I am assuming that the problem ask for the pressure in
the system. To be able to calculate this, we first assume that the system acts
like an ideal gas, then we can use the ideal gas equation to find for pressure
P.
P V = n R T
where,
P = Pressure (unknown)
V = 0.17 m^3
n = moles of lng / methane
R = gas constant = 8.314 Pa m^3 / mol K
T = 200 K
We find for the moles of lng. Molar mass of methane = 16
kg / kmol
n = 55 kg / 16 kg / kmol
n = 3.44 kmol CH4 = 3440 mol
Substituting all the values to the ideal gas equation:
P = 3440 mol * (8.314 Pa m^3 / mol K) * 200 K / 0.17 m^3
P = 33,647,247 Pa
<span>P = 33.6 MPa</span>
According to the conversation of mass, mass cannot be created or destroyed. This means whatever is done to one side, must be done to the other.
There are 4 Phosphorus atoms on the left, there must be 4 on the right. To do this, you must multiply the P2O3 by 2 to get 4 Phosphorus atoms and 6 Oxygen atoms. Now to balance the Oxygen atoms, you must multiply the oxygen atoms on the left by 3.
1 P4 + 3 O2 —-> 2 P2O3
Lastly, this equation type is synthesis (combination) because two reactants are becoming a single product.
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Answer:
B.C5H12 that is pentane
Explanation:
hydrocarbon has only hydrogen and carbon atoms
Answer:
- 13,150.6kJ
Explanation:
CH4 + 2 O2 ------> CO2 + 2 H2O ΔH= – 890 kJ
The ΔH is enthalpy change of combustion , which is the heat is either absorbed or released by the combustion of one mole of a substance.
ΔH=−890 kJ/mol (released in the combustion of one mole of methane)
using the molar mass (in grams )of methane to get moles of sample
(237g × 1 mole of CH4)/16.04g=14.776 moles of CH4
Since 1 mole produces 890 kJ of heat upon combustion, then 14.776 moles will produce
ΔH = 14.776moles of CH4 × 890kJ/1mole of CH4
=13,150.6kJ
Therefore ΔH = - 13,150.6kJ