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
Answer:
0.24M
Explanation:
The equation for the reaction is given below:
H2SO4 + 2KOH → K2SO4 + 2H2O
From the equation above, we obtained the following information:
nA (mole of acid) = 1
nB (mole of base) = 2
Data obtained from the question include:
Va (volume of the acid) = 12mL
Ca (concentration of the acid) =?
Vb (volume of the base) = 36mL
Cb (concentration of the base) = 0.16 M
The Ca (concentration of the acid) can be obtained as follow:
CaVa/CbVb = nA/nB
Ca x 12 / 0.16 x 36 = 1 /2
Cross multiply to express in linear form as shown below:
Ca x 12 x 2 = 0.16 x 36
Divide both side by 12 x 2
Ca = 0.16 x 36/ 12 x 2
Ca = 0.24M
Therefore, the concentration of the acid is 0.24M
Answer:
Metal (appropriate charge on metal) nonmetal-ide
Explanation:
1) Write the name of transition metal as shown on the Periodic Table.
2) Write the name and charge for the non-metal.
Use the total charge on the non-metal (or polyatomic ion) find the charge on the transition metal.
3) After the name for the metal, write its charge as a Roman Numeral in parentheses. Example: Iron (II) chloride.
Answer:
Lithium is used to treat mania that is part of bipolar disorder (manic-depressive illness). It is also used on a daily basis to reduce the frequency and severity of manic episodes.
A. is the answer i hope this helps
