Answer:
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Answer:
4HCl + 2Zn = 2H2 + 2ZnCl2
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Answer:
Carbon atoms in graphite and diamond are arranged in different ways. Hence, the two allotropes of carbon have different physical properties.
Explanation:
Both graphite and diamond are both made of only carbon atoms. However, their physical properties differ from each other. Hence, they are called allotropes. Think about how these carbon atoms are arranged in each of the allotropes.
<h3>Graphite</h3>
In graphite, each carbon atom is bonded to three other carbon atoms. These carbon atoms will be located in the same plane. A chunk of graphite can contain many of these planes.
Each carbon atom has four valence electrons. Three of these electrons will be used in the bonds. The other electron will be delocalized. These electrons would flow between the sheets of carbon atoms. That keeps the sheets separate and allow them to slide on top of each other.
<h3>Diamond</h3>
In diamond, each carbon atom is bonded to four other carbon atoms. These carbon atoms will form a tetrahedral network.
In graphite, there's a significant separation between two adjacent sheets of carbon atoms. The force between the two sheets is rather weak. When a piece of graphite is between two objects that move over one another, the layers in the graphite would also slide over one another. Since the attraction between two adjacent sheets isn't very strong, there wouldn't be much resistance. Hence the graphite acts as a lubricant.
In contrast, most of the carbon atoms in a piece of diamond would be connected to each other. Unlike the sheets in graphite, in a diamond there are almost no moving parts. Also, the forces between neighboring carbon atoms are very strong. When an external force acts on a chunk of diamond, the carbon atoms would barely move. Hence, the structure appears to be very rigid. That gives diamond its abrasive properties.
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Answer : The molarity of 
in the solution is 1.5 M.
Explanation : Given,
Moles of = 3.0 mol
Volume of solution = 2.00 L
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
Now put all the given values in this formula, we get:
Therefore, the molarity of in the solution is 1.5 M.
