<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
Answer:
1.16 moles CO₂
Explanation:
To find the moles of CO₂, you need to (1) convert grams C₈H₁₈ to moles (via the molar mass) and then (2) convert moles C₈H₁₈ to moles CO₂ (via the mole-to-mole ratio from equation coefficients). It is important that the conversions/ratios are arranged in a way that allows for the cancellation of units. The final answer should have 3 significant figures like the given value.
Molar Mass (C₈H₁₈): 8(12.011 g/mol) + 18(1.008 g/mol)
Molar Mass (C₈H₁₈): 114.232 g/mol
2 C₈H₁₈ + 25 O₂ -----> 16 CO₂ + 18 H₂O
^ ^
16.6 g C₈H₁₈ 1 mole 16 moles CO₂
-------------------- x ----------------- x ------------------------- = 1.16 moles CO₂
114.232 g 2 moles C₈H₁₈
Answer:
I believe your answer is B) make a prediction
Explanation:
<h3><u>Answer;</u></h3>
The value of Kc does not depend on starting concentrations.
<h3><u>Explanation;</u></h3>
- <em><u>At constant temperature, changing the equilibrium concentration does not affect the equilibrium constant, because the rate constants are not affected by the concentration changes. </u></em>
- When the concentration of one of the participants is changed, the concentration of the others vary in such a way as to maintain a constant value for the equilibrium constant.
