Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
Answer:However, unlike the outer core, the inner core is not liquid or even molten. The inner core's intense pressure—the entire rest of the planet and its atmosphere—prevents the iron from melting. The pressure and density are simply too great for the iron atoms to move into a liquid state.
Explanation:please give brainliest
The moles of O2 are needed to react completely with 1.00 mol of C₂H₂ is 5 mol , 0.23 moles of C₂H₂ are needed to form 0.46 mol of CO₂ .
<h3>What is a Balanced Reaction ?</h3>
A balanced reaction is a reaction in which the number of atoms present in the reactants is equal to the number of atoms in the products.
It is given in the question that
the combustion of Acetylene is given by
2 H―C≡C―H + 5 O₂ → 4 CO₂ + 2 H₂O
Given moles of Acetylene = 1 mol
Moles of oxygen needed = 5 mol
as the mole fraction of Acetylene to Oxygen is 1 :5
The mole of Acetylene needed to form 0.46 mol of CO₂ is ?
mole fraction of CO₂ to Acetylene is 4:2
Therefore 0.23 moles of CO₂ is required.
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Answer:
M
Explanation:
Concentration of
= 0.020 M
Constructing an ICE table;we have:
![Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%2B4NH_3_%7Baq%7D%20%5Crightleftharpoons%20%5BCu%28NH_3%29_4%5D%5E%7B2%2B%7D_%7B%28aq%29%7D)
Initial (M) 0.020 0.40 0
Change (M) - x - 4 x x
Equilibrium (M) 0.020 -x 0.40 - 4 x x
Given that: 
![K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}](https://tex.z-dn.net/?f=K_f%20%7D%20%3D%20%5Cfrac%7B%5BCu%28NH_3%29_4%5D%5E%7B2%2B%7D%7D%7B%5BCu%5E%7B2%2B%7D%5D%5BNH_3%5D%5E4%7D)

Since x is so small; 0.40 -4x = 0.40
Then:








M
To solve this, let's assume ideal gas behavior.
PV=nRT
Let's solve for n. Convert units to SI units first.
Pressure = 833 torr(101325 Pa/760 torr) = 111,057.53 Pa
Volume = 250 mL(1 L/1000 mL)(1 m³/1000 L) = 2.5×10⁻⁴ m³
Temperature = 42.4 + 273 = 315.4 K
n = (8,314 J/mol·K)(315.4 K)/(111057.53 Pa)(2.5×10⁻⁴ m³)
n = 94.45 mol
The molar mass of ammonia is 17.031 g/mol.
Mass = 94.45*17.031 = <em>1,608.51 g ammonia</em>