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>
1.2*104 is 12000 it is easy just try ur best
Answer:
The answer will be A
Explanation:
I will send a photo with it
Answer:
a. The second run will be faster.
d. The second run has twice the surface area.
Explanation:
The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.
The area of the 10.0 cm³-sphere is:
The area of each 1.25 cm³-sphere is:
The total area of the 8 1.25cm³-spheres is 8 × 5.61 cm² = 44.9 cm²
The ratio of 8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00
Since the surface area is doubled, the second run will be faster.
