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RUDIKE [14]
3 years ago
11

Ultraviolet radiation and radiation of shorter wavelengths can damage biological molecules because they carry enough energy to b

reak bonds within the molecules. A carbon-carbon bond requires 348 kJ/mol to break.What is the longest wavelength of radiation with enough energy to break carbon-carbon bonds?
Chemistry
1 answer:
Kamila [148]3 years ago
5 0

Answer:

344 nm is the longest wavelength of radiation with enough energy to break carbon-carbon bonds.

Explanation:

C-C(g)\rightarrow 2C(g) ,ΔH = 348 kJ/mol

Energy required to break 1 mole of C-C bond = 348 kJ

Energy required to break 1 C-C bond = E

E = \frac{348,000J}{6.022\times 10^{23}}=5.779\times 10^{-19} J

Energy related with the wavelength of light is given by Planck's equation:

E=\frac{hc}{\lambda }

\lambda =\frac{hc}{E}

=\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{5.779\times 10^{-19} J}

\lambda =3.44\times 10^{-7} m = 344 nm

1 m =10^9 nm

344 nm is the longest wavelength of radiation with enough energy to break carbon-carbon bonds.

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The percent composition of calcium is ?
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3 years ago
Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →
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Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = \frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0
3 years ago
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