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laila [671]
3 years ago
8

If an element contains 8 electrons how many electrons will be placed in the 2nd valence shell?

Chemistry
2 answers:
blondinia [14]3 years ago
4 0

Explanation:

As it is given that the element contains a total of 8 electrons. So, electronic distribution of this element will be 2, 6.

That is, there will be two electrons present in first shell and there will be six electrons present in second valence shell.

Here, we also get to know that this element contains 6 valence electrons hence, it is a group 16 element.

Thus, we can conclude that if an element contains 8 electrons then 6 electrons will be placed in the 2nd valence shell.

RideAnS [48]3 years ago
3 0
If an element contains 8 electrons, then there would be 6 electrons that will be placed in the 2nd valence shell. Each shell in an atom can only take up a fixed number of electrons. For the first shell, only two electrons can be found. For the second shell, it can hold up to 8 electrons. However, for this case only six electrons can be found since the others are found in the first shell.
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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 %

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Which title is most appropriate for this table?
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