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
Given mass of tungsten, W = 415 g
Molar mass of tungsten, W = 183.85 g/mol
Calculating moles of tungsten from mass and molar mass:
<u>Question:</u>
For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K
<u>Answer:</u>
The mean activity coefficient for HCl solution is 0.78.
<u>Explanation:</u>
Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is
As we know that 
= 0.22 V and E = 0.342 V, the equation will become
So, the mean activity coefficient is 0.78.
To solve this we assume
that the gas is an ideal gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant temperature and number of moles of the gas
the product of PV is equal to some constant. At another set of condition of
temperature, the constant is still the same. Calculations are as follows:
P1V1 =P2V2
<span>P2 = P1V1/V2</span>
<span>
</span>
<span>The correct answer is the first option. Pressure would increase. This can be seen from the equation above where V2 is indirectly proportional to P2.</span>
