Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures
Answer:
Answer E.
For a collision to be completely elastic, there must be NO LOSS in kinetic energy.
We can go through each answer choice:
A. Since the ball rebounds at half the initial speed, there is a loss in kinetic energy. This is NOT an elastic collision.
B. A collision involving sticking is an example of a perfectly INELASTIC collision. This is NOT an elastic collision.
C. A reduced speed indicates that there is a loss of kinetic energy. This is NOT elastic.
D. The balls traveling at half the speed after the collision indicates a loss of kinetic energy, making this collision NOT elastic.
E. This collision indicates an exchange of velocities, characteristic of an elastic collision. We can prove this:
Let:
m = mass of each ball
v = velocity
We have the initial kinetic energy as:
KE = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2KE=21mv2+0=21mv2
And the final as:
KE = 0 + \frac{1}{2}mv^2 = \frac{1}{2}mv^2KE=0+21mv2=21mv2
Chimps feed on seeds more
Explanation:
To answer this question, we'll need to use the Ideal Gas Law:
p
V
=
n
R
T
,
where
p
is pressure,
V
is volume,
n
is the number of moles
R
is the gas constant, and
T
is temperature in Kelvin.
The question already gives us the values for
p
and
T
, because helium is at STP. This means that temperature is
273.15 K
and pressure is
1 atm
.
We also already know the gas constant. In our case, we'll use the value of
0.08206 L atm/K mol
since these units fit the units of our given values the best.
We can find the value for
n
by dividing the mass of helium gas by its molar mass:
n
=
number of moles
=
mass of sample
molar mass
=
6.00 g
4.00 g/mol
=
1.50 mol
Now, we can just plug all of these values in and solve for
V
:
p
V
=
n
R
T
V
=
n
R
T
p
=
1.50 mol
×
0.08206 L atm/K mol
×
273.15 K
1 atm
= 33.6 L
this is not the answer but it will help you
do by the formula it is on the answer
