Answer: d
Explanation: I got it right
Apples turn brown when exposed to air because it undergoes aerial oxidation. Due to this, when the inside of the apple is exposed to the air containing oxygen and water, it turns brown. When apple is uncut, skin of the apple protects it from this process.
If the peeled apple in kept in the refrigerator, the oxidation reaction is greatly slowed down. This is because, rate of chemical reaction decreases with temperature. Hence, in refrigerator it would take several days for it to turn brown.
Answer:
The standard enthalpy of formation of this isomer of
is -220.1 kJ/mol.
Explanation:
The given chemical reaction is as follows.
![C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)](https://tex.z-dn.net/?f=C_%7B8%7DH_%7B18%7D%28g%29%2B%20%5Cfrac%7B25%7D%7B2%7DO_%7B2%7D%28g%29%20%5Crightarrow%208CO_%7B2%7D%28g%29%2B9H_%7B2%7DO%28g%29)
![\Delta H^{o}_{rxn}= -5104.1kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Brxn%7D%3D%20-5104.1kJ%2Fmol)
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)
![\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28H_%7B2%7DO%29%3D%20-241.8kJ%2Fmol)
![\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28CO_%7B2%7D%29%3D%20-393.5kJ%2Fmol)
![\Delta H^{o}_{f}(O_{2})= 0kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28O_%7B2%7D%29%3D%200kJ%2Fmol)
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)
![-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})](https://tex.z-dn.net/?f=-5104.1kJ%2Fmol%3D-5324.2kJ%2Fmol%20-%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29)
![\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%20%3D-5324.2kJ%2Fmol%20%2B5104.1kJ%2Fmol)
![=-220.1kJ/mol](https://tex.z-dn.net/?f=%3D-220.1kJ%2Fmol)
Therefore, The standard enthalpy of formation of this isomer of
is -220.1 kJ/mol.
Q6. 3
Q7. 3
Q8. pH
Q18. 3
Q19. 3
Q20. 4
Hope this helped??
I would say
change: Temp
Measure: Mass
Control: Volume