Q 13.3: Which of the following is the least stable radical choice? Tertiary radical. B : Allyl radical. C : Secondary radical. D : Methyl radical. E : Primary radical.

Answer:

1. This reaction is: B. Endothermic.

2. When the temperature is increased the equilibrium constant, K: A. Increases.

3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.

Explanation:

Hello,

In this case, considering the images, we can state that the red color at high temperature is due to the presence of nitrogen dioxide (product) and the lower coloring is due to the presence of dinitrogen tetroxide (reactant) at low temperature.

With the aforementioned, we can conclude that the chemical reaction:

[tex]N_2O_4(g) \rightleftharpoons 2 NO_2(g)[/tex]

Is endothermic since high temperatures favor the formation of the product and the low temperatures favor the consumption of the the reactant. thereby:

1. This reaction is: B. Endothermic.

2. When the temperature is increased the equilibrium constant, K: A. Increases. In this particular case, since the dinitrogen tetroxide has 1 molecule and nitrogen dioxide two molecules in the chemical reaction, the entropy change should be positive, therefore, by increasing the T, the Gibbs free energy of reaction becomes more negative:

[tex]G=H-TS[/tex]

As Gibbs free energy becomes more negative, the equilibrium constant becomes bigger given their relationship:

[tex]K=exp(-\frac{\Delta G}{RT} )[/tex]

3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.

Regards.

Answer:Liquid

Explanation:

Answer:

The empirical formulae for the compound is CH3.

Explanation:

a. Esters

b. Carboxylic acids

c. Esters (ethyl hexanoate smells like pineapple)

d. Carboxylic acids (produces a carboxylic salt)

For the given phrases the following description is better.

a. Esters

b. Carboxylic acids

c. Esters (ethyl hexanoate smells like pineapple)

d. Carboxylic acids (produces a carboxylic salt)

An ester is a synthetic compound got from a corrosive in which somewhere around one - OH hydroxyl bunch is supplanted by an - O-alkyl (alkoxy) bunch, as in the replacement response of a carboxylic acid and a liquor.

Carboxylic acid is any of a class of natural mixtures in which a carbon (C) particle is clung to an oxygen (O) molecule by a twofold bond and to a hydroxyl bunch (―OH) by a solitary bond.

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Answer: The number of moles of a solute per kilogram of solvent

Explanation:

Answer:

[tex]\Delta G=-359\frac{kJ}{mol}[/tex]

Explanation:

Hello,

In this case, we must remember that the Gibbs free energy is defined in terms of the enthalpy, temperature and entropy as shown below:

[tex]\Delta G=\Delta H -T\Delta S\\[/tex]

In such a way, for the given data, we obtain it, considering the conversion from J to kJ for the entropy in order to conserve the proper units:

[tex]\Delta G=-292\frac{kJ}{mol} -(298)(224\frac{J}{mol}*\frac{1kJ}{1000J} )\\\\\Delta G=-359\frac{kJ}{mol}[/tex]

Best regards.

Answer:

B- 358 kj

Explanation: I took the test

Answer:

Molar solubility is 1.12x10⁻⁴M

Explanation:

The dissolution of magnesium hydroxide is:

Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻

The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:

Mg(OH)₂(s) ⇄ X + 2X

Where X is solubility.

If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:

2X = 2.24x10⁻⁴M

X = 2.24x10⁻⁴M/2

X =1.12x10⁻⁴M

Answer:

d. Oxidation and reduction

Explanation:

For this question we have to remember the definition of each type of reaction:

-) Hydrogenation

In this reaction, we have the addition of hydrogen to a molecule. Usually, an alkene or alkyne. In the example, molecular hydrogen is added to a double bond to produce an alkane.

-) Alkylation

In this reaction, we have the addition of a chain of carbon to another molecule. In the example, an ethyl group is added to a benzene ring.

-) Hydrolysis

In this reaction, we have the breaking of a bond by the action of water. In the example, a water molecule can break the C-O bond in the ester molecule.

-) Halogenation

In this reaction, we have the addition of a halogen (atoms on the VIIIA group). In the example, "Cl" is added to the butene.

-) Ammoniation

In this reaction, we have the addition of the ammonium ion ([tex]NH_4^+[/tex]). In the example, the ammonium ion is added to an acid.

-) Oxidation and reduction

In this reaction, we have opposite reactions. The oxidation is the loss of electrons and the reduction is the gain of electrons. For example:

[tex]Ag^+~+~e^-~->~Ag[/tex] Reduction

[tex]Al~->~Al^+^3~+~3e^-[/tex] Oxidation

Answer:

1 L

Explanation:

ppm means parts per million. Generally the relationship between mass and litre is given as;

1 ppm = 1 mg/L

This means that 1 ppm is equivalent to 1 mg of a substance dissolved in 1 L of water.

Answer:

226Ra88→222Rn86+4He2

Explanation:

An α-particle usually consists of a helium nucleus which indicates the type of decay that was undergone in this radioactive process.

During α-decay(alpha decay), an atomic nucleus emits an alpha particle.

Answer:

The frequency of the photon is 7.41*10¹⁶ Hz

Explanation:

Planck states that light is made up of photons, whose energy is directly proportional to the frequency of radiation, according to a constant of proportionality, h, which is called Planck's constant. This is expressed by:

E = h*v

where E is the energy, h the Planck constant (whose value is 6.63*10⁻³⁴ J.s) and v the frequency (Hz or s⁻¹).

So the frequency will be:

[tex]v=\frac{E}{h}[/tex]

Being E= 4.91*10⁻¹⁷ J and replacing:

[tex]v=\frac{4.91*10^{-17} J}{6.63*10^{-34} J.s}[/tex]

You can get:

v= 7.41*10¹⁶ [tex]\frac{1}{s}[/tex]= 7.41*10¹⁶ Hz

The frequency of the photon is 7.41*10¹⁶ Hz

Answer:

3

Explanation:

Resonance is a valence bond concept put forward by Linus Pauling to explain the fact that the observed properties of a molecule may be as a result of the fact that its actual structure lie somewhere between a given number of structural extremes called canonical structures or resonance structures.

There are three resonance structures for SO3 that obey the octet rule. All the S-O bonds in SO3 are equivalent in these resonance structures.

Seven equivalent resonance structures for the molecular of SO3 can be drawn without breaking the octet rule.

We can arrive at this answer because:

In an SO3 molecule, electrons are shared between atoms. This sharing can be done with seven resonance structures.

These structures are shown in the figure below.

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Answer:

Ca²⁺ and Cl⁻

Explanation:

In a chemical reaction, spectator ions are ions that are not involved in the reaction, that means are the same before and after the reaction.

In water, calcium hydroxide, Ca(OH)₂ is dissociated in Ca²⁺ and OH⁻. Also, hydrochloric acid, HCl, dissociates in H⁺ and Cl⁻. The reaction is:

Ca²⁺ + 2OH⁻ + 2H⁺ + 2Cl⁻ → 2H₂O + Ca²⁺ + 2Cl⁻

The ions that react are H⁺ and OH⁻ (Acid and base producing water)

And the ions that are not reacting, spectator ions, are:

Answer:

n = 0.207 mole

Explanation:

We have,

P = 1 atm

V = 5 liter

R = 0.0821 L.atm/mol.K

T = 293 K

We need to find the value of n. The relation is as follows :

PV = nRT

Solving for n,

[tex]n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 5}{0.0821 \times 293}\\\\n=0.207\ \text{mol}[/tex]

So, the value of n is 0.207 mol.

Answer:

[tex]r_A=-1\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, given the rate of production of C, we can compute the rate of consumption of A by using the rate relationships which include the stoichiometric coefficients at the denominators (-1 for A and 2 for C) as follows:

[tex]\frac{1}{-1} r_A=\frac{1}{2}r_C[/tex]

In such a way, solving the rate of consumption of A, we obtain:

[tex]r_A=-\frac{1}{2} r_C=-\frac{1}{2}*2\frac{M}{s}\\ \\r_A=-1\frac{M}{s}[/tex]

Clearly, such rate is negative which account for consumption process.

Regards.

Answer:

1. pH = 2.98

2. pH = 4.02

3. pH = 8.12

Explanation:

1. Initial molarity of benzoic acid (Molar mass: 122.12g/mol; Ka = 6.14x10⁻⁵) is:

0.235 ₓ (1mol / 122.12g) = 1.92x10⁻³ moles / 0.100L = 0.01924M

The equilibrium of benzoic acid with water is:

C6H5CO2H(aq) + H2O(l) → C6H5O-(aq) + H3O+(aq)

And Ka is defined as the ratio between equilibrium concentrations of products over reactants, thus:

Ka = 6.14x10⁻⁵ = [C6H5O⁻] [H3O⁺] / [C6H5CO2H]

The benzoic acid will react with water until reach equilibrium. And equilibrium concentrations will be:

[C6H5CO2H] = 0.01924 - X

[C6H5O⁻] = X

[H3O⁺] = X

Replacing in Ka:

6.14x10⁻⁵ = [X] [X] / [0.01924 - X]

1.1815x10⁻⁶ - 6.14x10⁻⁵X = X²

1.1815x10⁻⁶ - 6.14x10⁻⁵X - X² = 0

Solving for X:

X = -0.0010→ False solution. There is no negative concentrations

X = 0.0010567M → Right solution.

pH = - log [H3O⁺] and as [H3O⁺] = X:

pH = - log [0.0010567M]

2.

pH of a buffer is determined using H-H equation (For benzoic acid:

pH = pka + log [C6H5O⁻] / [C6H5OH]

pKa = -log Ka = 4.21 and [] could be understood as moles of each chemical

The benzoic acid reacts with NaOH as follows:

C6H5OH + NaOH → C6H5O⁻ + Na⁺ + H₂O

That means NaOH added = Moles C6H5O⁻ And C6H5OH = Initial moles (1.92x10⁻³ moles - Moles NaOH added)

7.00mL of NaOH 0.108M are:

7x10⁻³L ₓ (0.108 mol / L) = 7.56x10⁻⁴ moles NaOH = Moles C₆H₅O⁻

And moles C6H5OH = 1.92x10⁻³ moles - 7.56x10⁻⁴ moles = 1.164x10⁻³ moles C₆H₅OH

Replacing in H-H equation:

pH = 4.21 + log [7.56x10⁻⁴ moles] / [ 1.164x10⁻³ moles]

3. At equivalence point, all C6H5OH reacts producing C6H5O⁻. The moles are 1.164x10⁻³ moles

Volume of NaOH to reach equivalence point:

1.164x10⁻³ moles ₓ (1L / 0.108mol) = 0.011L. As initial volume was 0.100L, In equivalence point volume is 0.111L and concentration of C₆H₅O⁻ is:

1.164x10⁻³ moles / 0.111L = 0.01049M

Equilibrium of  C₆H₅O⁻ with water is:

C₆H₅O⁻(aq) + H₂O(l) ⇄  C₆H₅OH(aq) + OH⁻(aq)

Kb = [C₆H₅OH] [OH⁻]/ [C₆H₅O⁻]

Kb = kw / Ka = 1x10⁻¹⁴ / 6.14x10⁻⁵ = 1.63x10⁻¹⁰

Equilibrium concentrations of the species are:

C₆H₅O⁻ = 0.01049M - X

C₆H₅OH = X

OH⁻ = X

Replacing in Kb expression:

1.63x10⁻¹⁰ = X² / 0.01049- X

1.71x10⁻¹² - 1.63x10⁻¹⁰X - X² = 0

Solving for X:

X = -1.3x10⁻⁶ → False solution

X = 1.3076x10⁻⁶ → Right solution

[OH⁻] =  1.3076x10⁻⁶

as pOH = -log [OH⁻]

pOH = 5.88

And pH = 14 - pOH

Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

(1) [tex]2SO_2(g)+O_2(g)\rightarrow SO_3(g)[/tex]

3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.

2) [tex]H_2O(l)\rightarrow H_2O(s)[/tex]

1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.

3) [tex]Br_2(l)\rightarrow Br_2(g)[/tex]

1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.

4)  [tex]H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)[/tex]

1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.

Answer:

12.5g

Explanation:

Half life = 2.4 Minutes.

The half life of a compound is the time it takes to decay to half of it's original concentration or mass.

Time lapsed= 7.2 minutes. This is equivalent to 3 half lives ( 3 * 2.4)

Initial mass = 100g

First half life;

100g --> 50g

Second half life;

50g --> 25g

Third half life;

25g --> 12.5g

The amount of Zn-71 that remains after 7.2 mins has elapsed is 12.5 g

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 2.4 mins

Time (t) = 7.2 mins

[tex]n = \frac{t}{t_{1/2}} \\\\n = \frac{7.2}{2.4} \\\\[/tex]

Thus, 3 half-lives has elapsed.

Original amount (N₀) = 100 g

Number of half-lives (n) = 3

[tex]N = \frac{N_{0}}{2^{n}} \\\\N = \frac{100}{2^{3 }}\\\\N = \frac{100}{8}\\\\[/tex]

Thus, the amount of Zn-71 that remains after 7.2 mins is 12.5 g

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Explanation:

The spontaneity of a system is deduced by the sign of the gibbs free energy value. If it is negative, it means the process / reaction is spontaneous however a positive value indicates the such process is not spontaneous.

Gibbs free energy, enthalpy and entropy are related by the following equation;

ΔG = ΔH - TΔS

A positive value of enthalpy, H and entropy, S means that G would always be a negative value at all temperatures.

Answer:

[tex]x_{et}=0.6068[/tex]

Explanation:

Hello,

In this case, since the mole fraction of a compound, in this case ethanol, in a binary mixture, in this constituted by both water and ethanol, is mathematically defined as follows:

[tex]x_{et}=\frac{n_{et}}{n_{et}+n_{w}}[/tex]

Whereas [tex]n[/tex] accounts for the moles in the solution for each species, we must first compute the moles of both ethanol (density: 0.789 g/mL and molar mass: 46.07 g/mol) and water (density: 1g/mL and molar mass: 18.02 g/mol)

[tex]n_{et}=10.00mL\ et*\frac{0.789g\ et}{mL\ et} *\frac{1mol\ et}{46.07g\ et}=0.1713mol\ et\\ \\n_w=2.00mL\ w*\frac{1g\ w}{mL\ w} *\frac{1mol\ w}{18.02g\ w}=0.1110mol\ w[/tex]

Therefore, the mole fraction turns out:

[tex]x_{et}=\frac{0.1713mol}{0.1713mol+0.1110mol}\\\\x_{et}=0.6068[/tex]

Best regards.

Answer:

An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction

Answer:

[tex]1.25~mol~H_2O[/tex] and [tex]0.627~mol~N_2[/tex]

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ([tex]N_2H_4[/tex]) and oxygen ([tex]O_2[/tex]). So, we can start with the reaction between these compounds:

[tex]N_2H_4~+~O_2~->~N_2~+~H_2O[/tex]

Now we can balance the reaction:

[tex]N_2H_4~+~O_2~->~N_2~+~2H_2O[/tex]

In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for [tex]N_2H_4[/tex] and 31.99 g/mol for [tex]O_2[/tex]):

[tex]20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4[/tex]

[tex]20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2[/tex]

In the balanced reaction we have 1 mol for each reagent (the numbers in front of [tex]O_2[/tex] and [tex]N_2H_4[/tex] are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is [tex]N_2H_4[/tex].

With this in mind, we can calculate the number of moles for each product. In the case of [tex]N_2[/tex] we have a 1:1 molar ratio (1 mol of [tex]N_2[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:

[tex]0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2[/tex]

We can follow the same logic for the other compound. In the case of [tex]H_2O[/tex] we have a 1:2 molar ratio (2 mol of [tex]H_2O[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:

[tex]0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O[/tex]

I hope it helps!

Answer:

The rate would have doubled

Explanation:

Answer:

I hope it works

Explanation:

As we know that

w=m*g

given m=0.145 , g=9.8

hence we get

w= (9.8)*(0.145)

w=1.421 m/sec 2

if its help-full thank hit the stars and brain-list it thank you

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

[tex]MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-[/tex]

And the undergoing chemical reaction:

[tex]MgCl_2+2NaF\rightarrow MgF_2+2NaCl[/tex]

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

[tex]n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2[/tex]

Next, the moles of magnesium chloride consumed by the sodium fluoride:

[tex]n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2[/tex]

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

[tex]n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2[/tex]

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[tex][Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M[/tex]

[tex][F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M[/tex]

Thereby, the reaction quotient is:

[tex]Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}[/tex]

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

Answer:

7.4 × 10⁻⁹ M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷

Concentration of lithium ion: 0.0020 M

Step 2: Write the reaction for the solution of Li₃PO₄

Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

Step 3: Calculate the phosphate concentration required for a precipitate to occur

The solubility product constant is:

Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]

[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³

[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³

[PO₄³⁻] = 7.4 × 10⁻⁹ M

Answer:

b) the Ka of NH₄⁺ to find the hydronium concentration

Explanation:

The equilbrium of NH₄⁺ (The conjugate acid of NH₃, a weak base), is:

NH₄⁺ ⇄ NH₃ + H⁺

Where Ka of the conjugate acid is:

Ka = [NH₃] [H⁺] / [NH₄⁺]

Thus, if you know Ka of NH₄⁺ and its molar concentration you can calculate  [H⁺], the hydronium concentration, to find pH (Because pH =  -log [H⁺])

Thus, right option is:

Answer:

NaN3 = 47.2 g

Explanation:

Given:

2 NaN3 ⇒ 2 Na  + 3 N2

Find:

Amount of NaN3

Computation:

N2 moles = Product of N2 / molar mass of N2

N2 moles =30.5/28

N2 moles = 1.0893

2NaN3 makes 3(N2 )

So,

NaN3 moles = (2/3) moles of N2  

NaN3 moles = ( 2/3) × 1.0893

NaN3 moles = = 0.7262

NaN3 mass = 0.7262 x 65

NaN3 = 47.2 g

Answer:

NaN3 = 47.2 g

Explanation:

Given:

2 NaN3 ⇒ 2 Na  + 3 N2

Find:

Amount of NaN3

Computation:

N2 moles = Product of N2 / molar mass of N2

N2 moles =30.5/28

N2 moles = 1.0893

2NaN3 makes 3(N2 )

So,

NaN3 moles = (2/3) moles of N2  

NaN3 moles = ( 2/3) × 1.0893

NaN3 moles = = 0.7262

NaN3 mass = 0.7262 x 65

NaN3 = 47.2 g

Explanation:

Answer:

I believe the answer The case study was influenced by bias, and led to incorrect conclusions being drawn. plz correct me if I am wrong

Explanation:

Answer: options B

Explanation:

Answer:

Triacontyl palmitate

Explanation:

In this case, we have a reaction between an acid and an alcohol. When we put together these kind of compounds an ester is produced. This reaction is called "esterification".

In our case, the alcohol is a structure with 30 carbon in which the "OH" group is bonded on carbon 1. The name of this compound is "n-triacontanol". The acid is a structure in which we have 16 carbon in which the "COOH" group is placed on carbon 1. The name of this compound is "palmitic acid". The ester produced by the acid and the alcohol is "Triacontyl palmitate".

See figure 1.

I hope it helps!