Draw the molecules for the following reaction:

An element with the valence electron configuration 4s1 would form a monatomic ion with a charge of_______ . In order to form thi

qwelly [4]

Answer:

+1, lose, 1, 4s, 4s and 3d

Explanation:

An element with the valence electron configuration 4s¹ would form a monatomic ion with a charge of +1. In order to form this ion, the element will lose 1 electron from the 4s subshell.

The corresponding oxidation reaction is:

K ⇒ K¹⁺ + 1 e⁻

[Ar] 4s¹ ⇒ [Ar]

If an element with the valence configuration 4s² 3d⁶ loses 3 electrons, these electrons would be removed from the 4s and 3d subshell(s).

The corresponding oxidation reaction is:

Fe ⇒ Fe³⁺ + 3 e⁻

[Ar] 4s² 3d⁶ ⇒ [Ar] 4s⁰ 3d⁵

3 0

3 years ago

Refer to the following illustration to answer this Question.

ad-work [718]

Answer:

Option (B) It separates the hydrogen ion (H+) from the chlorine ion (Cl–).

Explanation:

The reaction above simply illustrate the dissociation of HCl when dissolve is water.

When HCl is dissolved in H2O, it will separate into the hydrogen ion, H+ and chloride ion Cl- as shown below

HCl —> H+ + Cl-

The water help to separate the H+ from the Cl-

4 0

3 years ago

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.892 M and [Fe2 ] = 0.0150

Effectus [21]

Given:

Concentration of Cr2+ = 0.892 M

Concentration of Fe2+ = 0.0150 M

To determine:

The cell potential, Ecell

Explanation:

The half cell reactions for the given cell are:

Anode: Oxidation

Cr(s) ↔ Cr2+(aq) + 2e⁻ E⁰ = -0.91 V

Cathode: Reduction

Fe2+ (aq) + 2e⁻ ↔ Fe (s) E⁰ = -0.44 V

------------------------------------------

Net reaction: Cr(s) + Fe2+(aq) ↔ Cr2+(aq) + Fe(s)

E°cell = E°cathode - E°anode = -0.44 - (-0.91) = 0.47 V

The cell potential can be deduced from the Nernst equation as follows:

Ecell = E°cell - (0.0591/n)log[Cr2+]/[Fe2+]

Here, n = number of electrons = 2

Ecell = 0.47 - 0.0591/2 * log[0.892]/[0.0150] = 0.418 V

Ans: The cell potential is 0.418 V

8 0

3 years ago

Read 2 more answers

Metal plating is done by passing current through a metal solution. For example, an item can become gold plated by attaching the

Norma-Jean [14]

Answer:

9.18g

Explanation:

Step 1: Write the reduction half-reaction

Au³⁺(aq) + 3 e⁻ ⇒ Au(s)

Step 2: Calculate the mass of gold is produced when 15.0A of current are passed through a gold solution for 15.0min

We will use the following relationships:

1 min = 60 s

1 A = 1 C/s

1 mole of electrons has a charge of 96486 C (Faraday's constant).

1 mole of Au is produced when 3 moles of electrons circulate.

The molar mass of Au is 196.97 g/mol.

The mass of gold produced is:

$15.0 min \times \frac{60s}{1 min} \times \frac{15.0C}{1s} \times \frac{1 mol e^{-} }{96486C} \times \frac{1molAu}{3 mol e^{-} } \times \frac{196.97gAu}{1molAu} = 9.18gAu$

3 0

3 years ago

The phenolic indicator (In-OH)has approximately the same pKa as a carboxylic acid. Which H is the most acidic proton in In-OH? C

Sidana [21]

The –OH+ group is most acidic proton in ln-OH as shown in figure (a). The proton is circled in the figure.

The stabilisation of the conjugate base produced is stabilises due to resonance factor. The possible resonance structures are shown in figure (b).

The acidity of a protonated molecule depends upon the stabilisation of the conjugate base produced upon deprotonation. The conjugate base of ln-OH is shown in figure (a).

The possible resonance structures are shown in figure (b). As the number of resonance structures of the conjugate base increases the stabilisation increases. Here the unstable quinoid (unstable) form get benzenoid (highly stable) form due to the resonance which make the conjugate base highly stabilise.

Thus the most acidic proton is assigned in ln-OH and the stability of the conjugate base is explained.

4 0

3 years ago