Answer:
So sorry if I was wrong but I think it's B. Because from the source states,
https://socratic.org/questions/which-group-on-the-periodic-table-is-the-least-reactive-why
"The least reactive elements are those who have a full outermost valence shell ie they have 8 electrons in the outer shell so elements such as helium, neon, radon or the transition elements."
Answer:

Explanation:
The unbalanced nuclear equation is

Let's write the question mark as a nuclear symbol.

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.
Then
60 = 0 + A, so A = 60 - 0 = 60, and
27 = -1 + Z, so Z = 27 + 1 = 28
Your nuclear equation becomes

Element 28 is nickel, so the balanced nuclear equation is

1) You need to use the atomic mass of copper.
You can find it in a periodic table. It is 63.546 amu.
2) The atomic mass is the weigthed mass of the different isotopes.
This is, the atomic mass of one element is the atomic mass of each isotope times its corresponding abundance:
=> atomic mass of the element = abundance isotope 1 * atomic mass isotope 1 + abundance isotope 2 * atomic mass isotope 2 + ....+abundance isotope n * atomic mass isotope n.
3) The statement tells there are two isotopes so the abundance of one is x and the abundance of the other is 1 - x
=> 63.546 amu = x * 62.9296 amu + (1-x)*64.9278
=> 63.546 = 62.9296x + 64.9278 - 64.9278x
=> 64.9278x - 62.9296 = 64.9278 - 63.546
=> 1.9982x = 1.3818
=> x = 1.3818 / 1.9982 = 0.6915 = 69.15%
=> 1 - x = 1 - 0.6915 = 0.3085 = 30.85%
Answer:
Cu-63 69.15%;
Cu-65 : 30.85%
Answer:
Answer A
Explanation:
atmosphere is made up of approx 78% N and 21% O2
Answer:
Work done by the system = 4545 J
Explanation:
The expression for the calculation of work done is shown below as:
Where, P is the pressure
is the change in volume
From the question,
= 45 - 15 L = 30 L
P = 1.5 atm
Also, 1 atmL = 101 J
So,
(negative sign implies work is done by the system)
<u>Work done by the system = 4545 J</u>