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motikmotik
3 years ago
9

1. As Z stays constant and the number of electrons increases, the electron-electron repulsions _____, and the anion becomes larg

er.
2. The reverse is true for the cation, which becomes ____ than the neutral atom.
Chemistry
2 answers:
Ray Of Light [21]3 years ago
3 0

Answer:

1) increases

2) smaller

Explanation:

Generally, as electron- electron repulsion increases and more electrons are added to the atom while Z is held constant, the electron cloud size is increased. The size of the anion formed is usually measured as the size of this extended electron cloud. Hence the larger electron cloud means a larger anion size compared to the size of the neutral atom.

For a cation, the converse is true and the cation is found to be smaller than the neutral atom.

PilotLPTM [1.2K]3 years ago
3 0

This question is incomplete, I got the complete one from google as below:

I−>I>I+

1. As Z stays constant and the number of electrons increases, the electron-electron repulsions ______ (increases or decreases), and the anion becomes larger.

2. The reverse is true for the cation, which becomes ____ (smaller or larger) than the neutral atom.

Answer:

1. As Z stays constant and the number of electrons increases, the electron-electron repulsion increases, and the anion becomes larger.

2.The reverse is true for the cation, which becomes smaller than the neutral atom.

Explanation:

1. As Z stays constant and the number of electrons increases, the electron-electron repulsion increases, and the anion becomes larger. This is because in anions of the same atoms, the net force of attraction on electrons decreases.

2. The reverse is true for the cation, which becomes smaller than the neutral atom. This is because in cations of the same atoms, the net force of attraction on electrons increases.

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4 0
3 years ago
Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing
Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of AgNO_3 = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of AgNO_3 = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and AgNO_3.

\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles

\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag

From the balanced reaction we conclude that

As, 2 mole of AgNO_3 react with 1 mole of Zn

So, 0.1479 moles of AgNO_3 react with \frac{0.1479}{2}=0.07395 moles of Zn

From this we conclude that, Zn is an excess reagent because the given moles are greater than the required moles and AgNO_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of Ag

From the reaction, we conclude that

As, 2 mole of AgNO_3 react to give 2 mole of Ag

So, 0.1479 moles of AgNO_3 react to give 0.1479 moles of Ag

Now we have to calculate the mass of Ag

\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag

\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g

Theoretical yield of Ag = 15.95 g

Experimental yield of Ag = 13.32 g

Now we have to calculate the percent yield.

\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100

\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%

Therefore, the percent yield is, 83.51 %

7 0
3 years ago
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