Answer:
140 K
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 3 atm
- Initial temperature of the gas (T₁): 280 K
- Final pressure of the gas (P₂): 1.5 atm
- Final temperature of the gas (T₂): ?
Step 2: Calculate the final temperature of the gas
We have a gas whose pressure is reduced. If we assume an ideal behavior, we can calculate the final temperature of the gas using Gay-Lussac's law.
T₁/P₁ = T₂/P₂
T₂ = T₁ × P₂/P₁
T₂ = 280 K × 1.5 atm/3 atm = 140 K
Density is Mass divided by volume.
Answer:

Explanation:
Hello there!
In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

In terms of mass, specific heat and temperature change is:

Now, solve for the final temperature, as follows:

Then, plug in the masses, specific heat and temperatures to obtain:

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.
Regards!
Answer:density
Explanation:
it’s how’s how dens the ball is