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Anon25 [30]
2 years ago
15

It takes 839./kJmol to break a carbon-carbon triple bond. Calculate the maximum wavelength of light for which a carbon-carbon tr

iple bond could be broken by absorbing a single photon.
Round your answer to 3 significant digits in nm.
Chemistry
1 answer:
tresset_1 [31]2 years ago
6 0

Answer:

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

Explanation:

It takes 839 kJ/mol to break a carbon-carbon triple bond.

Energy required to break 1 mole of carbon-carbon triple bond = E = 839 kJ

E = 839 kJ/mol = 839,000 J/mol

Energy required to break 1 carbon-carbon triple bond = E'

E'=\frac{ 839,000 J/mol}{N_A}=\frac{839,000 J}{6.022\times 10^{23} mol^{-1}}=1.393\times 10^{-18} J

The energy require to single carbon-carbon triple bond will corresponds to wavelength which is required to break the bond.

E'=\frac{hc}{\lambda } (Using planks equation)

\lambda =\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{1.393\times 10^{-18} J}

\lambda =1.427\times 10^{-7} m =142.7 nm = 143 nm

(1 m = 10^9 nm)

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

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2 years ago
Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −
kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)  ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca)  = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207  KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH  of the respective molecules given above,

ΔH  = -1270 + (-393.5) - (-1207)

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3 0
2 years ago
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<u>Answer:</u> The above reaction is non-spontaneous.

<u>Explanation:</u>

For the given chemical reaction:

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Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

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Relationship between standard Gibbs free energy and standard electrode potential follows:

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2 years ago
Calculate the number of grams of solute in 500.0 mL of 0.189 M KOH.
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Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

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Volume of solution = 500 ml

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Formula used :

Molarity=\frac{\text{Mass of KOH}\times 1000}{\text{Molar mass of KOH}\times \text{Volume of solution in ml}}

Now put all the given values in this formula, we get the mass of solute KOH.

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Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

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