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Rainbow [258]
2 years ago
8

If you are looking for a one specific electron you will have....% chance of finding it in an area within the atom electron cloud

s?
90%
50%
100%
10%
Chemistry
1 answer:
Vika [28.1K]2 years ago
7 0

If you are looking for a one specific electron you will have 90% chance of finding it in an area within the atom electron clouds.

According to the wave mechanical or electron cloud model of the atom, postulates that the atomic orbital is a region in space where there is a very high probability of finding an electron.

The term "electron cloud" is just another term used for "atomic orbital".

Since electrons reside in orbitals, if you are looking for a one specific electron you will have 90% chance of finding it anywhere within the electron clouds. This would make mathematical sense as the probability of finding an electron within the electron cloud can never be 100%.

Learn more: brainly.com/question/1869903

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Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
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Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



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