Answer:
113 g NaCl
Explanation:
The Ideal Gas Law equation is:
PV = nRT
In this equation,
> P = pressure (atm)
> V = volume (L)
> n = number of moles
> R = 8.314 (constant)
> T = temperature (K)
The given values all have to due with the conditions fo F₂. You have been given values for all of the variables but moles F₂. Therefore, to find moles F₂, plug each of the values into the Ideal Gas Law equation and simplify.
(1.50 atm)(15.0 L) = n(8.314)(280. K)
2250 = n(2327.92)
0.967 moles F₂ = n
Using the Ideal Gas Law, we determined that the moles of F₂ is 0.967 moles. Now, to find the mass of NaCl that can react with F₂, you need to (1) convert moles F₂ to moles NaCl (via the mole-to-mole ratio using the reaction coefficients) and then (2) convert moles NaCl to grams NaCl (via molar mass from periodic table). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator).
1 F₂ + 2 NaCl ---> Cl₂ + 2NaF
Molar Mass (NaCl): 22.99 g/mol + 35.45 g/mol
Molar Mass (NaCl): 58.44 g/mol
0.967 moles F₂ 2 moles NaCl 58.44 g
---------------------- x ----------------------- x ----------------------- = 113 g NaCl
1 mole F₂ 1 mole NaCl
A3B2 because the oxidation numbers are the same as ionic charge just switch symbol and number. Then use the cross cross method and you get A3B2.
Answer:
Explanation:
The contact between the sheet of gold and the sheet of iron allows a heat transfer until thermal equilibrium is done, which means that both sheets have the same temperature:
The final temperature is:
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>
If you are given the
standard potential for the reduction of X^2+ is +0.51 V, and the standard
potential for the reduction of A^2+ is -0.33, just add the two. The standard
potential for an electrochemical cell with the cell is 0.18V
