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Vadim26 [7]
3 years ago
6

Which statement describes the 3d, 4s, and 4p orbitals of Arsenic (As) based on its electronic configuration and position in the

periodic table?
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The 3d orbital is completely filled, and the 4s and 4p orbitals are partially filled.
The 3d, 4s, and 4p orbitals are completely filled.
The 3d, 4s, and 4p orbitals are partially filled.
Chemistry
1 answer:
tekilochka [14]3 years ago
8 0

Answer:

The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

Explanation:

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PLEASE HELP!!!!!!!
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Answer: 1. 2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu

2. 3 moles of CuCl_2 : 2 moles of Al

3. 0.33 moles of CuCl_2 : 0.92 moles of Al

4. CuCl_2 is the limiting reagent and Al is the excess reagent.

5. Theoretical yield of AlCl_3 is 29.3 g

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles

\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles

The balanced chemical equation is:

2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu  

According to stoichiometry :

3 moles of CuCl_2 require = 2 moles of Al

Thus 0.33 moles of CuCl_2 will require=\frac{2}{3}\times 0.33=0.22moles  of Al

Thus CuCl_2 is the limiting reagent as it limits the formation of product and Al is the excess reagent.

As 3 moles of CuCl_2 give = 2 moles of AlCl_3

Thus 0.33 moles of CuCl_2 give =\frac{2}{3}\times 0.33=0.22moles  of AlCl_3

Theoretical yield of AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3

Thus 29.3 g of aluminium chloride is formed.

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