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This problem is describing the state two gases have when separated and together as shown on the attached picture. First of all, diagram 1 shows how they are separated in two containers with apparently equal volumes, whereas diagram 2 shows the removal of the barrier so that they get mixed together.
In this case, we can analyze that each gas has its own pressure and due to the removal of the barrier, both pressure and volume undergo a change. Thus, we can infer that the final volume is doubled with respected to the initial one for each gas, causing the pressure of each gas to be halved and the total pressure the half of the added ones, in agreement to the Boyle's law (inversely proportional relationship between pressure and temperature).
Therefore, the correct choice is:
C. The partial pressure of each gas in the mixture is half its initial pressure; the final total pressure is half the sum of the initial pressures of the two gases.
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Answer:
∆H > 0
∆Srxn <0
∆G >0
∆Suniverse <0
Explanation:
We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.
Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.
The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.
Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.
Answer:
0.047 %
Explanation:
Step 1: Given data
- Partial pressure of ozone (pO₃): 0.33 torr
- Total pressure of air (P): 695 torr
Step 2: Calculate the %v/v of ozone in the air
Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.
<em>%v/v = pO₃/P × 100%</em>
%v/v = 0.33 torr/695 torr × 100%
%v/v = 0.047 %
Li2O is the formula for <span> lithium oxide</span>
