What is the geometry name for PF5? Is PF5 polar or non-polar?

How many total moles of ions are released when the following sample dissolves completely in water?

pogonyaev

Answer:

8.55 × 10²⁴ Ions

Explanation:

Ammonium Chloride is an ionic compound which contains a monatomic anion (Cl⁻ ; Chloride) and a polyatomic cation (NH₄⁺ ; Ammonium).

Hence, when added in water Ammonium Chloride ionizes as;

NH₄Cl → NH₄⁺ + Cl⁻

Hence, we can say that it produces two ions when dissolved in water.

Also,

We know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of ions contained by 7.1 moles of NH₄Cl, we will use following relation to first calculate the number of molecules as;

Moles = Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Molecules,

Number of Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting values,

Number of Molecules = 7.1 mol × 6.022 × 10²³

Number of Molecules = 4.27 × 10²⁴ Molecules

So,

As,

1 Molecule of NH₄Cl contained = 2 Ions

So,

4.27 × 10²⁴ Molecules of NH₄Cl will contain = X ions

Solving for X,

X = 2 Ions × 4.27 × 10²⁴ Molecules / 1 Molecule

X = 8.55 × 10²⁴ Ions

8 0

3 years ago

Atoms of arsenic (As) are often added to silicon (Si) in a process called doping to change the conductivity of the silicon. How

Basile [38]

Answer:

B.) An atom of arsenic has one more valence electron and more electron shells than an atom of silicon, so the conductivity decreases because the arsenic atom loses the electron.

Explanation:

Silicon is located in the 3rd row and 14th column in the periodic table. Arsenic is located in the 4th row and 15th column in the periodic table. This means that arsenic has one more valence electron than silicon. Since arsenic is located one row down from silicon, its valence electrons occupy higher energy orbitals.

Silicon maintains a crystal-like lattice structure. Each silicon atom is covalently connected to assume this shape. When silicon gains one extra electron from arsenic, it experiences n-type doping. This new electron is not tightly bound in the lattice structure. This allows it to move more freely and conduct more electricity. This can also be explained using band gaps. Silicon, which previously had an empty conduction band, now has one electron in this band. This lowers the band gap between the conduction and valence bands and increases conductivity.

5 0

2 years ago

In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

Answer: The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

Explanation:

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$ .......(1)

Calculating for first dilution:

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

Calculating for second dilution:

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

8 0

3 years ago

Balance the following equation in acidic conditions. Phases are optional. Cu NO3,

Gwar [14]

If your equation is Cu+NO^- 3-->Cu^2+NO, then the answer is
2 Cu + 1 NO3{-} → 1 Cu^{2+} + 3 NO

To check if it is balance, this is the solution:
2- Cu- 2
3- N -3
3- O -3

5 0

3 years ago

Explain why the amino group of p-aminobenzoic acid does not participate in the reaction

lubasha [3.4K]

This is because amino group of p-aminobenzoic acid is an aniline and is less electrophilic than an alkyl amine.

<h3>What is an Aniline?</h3>

This is an aromatic amine which consists of a phenyl group attached to an amino group.

The amino group of p-aminobenzoic acid being an aniline makes it less electrophilic which is why an alkyl amine participates in the reaction instead.

Read more about Aniline here brainly.com/question/9982058

5 0

2 years ago