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zalisa [80]
2 years ago
6

An atom with high ionization energy will form a positive ion more easily than an atom with low ionization energy. True False

Chemistry
1 answer:
andrew11 [14]2 years ago
8 0

The statement "An atom with high ionization energy will form a positive ion more easily than an atom with low ionization energy" is false.

In this context , we will define ionization energy as the minimum energy required to remove a valence electron from a neutral atom in it's gaseous state.  In a sense the ionization energy is a measure the amount of 'difficulty' of making something an ion. A high ionization energy means that it takes a lot of energy to remove a valence electron from that atom. A low ionization energy means that it is easy to remove a valence electron from the atom. It is known that group 1 elements generally have a low ionization energy.  On the other hand, it is harder for noble gasses and group 7 atoms to loose electrons because they have higher ionization energy.

To form a positive ion, you have to remove an electron. When an electron is removed from an atom, there ion formed has more positive charges than negative charges in it, making it net positive. We have established that atoms with low ionization energy loose elections much more easily. We have also established that atoms with high ionization energy do not loose electrons easily. From this we can gather that the statement is false. An atom with high ionization energy will not form a positive ion more easily that an atom with low ionization energy.

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Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta
shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]

\Delta H=-5314.8kJ

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = \frac{5314.8}{2}\times 1=2657.4kJ

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Explanation:

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Answer: option (1) an electron.


Justification:


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2) The next model of the atom, developed by the scientist Ernest Rutherford, depicted the atom a mostly empty space with a small dense positively charged nucleous and the electrons surrounding it.


3) Then, Niels Bhor came out with the model of electrons in fixed orbits around the nucleous, just like the planets orbit the Sun. So, the path followed by the electrons were orbits.


4) The quantum model of the atom did not place the electrons in fixed orbits around the nucleous but in regions around the nucleous. Those regions were named orbitals. And they are regions were it is most probable to find the electron, since it is not possible to tell the exact position of an electron.


As per this model, the electron has a wave function associated. The scientist Schrodinger developed the wave equation which predicts the location of the electron as a probability.


The orbitals are those regions were it is most likely to find the electron. Those regions are thought as clouds of electrons.

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I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y, for the following reason: 

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Now, you suppose to prepare each of these two compounds, starting with the same fixed mass of element Y ( I will choose 12g of Y for an easy calculation!) 

The first compound will then contain 4g of X and 12g of Y 
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<span>The ratio which combined the masses of X and the fixed mass (12g) of Y
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<span>or 2 : 3 </span>

So, the ratio of MOLES of X which combined with the fixed amount of Y in the two compounds is also = 2 : 3 </span>

The two compounds given with the plausible formula must therefore contain the same ratio.

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