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guajiro [1.7K]
3 years ago
5

Summarize the relationship between an electron's energy level and its location in an atom

Chemistry
1 answer:
FromTheMoon [43]3 years ago
6 0

Answer:

The further an electron is from the nucleus. the greater its energy level.

Explanation:

When an electron is close to the nucleus, it is at as low an energy level as it can get.

We must put energy into an electron to pull it away from the attraction of a nucleus.

So, electrons that are further from the nucleus are at higher energy levels.

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Calculate the standard enthalpy of formation of liquid methanol, CH3OH(l), using the following information: C(graphite) + O2 --&
Darya [45]

Answer : The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation :

Standard formation of reaction : It is a chemical reaction that forms one mole of a substance from its constituent elements in their standard states.

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C(graphite)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.5kJ/mole

(2) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.8kJ/mole

(3) CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-726.4kJ/mole

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, we get :

(1) C(graphite)+O_2(g)\rightarrow CO_2(g)    \Delta H_1=-393.5kJ/mole

(2) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol

(3) CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g)     \Delta H_3=726.4kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

Therefore, the standard enthalpy of formation of methanol is, -238.7 kJ/mole

6 0
3 years ago
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