• Animals such as prairie dogs, can cause physical weathering by digging burrows.
The answer is c. hg (mercury)
Answer:
Burn, able to rust, and sour.
Explanation:
I actually couldn't guess at first what the question is. But looking closely at the statements, I deduced that some are correct and some are not. So, I think this is a true or false problem. So,
<span>The 1s orbital(s) do(es) not have any nodes. - This is false. Nodes are the planes that the orbitals do not fill. The formula for the number of nodes is:
N = n - l
where
n is the energy level
l is 0 for s subshell, 1 for p subshell, 2 for d subshell, 3 for f subshell; l also signifies the number of angular nodes.
Thus,
N = 1 - 0 = 1 node
</span><span>The 1s orbital(s) has(have) a node at the nucleus. Since this is the opposite of the first statement, this is true.
</span><span>The 3d orbital(s) has(have) a cloverleaf shape, with four lobes of electron density around the nucleus and two perpendicular nodal planes.
This is true. The shape of d subshell is cloverleaf, and all have four lobes. Since l=2, there are 2 perpendicular or angular nodes.
</span><span>The f orbitals are even more complex. This is true. The f subshell is the last subshell. It has complex shapes and it rarely comes up in chemistry.
</span><span><em>The number of nodes (and nodal planes) depends on the specific orbital, but there will be more than for s, p, or d orbitals. </em><em />This is false. In fact, f orbitals have more nodes because l = 3. That means they always have 3 angular nodes, which is greater than the other subshells.</span>