 * 10^24
1 mole of Barium Acetate Contains 6.02*10^23 particles.
There are 4 moles of carbon to every mole of Barium Acetate.
1.98 * 10^24 atoms / (4*6.02*10^23)
0.8223 moles of Ba(C2H3O2)2
Ba = 137
4C = 4*12 48
6H = 6*1 6
4O = 4*16 64
1mole 255 grams
0.8223 * 255 = 209.68 grams
I have used rounded masses for these elements depending on the periodic table you use. Go through the question with your masses to get a more accurate answer. My answer will not differ by much. It is a guide.
