Answer:
M.Mass = 120 g/mol
Explanation:
Data Given:
Volume = V = 0.0650 L
Temperature = T = 547 °C = 820.15 K
Pressure = P = 70.5 kPa = 0.695 atm
Gas Constant = R = 0.082057 L.atm.mol⁻¹.K⁻¹
Formula Used:
Assuming that the gas is ideally then according to ideal gas equation,
P V = n R T
Solving for n,
n = P V / R T
Putting Values,
n = (0.695 atm × 0.0650 L) ÷ (0.082057 L.atm.mol⁻¹.K⁻¹ × 820.15 K)
n = 6.71 × 10⁻⁴ moles
Now, Knowing that,
Moles = Mass / M.Mass
Or,
M.Mass = Mass / Moles
Putting values,
M.Mass = 8.06 × 10⁻² g / 6.71 × 10⁻⁴ mol
M.Mass = 120 g/mol
Answer:
Stratosphere
Explanation:
Small storms would be residing in the troposphere but stronger thunderstorms and hurricanes reside in the stratosphere.
Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is 
.
Explanation:
When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.
For example, chemical equation for oxidation of methane is as follows.
Number of atoms present on reactant side are as follows.
Number of atoms present on product side are as follows.
To balance this equation, multiply 
by 2 on reactant side. Also, multiply 
by 2 on product side. Hence, the equation can be rewritten as follows.
Now, the number of atoms present on reactant side are as follows.
Number of atoms present on product side are as follows.
Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.
Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is .
Air is mainly composed of N2 (78%), O2 (21%) and other trace gases. Now, the total pressure of air is the sum of the partial pressures of the constituent gases. The partial pressure of each gas, for example say O2, can be expressed as:
p(O2) = mole fraction of O2 * P(total, air) ----(1)
Thus, the partial pressure is directly proportional to the total pressure. If we consider a sealed container then, as the temperature of air increases so will its pressure. Based on equation (1) an increase in the pressure of air should also increase the partial pressure of oxygen.
