Answer:
The 3rd answer down.
Na²O (sodium oxide) will be a base when exposed to water H²O
Explanation:
Sodium Oxide Na²O, will become Sodium Hydroxide after being exposed to water (at 80% I believe).
The oxygen ion in Na²O has 2 extra electrons which makes it highly charged and very attractive to hydrogen ions. The attraction is so strong that when Na²O comes in contact with H²O, the O(-2) strips off a hydrogen from water, forming 2 x OH ions which of course are still strongly basic.
Answer:
atomic number of Be =4
electronic configuration :1s²2s²
but for Be2+ atomic number = 4-2 =2
electronic configuration : 1s²
Answer: The heat of reaction (ΔHrxn) for the reaction is -164.9kJ
Explanation:
The given balanced chemical reaction is,
To calculate the enthalpy of reaction 
.
![\Delta H^o=[n_{CaCl_2}\times \Delta H_f^0_{(CaCl_2)}+n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{CaCO_3}\times \Delta H_f^0_{(CaCO_3)+n_{HCl}\times \Delta H_f^0_{(HCl)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BCaCl_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28CaCl_2%29%7D%2Bn_%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_%7BCaCO_3%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28CaCO_3%29%2Bn_%7BHCl%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28HCl%29%7D%5D)
where,
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times -877.1)+(1\times -393.51)+(1\times -285.8)]-[(1\times -1206.9)+(2\times -92.30)]=-164.9kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20-877.1%29%2B%281%5Ctimes%20-393.51%29%2B%281%5Ctimes%20-285.8%29%5D-%5B%281%5Ctimes%20-1206.9%29%2B%282%5Ctimes%20-92.30%29%5D%3D-164.9kJ)
Therefore the heat of reaction (ΔHrxn) for the reaction is -164.9kJ
