Answer:
27 g
Explanation:
M(C6H12O6) = 6*12 + 12*1 + 6*16 = 180 g/mol
100 mL = 0.1 L solution
1.5 M = 1.5 mol/L
1.5 mol/L * 0.1 L = 0.15 mol C6H12O6
0.15 mol * 180 g/1 mol = 27 g C6H12O6
Answer:
Solid
Explanation:
The reason is because the liquid aspirin has a plastic cover with a liquid inside and it doesn't take long for out stomach acid to dissolve the cover unlike the Solid aspirin where is slowly dissolves.
<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:
V₂ → 106.6 mL
Explanation:
We apply the Ideal Gases Law to solve the problem. For the two situations:
P . V = n . R . T
Moles are still the same so → P. V / R. T = n
As R is a constant, the formula to solve this is: P . V / T
P₁ . V₁ / T₁ = P₂ .V₂ / T₂ Let's replace data:
(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C
((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂
58.66 mL.atm = 0.55 atm . V₂
58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL
