Answer:
Any two metal which are not ductile are iron and zinc ..
it is a molecule* that can be joined with other molecules that are identical to form a polymer*
key words :
a molecule:
a group of atoms bonded together, representing the smallest fundamental unit of a chemical compound that can take part in a chemical reaction.
a polymer:
a substance that has a molecular structure consisting chiefly or entirely of a large number of similar units bonded together
hope this helped, good luck in future studies !
-A
Answer:
Option C = same period.
Explanation:
All these elements belongs to second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.
Electronic configuration of lithium:
Li₃ = [He] 2s¹
Electronic configuration of beryllium:
Be₄ = [He] 2s²
Electronic configuration of boron:
B₅ = [He] 2s² 2p¹
Electronic configuration of carbon:
C₆ = [He] 2s² 2p²
Electronic configuration of nitrogen:
N₇ = [He] 2s² 2p³
Electronic configuration of oxygen:
O₈ = [He] 2s² 2p⁴
Electronic configuration of fluorine:
F₉ = [He] 2s² 2p⁵
Electronic configuration of neon:
Ne₁₀ = [He] 2s² 2p⁶
All these elements present in same period having same electronic shell.
However their families, valance electrons and group are different. Boron have three valance electrons and belongs to group 3A. Carbon belongs to group 4A and have 4 valance electrons. Nitrogen belongs to group 5A and have five valance electrons. Oxygen belongs to group 6A and have six valance electrons. Fluorine belongs to group 7A and have seven valance electrons.
Second orbital can hold 8 electrons
Answer:
At equilibrium, the concentration of
is going to be 0.30M
Explanation:
We first need the reaction.
With the information given we can assume that is:
+
⇄ 2
If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no
nor
present. Immediately,
and
are going to be produced until equilibrium is reached.
By the ICE (initial, change, equilibrium) analysis:
I: [
]=0 ; [
]= 0 ; [
]=0.60M
C: [
]=+x ; [
]= +x ; [
]=-2x
E: [
]=0+x ; [
]= 0+x ; [
]=0.60-2x
Now we can use the constant information:
![K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5Bproducts%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D%7B%5Breactants%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D)
= 
= 
= 




At equilibrium, the concentration of
is going to be 0.30M