Answer: Question 8) 0.45 mole NaCl
Question 9) 1.7 moles Fe2+
Question 10) 9.52 moles Fe2+
Explanation: solution attached
Hey there!
2 moles will be produced.
In N₂ there are 2 nitrogen atoms. In NH₃ there is 1 nitrogen atom.
So, there will be twice as many moles produced because there will be twice as many molecules.
Hope this helps!
The given chemical equation is:

On balancing the equation we get,

Calculating enthalpy of formation of this reaction from the standard heats of formation of the products and reactants:
Δ![H_{reaction}^{0}=[H_{f}^{0}(Al_{2}O_{3}(s)) + (3*H_{f}^{0}(H_{2}SO_{4}(aq))] - [H_{f}^{0}(Al_{2}SO_{4}(aq)) + (3*H_{f}^{0}(H_{2}O(l))]](https://tex.z-dn.net/?f=H_%7Breaction%7D%5E%7B0%7D%3C%2Fp%3E%3Cp%3E%3D%5BH_%7Bf%7D%5E%7B0%7D%28Al_%7B2%7DO_%7B3%7D%28s%29%29%20%2B%20%283%2AH_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DSO_%7B4%7D%28aq%29%29%5D%20-%20%20%20%5BH_%7Bf%7D%5E%7B0%7D%28Al_%7B2%7DSO_%7B4%7D%28aq%29%29%20%2B%20%283%2AH_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%28l%29%29%5D)
=[(-1669.8kJ/mol)+ {3* (-909.27 kJ/mol)}]-[(-3442kJ/mol)+{3*(-285.8 kJ/mol)}]
=[(-4397.61kJ/mol)]-[(-4299.4kJ/mol)]
=-98.21kJ/mol
Total enthalpy change when 15 mol of
reacts will be=

Answer:
I think so
Explanation:
There are atone in every particle so the more particles, the more atoms
Answer:
Are basic:
[OH⁻] = 3.13x10⁻⁷M and [H₃O⁺] = 9.55x10⁻⁹M
Explanation:
A solution is basic when pH = - log [H₃O⁺] is higher than 7.
It is possible to convert [OH⁻] to [H₃O⁺] using:
[H₃O⁺] = 1x10⁻¹⁴ / [OH⁻]
a. [OH⁻] = 3.13x10⁻⁷M
[H₃O⁺] = 1x10⁻¹⁴ / [3.13x10⁻⁷M]
[H₃O⁺] = 3.19x10⁻⁸M
pH = - log [H₃O⁺] = 7.50
[OH⁻] = 3.13x10⁻⁷M is basic
b. pH = -log [H₃O⁺] = - log 0.000747M = 3.13.
This solution is not basic
c. [H₃O⁺] = 9.55x10⁻⁹M
pH = 8.02
This solution is also basic.