Answer:
Democritus was an Ancient Greek pre-Socratic philosopher primarily remembered today for his formulation of an atomic theory of the universe. Democritus was born in Abdera, Thrace, around 460 BC, although there are disagreements about the exact
Explanation:
there are seven valence electrons<span> available of bonding in bromine(Br)</span>
Trainers use this substance because the reaction of the product on the skin is an exothermic one. This produces a feeling of warmth and heat. Heating an injury like a sprain with warmth or heat increases blood flow to the area and promotes healing as the flow of blood carries the necessary substances for healing.
MM Zn(NO₃)₂ = 189.36 g/mol
mass 1 mol Zn(NO₃)₂ = 189.36 g
mass hydrate = 100 / 63.67 x 189.36 = 297.409 g
mass 1 mol hydrate = 297.409 g
MM hydrate = 297.409 g/mol
MM hydrate = MM Zn(NO₃)₂ + MM xH₂O
297.409 = 189.36 + x(18)
x = 6
<u>Answer:</u> The amount of energy released per gram of 
is -71.92 kJ
<u>Explanation:</u>
For the given chemical reaction:
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_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_2O_3%28s%29%29%7D%29%2B%289%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28l%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_5H_9%28l%29%29%7D%29%2B%2812%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
Taking the standard enthalpy of formation:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20%281271.94%29%29%2B%289%5Ctimes%20%28-285.83%29%29%5D-%5B%282%5Ctimes%20%2873.2%29%29%2B%2812%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-9078.57kJ)
We know that:
Molar mass of pentaborane -9 = 63.12 g/mol
By Stoichiometry of the reaction:
If 2 moles of produces -9078.57 kJ of energy.
Or,
If 
of produces -9078.57 kJ of energy
Then, 1 gram of will produce = 
of energy.
Hence, the amount of energy released per gram of is -71.92 kJ
