Water can't cool at a single temperature. It must start at a higher temperature, and drop to a lower temperature in order to cool. Unless we know the other temperature, there is no way to calculate the amount of thermal energy released.
Answer:
1. Cu
2. Cu
3. 2 electrons.
Explanation:
Step 1:
The equation for the reaction is given below:
3Cu(s) + 8HNO3(aq) -> 2NO(g) 3Cu(NO3)2(aq) + 4H2O(l)
Step 2:
Determination of the change of oxidation number of each element present.
For Cu:
Cu = 0 (ground state)
Cu(NO3)2 = 0
Cu + 2( N + 3O) = 0
Cu + 2(5 + (3 x -2)) =0
Cu + 2 (5 - 6) = 0
Cu + 2(-1) = 0
Cu - 2 = 0
Cu = 2
The oxidation number of Cu changed from 0 to +2
For N:
HNO3 = 0
H + N + 3O = 0
1 + N + (3 x - 2) = 0
1 + N - 6 = 0
N = 6 - 1
N = 5
NO = 0
N - 2 = 0
N = 2
The oxidation number of N changed from +5 to +2
The oxidation number of oxygen and hydrogen remains the same.
Note:
1. The oxidation number of Hydrogen is always +1 except in hydride where it is - 1
2. The oxidation number of oxygen is always - 2 except in peroxide where it is - 1
Step 3:
Answers to the questions given above
From the above illustration,
1. Cu is oxidize because its oxidation number increased from 0 to +2 as it loses electron.
2. Cu is the reducing agent because it reduces the oxidation number of N from +5 to +2.
3. The reducing agent i.e Cu transferred 2 electrons to the oxidising agent HNO3 because its oxidation number increase from 0 to +2 as it loses its electrons. This means that Cu transfer 2 electrons.
Answer:
Q= 245 =2.5 * 10^2
Explanation:
ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK
R= 8,314 J/molK, T=298K
ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol
ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ
→ 5.5 = LnQ → Q= 245 =2.5 * 10^2
Answer:
C
Explanation:
because valence electrons are located at the last energy level
Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Explanation:
The given balanced chemical reaction is,
First we have to calculate the enthalpy of reaction 
.
![\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
We are given:
Putting values in above equation, we get:
![-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]](https://tex.z-dn.net/?f=-1.0940%5Ctimes%2010%5E4%3D%5B%2816%5Ctimes%20-393.5%29%2B%2818%5Ctimes%20-285.8%29%5D-%5B%2825%5Ctimes%200%29%2B%282%5Ctimes%20%5CDelat%20H_f%7BC_8H_%7B18%7D%28l%29%7D%5D)
Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol
