Let us assume that there is a 100g sample of Opal. The masses of each element will be:
29.2g Si
33.3g O
37.5g H2O
Now we divide each constituent's mass by its Mr to get the moles present
Si: (29.2 / 28) = 1.04
O: (33.3 / 16) = 2.08
H2O: (37.5 / 18) = 2.08
Now we divide by the smallest number and obtain:
Si: 1
O: 2
H2O: 2
Thus, the empirical formula of Opal is:
SiO2 . 2H2O
Answer : The enthalpy of combustion per mole of
is -2815.8 kJ/mol
Explanation :
Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
The equilibrium reaction follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%28n_%7B%28CO_2%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%29%7D%29%2B%28n_%7B%28H_2O%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%29%7D%29%5D-%5B%28n_%7B%28C_6H_%7B12%7DO_6%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_6H_%7B12%7DO_6%29%7D%29%2B%28n_%7B%28O_2%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%286%5Ctimes%20-393.5%29%2B%286%5Ctimes%20-285.8%29%5D-%5B%281%5Ctimes%20-1260%29%2B%286%5Ctimes%200%29%5D%3D-2815.8kJ%2Fmol)
Therefore, the enthalpy of combustion per mole of
is -2815.8 kJ/mol
Answer:
-1.42, -0.375, 32.5% (.325), 3/8 (.375), √4 (2.0), 3 (3.0), 2³ (8.0)
Answer:
ionization energy decreases from top to bottom in groups, and increases from left to right across a period. So technically, the noble games have the largest ionization energy.
Explanation: