A neutral sodium atom, for example, contains 11 protons and 11 electrons. By removing an electron from this atom we get a positively charged Na+ ion that has a net charge of +1. Atoms that gain extra electrons become negatively charged. A neutral chlorine atom, for example, contains 17 protons and 17 electrons.
Answer:
1.20 V
Explanation:
The standard cell potential is calculated from the expression
ε⁰ cell = ε⁰ oxidation + ε⁰ reduction
The species that will be reduced is the one with the higher standard reduction potential and the species that will be oxidized will be the one with the more negative reduction potential.
Thus for our question we will have
oxidation:
Pb(s) → Pb2+(aq) + 2 e- ε⁰ oxidation = - ε⁰ reduction
= - ( - 0.13 V ) = + 0.13 V
reduction
Br2(l) + 2 e- → 2 Br-(aq) ε⁰ reduction = +1.07 V
ε⁰ cell = ε⁰ oxidation + ε⁰ reduction = + 0.13 V + 1.07 V = 1.20 V
Yes they do. But a very small kinetic energy. They vibrate in fixed positions
Answer: The standard enthalpy of formation of this isomer of 
is -210.9 kJ
Explanation:
The given balanced chemical reaction is,
First we have to calculate the enthalpy of formation of 
.
![\Delta H^o=[n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{O_2}\times \Delta H_f^0_{(O_2)+n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28CO_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%2Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%7D%5D)
where,
We are given:
Putting values in above equation, we get:
![-511.3kJ/mol=[(8\times -393.5)+(9\times -241.8)]-[(\frac{25}{2}\times 0)+(1\times \Delta H_f^0_{(C_8H_{18})}](https://tex.z-dn.net/?f=-511.3kJ%2Fmol%3D%5B%288%5Ctimes%20-393.5%29%2B%289%5Ctimes%20-241.8%29%5D-%5B%28%5Cfrac%7B25%7D%7B2%7D%5Ctimes%200%29%2B%281%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%7D)
