Particles are measured in atomic mass units, or AMU.
This is an exercise in the general or combined gas law.
To start solving this exercise, we must obtain the following data:
<h3>Data:</h3>
- V₁ = 4.5 l
- T₁ = 33 °C + 273 = 306 k
- P₁ = 6.54 atm
- T₂ = 94 °C + 273 = 367 k
- V₂ = 2.3 l
- P₂ = ¿?
We use the following formula:
- P₁V₁T₂ = P₂V₂T₁ ⇒ General Formula
Where
- P₁ = Initial pressure
- V₁ = Initial volume
- T₂ = Initial temperature
- P₂ = Final pressure
- V₂ = Final volume
- T₁ = Initial temperature
We clear the general formula for the final pressure.

We solve by substituting our data in the formula:



If I raise the temperature to 94°C and decrease the volume to 2.3 liters, the pressure of the gas will be 15,346 atm.
Answer:
H2O acts as an oxidizing agent
Answer:
The pH of the solution is 9.43
Explanation:
sodium acetate = CH3COONa
after removing the spectator ion Na
CH3COO⁻ + H2O ⇄ CH3COOH + H3O⁺
since sodium acetate is a weak base, it will not react completely with water. Thus a value "x" will be used to make the products.
At equilibrium:
[CH3COO⁻] = 1.288 - x
[CH3COOH] = x
[H3O⁺] = x
Since a base is reacting with H2O, Kb is used at equilibrium: Kb=(1*10^-14)/(1.8*10^-5) = 5.6*10^-10)
Kb = [CH3COOH][H3O⁺]/[CH3COO⁻]
Kb = (x²)/(1.288-x) since the Kb is significantly lower than 1.288 we can make a mathematical assumption that x is significantly lower than 1.288.
5.6*10⁻¹⁰ = x²/1.288
x = [OH⁻] = 2.675*10⁻⁵ M
pOH = -log[OH⁻] = 4.57
pH=14-pOH = 14 - 4.57 = 9.43