<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
Explanation:
A ____Chemical Reaction_______________________ is a well defined example of a chemical change. A chemical ___ _____chemical equation___________________ can be used to show the changes that occur in a chemical reaction. In a chemical reaction, the substance on the left side of the arrow are the starting substance. These substances are called ___Reactants________________________. The substances on the right side of the arrow are the substances that result from the reaction. These substances are called ____________Products_______________. The arrow is read as either produces or ______yields_____________________. According to the law of conservation of __________mass_________________, atoms are neither lost nor gained during a chemical reaction. This law is illustrated when a chemical equation is ________Balanced___________. When this is done, there will be the same number of ___________atoms________________ of each kind on both sides of the equation. In a chemical equation, the numbers that are placed in front of the symbols and the formulas are called ______________coefficients_____________. They are necessary to keep the ___________________________ of atoms in balance. There are several rules for balancing an equation. First, write the correct ____________(not so sure)_____________ for each reactant and product. Next, choose the coefficients that make the number of atoms of each _______elements(not so sure)________________ on each side of the equation equal. The correctly written formula should not be changed. If you change the formula of a substance, the equation is no longer ___________correct_____________. Changing a formula will indicate a ________Substance___________________ different than the one intended. To balance the equation Mg + O2 à MgO, first choose coefficients to make the number of atoms of each element on each side of the equation equal. You would need to place a coefficient of _________two___________
Stratus clouds reside below the 6,000ft range.
Altocumulus clouds reside in the 6,500-20,000ft range.
Cirrus clouds reside above the 20,000ft range.
Ordered from highest to lowest, the sequence of clouds will be:
Cirrus
Altocumulus
Stratus
It becomes a liquid with the water
