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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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What is 17.8 + 12.115 in sig fig
iogann1982 [59]

Answer:

Explanation:

17.8 + 12.115 = 29.9

Sig Figs

3

29.9

Decimals

1

29.9

Scientific Notation

2.99 × 10 1

E-Notation

2.99e+1

Words

twenty-nine point nine

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3 years ago
The sum of the number of protons and neutrons in the nucleus of an atom is called the
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4 years ago
The entropy of a system at 337 K increases by 221.7 J/mol•K. The free energy value is found to be –717.5 kJ/mol. Calculate the c
OlgaM077 [116]

<u>Answer:</u> The change in enthalpy for the given system is -642.8 kJ/mol

<u>Explanation:</u>

To calculate the change in enthalpy for given Gibbs free energy, we use the equation:

\Delta G=\Delta H-T\Delta S

where,

\Delta G = Gibbs free energy = -717.5 kJ/mol = -717500 J/mol    (Conversion factor: 1 kJ = 1000 J)

\Delta H = change in enthalpy = ?

T = temperature = 337 K

\Delta S = change in entropy = 221.7 J/mol.K

Putting values in above equation, we get:

-717500J/mol=\Delta H-(337K\times 221.7J/mol.K)\\\\\Delta H=-642787J/mol=-642.8kJ/mol

Hence, the change in enthalpy for the given system is -642.8 kJ/mol

4 0
4 years ago
True or False: The atmosphere is the collection of gases surrounding the earth trapped by earth's gravity.
Vladimir79 [104]

Answer:

I think its False

Explanation:

ozone layer surroundes the atmosphere and traps the heat

4 0
4 years ago
Will these changes increase, decrease, or have no effect on the mean free path of the molecules in a gas sample?
lisabon 2012 [21]
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l = μ/P * √(πkT/2m), where
l is the mean free path
μ is the viscosity of the fluid
P is the pressure
k is the Boltzmann's constant
T is the absolute temperature
m is the molar mass

So, here are the general effects of the factors on the mean free path:

Mean free path increases when:
1. The fluid is viscous (↑μ)
2. At low pressures (↓P)
3. At high temperatures (↑T)
4. Very light masses (↓m)

The opposite is also true for when the mean free path decreases. Factors that are not found here have little or no effect.
6 0
4 years ago
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