Answer:
Explanation:
The compound 
is .
The reaction included are:
(because bromine is less reactive than chlorine)
But , because of hexane the solution get dilute and its color changes to orange.
Now, NaI is added and we know Br is more reactive than I.
Therefore it replace it.
Reaction: 
Purple color develop due to formation of Iodine.
Answer:
A. Interactions between the ions of sodium chloride (solute-solute interactions).
B. Interactions involving dipole-dipole attractions (solvent-solvent interactions).
C. Interactions formed during hydration (solute-solvent interactions).
D. Interactions involving ion-ion attractions (solute-solute interactions).
E. Interactions associated with an exothermic process during the dissolution of sodium chloride (solute-solvent interactions).
F. Interactions between the water molecules (solvent-solvent interactions).
G. Interactions formed between the sodium ions and the oxygen atoms of water molecules (solute-solvent interactions).
Explanation:
The solution process takes place in three distinct steps:
- Step 1 is the <u>separation of solvent molecules.
</u>
- Step 2 entails the <u>separation of solute molecules.</u>
These steps require energy input to break attractive intermolecular forces; therefore, <u>they are endothermic</u>.
- Step 3 refers to the <u>mixing of solvent and solute molecules.</u> This process can be <u>exothermic or endothermic</u>.
If the solute-solvent attraction is stronger than the solvent-solvent attraction and solute-solute attraction, the solution process is favorable, or exothermic (ΔHsoln < 0). If the solute-solvent interaction is weaker than the solvent-solvent and solute-solute interactions, then the solution process is endothermic (ΔHsoln > 0).
In the dissolution of sodium chloride, this process is exothermic.
Answer
Electrons live inside the nucleus and carry a neutral charge.
Explanation:
<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
