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skelet666 [1.2K]
3 years ago
6

State hess's law.how is it used

Chemistry
2 answers:
Vesnalui [34]3 years ago
5 0
 <span>Hess' Law states that the enthalpy change in a reaction can be calculated from the enthalpy changes of reactions that, when combined, result in the desired reaction. 

For example, to check the enthalpy change that occurs when benzene undergoes incomplete combustion to water and carbon monoxide is not an easy task, because the products invariably contain CO2. However, by combining the reactions of the complete combustion of benzene and the combustion of CO, you can get the reaction you want. 

Reaction wanted: 2C6H6 + 9O2 → 12CO + 6H2O 
Reactions provided: 2C6H6 + 15O2 → 12CO2 + 6H2O and 2CO + O2 → 2CO2, and their associated ΔH. 

Rearrange the reactions so that, when they add up, they result in the wanted reaction. 

2C6H6 + 15O2 → 12CO2 + 6H2O (leave as is; no changes to ΔH) 
12CO2 → 12CO + 6O2 (reverse and multiply by 6; this changes the sign of ΔH and multiplies it by 6) 

Added up, it will result in 2C6H6 + 9O2 → 12CO + 6H2O. Add up the ΔH values for the rearranged reactions to find ΔH for this particular reaction.</span>
PSYCHO15rus [73]3 years ago
5 0

Answer: Hess's Law says the total enthalpy change does not rely on the path taken from beginning to end. Enthalpy can be calculated in one grand step or multiple smaller steps.

To solve this type of problem, organize the given chemical reactions where the total effect yields the reaction needed. There are a few rules that you must follow when manipulating a reaction.

The reaction can be reversed. This will change the sign of ΔHf.

The reaction can be multiplied by a constant. The value of ΔHf must be multiplied by the same constant.

Any combination of the first two rules may be used.

Finding a correct path is different for each Hess's Law problem and may require some trial and error.

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True or false<br><br> The central core of an atom is called the hub.
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What is the molarity of a 4.447 L solution
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Answer:52M

Explanation:

use S= 1000w/ MV

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4 0
3 years ago
Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in air to be
amid [387]

Answer:

X(O₂) = 0.323

X(N₂) = 0.677

Explanation:

We have the partial pressures of oxygen (O₂) and nitrogen (N₂):

P(O₂) = 0.20 atm

P(N₂) = 0.80 atm

In order to solve the problem, you need the solubilities of each gas in water at 298 K. We can consider 1.3 x 10⁻³ mol/(L atm) for oxygen (O₂) and 6.8 x 10⁻⁴mol/(L atm) for nitrogen (N₂) from the bibliography.

s(O₂) = 1.3 x 10⁻³ mol/(L atm)

s(N₂) = 6.8 x 10⁻⁴mol/(L atm)

So, we calculate the concentration (C) of each gas as the product of its partial pressure (P) and the solubility (s):

C(O₂) = P(O₂) x s(O₂) = 0.20 atm x 1.3 x 10⁻³ mol/(L atm) = 2.6 x 10⁻⁴mol/L

C(N₂) = P(N₂) x s(N₂) = 0.80 atm x 6.8 x 10⁻⁴mol/(L atm) = 5.44 x 10⁻⁴ mol/L

In 1 liter of water, we have the following number of moles (n):

n(O₂) = 2.6 x 10⁻⁴ mol

n(N₂) = 5.44 x 10⁻⁴ mol

Thus, the total number of moles (nt) is calculated as the sum of the number of moles of the gases in the mixture:

nt = n(O₂) + n(N₂) = 2.6 x 10⁻⁴ mol + 5.44 x 10⁻⁴ mol = 8.04 x 10⁻⁴ mol

Finally, the mole fraction of each gas is calculated as the ratio between the number of moles of each gas and the total number of moles:

X(O₂) = n(O₂)/nt = 2.6 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.323

X(N₂) = n(N₂)/nt = 5.44 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.677

5 0
3 years ago
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