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3 years ago

Which statement correctly describes diamond and graphite, which are different forms of solid carbon?

Chemistry

2 answers:

alukav5142 [94]3 years ago

8 0

Answer: Option (3) is the correct answer.

Explanation:

When an element exists in two or more different physical forms then they are known as allotropes.

For example, diamond and graphite are both allotropes of carbon.

In graphite, carbon atoms are joined together in sheets of six sided lattice. Whereas in diamond, carbon atoms are joined together in four cornered lattice.

Therefore, as these allotropes are made up of same element so they have similar chemical properties but different physical properties like melting point, boiling point etc.

Also, both of them have different molecular structure.

Thus, we can conclude that the statement the differ in their molecular structure and properties correctly describes diamond and graphite.

mart [117]3 years ago

5 0

They differ in their molecular structures and properties.

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2 years ago

The equilibrium constant for the dissolution of silver chloride (AgCl(s) Ag+(aq) + Cl–(aq)) has a value of 1.79 × 10–10. Which s

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Option: b

Explanation:

As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant $\left(\boldsymbol{K}_{s p}\right)$ for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures. Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.

To calculate $\left(\boldsymbol{K}_{s p}\right)$ firstly molarity of ions are needed to be found with formula: $\text { Molarity of ions }=\frac{\text { number of moles of solute }}{\text { Volume of solution in litres }}$

Then at equilibrium cations and anions concentration is considered same hence:

$\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }$

Hence from above data $\left(\boldsymbol{K}_{s p}\right)$ can be calculated by: $\left(\boldsymbol{K}_{s p}\right)$ = $\left[\mathbf{A} \mathbf{g}^{+}\right] \cdot[\mathbf{C} \mathbf{I}]$

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3 years ago