An experimental design is used to assign variables for testing. In contrast to a control design where nothing is changed, the experimental design allows you to test various new inputs to see how they would vary from the original results.
Answer:
In chemical compounds, atoms tends to have the electron configuration of a noble gas.
Explanation:
The noble gases are unreactive because of their electron configurations. This noble gas neon has the electron configuration of 1s22s22p6 . It has a full outer shell and cannot incorporate any more electrons into the valence shell.
The octet rule states that atoms tend to form compounds in ways that give them eight valence electrons and thus the electron configuration of a noble gas. An exception to an octet of electrons is in the case of the first noble gas, helium, which only has two valence electrons.
Answer:
-320 μJ.
Explanation:
Consider a point with an electrical charge of
. Assume that
is the electrical potential at the position of that charge. The electrical potential of that point charge will be equal to:
.
Keep in mind that since both
and
might not be positive, the size of the electrical potential energy might not be positive, either.
For this point charge,
; (that's -8.0 microjoules, which equals to
)
.
Hence its electrical potential energy:
.
Why is this value negative? The electrical potential energy of a charge is equal to the work needed to bring that charge from infinitely far away all the way to its current position. Also, negative charges are attracted towards regions of high electrical potential. Bringing this
negative charge to the origin will not require any external work. Instead, this process will release 320 μJ of energy. As a result, the electrical potential energy is a negative value.
Answer:
The final temperature of both objects is 400 K
Explanation:
The quantity of heat transferred per unit mass is given by;
Q = cΔT
where;
c is the specific heat capacity
ΔT is the change in temperature
The heat transferred by the object A per unit mass is given by;
Q(A) = caΔT
where;
ca is the specific heat capacity of object A
The heat transferred by the object B per unit mass is given by;
Q(B) = cbΔT
where;
cb is the specific heat capacity of object B
The heat lost by object B is equal to heat gained by object A
Q(A) = -Q(B)
But heat capacity of object B is twice that of object A
The final temperature of the two objects is given by

But heat capacity of object B is twice that of object A

Therefore, the final temperature of both objects is 400 K.