<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ
<u>Explanation:</u>
Enthalpy change 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_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28MgCl_2%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28Mg%28OH%29_2%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCl%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-641.8%29%29%2B%282%5Ctimes%20%28-241.8%29%29%5D-%5B%281%5Ctimes%20%28-924.5%29%29%2B%282%5Ctimes%20%28-92.30%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-16.3kJ)
Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ
Helium
Remember: electron filling of atomic shell ...
The element which have electron in the lowest quantum shell will have the smallest atomic radius\
He:1s1 (Helium)
H:1s2 (hydrogen)
Answer:
Niels Bohr The electron in a Hydrogen atoms can only circle the nucleus in certain paths or orbits.
The empirical formula is Ca₃P₂O₈.
<em>Assume</em> that you have 100 g of the compound.
Then you have 38.76 g Ca, 19.97 g P, and 41.28 g O.
Now, we must convert these <em>masses to moles</em> and <em>find their ratio</em>s.
If the number in the ratio are not close to integers, you <em>multiply them by a numbe</em>r that makes them close to integers.
From here on, I like to summarize the calculations in a table.
<u>Element</u> <u>Mass/g</u> <u> Moles </u> <u> </u><u>Ratio </u> <u> ×2 </u> <u>Integers</u>
Ca 38.76 0.967 07 1.4998 2.9995 3
P 19.97 0.644 82 1 2 2
O 41.28 2.580 0 4.0011 8.0023 8
The empirical formula is Ca₃P₂O₈.
