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blsea [12.9K]
3 years ago
15

How much energy is released when 0.40 mol C6H6(g) completely reacts with oxygen?

Chemistry
2 answers:
podryga [215]3 years ago
6 0

Answer:

The right answer is -1268 kJ

Explanation:

Vsevolod [243]3 years ago
4 0
 <span>This question asksyou to apply Hess's law. 
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C. 
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3 
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12 
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation. 
Then, 
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ 
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ 
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ 
______________________________________... 
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ 
I hope this helps and my answer is right.</span>
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Answer:1.

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3 years ago
Which statement is false? a) When reactants become products, they do so through an intermediate transition state. b) At equilibr
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Answer:

b) At equilibrium, equal amounts of products and reactants are present.

Explanation:

At equilibrium , the ratio of product of concentration of products  and product of concentration of  reactants is constant .

                                             A + B ⇄ C + D

                                 [C] [ D]  / [ A ] [ B ] = Constant

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3 years ago
The value of delta for the [C_rF_6]^3- complex is 182 kJ/mol. Calculate the expected wavelength of the absorption corresponding
kirza4 [7]

Answer:  Yes the absorb in the visible range.

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

E=\frac{Nhc}{\lambda}

where,

E = energy of the wave  = 182 kJ/mol  = 182000 J/mol

N = avogadro's number =  6.023\times 10^{23}

h = plank constant = 6.6\times 10^{-34}Js^{-1}

c = speed of light = 3\times 10^8m/s

\lambda = wavelength of the wave = ?

Putting all the values:

182000=\frac{6.023\times 10^{23}\times 6.6\times 10^{-34}\times 3\times 10^8m/s}{\lambda}

\lambda=0.65\times 10^{-6}m=650nm    (1nm=10^{-9}m)

The wavelength range for visible rays is 400 nm to 750 nm, thus the complex absorb in the visible range.

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Answer:

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