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GarryVolchara [31]
3 years ago
7

If the electron is more attracted to the nucleus why does that mean the IE is greater

Chemistry
1 answer:
Nostrana [21]3 years ago
8 0

Answer:The bigger the atom the lesser the ability of the atom to hold on to its valence electrons.

Explanation:

The larger the atom, the lesser the tendency for the atom to withhold its valence electrons.

The size of an atom is usually estimated in terms of its atomic radius. Across the period in the periodic table, the atomic radius decreases progressively from left to right. This is due to the progressive increase in the nuclear charge (i.e., the number of protons in the nucleus) without an attendant increase in the number of electron shells.

Similarly, down the group, the atomic radius increases progressively from top to bottom due to the successive number of electrons' shells.

Thus, the larger the atom's radius, the smaller the Ionization energy because of the increasing shielding effect of inner shell electrons on the electron to be removed from the nuclear attraction.

Finally, we can conclude that as the atom becomes larger, the tendency to withhold the electrons on the outer shell decreases due to distance and the shielding/ screening effect.

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Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze
Rashid [163]

Answer:

a. ΔH^0_{rxn} = -108.0\frac{kJ}{mol}

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate ΔH^0_{rxn} , for the gas-phase reaction )  using the Hess's Law.

ΔH^0_{rxn} =  E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}

Given from the question, the table below shows the corresponding  ΔH^0_{t}(kJ/mol) for each compound.

Compound                    H^0_{t}(kJ/mol)

Liquid EO                       -77.4

CH_4_(g_)                            -74.9                

CO_(g_)                              -110.5

If we incorporate our data into the above previous equation; we have:

ΔH^0_{rxn} = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

          =   -108.0 \frac{kJ}{mol}

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

T_{initial} = 93.0°C   &

the  enthalpy of vaporization  (ΔH^0_{vap}) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as =  \frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}

Let's not forget as well, that  ΔH^0_{vap} = \frac{q}{mass}

If we substitute  ΔH^0_{vap}  for  \frac{q}{mass} in the above equation, we have;

specific heat capacity (c) = \frac{deltaH^0_{vap}}{T_{final}-T_{initial}}

Making (T_{final}- T_{initial}) the subject of the formula; we have:

T_{final}- T_{initial}  = \frac{delat H^0_{vap}}{specificheat capacity}

(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}

T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C

         = 227.76°C +93.0°C

          = 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C                

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In what way did Rannan make an error in his scientific process? He should not use a systemic process. He should not use his opin
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