In drawing the Lewis structure for C2H2 (also called ethyne) you'll find that you don't have enough valence electrons available to satisfy the octet for each element (if you use only single bonds). The solution is to share three pairs of valence electrons and form a triple bond between the Carbon atoms in C2H2
A 25.0 mL solution of Sr(OH)₂ is neutralized with 31.6 mL of 0.150 M HBr, then the concentration of the original Sr(OH)₂ solution is 0.189M.
<h3>How do we calculate the concentration?</h3>
Concentration of the solution will be calculated by using the below chemical reaction as:
M₁V₁ = M₂V₂, where
M₁ & V₁ is the molarity and volume of HBr solution and M₂ & V₂ is the molarity and volume of original Sr(OH)₂ solution.
On putting values on above equation by taking from question, we get
M₂ = (0.15)(31.6) / (25) = 0.189 M
Hence required molarity of Sr(OH)₂ solution is 0.189M.
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<u>Answer:</u> The standard enthalpy change of the reaction is -1076.82kJ
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
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%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CO_2)})+(2\times \Delta H^o_f_{(SO_2)})]-[(1\times \Delta H^o_f_{(CS_2)})+(3\times \Delta H^o_f_{(O_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CS_2%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-393.52))+(2\times (-296.8))]-[(1\times (89.70))+(3\times (0)]\\\\\Delta H^o_{rxn}=-1076.82kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-393.52%29%29%2B%282%5Ctimes%20%28-296.8%29%29%5D-%5B%281%5Ctimes%20%2889.70%29%29%2B%283%5Ctimes%20%280%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-1076.82kJ)
Hence, the standard enthalpy change of the reaction is -1076.82kJ
No. The mass is released into the air.
For the concentration of Na in M when it begins to crystallize out of the solution is mathematically given as
Na+ = 0.125M
<h3>what is the concentration of Na in M when it begins to crystallize out of the solution?</h3>
Generally, the equation for the Ksp of the solution is mathematically given as
Na⁺+CL⁻
Ksp = SxS
53.9 = s^2
S = 7.34
Molar mass Nacl 58,44
Solubility = 7.34/58.44
Solubility = 0.125M
In conclusion, the concentration of Na in M when it begins to crystallize out of the solution
Na+ = 0.125M
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