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3 years ago

Why do graphite and diamond appear to be different substances if they are composed of the same atom?

Chemistry

1 answer:

Alekssandra [29.7K]3 years ago

6 0

Answer:

Diamond and graphite are allotropes of carbon.

Explanation:

Diamond has a close packed crystal structure which is responsible for its extreme hardness. In it, each carbon atom is sp³ hybridised and bonded to four other carbon atoms in a tetrahedral arrangement. Diamond has a hardness of 10 on mohs scale with a cubic crystal form.

Note: allotropes of an element have different molecular structure.

Graphite on the other hand is made up of layers of hexagonal structure that are weakly bonded by van-der-waals forces. This layered arrangement explains its softness/slippery feeling and hence it is used as a lubricant. In each layer, each carbon atom is sp² hybridised and bonded to three other carbon atoms in a trigonal planar arrangement.

The presence of $\pi$ electrons in the layers accounts for its ability to conduct electricity.

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What reagent is used to test for starch

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Answer:

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3 years ago

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eimsori [14]

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2 and 4

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3 years ago

When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem

masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

Step 1: Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

Step 2: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

Step 3: Define the formula

Pressure is constant. → the heat evolved from the reaction is equivalent to the enthalpy of reaction.

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for water (1g/mL and 4.184 J/g°C)

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3 years ago