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
1.27moles of K3PO4 can be formed.
Given:
<span>CS2 + 3O2 → CO2 + 2SO2
</span><span>114 grams of CS2 are burned in an excess of O2
</span>
moles CS2 = 114 g/76.143 g/mol → 114g * mol/76.143 g = 1.497 mol
<span>the ratio between CS2 and SO2 is 1 : 2 </span>
moles SO2 formed = 1.497 x 2 = 2.994 moles → 2nd option
Answer:
<h3>C6H5CO2H (aq) + H2O (l) _C6H5CO2<em>-</em><em> </em><em>+</em><em> </em><em>H3O</em></h3>
Answer:
they are in group II and they have two electrons in their outermost energy levels ( valency)...is this helpful?
