Answer:
Explanation has been given below.
Explanation:
- Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
- Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
- First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
- Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
- Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
- Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
Calcite can be either dissolved by groundwater or precipitated by groundwater, depending on several factors including the water temperature, pH, and dissolved ion concentrations. Although calcite is fairly insoluble in cold water, acidity can cause dissolution of calcite and release of carbon dioxide gas.
Answer: A. -396 kJ
Explanation:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
Reversing the reaction, changes the sign of 
On multiplying the reaction by 2, enthalpy gets multiplied by 2:
Thus the enthalpy change for the reaction is -396 kJ.
2H3PO4 + 3Ca(OH)2 ➡ Ca3(PO4)2 + 6H2O.
Hence salt formed is Ca3(PO4)2.
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