Spontaneous at low temperatures.
The reaction for the combustion of methane can be expressed as follows.
CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
moles of CO2 = (5.6 L) / (22.4 L/1 mole)
moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
= 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane.
Answer:
Partial pressure SO₂ → 0.440 atm
Explanation:
We apply the mole fraction concept to solve this:
Moles of gas / Total moles = Partial pressure of the gas / Total pressure
Total moles = 0.3 moles of CO₂ + 0.2706 moles of SO₂ + 0.35 moles H₂O
Total moles = 0.9206 moles
Mole fraction SO₂ = 0.2706 moles / 0.9206 moles → 0.29
Now, we can know the partial pressure:
0.29 = Partial pressure SO₂ / Total pressure
0.29 = Partial pressure SO₂ / 1.5 atm
0.29 . 1.5atm = Partial pressure SO₂ → 0.440 atm
Answer:
Explanation:
Density is
mass / volume = d
Mass:
840g
Volume:
7 cm x 4 cm x 10 cm = 280 cm^3
840g / 280 cm^3 = 3 g/cm^3
