Answer : The final volume of gas will be, 26.3 mL
Explanation :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 0.974 atm
= final pressure of gas = 0.993 atm
= initial volume of gas = 27.5 mL
= final volume of gas = ?
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:
Therefore, the final volume of gas will be, 26.3 mL
Answer:
1.4 × 10² mL
Explanation:
There is some info missing. I looked at the question online.
<em>The air in a cylinder with a piston has a volume of 215 mL and a pressure of 625 mmHg. If the pressure inside the cylinder increases to 1.3 atm, what is the final volume, in milliliters, of the cylinder?</em>
Step 1: Given data
- Initial volume (V₁): 215 mL
- Initial pressure (P₁): 625 mmHg
- Final pressure (P₂): 1.3 atm
Step 2: Convert 625 mmHg to atm
We will use the conversion factor 1 atm = 760 mmHg.
625 mmHg × 1 atm/760 mmHg = 0.822 atm
Step 3: Calculate the final volume of the air
Assuming constant temperature and ideal behavior, we can calculate the final volume of the air using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 0.822 atm × 215 mL / 1.3 atm = 1.4 × 10² mL
Answer:
The average velocity of the airplane for this trip is 1684.21 km/h
Explanation:
Average velocity is the rate of change of displacement with time. That is,
Average velocity = 
= Δx / Δt = 
Now we will calculate the time taken by the airplane for the first motion before it encounters a wind.
From,
Velocity = 
Time = 
Therefore, Time = 
Time = 2.1h
This is the time taken before the airplane encounters a wind.
Hence, t1 = 2.1h
Now, For the time taken by the airplane when it encounters a wind
Also from,
Velocity =
Time =
Therefore, Time = 
Time = 1.625h
Hence, t2 = 1.625h
Now, to calculate the average velocity
Average velocity =
x1= 2100, x2= 1300, t1= 2.1h and t2= 1.625h
Hence, Average velocity = 
Average velocity = 1684.21 km/h
The correct answer of the given question above would be option B. IRON 0.449. Based on the given details above about an unknown substance that has a mass of 14.7 g and the substance absorbs 1.323×102 J of heat, the temperature of the substance is raised from 25.0 ∘C to45.0 ∘C, most likely, the substance is IRON. Hope this answers the question.
