Answer:
If you mix equal amounts of a strong acid and a strong base, the two chemicals essentially cancel each other out and produce a salt and water. Mixing equal amounts of a strong acid with a strong base also produces a neutral pH (pH = 7) solution.
<u>Answer:</u> The value of 
of the reaction is 28.38 kJ/mol
<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}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2Cl_2%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Cl_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-364%29%29%5D-%5B%281%5Ctimes%20%28-296.8%29%29%2B%281%5Ctimes%200%29%5D%3D-67.2kJ%2Fmol%3D-67200J%2Fmol)
- The equation used to calculate entropy change is of a reaction is:
![\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the entropy change of the above reaction is:
![\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28SO_2Cl_2%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28Cl_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20311.9%29%5D-%5B%281%5Ctimes%20248.2%29%2B%281%5Ctimes%20223.0%29%5D%3D-159.3J%2FKmol)
To calculate the standard Gibbs's free energy of the reaction, we use the equation:
where,
= standard enthalpy change of the reaction =-67200 J/mol
= standard entropy change of the reaction =-159.3 J/Kmol
Temperature of the reaction = 600 K
Putting values in above equation, we get:
Hence, the value of of the reaction is 28.38 kJ/mol
Answer:
The minimum number of boxes of pencils to be ordered is 630 boxes.
Explanation:
Since a pupil uses averagely 9.3 pencils
and a box contains 12 pencils,
the school enrollment is also 812
school's enrollment x average use of pencil per student
__________________________________________
number of pencils in a box
812 x 9.3 = 7551.6
7551.6 /12 = 629.3
Having a total number of 630 boxes of pencils to be ordered.
Answer:
Average atomic mass = 51.9963 amu
Explanation:
Given data:
Abundance of Cr⁵⁰ with atomic mass= 4.34%
, 49.9460 amu
Abundance of Cr⁵² with atomic mass = 83.79%, 51.9405 amu
Abundance of Cr⁵³ with atomic mass =9.50%, 52.9407 amu
Abundance of Cr⁵⁴ with atomic mass = 2.37%, 53.9389 amu
Average atomic mass = 51.9963 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass +....n) / 100
Average atomic mass = (4.34×49.9460)+(83.79×51.9405) +(9.50×52.9407)+ (2.37×53.9389) / 100
Average atomic mass = 216.7656 + 4352.0945 + 502.9367 +127.8352 / 100
Average atomic mass = 5199.632 / 100
Average atomic mass = 51.9963 amu
Answer:
Option (D)
Explanation:
<u>Eskers are the long ridges that are comprised of rocks, sands and clay particles and are deposited towards the end of the glaciers</u>. These are fluvioglacial depositional features. These particles are exposed after the glaciers recede. These ridges are formed parallel to the earlier flow direction of ice. The size of eskers is generally smaller as it carries smaller particles such as rocks, sands, and gravels, in comparison to the different type of moraines. It is because the flow velocity decreases as the glaciers melt. So, these eskers are formed at the end of the glaciers.
Thus, the correct answer is option (D).
