Answer : The concentration of
is, 0.12 M
Explanation :
Using Henry's law :

where,
= concentration of
= ?
= partial pressure of
= 4.5 atm
= Henry's law constant = 
Now put all the given values in the above formula, we get:


Thus, the concentration of
is, 0.12 M
Mark Brainliest please
Answer :
In my opinion , answer is A
In HCl-HCl, the hydrogen-chlorine link is a polar covalent bond. It is produced when two atoms share an electron pair.
When atoms with various electronegativities share electrons in a covalent link, the result is a polar covalent bond. Think about the molecule of hydrogen chloride (HCl). In order to generate an inert gas electron configuration, each atom of HCl needs an additional electron. Despite having a stronger electronegativity than hydrogen, the chlorine atom cannot remove an electron from hydrogen due to its inability to attract electrons. As a result, a polar covalent bond in hydrogen chloride has an unbalanced distribution of bonding electrons.
Learn more about electronegativity here-
brainly.com/question/17762711
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<u>Answer:</u> The
for the reaction is 54.6 kJ/mol
<u>Explanation:</u>
For the given balanced chemical equation:

We are given:

- To calculate
for the reaction, we use the equation:
![\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G_f%28reactant%29%5D)
For the given equation:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28COCl_2%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28CO_2%29%7D%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28CCl_4%29%7D%29%5D)
Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-204.9%29%29-%28%281%5Ctimes%20%28-394.4%29%29%2B%281%5Ctimes%20%28-62.3%29%29%29%5D%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D46.9kJ%3D46900J)
Conversion factor used = 1 kJ = 1000 J
- The expression of
for the given reaction:

We are given:

Putting values in above equation, we get:

- To calculate the Gibbs free energy of the reaction, we use the equation:

where,
= Gibbs' free energy of the reaction = ?
= Standard gibbs' free energy change of the reaction = 46900 J
R = Gas constant = 
T = Temperature = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
= equilibrium constant in terms of partial pressure = 22.92
Putting values in above equation, we get:

Hence, the
for the reaction is 54.6 kJ/mol