A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a stock solution whose concentration is known.
Answer: +178.3 kJ
Explanation:
The chemical equation follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CaO%28s%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5E0f_%7BCO_2%7D%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CaCO_3%28s%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-635.1%29%29%2B%281%5Ctimes%20%28-393.5%29%29%5D-%5B%281%5Ctimes%20%28-1206.9%29%29%5D)
The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ
Explanation:
Both melting and boiling point of water are much higher that that of H2S.
The MP and BP of water are 0°C and 100°C respectively whereas the MP and BP of H2S are -82°C and -60°C respectively.
Clearly, the intermolecular forces in H2O are much stronger than in H2S because, H2O has hydrogen bonding in it, which is both inter and intra. Whereas H2S has only weak van der waal force between their molecules. This is why boiling and melting points of H2O is more than that of H2S.
Hydrogen bonding is a particular type of dipole-dipole association which occurs between atoms of hydrogen bound to strongly electronegative atoms that are either fluorine, oxygen or nitrogen.
The hydrogen atom's partially positive side is drawn to the partially negative end of these atoms found in some other molecule. It is a strong binding mechanism between the molecules.
