Based on experiment 1:
Mass of Hg = 1.00 g
Mass of sulfide = 1.16 g
Mass of sulfur = 1.16 - 1.00 = 0.16 g
# moles of Hg = 1 g/200 gmol-1 = 0.005 moles
# moles of S = 0.16/32 gmol-1 = 0.005 moles
The Hg:S ratio is 1:1, hence the sulfide is HgS
Based on experiment 2:
Mass of Hg taken = 1.56 g
# moles of Hg = 1.56/200 = 0.0078
Mass of S taken = 1.02 g
# moles of S = 1.02/32 = 0.0319
Hence the limiting reagent is Hg
# moles of Hg reacted = # moles of HgS formed = 0.0078 moles
Molar mass of HgS = 232 g/mol
Therefore, mass of HgS formed = 0.0078 * 232 = 1.809 g = 1.81 g
Paraffin wax at room temperature is solid.
The ideal gas law states that the relationship between partial pressure and volume is inverse. Additionally, it relates temperature and moles in a direct proportion. The partial pressures of the gases are found to be PN2 = 0.094 atm, PH2 = 0.039 atm, and PNH3 = 0.003 atm in the following reaction at equilibrium at room temperature.
<h3>What is ideal gas law ?</h3>
PV = NkT, where P is a gas's absolute pressure, V is the volume it occupies, N is the quantity of atoms and molecules in the gas, and T is its absolute temperature, is the formula for the ideal gas law.
Equal volumes of two gases with the same temperature and pressure will contain an equal number of molecules, according to the law. The ideal gas law is the result of the union of all these relationships.
To learn more about ideal gas law from the given link:
brainly.com/question/28257995
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Answer: 20L of H2O
Explanation:
C3H8 + 5O2 → 3CO2 + 4H2O
Recall 1mole of a gas contains 22.4L at stp
5moles of O2 contains = 5 x 22.4 = 112L
4moles of H2O contains = 4 x 22.4 = 89.6L
From the equation,
112L of O2 produced 89.6L H2O
There for 25L of O2 will produce XL of H2O i.e
XL of H2O = (25 x 89.6)/112 = 20L
