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:
C. 100.7 amu
Explanation:
Isotopes of an element are atoms of an element with the same atomic number but different atomic masses. Each atomic mass of an isotope is known as an isotopic mass. An element that exhibits isotope, that is, that have two or more isotopes has a relative atomic mass that is not a whole number.
Relative atomic mass of X is the sum of the products of the relative abundances of each isotope and its isotopic mass.
For Isotope ¹⁰⁰X: 30% × 100 = 30 amu
For Isotope ¹⁰¹X: 70% × 101 = 70.7 amu
Relative atomic mass of X = (30 + 70.7) amu = 100.7 amu
Therefore, the approximate atomic mass of X is 100.7 amu
You spelled bored wrong lol
Answer:
FORMULA: HCl PESO MOLECULAR: 36.46 g/mol COMPOSICION: Cl: 97.23 % y H: 2.76 %. GENERALIDADES: El ácido clorhídrico es una disolución acuosa de cloruro de hidrógeno.
