Answer:
pH = 4.27. Porcentaje de disociación: 0.03%
Explanation:
El pH de un ácido débil, HX, se obtiene haciendo uso de su equilibrio:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
Donde la constante de equilibrio, Ka, es
Ka = 1.65x10⁻⁸ = [H⁺] [X⁻] / [HX]
Como los iones H⁺ y X⁻ vienen del mismo equilibrio podemos decir:
[H⁺] = [X⁻]
[HX] es:
20g * (1mol/55g) = 0.3636moles / 2.100L = 0.1732M
Reemplazando es Ka:
1.65x10⁻⁸ = [H⁺] [H⁺] / [0.1732M]
2.858x10⁻⁹ = [H⁺]²
5.35x10⁻⁵M = [H⁺]
pH = -log[H⁺]
<h3>pH = 4.27</h3>
El porcentaje de disociacion es [X⁻] / [HX] inicial * 100
Reemplazando
5.35x10⁻⁵M / 0.1732M * 100
<h3>0.03%</h3>
<h3>
Answer:</h3>
0.10 L
<h3>
Explanation:</h3>
The concentration of glucose is given as 180 g/L
The mass of glucose is 18 g
- Concentration in g/L is calculated by dividing mass of the solute by the volume of the solution.
- When calculating molarity on the other hand, we divide number of moles of the solute by the volume of the solution.
- Concentration in g/L = Mass of solute ÷ Volume
Rearranging the formula,
Volume = Mass of the solute ÷ concentration
= 18 g ÷ 180 g/L
= 0.10 L
Therefore, volume of water is 0.10 L
<u>Answer:</u> The equilibrium concentration of water is 0.597 M
<u>Explanation:</u>
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For a general chemical reaction:
The expression for 
is written as:
![K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
The concentration of pure solids and pure liquids are taken as 1 in the expression.
For the given chemical reaction:
The expression of 
for above equation is:
![K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2S%5D%5E2%5Ctimes%20%5BO_2%5D%7D)
We are given:
![[H_2S]_{eq}=0.671M](https://tex.z-dn.net/?f=%5BH_2S%5D_%7Beq%7D%3D0.671M)
![[O_2]_{eq}=0.587M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D0.587M)
Putting values in above expression, we get:
![1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}](https://tex.z-dn.net/?f=1.35%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%280.671%29%5E2%5Ctimes%200.587%7D)
![[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M](https://tex.z-dn.net/?f=%5BH_2O%5D%3D%5Csqrt%7B%281.35%5Ctimes%200.671%5Ctimes%200.671%5Ctimes%200.587%29%7D%3D0.597M)
Hence, the equilibrium concentration of water is 0.597 M
Oxygen (6O2) and Glucose (C6H12O6)
<span>Reference: 6CO2 + 6H2O + light energy = C6H12O6 + 6O2.</span>
Melting point- the temperature at which a substance has changed from a solid to a liquid
freezing- the temperature at which a substance chanes from liquid to a solid
boiling point- the temperature at which a substance changes from a liquid to a gas phase
