Answer:
Explanation:
MM: 30.01
N₂ + O₂ ⟶ 2NO; ΔH = +43 kcal/mol
m/g: 147
Treat the heat as if it were a reactant in the reaction. Then you can write
N₂ + O₂ + 43 kcal ⟶ 2NO
The conversion factor is then 43 kcal/2 mol NO.
1. Moles of NO
2. Amount of heat
Answer:
4.31 × 10²
Explanation:
Equation of the reaction;
⇌
The ICE Table is shown as follows:
⇌
Initial 3.10 2.50 0
Change - x -x + 2x
Equilibrium (3.10 - x) 0.0800 2x
From ;
We can see that 2.50 - x = 0.0800
So; we can solve for x;
x = 2.50 - 0.0800
x = 2.42
which = (3.10 -x) will be :
= 3.10 - 2.42
= 0.68
= 2x
= 2 (2.42)
= 4.84
430.62
≅ 431
= 4.31 × 10²
Explanation:
The given data is as follows.
= 85.0 ml,
= 0.9 M
= 85.0 ml,
= 0.9 M
Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.
So, No. of moles = Molarity × Volume
= (as 1 L = 1000 ml so, 85 ml = 0.085 L)
= 0.076 mol
As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.
= 4.271 kJ
or, = 4271 J (as 1 kJ = 1000 J)
Therefore, heat released = - heat of gained by calorimeter
Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.
Hence, mass of HCl = density × Volume of HCl
= 1.00 g/ml × 85.0 ml
= 85 g
Similarly, mass of KOH = = density × Volume of HCl
= 1.00 g/ml × 85.0 ml
= 85 g
Hence, total mass of the solution = 85 g + 85 g
= 170 g
Also, q =
4271 J =
0.0773 =
=
Thus, we can conclude that final temperature of the mixed solution is .