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alekssr [168]
2 years ago
5

How many electrons may be accommodated in the first three energy levels?

Chemistry
2 answers:
viktelen [127]2 years ago
3 0
Each shell can contain only a fixed number of electrons: The first shell can hold up to two electrons, the second shell can hold up to eight (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on. Some points will be nice
ollegr [7]2 years ago
3 0
The first energy level can hold 2 electrons after that all other energy levels hold 8 electrons which means 3 energy levels can hold 18 electrons 2+8+8=18
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Theoretically, what mass of [Co(NH3)4(H2O)2]Cl2 could be produced from 4.00 g of CoCl2•6H2O starting material. If 1.20 g of [Co(
Romashka [77]

<u>Answer:</u> The theoretical yield and percent yield of [Co(NH_3)_4(H_2O)_2]Cl_2 is 3.93 g and 30.53 % respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of CoCl_2.6H_2O = 4.00 g

Molar mass of CoCl_2.6H_2O = 238 g/mol

Putting values in equation 1, we get:

\text{Moles of }CoCl_2.6H_2O=\frac{4.00g}{238g/mol}=0.0168mol

The chemical equation for the reaction of CoCl_2.6H_2O to form  [Co(NH_3)_4(H_2O)_2]Cl_2 follows:

CoCl_2.6H_2O+4NH_3\rightarrow [Co(NH_3)_4(H_2O)_2]Cl_2+4H_2O

By Stoichiometry of the reaction:

1 mole of CoCl_2.6H_2O produces 1 mole of [Co(NH_3)_4(H_2O)_2]Cl_2

So, 0.0168 moles of CoCl_2.6H_2O will produce = \frac{1}{1}\times 0.0168=0.0168mol of [Co(NH_3)_4(H_2O)_2]Cl_2

Now, calculating the mass of [Co(NH_3)_4(H_2O)_2]Cl_2 from equation 1, we get:

Molar mass of [Co(NH_3)_4(H_2O)_2]Cl_2 = 234 g/mol

Moles of [Co(NH_3)_4(H_2O)_2]Cl_2 = 0.0168 moles

Putting values in equation 1, we get:

0.0168mol=\frac{\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2}{234g/mol}\\\\\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2=(0.0168mol\times 234g/mol)=3.93g

To calculate the percentage yield of [Co(NH_3)_4(H_2O)_2]Cl_2, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of [Co(NH_3)_4(H_2O)_2]Cl_2 = 1.20 g

Theoretical yield of [Co(NH_3)_4(H_2O)_2]Cl_2 = 3.93 g

Putting values in above equation, we get:

\%\text{ yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=\frac{1.20g}{3.93g}\times 100\\\\\% \text{yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=30.53\%

Hence, the theoretical yield and percent yield of [Co(NH_3)_4(H_2O)_2]Cl_2 is 3.93 g and 30.53 % respectively

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