Answer:
M.Mass = 3.66 g/mol
Data Given:
M.Mass = M = ??
Density = d = 0.1633 g/L
Temperature = T = 273.15 K (Standard)
Pressure = P = 1 atm (standard)
Solution:
Let us suppose that the gas is an ideal gas. Therefore, we will apply Ideal Gas equation i.e.
P V = n R T ---- (1)
Also, we know that;
Moles = n = mass / M.Mass
Or, n = m / M
Substituting n in Eq. 1.
P V = m/M R T --- (2)
Rearranging Eq.2 i.e.
P M = m/V R T --- (3)
As,
Mass / Volume = m/V = Density = d
So, Eq. 3 can be written as,
P M = d R T
Solving for M.Mass i.e.
M = d R T / P
Putting values,
M = 0.1633 g/L × 0.08205 L.atm.K⁻¹.mol⁻¹ × 273.15 K / 1 atm
M = 3.66 g/mol
Answer:
it will option B ,hope it helps
Answer:
Iodine
Explanation:
It's in the same group as chlorine.
4. The pressure of the inner core is higher than the outer core
5. The coolest layers are farthest from the core
<u>Given:</u>
Moles of Al = 0.4
Moles of O2 = 0.4
<u>To determine:</u>
Moles of Al2O3 produced
<u>Explanation:</u>
4Al + 3O2 → 2Al2O3
Based on the reaction stoichiometry:
4 moles of Al produces 2 moles of Al2O3
Therefore, 0.4 moles of Al will produce:
0.4 moles Al * 2 moles Al2O3/4 moles Al = 0.2 moles Al2O3
Similarly;
3 moles O2 produces 2 moles Al2O3
0.4 moles of O2 will yield: 0.4 *2/3 = 0.267 moles
Thus Al will be the limiting reactant.
Ans: Maximum moles of Al2O3 = 0.2 moles
