Answer:
2Li + F₂ → 2LiF
Explanation:
The reaction expression is given as:
Li + F₂ → LiF
We are to balance the expression. In that case, the number of atoms on both sides of the expression must be the same.
Let use a mathematical approach to solve this problem;
Assign variables a,b and c as the coefficients that will balance the expression:
aLi + bF₂ → cLiF
Conserving Li: a = c
F: 2b = c
let a = 1, c = 1 and b = 
Multiply through by 2;
a = 2, b = 1 and c = 2
2Li + F₂ → 2LiF
Answer:
s = 4.41 g/L.
Explanation:
¡Hola!
En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:
Lo cual hace que la expresión de equilibrio se calcule como:
![Ksp=[Pb^{2+}][Cl^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5BCl%5E-%5D%5E2)
Y que en términos de la solubilidad molar, s, se resuelve como:
![1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L](https://tex.z-dn.net/?f=1.6x10%5E%7B-5%7D%3Ds%282s%29%5E2%5C%5C%5C%5C1.6x10%5E%7B-5%7D%3D4s%5E3%5C%5C%5C%5Cs%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B1.6x10%5E%7B-5%7D%7D%7B4%7D%20%7D%20%5C%5C%5C%5Cs%3D0.0159molPbCl_2%2FL)
Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):
¡Saludos!
C. HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O
Answer:
molar mass M(s) = 65.326 g/mol
Explanation:
- M(s) + H2SO4(aq) → MSO4(aq) + H2(g)
∴ VH2(g) = 231 mL = 0.231 L
∴ P atm = 1.0079 bar
∴ PvH2O(25°C) = 0.03167 bar
Graham´s law:
⇒ PH2(g) = P atm - PvH2O(25°C)
⇒ PH2(g) = 1.0079 bar - 0.03167 bar = 0.97623 bar = 0.9635 atm
∴ nH2(g) = PV/RT
⇒ nH2(g) = ((0.9635 atm)(0.231 L))/((0.082 atmL/Kmol)(298 K))
⇒ nH2(g) = 9.1082 E-3 mol
⇒ n M(s) = ( 9.1082 E-3 mol H2(g) )(mol M(s)/mol H2(g))
⇒ n M(s) = 9.1082 E-3 mol
∴ molar mass M(s) [=] g/mol
⇒ molar mass M(s) = (0.595 g) / (9.1082 E-3 mol)
⇒ molar mass M(s) = 65.326 g/mol
According to boiling point elevation equation:
Δ T = i Kb m
when ΔT (change in boiling point) = 7.10 C°
and i (van't Hoff factor)= 1
and Kb = 0.520
so, by substitution:
7.10 = 1*0.520 *m
m = 7.1 / 0.52 = 13.65 m
