How many total electrons does the P^3- ion have? O a. 3 O b. 31 O c.1 O d. 15 O e. 18

Chemistry

1 answer:

STALIN [3.7K]2 years ago

7 0

Answer:

e. 18

Explanation:

A neutral P atom has an atomic number of 15, which means there are 15 protons in the atom. In order to be neutral, the P atom must also have 15 electrons.

The P³⁻ anion has 3 electrons more than the neutral P atom since it has a charge of -3.

Thus, the total number of electrons are 15 + 3 = 18 electrons.

You might be interested in

You are given a solution of hcooh (formic acid) with an approximate concentration of 0.20 m and you will titrate this with a 0.1

balu736 [363]

V(NaOH)*M(NaOH)=V(HCOOH)M(HCOOH)
35.43ml*0.1150m=20.00ml*M(HCOOH)
M(HCOOH)=0.2037m

6 0

2 years ago

Read 2 more answers

What are the most common elements in the canyon

katen-ka-za [31]

The most common element found in crayons would be carbon and hydrogen. Since crayons are made up of paraffin and other chemical pigments. Paraffin is wax like material, which is derived from petroleum, wood, or coal. All three contain large amounts of hydrocarbons, which is carbon and hydrogen compounds.

4 0

2 years ago

90 POINTSSSS!!!

Katarina [22]

Answer:

$\large \boxed{\text{-2043.96 kJ/mol}}$

Explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn: C₃H₈(g) + O₂(g) ⟶ CO₂(g) + H₂O(g)

Balanced eqn: C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

(a) Table of enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}$

(b)Total enthalpies of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\= \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}$