g n the following three compounds(1,2,3) arrange their relative reactivity towards the reagent CH3Cl / AlCl3. Justify your order

Answers

Answer 1

Answer:

3 > 2> 1

Explanation:

Aromatic compounds undergo electrophilic substitution reaction which passes through a positively charged intermediate to yield the product.

Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene molecule depending on the nature of the substituent.

Certain substituents increase the ease of reaction of benzene towards aromatic substitution.

If we look at the compounds closely, we will notice that toluene reacts readily with CH3Cl / AlCl3. This is because, the methyl group is electron donating hence it stabilizes the positively charged intermediate produced in the reaction.

Carbonyl compounds are electron withdrawing substituents hence they decrease the magnitude of the positive charge and hence decrease the rate of electrophilic aromatic substitution.

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The _______ are found on the left side of the arrow in a chemical reaction. A. coefficients B. reactants C. products D. subscripts

Answers

B reactants then C is on the right side! If you take h2O and air (the reactants) then the product would be a snowflake. happy holidays!!

Answer:

Reactants

Explanation:

Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V

An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:

Ni(s)VO2+(aq,0.083M)+2H+(aq,1.1M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l)

Calculate the cell potential under these nonstandard concentrations.

Express the cell potential to two decimal places and include the appropriate units.

Answers

Answer:

Cell potential under non standard concentration is 4.09 v

Explanation:

To solve this problem we need to use Nernst Equation because concentrations of the components of the chemical reactions are differents to 1 M (normal conditions: 1 M , 1 atm).

Nernst equation at 25ºC is:

$E = E^(0) - [((0.0592)/(n) · log Q)]$

where

E: Cell potential (non standard conditions)

$E^(0)$ = Cell potential (standard conditions)

n: Number of electrons transfered in the redox reaction

Q: Reaction coefficient (we are going to get it from the balanced chemical reaction)

For example, consider the following general chemical reaction:

aA + bB --> cC + dD

where

a, b, c, d: coefficient of balanced chemical reaction

A,B,C,D: chemical compounds in the reaction.

Using the previous general reaction, expression of Q is:

$Q = (C^(c) * D^(d) )/(A^(a)*B^(b) )$

Previous information is basic to solve this problem. Let´s see the Nernst equation, we need to know: $E^(0)$ , n and Q

Let´s calcule potential in nomal conditions ( $E^(0)$ ):

1. We need to know half-reactions (oxidation and reduction), we take them from the chemical reaction given in this exercise:

Half-reactions: Eo (v):

$(VO_(2))^(2+)$ + e- --> $(VO_(2))^(+)$ -0.23

$Ni --> Ni^(2+)$ + 2 e- +0.99

Balancing each half-reaction, first we are going to balance mass and then we will balance charge in each half-reaction and then charge between half-reactions:

Half-reactions: Eo (v):

2 * [ $(VO_(2))^(2+)$ + e- --> $(VO_(2))^(+)$ ] -0.23

1 * [ $Ni --> Ni^(2+)$ + 2 e- ] +0.99

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

2 $(VO_(2))^(2+)$ + 2e- + Ni --> 2 $(VO_(2))^(+)$ + $Ni^(2+)$ + 2e- 0.76 v

Then global balanced chemical reaction is:

2 $(VO_(2))^(2+)$ + Ni --> 2 $(VO_(2))^(+)$ + $Ni^(2+)$

and the potential in nomal conditions is:

$E^(0)$ = 0.76 v

Also from the balanced reaction, we got number of electons transfered:

n = 2

2. Calculate Q:

Now using previous information, we can establish Q expression and we can calculate its value:

$Q = ([(VO_(2)+]^(2)* [Ni^(2+) )/([VO_(2+)]^(2) * Ni )]$

From the exercise we know:

$[VO_(2) ^(2+)] = 2.5 M$

$[VO_(2)+] = 0.083 M$

$[Ni^(2+)] = 2.5 M$

$[Ni] = 1 M, it is a pure solid, so its activity in Q is unit (1). It is also applied for pure liquids.$

$Q = ((2.5)^(2)* 2.5 )/((0.083)^(2) * 1) = 2,268.11$

3. Use Nernst equation:

Finally, we replace all these results in the Nernst equation:

$E = E^(0) - ((0.0592)/(n) - log Q)\n \nE = 0.76 - ((0.0592)/(2)-log (2,268.11) \nE = 0.76 - (0.0296 - 3.36)\nE = 4.09 v$

Cell potential under non standard concentration is 4.09 v

Final answer:

To calculate the cell potential under nonstandard conditions, we need to apply the Nernst Equation. This involves finding the reaction quotient (Q) from the given concentrations and then subtracting a value derived from Q and the number of electrons transferred, from the cell potential under standard conditions.

Explanation:

For calculating the cell potential under nonstandard conditions for an electrochemical cell, we need to use the Nernst equation. In this case, the Nernst Equation is Ecell = E∘cell - (0.0592/n) * logQ, where Q, the reaction quotient, is the ratio of the concentrations of the products to the reactants raised to their stoichiometric coefficients.

Given the half-cell reduction potentials, we can calculate the cell potential under standard conditions (E°cell) by subtracting the potential of the anode from the potential of the cathode (E°cell = Ecathode - Eanode = 0.99V - (-0.23V), resulting in E°cell = 1.22V.

Next, Q = [Ni2+]/([VO2+]×[H+]²), substituting the given concentrations, Q = (2.5)/(0.083×1.1²).

After calculating Q, we substitute all known values into the Nernst Equation and solve for Ecell. Hence, the cell potential under these nonstandard conditions can be calculated.

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Consider the neutralization reaction 2HNO3(aq) + Ba(OH)2 ( aq ) ⟶ 2H2O ( l ) + Ba ( NO3)2 ( aq ). A 0.125 L sample of an unknown HNO 3 solution required 32.3 mL of 0.200 M Ba ( OH ) 2 for complete neutralization. What is the concentration of the HNO 3 solution?

Answers

Answer:

The concentration of the HNO3 solution is 0.103 M

Explanation:

Step 1: Data given

Volume of the unknow HNO3 sample = 0.125 L

Volume of 0.200 M Ba(OH)2 = 32.3 mL = 0.0323 L

Step 2: The balanced equation

2HNO3(aq) + Ba(OH)2 ( aq ) ⟶ 2H2O ( l ) + Ba( NO3)2 (aq)

Step 3:

n2*C1*V1 = n1*C2*V2

⇒ n2 = the number of moles of Ba(OH)2 = 1

⇒ C1 = the concentration of HNO3 = TO BE DETERMINED

⇒ V1 = the volume of the HNO3 solution = 0.125 L

⇒ n1 = the number of moles of HNO3 = 2

⇒ C2 = the concentration of Ba(OH)2 = 0.200 M

⇒ V2 = the volume of Ba(OH)2 = 0.0323 L

1*C1 * 0.125 L = 2*0.200M * 0.0323 L

C1 = (2*0.200*0.0323)/0.125

C1 = 0.103 M

The concentration of the HNO3 solution is 0.103 M

Of the following elements, which one has the lowest first ionization energy?boron carbon aluminum silicon

Answers

Answer:

Boron

Explanation:

Because it has a complete 2s orbital and therefore, an increased shielding of the 2s orbital will reduce the ionisation energy.

Final answer:

Among boron, carbon, aluminum, and silicon, aluminum has the lowest first ionization energy due to its position on the periodic table, which is further to the left and in a higher period than the other elements.

Explanation:

Ionization energy refers to the energy necessary to remove an electron from an atom in its gaseous state. The element with the lowest first ionization energy among boron, carbon, aluminum, and silicon is aluminum. Ionization energy increases from left to right across a period in the periodic table and from bottom to top in a group. Thus, aluminum, being to the left of boron, carbon, and silicon, has the lowest first ionization energy. Furthermore, aluminum is in the third period, which is below boron and carbon's second period.

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A voltaic cell with an aqueous electrolyte is based on the reaction between Cd2 (aq) and Mg(s), producing Cd(s) and Mg2 (aq). Write half-reactions for the anode and cathode and then write a balanced cell reaction. Please include the states of matter in the equations.

Answers

Answer:

Cd²⁺(aq) + 2 e⁻ → Cd(s)

Mg(s) → Mg²⁺(aq) + 2 e⁻

Cd²⁺(aq) + Mg(s) → Cd(s) + Mg²⁺(aq)

Explanation:

A voltaic cell is composed of two half-reactions:

Reduction (cathode): Cd²⁺(aq) + 2 e⁻ → Cd(s)

Oxidation (anode): Mg(s) → Mg²⁺(aq) + 2 e⁻

Cd²⁺ reduces to Cd (the oxidation number decreases from 2 to 0), whereas Mg oxidizes to Mg²⁺ (the oxidation number increases from 0 to 2). We can get the overall cell reaction by adding both half-reactions:

Cd²⁺(aq) + 2 e⁻ + Mg(s) → Cd(s) + Mg²⁺(aq) + 2 e⁻

Cd²⁺(aq) + Mg(s) → Cd(s) + Mg²⁺(aq)

What are four properties of water?

Answers

Answer:

Polarity

Cohesion

Adhesion

High Specific Heat

Explanation:

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