<span>These are inert gases, so we can assume they don't react with one another. Because the two gases are also subject to all the same conditions, we can pretend there's only "one" gas, of which we have 0.458+0.713=1.171 moles total. Now we can use PV=nRT to solve for what we want.
The initial temperature and the change in temperature. You can find the initial temperature easily using PV=nRT and the information provided in the question (before Ar is added) and solving for T.
You can use PV=nRT again after Ar is added to solve for T, which will give you the final temperature. The difference between the initial and final temperatures is the change. When you're solving just be careful with the units!
SIDE NOTE: If you want to solve for change in temperature right away, you can do it in one step. Rearrange both PV=nRT equations to solve for T, then subtract the first (initial, i) from the second (final, f):
PiVi=niRTi --> Ti=(PiVi)/(niR)
PfVf=nfRTf --> Tf=(PfVf)/(nfR)
ΔT=Tf-Ti=(PfVf)/(nfR)-(PiVi)/(niR)=(V/R)(Pf/nf-Pi/ni)
In that last step I just made it easier by factoring out the V/R since V and R are the same for the initial and final conditions.</span>
Density=mass/volume
5.45g/ml=65g/V
V=65g/5.42g/ml
V=11.92ml
Answer:
0 kg m/s before and after collision
Explanation:
Let m, v be the mass and speed of the 2 balls, respectively, before the collision. Since they have the same mass and same speed but in opposite direction, the total momentum of the system would be:
P = mv - mv = 0 kg m/s
As the collision is elastic. The total momentum after the collision is the same as the total momentum before the collision, which is 0.
Since the law of gravitation is an inverse square law if you
quadruple the radius the f will drop by a factor of 16 SO the object would
weigh 200/16 = 12.5N
In other words, as the distance, or radius, quadruples the
weight becomes 1/16 of the original weight. Just plug in 4 for r and when you
square it you get 16. The numerator is 1 so that is how the weight becomes
1/16.
