Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
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2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
Thermodynamic quantity equivalent to the total heat content of a system It is equal to the internal energy of the system plus the product of pressure and volume
It is called permafrost :)
Answer:
Supersaturated solution.
Explanation:
Hello!
In this case, according to the types of solution in terms of the relative amounts of solute and solvent, we can define a point called solubility at which the amount of solute is no longer dissolved in the solvent; thus, a value of solute/solvent less than the solubility is related to unsaturated solutions, equal to the solubility is related to the saturated solutions and more than the solubility to supersaturated solutions.
Thus, since solubility is temperature-dependent, at 30 °C the solubility of sodium chloride is 36.09 g per 100 mL of water; which means that, since the solution has 50 g of sodium chloride, more than 36.09 g, we infer this is a supersaturated solution.
Best regards!
Rutherford's atomic structure model was revolutionary. Contrary to J.J. Thompson's "plum pudding" model (which consisted of a solid, even mixture of protons and electrons), Rutherford's model consisted of one small, positively charged, dense nucleus, a layer of empty space, and a layer of negatively charged electrons. He came to this conclusion through his gold-foil experiment. He shot a ray of alpha particles towards the thin gold foil, and to Rutherford's surprise, some of the rays reflected back instead of going straight through the foil as he originally thought.
