<span>128 g/mol
Applying Graham's law of effusion, we can utilize the formula:
r1/r2 = sqrt(m2/m1)
where
r1 = effusion rate of gas 1
r2 = effusion rate of gas 2
m1 = molar mass of gas 1
m2 = molar mass of gas 2
Given that the atomic weight of oxygen is 15.999, the molar mass of O2 = 2 * 15.999 = 31.998.
We can now insert the known values into Graham's equation to find m2.
r1/r2 = sqrt(m2/m1)
2/1 = sqrt(m2/31.998)
4/1 = m2/31.998
Thus, we find m2 to be 127.992.
Rounding to three significant figures yields 128 g/mol</span>
At a pressure of 1.00 × 10⁻¹⁰ mmHg and a temperature of 273.15 K, the volume occupied by the 1.00 × 10⁶ moles of gas is 1.70 × 10²³ millilitres. This is derived from the universal gas equation PV = nRT, where V is the volume, n is the number of moles (1.00 × 10⁶), R is the universal gas constant (62.363 mmHg·L/(mol·K)), T is the temperature (273.15 K), and P is the pressure (1.00 × 10⁻¹⁰ mmHg). By substituting these values into the equation, we find the volume in millilitres equals 1.703 × 10²⁰ L converted to millilitres equals 1.703 × 10²³ millilitres.
The reaction will yield 2 mol of H₂O. The balanced chemical equation for this process is: c3h8 (g) + 5o2 (g) → 3co2 (g) + 4h2o (g). Since 5 moles of O₂ create 4 moles of H₂O, 2.5 mol of O₂ will produce: 2.5 x 4/5 = 2 mol of H₂O.
Water is made up of H2 gas and O2 gas in a 2:1 proportion, and the bubbles seen at the bottom of the pot represent the transition of water from a liquid state to a gaseous state. I'm fairly certain that the substance released from the pot is water vapor. Hope this clarifies things!???:)
