Answer:
Lewis structure in attachment.
Explanation:
Atoms of elements in and beyond the third period of the periodic table form some compounds in which more than eight electrons surround the central atom. In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbitals that can be used in bonding. These orbitals enable an atom to form an <u>expanded octet</u>.
The central Xe atom in the XeF₄ molecule has <u>two</u> unbonded electron pairs and <u>four</u> bonded electron pairs in its valence shell.
The periodic table is arranged by the outermost electrons of the elements and this is the statement among the given statements that explains how the periodic table is arranged. The correct option among all the options that are given in the question is the second option or option "B".
Answer:
Explanation:
Matter is anything that has weight and occupies space.
To prove that a bicycle is a matter, we need to show that it has weight and will occupy space.
When you put the bicycle on a weighing scale, you will see the weight of the bicycle. This a proof that bicycle is a matter.
To show that the bicycle can occupy space, place the device in a tank full of water. From the tank, the bicycle will displace some water. Substances that cannot occupy space will not behave in such manner.
Answer:
It is important to collect all data first, or else your guesses could purely be the opposite of the right answer. If you make inferences of what might happen, your guesses may be purely fictional, and totally off-topic. During experiments, this step is important.
The number of liters of 3.00 M lead (II) iodide : 0.277 L
<h3>Further explanation</h3>
Reaction(balanced)
Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)
moles of KI = 1.66
From the equation, mol ratio of KI : PbI₂ = 2 : 1, so mol PbI₂ :
Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution
Where
M = Molarity
n = Number of moles of solute
V = Volume of solution
So the number of liters(V) of 3.00 M lead (II) iodide-PbI₂ (n=0.83, M=3):
