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3 years ago

What is the total number of p electrons in a single phosphorus atom in its ground state?

Chemistry

2 answers:

Law Incorporation [45]3 years ago

7 0

Phosphorus has an electron configuration of [Ne} 3s2 3p3. Phosphorus has 9 electrons in the p orbital.

alekssr [168]3 years ago

7 0

Answer:

9 p-electrons

Explanation:

Phosphorus is a reactive chemical element that belongs to the group 15 and period 3 of the periodic table. It is a p-block element having atomic number 15.

The electronic configuration of phosphorus in the ground state is 1s² 2s² 2p⁶ 3s² 3p³.

Since, 6 electrons are present in the 2p subshell and 3 electrons are present in the 3p subshell.

Therefore, there are total 9 p-electrons in a phosphorus atom present in its ground state.

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Samira has two pens. She wants to know which pen was used to write a note. She labels her pens A and B. She puts a sample of ink

Ann [662]

Answer:

Explanation:

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3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

Given the chemical formulas of the following compounds, name each compound and state the rules you used to determine each name.

Svetlanka [38]

When naming an ionic compound, write the name of the cation, which is the metal first. Then, write the name of the anion, which is the nonmetal. However, you remove the last 2-3 letters and replace suffixes.

1. RbF --> Rubidium Fluoride
Change fluorine to fluoride
2. CuO --> Copper (II) Oxide
Change oxygen to oxide. Oxide has a charge of -2. Since no subscripts are written, it means they have the same opposite charge. So, we use Copper (II).
3. (NH₄)₂C₂O₄ ---> Ammonium Oxalate
NH₄ is ammonia, but we change it to ammonium for polyatomic ions.

5 0

3 years ago

What is the mass of a sample of water that takes 2000 kJ of energy to boil into steam at 373 K. The latent heat of vaporization

zzz [600]

Answer:

$\boxed{\text{889 g}}$

Explanation:

The formula relating the mass m of a sample and the heat q to vaporize it is

q = mL, where L is the latent heat of vaporization.

$\begin{array}{rcl}2000 \times 10^{3} \text{ J} & = & m \times \dfrac{2.25 \times 10^{6} \text{ J}}{\text{1 kg}}\\\\m & = & \dfrac{2000 \times \times 10^{3}\text{ kg}}{2.25 \times 10^{6}}\\ & = & \text{0.889 kg}\\\\ & = & \text{889 g}\\\end{array}\\\text{The mass of water is $\boxed{\textbf{889 g}}$}$

5 0

3 years ago

an amber alert object is placed in 50 mL of water in the water level rises 72 ml. What is the volume of the object?

Ilya [14]

Answer:

72 ml-50 ml =22 ml is the volume of the object

Explanation:

7 0

3 years ago