At constant temperature, if the pressure is compressed to the given value, the volume of the nitrogen gas increases to 23.2L.
<h3>What is Boyle's law?</h3>
Boyle's law simply states that "the volume of any given quantity of gas is inversely proportional to its pressure as long as temperature remains constant.
Boyle's law is expressed as;
P₁V₁ = P₂V₂
Where P₁ is Initial Pressure, V₁ is Initial volume, P₂ is Final Pressure and V₂ is Final volume.
Given that;
- Initial volume of the gas V₁ = 22.5L
- Initial pressure of the gas P₁ = 0.98atm
- Final pressure of the gas P₂ = 0.95atm
- Final volume of the gas V₂ = ?
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂
V₂ = (0.98atm × 22.5L) / 0.95atm
V₂ = 22.05Latm / 0.95atm
V₂ = 23.2L
Therefore, at constant temperature, if the pressure is compressed to the given value, the volume of the nitrogen gas increases to 23.2L.
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Answer : Al
Explanation :
- For a given redox reaction there must be an oxidizing agent/(s) and a reducing agent/(s).
- the oxidizing agent is the substance that undergoes reduction process (gaining electrons or loss of oxygen atoms), meanwhile, the reducing agent is the substance that undergoes oxidation process (loss of electrons or gaining of oxygen atoms).
- In the reaction above, the oxidation number of (Al) in AlCl3 is (3+). However, the oxidation number of (Al) in the products is zero because it exists as a single element.
Therefore, changing from (3+) to zero means gaining of (3) electrons to neutralize the previously existing (3) protons on (Al) in AlCl3.
So Al is the oxidizing agent..
If the change in entropy of the surroundings for a process at 451 k and constant pressure is -326 j/k, then heat flow absorbed (in kj) by the system is -147.026kJ.
<h3>What is entropy? </h3>
The entropy of particle is defined as how random it move. It shows the randomness of the system or may be disorders of the system. It is used to measure the unavailable energy for performing useful work.
Unit of entropy = J/K
<h3>Formula:</h3>
∆s = ∆Q/T
where,
∆s = change in entropy of the surrounding = -326J/K
∆Q = heat absorbed from surrounding
T = Temperature = 451K
∆Q = ∆s × T
∆Q = -326 × 451
∆Q = 147,026 J
∆Q = 147.026 kJ
Thus we find that the heat absorbed by the system is 147.026 kJ.
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