Hello!
Your answer would be polar covalent.
Covalent bonds are where two atoms come together, and share electrons between each other, and are therefore, bonded.
In some cases of molecules that are bonded with a covalent bond, one of the atoms is more, you could call it selfish, and takes more of the electrons. A prime example of this is H20, or water. One of the atoms takes the electrons for longer, and therefore has a more negative charge because electrons are counted as negative charges.
This bond where an atom "hogs" electrons, is called a polar covalent bond, respective to the changing charges for the atoms.
So your answer is d.
Hope this helped!
The ions of Noble gases, <em>group VIII</em> elements have a full octet configuration on their outermost shell and as such are highly stable.
The periodic table is a systematic arrangement of elements in order of their atomic numbers into a set of 8 columns each called groups and a set of 7 rows each called a period.
Elements are arranged in different groups according to the number of Valence electrons they have.
- For instance, elements in the group I of the periodic table are highly electropositive and as such are highly reactive.
The same is evident in group 7 elements are highly electronegative and have high electron affinity and as such are unstable and reactive.
- However, Noble gases, <em>group VIII</em> elements have a full octet configuration on their outermost shell and as such are highly stable.
Consequently, the <em>Noble gases ion</em> has a stable Valence electron configuration.
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The one in the middle since when ur going down it’s more quicker anc faster .
Answer:
1.62
Explanation:
From the given information:
number of moles of benzamide 
= 0.58 mole
The molality = 
= 0.6837
Using the formula:
where;
dT = freezing point = 27
l = Van't Hoff factor = 1
kf = freezing constant of the solvent
∴
2.7 °C = 1 × kf × 0.6837 m
kf = 2.7 °C/ 0.6837m
kf = 3.949 °C/m
number of moles of NH4Cl = 
= 1.316 mol
The molality = 
= 1.5484
Thus;
the above kf value is used in determining the Van't Hoff factor for NH4Cl
i.e.
9.9 = l × 3.949 × 1.5484 m
l = 1.62
