Answer:
- <u>Option b. Atom P has an estimated Zeff of 7 and is therefore to the right of Atom Q, which has a Zeff of 6.</u>
Explanation:
Please, find attached the figures of both atom Q and atom P corresponding to this question.
The <u>features of atom Q are</u>:
- Each <em>black sphere</em> represents an electron
- In total this atom has 8 electrons: 2 in the inner shell and 6 in the outermost shell.
- Since it is assumed that the atom is neutral, it has 8 protons: one positive charge of a proton balances one negative charge of an electron. Thus, the atomic number of this atom is 8.
- Since only two shells are ocuppied, you can assert that the atom belongs to the period 2 (which is confirmed looking into a periodic table with the atomic number 8).
- <em>Zeff </em>is the effective nuclear charge of the atom. It accounts for the net positive charge the valence electrons experience. And may, in a very roughly way, be estimated as the number of protons less the number of electrons in the inner shells. Thus, for this atom, an estimated Z eff = 8 - 2 = 6.
The <u>features of atom P</u> are:
- Again, each black sphere represents an electron
- In total this atom has 9 electrons: 2 in the inner shell and 7 in the outermost shell.
- Since it is assumed that the atom is neutral, it has 9 protons.
- The atomic number of this atom is 9.
- Using the same reasoning used for atom Q, this atom is also in the period 2.
- Estimated Z eff = 9 - 2 = 7.
Then, since atom P has a greater Z eff than atom Q (an estimated Zeff of 7 for atom P against an estimated Z eff of 6 for atom Q), and both atoms are in the same period, you can affirm that <em>atom P</em> has a greater atomic number and<em> is therefore to the right of atom Q</em>.
It would be a positive charge because it lost two electrons, if the charge was neutral it would be the same amount of protons and electrons, if the charge was negative the electrons would be 20 instead of 18. So in this case it is positive.
Answer: the answer is D heat and pressure
Explanation: I just took the test.
Answer:
calcium ca
Explanation:
We can see here that the only element that is on the same group (column) as Ba (Barium) is Calcium (Ca).
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