<u>Given:</u>
Moles of CS2 (it cannot be CO2 as mentioned in the question, since O2 reacts with CS2 and not CO2) = 34.5 mol
<u>To determine:</u>
Moles of O2 undergoing the reaction
<u>Explanation:</u>
The reaction is-
CS2 + 3O2 → CO2 + 2SO2
Based on the stoichiometry: 3 moles of O2 reacts with 1 mole of CS2
therefore the moles of O2 that would combine with 34.5 moles of CS2 are
= 3 moles O2 * 34.5 moles CS2/1 mole CS2 = 103.5 moles
Ans: Around 104 moles of O2 would react with 34.5 moles of CS2
We have 25cm^3 of 0.1mol AgNO3.
25cm^3 = 0.025L, so we have 0.025 x 0.1 = 0.0025mol AgNO3, so
0.0025AgNO3 + 0.0025NaCl = 0.0025AgCl + 0.0025NaNO3
Change in Free Energy: ΔG(20C) = -0.064kJ (negative, so the reaction runs)
Change in Enthalpy: ΔH(20C) = -0.110kJ (negative, so the reaction is exothermic)
This reaction produces 0.358g of AgCl and 0.213g of NaNO3
Les Mclean PhD
25,000 buts you in the 30% bracet, so your answer would be, 30% or $7,500
Answer:
- The room mantained at a lower temperature will contain more air molecules.
Explanation:
1) Since the two rooms are <em>connected by an open door</em>, you assume pressure equilibrium: the pressure on the two rooms is the same.
2) Since you consider <em>two equal size rooms</em>, both volumes are equal.
3) Assuming ideal gas behavior, pressure (P), temperature (T), volume (V) and number of moles (n) are related by the equation PV = nRT
4) Naming T₁ the lower temperature, T₂ the higher temperature, n₁ the number of moles of air in the room at lower temperature, and n₂ the number of moles of air in the room at higher temperature, you get:
- n₁ T₁ = n₂ T₂, or n₁ / n₂ = T₂ / T₁
5) That means that the amount of molecules (number of moles) is inversely related to the temperature: the higher the temperature the lower the number of moles, and the lower the temperature the greater the number of moles.
Hence, the answer is that <em>the room that contains more air molecules is the room mantained at a lower temperature.</em>
