Why isn't hess's law helpful for calculating the heat of reaction involved in the conversion of diamond to graphite?

2 answers:

7 0

Hess' Law of Constant Heat Summation was developed by Germain Henri Hess who lived from 1802 - 1850. This law became important primarily for his thermochemical studies. This law states that the heat released or absorbed in the chemical reaction is unchanged or the same irregardless of how many steps the reactants take in order to convert to products.

This law is not applicable or not helpful in the conversion of diamond to graphite because both are just allotropes of carbon along with C60 (Buckminsterfullerene or buckyball), C540, C70, amorphous carbon, and carbon nanotube.

IceJOKER [234]2 years ago

3 0

Hess's law is applicable when the reactants are different than the products. In a chemical reaction there are rearrangement of atoms such that the chemical property of the reactant is different than that of the product.

Diamond and graphite are purely made up of C atoms with different arrangements of C atoms thus making them allotropes. Hess's law is not applicable for allotropes.

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2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'

Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

$\displaystyle M_1V_1= M_2V_2$

Where M represents molarity and V represents volume.

Let the initial concentration and unknown volume be M₁ and V₁, respectively. Let the final concentration and required volume be M₂ and V₂, respectively. Solve for V₁:

$\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}$

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
$\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}$

Convert this value to mL:
$\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}$

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

8 0

2 years ago

Yasmin's teacher asks her to make a supersaturated saline solution. Her teacher tells her that the solubility of the salt is 360

Aleks [24]

Answer:

She can add 380 g of salt to 1 L of hot water (75 °C) and stir until all the salt dissolves. Then, she can carefully cool the solution to room temperature.

Explanation:

A supersaturated solution contains more salt than it can normally hold at a given temperature.

A saturated solution at 25 °C contains 360 g of salt per litre, and water at 70 °C can hold more salt.

Yasmin can dissolve 380 g of salt in 1 L of water at 70 °C. Then she can carefully cool the solution to 25 °C, and she will have a supersaturated solution.

B and D are wrong. The most salt that will dissolve at 25 °C is 360 g. She will have a saturated solution.

C is wrong. Only 356 g of salt will dissolve at 5 °C, so that's what Yasmin will have in her solution at 25 °C. She will have a dilute solution.

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2 years ago

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777dan777 [17]

Answer:

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