<u>Answer:</u> The value of 
for the given reaction is 1.435
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
Given mass of 
= 9.2 g
Molar mass of = 92 g/mol
Volume of solution = 0.50 L
Putting values in above equation, we get:
For the given chemical equation:
<u>Initial:</u> 0.20
<u>At eqllm:</u> 0.20-x 2x
We are given:
Equilibrium concentration of = 0.057
Evaluating the value of 'x'
The expression of for above equation follows:
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![[NO_2]_{eq}=2x=(2\times 0.143)=0.286M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.143%29%3D0.286M)
![[N_2O_4]_{eq}=0.057M](https://tex.z-dn.net/?f=%5BN_2O_4%5D_%7Beq%7D%3D0.057M)
Putting values in above expression, we get:
Hence, the value of for the given reaction is 1.435
Answer:
F. 2NO + 02 —> 2NO
H. 4NH3 + 502 —> 4NO + 6H20
Explanation:
The law of conservation of mass states that matter can neither be created nor destroyed during a chemical reaction but can be convert from one form to another.
2NO + 02 —> 2NO
From the above, the total number of N on the left balance the total number on the right i.e 2 atoms of N on both side of the equation.
The total number of O on the left balance the total number on the right i.e 2 atoms of O on both side of the equation. This is certified by the law of conservation of mass.
4NH3 + 502 —> 4NO + 6H20
From the above, the total number of N on the left balance the total number on the right i.e 4 atoms of N on both side of the equation.
The total number of O on the left balance the total number on the right i.e 10 atoms of O on both side of the equation.
The total number of H on the left balance the total number on the right i.e 12 atoms of O on both side of the equation.
This is certified by the law of conservation of mass.
The rest equation did not conform to the law of conservation of mass as the atoms on the left side did not balance those on the right side
Answer:
Oxidation–reduction or redox reactions are reactions that involve the transfer of electrons between chemical species (check out this article on redox reactions if you want a refresher!). The equations for oxidation-reduction reactions must be balanced for both mass and charge, which can make them challenging to balance by inspection alone. In this article, we’ll learn about the half-reaction method of balancing, a helpful procedure for balancing the equations of redox reactions occurring in aqueous solution.
Explanation:
Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)
The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)
ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa
