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3 years ago

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The oxygen atom of a ketone (such as cyclohexanone) contains 2 lone pairs of electrons. These pairs of electrons most likely res

ide in what type of orbital

1 answer:

skelet666 [1.2K]3 years ago

5 0

Answer: The given pairs of electrons most likely reside in $sp^{2}$ type of orbital.

Explanation:

As it is given that two lone pair of electrons are present on the oxygen atom of ketone (such as cyclohexanone).

Also, there will be one bond pair between carbon and oxygen atom.

Hence, total electrons present in the domain are as follows.

2 lone pairs + 1 bon pair of electron = 3 electron domains

This means that there will be $sp^{2}$ type of orbital present.

Thus, we can conclude that given pairs of electrons most likely reside in $sp^{2}$ type of orbital.

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Which soup would likely be more flavorful, a nicely heated but not too hot to eat bowl of soup or a bowl of leftover soup direct

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The soup that is not too hot to eat would be better

Explanation:

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3 years ago

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You burn a 15g jellybean to warm 50 mL of water, which increases the temperature of the water 25 °C. How many calories of heat a

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Answer:

$Q = 375\,cal$

Explanation:

The quantity of heat transfered from the jellybean to the water is:

$Q = \rho\cdot V \cdot c\cdot \Delta T$

$Q = \left(1\,\frac{g}{cm^{3}}\right)\cdot (15\,cm^{3})\cdot \left(1\,\frac{cal}{g\cdot ^{\circ} C} \right)\cdot (25\,^{\circ}C )$

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3 years ago

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How much energy is needed to completely boil a 5.05g sample of water?

Keith_Richards [23]

Given what we know, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.

<h3>Why does it take this much energy to boil the water?</h3>

We arrive at this number by taking into account the energy needed to boil 1g of water to its vaporization point. This results in the use of 2260 J of heat energy. We then take this number and multiply it by the total grams of water being heated, in this case, 5.05g, which gives us our answer of 11.4 kJ of energy required.

Therefore, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.

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2 years ago

Can someone help me please

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3 years ago

Consider the reaction FeO (S) + CO(g) <-----> Fe(s) + CO2(g) for which KP is found to have the following values:

Svetach [21]

Explanation:

$\Delta G^o=-RT\ln K_1$

where,

R = Gas constant = $8.314J/K mol$

T = temperature = $600^oC=[273.15+600]K=873.15 K$

$K_p$ = equilibrium constant at 600°C = 0.900

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/Kmol)\times 873.15 K\times \ln (0.900 )$

$\Delta G^o=764.85 J/mol$

The ΔG° of the reaction at 764.85 J/mol is 764.85 J/mol.

Equilibrium constant at 600°C = $K_1=0.900$

Equilibrium constant at 1000°C = $K_2=0.396$

$T_1=[273.15+600]K=873.15 K$

$T_2=[273.15+1000]K=1273.15 K$

$\ln \frac{K_2}{K_1}=\frac{\Delta H^o}{R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

$\ln \frac{0.396}{0.900}=\frac{\Delta H^o}{8.314 J/mol K}\times [\frac{1}{873.15 K}-\frac{1}{1273.15 K}]$

$\Delta H^o=-18,969.30 J/mol$

The ΔH° of the reaction at 600 C is -18,969.30 J/mol.

ΔG° = ΔH° - TΔS°

764.85 J/mol = -18,969.30 J/mol - 873.15 K × ΔS°

ΔS° = -22.60 J/K mol

The ΔS° of the reaction at 600 C is -22.60 J/K mol.

$FeO (s) + CO(g)\rightleftharpoons Fe(s) + CO_2(g)$

Partial pressure of carbon dioxide = $p_1=P\times \chi_1$

Partial pressure of carbon monoxide = $p_2=P\times \chi_2$

Where $\chi_1\& \chi_2$ mole fraction of carbon dioxide and carbon monoxide gas.

The expression of $K_p$ is given by:

$K_p=\frac{p_1}{p_2}=\frac{P\times \chi_1}{P\times \chi_2}$

$0.900=\frac{\chi_1}{\chi_2}$

$\chi_1=0.900\times \chi_2$

$\chi_1+\chi_2=1$

$0.9\chi_2+\chi_2=1$

$1.9\chi_2=1$

$\chi_2=\frac{1}{1.9}=0.526$

$\chi_1=1-\chi_2=1-0.526=0.474$

Mole fraction of carbon dioxide at 600°C is 0.474.

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3 years ago