Answer:
C is the element thats has been oxidized.
Explanation:
MnO₄⁻ (aq) + H₂C₂O₄ (aq) → Mn²⁺ (aq) + CO₂(g)
This is a reaction where the manganese from the permanganate, it's reduced to Mn²⁺.
In the oxalic acid, this are the oxidation states:
H: +1
C: +3
O: -2
In the product side, in CO₂ the oxidation states are:
C: +4
O: -2
Carbon from the oxalate has increased the oxidation state, so it has been oxidized.
The characteristics of an Arrhenius acid was the acid donates H+ when it is in the aqueous solution, whether the solution is acid or base. If the solution is a acid, then be H+ builds up and increase its thing. The pH level for acid solution is the scale from 1-6. Even though the ph scale is lower than the base, the color also varies. It also applies to the base, too.
Here we have to get the moles of hydrogen (H₂) consumed to form water (H₂O) from 1.57 moles of oxygen (O₂)
In this process 3.14 moles of H₂ will be consumed.
The balanced reaction between oxygen (O₂) and hydrogen (H₂); both of which are in gaseous state to form water, which is liquid in nature can be written as-
2H₂ (g) + O₂ (g) = 2H₂O (l).
Thus form the equation we can see that 1 mole of oxygen reacts with 2 moles of hydrogen to form 2 moles of water.
So, 1.57 moles of oxygen will consume (1.57×2) = 3.14 moles of hydrogen to form water.
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
Answer:
59.2 grams
Explanation:
We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:
There are 59.2 grams of water in this solution.
