I believe it would be A. Equal out in both directions.
The balanced chemical equation for the combustion of butane is:
Δ
= Σ
Δ
-Σ
Δ
= ![[{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]](https://tex.z-dn.net/?f=%5B%7B8%2A%28-393.5kJ%2Fmol%29%7D%2B%7B10%2A%28-241.82kJ%2Fmol%29%7D%5D-%5B%7B2%2A%28-125.6kJ%2Fmol%29%7D%2B13%2A%280%20kJ%2Fmol%29%7D%5D)
=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]
= -5315 kJ/mol
Calculating the enthalpy of combustion per mole of butane:
Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol
Correct answer: -2657.5 kJ/mol
Equation: NH4HS(s) <=> NH3 (g) + H2S(g)
Kc = [NH3]eq[H2S]eq; where eq subscripts are used to indicate that those are the concentrations at equilibrium
Kc= 0.272 * 0.365 =0.09928
Answer:
Mass = 1.84 g
Explanation:
Given data:
Total pressure = 3.44 atm
Volume of flask = 9.62 L
Temperature = 63°C = 63+273.15K = 336.15 K
Mass of CO₂ = 12.5 g
Mass of hydrogen = ?
Solution:
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 12.5 g/ 44 g/mol
Number of moles = 0.28 mol
Pressure of CO₂:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
P = 0.28 mol × 0.0821 atm.L/mol.K × 336.15 K / 9.62 L
P = 7.73 atm.L/ 9.62 L
P = 0.8 atm
Total pressure = Partial pressure of CO₂ + Partial pressure of H₂
3.44 atm = 0.8 atm + P(H₂)
P(H₂) = 3.44 atm - 0.8 atm
P(H₂) = 2.64 atm
Number of moles of hydrogen:
PV = nRT
2.64 atm × 9.62 L = n × 0.0821 atm.L/mol.K × 336.15 K
25.39 atm.L = n × 27.59atm.L/mol
n = 25.39 atm.L / 27.59atm.L/mol
n = 0.92 mol
Mass of hydrogen:
Mass = number of moles × molar mass
Mass = 0.92 mol × 2 g/mol
Mass = 1.84 g
Answer:
12.044 ×10²³ molecules of HCl
Explanation:
Given data:
Number of moles of MgO = 1 mol
Number of molecules of HCl react = ?
Solution:
Chemical equation:
MgO + 2HCl → MgCl₂ + H₂O
with 1 mole of MgO 2 moles of HCl are react.
Number of molecules of HCl react:
1 mole contain 6.022×10²³ molecules
2 mol × 6.022×10²³ molecules / 1 mol
12.044 ×10²³ molecules
