Answer:
0.18 moles
Explanation:
Applying,
PV = nRT................... Equation 1
Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = temperature.
make n the subject of the equation
n = PV/RT............... Equation 2
Given: V = 5.3 L, T = 22 °C = (22+272) K = 295 K, P = 632 mmHg = (0.00131579×632) = 0.8316 atm, R = 0.083 L.atm/K.mol
Substitute these values into equation 2
n = (0.8316×5.3)/(0.083×295)
n = 0.18 moles
Answer: KMnO4-
Explanation:
You're looking at one potassium plus a polyatomic ion.
So K plus MnO4, equals:
KMnO4-
It also has a molar mass of 158.04 g/mol, I don't know if you need that, but I thought it would be nice to include it.
<span>In the Bohr model electrons in atoms can occupy allowed orbits where they do not emit energy. Exchange of energy with the surrounding environment occurs only when an electron "jumps" from an orbit to another. Hope this answers the question. Have a nice day.</span>
Answer: when reactants and products are gases at STP.
Justification:
1) STP stands for standard temperature (0°) and pressure (1 atm).
2) According to the kinetic molecular theory of the gases, and as per Avogadro's principle, equal volumes of gases, at the same temperature and pressure, have the same number of molecules.
3) Since the coefficients in a balanced chemical equation represent number of moles, when reactants and products are gases at the same temperature and pressure, the mole ratios are the same that the volume ratios, and then the coefficients of the chemical equation represent the volume ratios.
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