Hydrogen bonding is a type of intermolecular forces of attraction in which hydrogen atom is bonded to one of the most electronegative atoms. This gives a partial positive charge to hydrogen atom and a partial negative charge to the electronegative atom involved in the bonding. The electronegative atoms that can form hydrogen bonding are fluorine (F), nitrogen (N), and oxygen (O).
Therefore the correct option is,
A) NH3
1, Read the entire lab procedure through and make sure it is understandable. 2, Put on safety goggles and an apron. 3, Check the flask for chips and cracks. 4, <span> Use heat-resistant gloves or tongs to swirl the flask.</span>
<u>Answer:</u> The 
for the reaction is 72 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The given chemical reaction follows:
The intermediate balanced chemical reaction are:
(1) 
(2) 
( × 2)
(3) 
( × 2)
The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B1%5Ctimes%20%28%5CDelta%20H_1%29%5D%2B%5B2%5Ctimes%20%28-%5CDelta%20H_2%29%5D%2B%5B2%5Ctimes%20%28%5CDelta%20H_3%29%5D)
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-1184%29%29%2B%282%5Ctimes%20-%28-234%29%29%2B%282%5Ctimes%20%28394%29%29%5D%3D72kJ)
Hence, the for the reaction is 72 kJ.
When many atoms share electrons together.
Convert 55.0g Ca(OH)2 to mols.
55.0g Ca(OH)2 = 0.742 mols Ca(OH)2
0.742mol Ca(OH)2/ 0.680M Ca(OH)2 = 1.09L Ca(OH)2
Neglecting the volume of the Ca(OH)2 itself, since it is minimal and its density wasn't provided, 1.09L would be the total volume of a 0.680M solution produced by dissolving 55.0g of Ca(OH)2 in enough water.
