1. The weight loss from 45.850g to 45.351g is the mass of oxygen, O2, gas released. The molar mass of O2 is 16.00x2 = 32.00g. This information should enable you get the moles of O2.
2. The volume of O2 = volume of water displaced.
volume = mass/density.
The density of water is very close to 1.00 g/mL
3. P(dry O2) = 754.2torr - P(water vapor)
Look up the vapor pressure of H2O at 23 deg C for the above calculation.
4. The Ideal Gas Law is PV = nRT
Convert the pressure of dry O2 to atmospheres.
Convert the volume of O2 to liters.
Convert 23.0 deg C to deg K
n is the number of moles of O2 from a previous calculation.
Substitute the above values into PV = nRT and solve for R.
KClO3 decomposes in the presence of MnO2 to form oxygen gas, collected over H2O at 23.0 degrees C and a barometric pressure of 754.2 torr, as described in the experimental description. After the pressure of the oxygen gas is equalized with atmospheric pressure, the mass of water collected is 386.2g. If the test tube plus KCLO3 initially weighed 45.850g and had a final mass of 45.351g, answer the following questions.
1. Calculate the number of moles of oxygen generated.
2. Calculate the volume of the gas generated, from the mass and density of water displaced.
3. Calculate the pressure of dry oxygen using Dalton's law of partial pressures.
4. Calculate the value of R, the gas constant, in L atm/mole K, and in mL torr/mole K.
Please help me!
2 answers
no this is not correct because you guys do not have a brain
1 answer