RCOOH + NaOH → RCOONa + H₂O (salt and water)
RCOOH + OH⁻ → RCOO⁻ + H₂O
It has 8 bonding electrons
The water molecules received enough energy through the heat for them to change to a liquid state by unfixing themselves from their fixed positions.
The equilibrium constant is 1.3 considering the reaction as written in the question.
<h3>Equilibrium in chemical reactions</h3>
In a chemical reaction, the equilibrium constant is calculated based on the equilibrium concentration of each specie. The equation of this reaction is;
A (g) + 2B (g) ⇌ 3C (g).
The initial concentration of each specie is;
- A - 9.22 M
- B - 10.11 M
- C - 27.83 M
The equilibrium concentration of B is 18.32 M
We now have to set up the ICE table as follows;
A (g) + 2B (g) ⇌ 3C (g)
I 9.22 10.11 27.83
C -x -x +x
E 9.22 - x 10.11 - x 27.83 + x
The equilibrium concentration of B is 18.32 M hence;
10.11 - x = 18.32
x = 10.11 - 18.32 = -8.21
Hence;
Equilibrium concentration of A = 9.22 - (-8.21) = 17.43
Equilibrium concentration of C = 27.83 + (-8.21) = 19.62
Equilibrium constant K = [19.62]^3/[17.43] [18.32]^2
K = 1.3
Learn more about equilibrium constant: brainly.com/question/17960050
Answer:
The heat required to raise the temperature of 12g of water from 16 C to 21 C is 60 cal.
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case, you know:
- Q=?
- c= 4.186

- m= 12 g
- ΔT=Tfinal - Tinitial= 21 °C - 16°C= 5 °C
Replacing:
Q= 4.186
*12 g *5 °C
Solving:
Q=251.16 J
Since 1 J is equal to 0.2388 cal, then the following rule of three can be applied: if 1 J is equal to 0.2388 cal, then 251.16 J to how many cal are?

cal= 59.98 ≅ 60
<u><em>The heat required to raise the temperature of 12g of water from 16 C to 21 C is 60 cal.</em></u>