Answer:
Mass = 99.8 g
Explanation:
Given data:
Mass of potassium nitride = ?
Mass of nitrogen produced = 10.65 g
Solution:
Chemical equation:
2K₃N→ 6K + N₂
Moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 10.65 g / 28 g/mol
Number of moles = 0.38 mol
Now we will compare the moles of nitrogen with potassium nitride.
N₂ ; K₃N
1 : 2
0.38 : 2×0.38 =0.76
Mass of potassium nitride:
Mass = molar mass × number of moles
Mass = 131.3 g/mol × 0.76 mol
Mass = 99.8 g
Answer:
K(48.5°C) = 1.017 E-8 s-1
Explanation:
- CH3Cl + H2O → CH3OH + HCl
at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1
at T2 = 48.5°C (321.5 K) ⇒ K2 = ?
Arrhenius eq:
- K(T) = A e∧(-Ea/RT)
- Ln K = Ln(A) - [(Ea/R)(1/T)]
∴ A: frecuency factor
∴ R = 8.314 E-3 KJ/K.mol
⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)
⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)
(1)/(2):
⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)
⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)
⇒ Ln (K1/K2) = - 3.422
⇒ K1/K2 = e∧(-3.422)
⇒ (3.32 E-10 s-1)/K2 = 0.0326
⇒ K2 = (3.32 E-10 s-1)/0.0326
⇒ K2 = 1.017 E-8 s-1
Answer:
Most likely true
Explanation:
because the speed of all ocean waves is controlled by gravity, wavelength, and water depth. ... Waves moving through water deeper than half their wavelength are known as deep-water waves. On the other hand, the orbits of water molecules in waves moving through shallow water are flattened by the proximity of the sea surface bottom.
Answer:
[Ag⁺] = 0.0666M
Explanation:
For the addition of Ag⁺ and CN⁻, the (Ag(CN)₂⁻ is produced, thus:
Ag⁺ + 2CN⁻ ⇄ Ag(CN)₂⁻
Kf = 1x10²¹ = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺]
As initial concentrations of Ag⁺ and CN⁻ are:
[Ag⁺] = 0.110L × (3.0x10⁻³mol / L) = 3.3x10⁻⁴mol / (0.110L + 0.230L) = 9.7x10⁻⁴M
[CN⁻] = 0.230L × (0.1mol / L) = 0.023mol / (0.110L + 0.230L) = 0.0676M
The equilibrium concentrations of each compound are:
[CN⁻] = 9.7x10⁻⁴M - x
[Ag⁺] = 0.0676M - x
[Ag(CN)₂⁻] = x
<em>Where x is reaction coordinate</em>
Replacing in Kf formula:
1x10²¹ = [x] / [9.7x10⁻⁴M - x]² [0.0676M - x]
1x10²¹ = [x] / 6.36048×10⁻⁸ - 0.000132085 x + 0.06954 x² - x³
-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = x
-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = 0
Solving for x:
X = 9.614x10⁻⁴M
Thus, equilibrium concentration of Ag⁺ is:
[Ag⁺] = 0.0676M - 9.614x10⁻⁴M = <em>0.0666M</em>
