I think its A.
if one force cannot overcome the other, the object remains stationary.
Answer:
ΔG°rxn = -69.0 kJ
Explanation:
Let's consider the following thermochemical equation.
N₂O(g) + NO₂(g) → 3 NO(g) ΔG°rxn = -23.0 kJ
Since ΔG°rxn < 0, this reaction is exergonic, that is, 23.0 kJ of energy are released. The Gibbs free energy is an extensive property, meaning that it depends on the amount of matter. Then, if we multiply the amount of matter by 3 (by multiplying the stoichiometric coefficients by 3), the ΔG°rxn will also be tripled.
3 N₂O(g) + 3 NO₂(g) → 9 NO(g) ΔG°rxn = -69.0 kJ
Explanation:
the pH of the solution defined as negatuve logarithm of
ion concentration.
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
1. Hydrogen ion concentration when pH of the solution is 11.
![11=-\log[H^+]](https://tex.z-dn.net/?f=11%3D-%5Clog%5BH%5E%2B%5D)
..(1)
At pH = 11, the concentration of
ions is
.
2. Hydrogen ion concentration when the pH of the solution is 6.
![6=-\log[H^+]'](https://tex.z-dn.net/?f=6%3D-%5Clog%5BH%5E%2B%5D%27)
..(2)
At pH = 6, the concentration of
ions is
.
3. On dividing (1) by (2).
![\frac{[H^+]}{[H^+]'}=\frac{1\times 10^{-11} mol/L}{1\times 10^{-6} mol/L}=1\times 10^{-5}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5BH%5E%2B%5D%27%7D%3D%5Cfrac%7B1%5Ctimes%2010%5E%7B-11%7D%20mol%2FL%7D%7B1%5Ctimes%2010%5E%7B-6%7D%20mol%2FL%7D%3D1%5Ctimes%2010%5E%7B-5%7D%20)
The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is
.
4. Difference between the
ions at both pH:

This means that Hydrogen ions in a solution at pH = 7 has
ions fewer than in a solution at a pH = 6
Answer:
The standard enthalpy of formation of this isomer of
is -220.1 kJ/mol.
Explanation:
The given chemical reaction is as follows.


The expression for the entropy change for the reaction is as follows.
![\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Brxn%7D%3D%5B8%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28CO_%7B2%7D%29%20%2B9%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28H_%7B2%7DO%29%5D-%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%2B%20%5Cfrac%7B25%7D%7B2%7D%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28O_%7B2%7D%29%5D)



Substitute the all values in the entropy change expression.
![-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]](https://tex.z-dn.net/?f=-5104.1kJ%2Fmol%3D%5B8%28-393.5%29%2B9%28-241.8%29kJ%2Fmol%5D-%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%2B%20%5Cfrac%7B25%7D%7B2%7D%280%29kJ%2Fmol%5D)



Therefore, The standard enthalpy of formation of this isomer of
is -220.1 kJ/mol.
I would believe it to be 55g. A -> B YIELDS AB. So, 10g + 45g = 55g.