The concentration of NO2 is 0.054 M and that of N2O4 is 0.003 M.
<h3>Equilibrium constant</h3>
The equilibrium constant shows the extent to which reactants are converted into products.
We have the reaction; N2O4⇌2NO2 and Kp = 4.5. Setting up the ICE table, we have;
N2O4 ⇌ 2NO2
I 0. 0550 0
C -x +x
E 0. 0550 - x x
Given that;
Kp = [NO2]^2/[N2O4]
4.5= (x)^2/0. 0550 - x
4.5(0. 0550 - x) = x^2
0.2475 - 4.5x = x^2
x^2 + 4.5x - 0.2475 = 0
x=0.054 M
The concentration of the required NO2 gas at equilibrium is 0.054 M and that of N2O4 is 0.003 M.
Learn more about equilibrium constant: brainly.com/question/17960050
<h2>Answer:</h2>
The correct answer is option C which is, "Electrons in the orbit closest to the nucleus have the least amount of energy".
<h3>
Explanation:</h3>
- There are different orbitals around the nucleus on which the electrons moves around the nucleus.
- These orbitals have a specific energy, due to which they are known as energy levels.
- The energy level near to the nucleus has least amount of the energy and the energy of the orbitals increase as the distance of the orbitals increase to the nucleus.
I will try to do it but not more than 70.4
Answer:
This question is incomplete, the complete question is; assuming that the solution has a specific heat of 4.18 J/g°C
The answer is 381.67 J/g
Explanation:
Enthalpy change denoted by ΔH can be calculated using the formula;
ΔH = m × c × ΔT
Where; m= mass of reactants
c= specific heat
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 32.76 - 19.3
ΔT = 13.46 °C
mass of reactants= 85.6 + 14.8 = 100.4g, c = 4.18J/g°C
Hence; ΔH = m × c × ΔT
ΔH = 100.4 × 4.18 × 13.46
ΔH = 5648.78J
Enthalpy change per gram of potassium hydroxide dissolved in the water is;
ΔH = 5648.78/14.8
ΔH = 381.67 J/g
Explanation:
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