A 0.10 M solution of a weak monoprotic acid has a pH of 3.40 at 25°C. What is the acid-ionization
constant, Ka, for this acid?
A) 1.6 x 10-6
B) 4.0 x 10-4
C) 3.4 x 10-5
D) 1.2 x 10-3
E) 1.8 x 10-7

Explanation:

In the Figure can be seen the structure of a soap surfactant. It has two key parts: the hydrophobic tail and the hydrophilic head.

Surfactant molecules bond with each other forming miscelles, expossing the part that is compatible with the solvent and protecting the other one.

This means thar the polar head enables the surfactant to dissolve in water and the non-polar tail enables it to dissolve in non polar solvents such as oil mixtures.

Answer:

The anode half reaction is : [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}[/tex]

Explanation:

In electrochemical cell, oxidation occurs in anode and reduction occurs in cathode.

In oxidation, electrons are being released by a species. In reduction, electrons are being consumed by a species.

We can split the given cell reaction into two half-cell reaction such as-

Oxidation (anode): [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}[/tex]

Reduction (cathode): [tex]Sn^{4+}(aq.)+2e^{-}\rightarrow Sn^{2+}(aq.)[/tex]

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

overall: [tex]Fe(s)+Sn^{4+}(aq.)\rightarrow Fe^{2+}(aq.)+Sn^{2+}(aq.)[/tex]

So the anode half reaction is : [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}[/tex]

In the electrochemical cell with redox reaction, Sn4+ (aq) + Fe (s) → Sn2+ (aq) + Fe2+ (aq), the anode half-reaction is Fe (s) → Fe2+ (aq) + 2e-, as Fe is oxidized from 0 to +2.

In the given chemical reaction, Sn4+ (aq) + Fe (s) → Sn2+ (aq) + Fe2+ (aq), we see that tin (Sn) and iron (Fe) change their oxidation states. Tin is being reduced from an oxidation state of +4 to +2, and iron is being oxidized from 0 to +2.

Type of reaction happening at anode is always an oxidation. Oxidation is defined as a reaction where a substance loses electrons. Since Fe goes from Fe to Fe2+, it loses 2 electrons in the process. Therefore, the anode half-reaction for this electrochemical cell is: Fe (s) → Fe2+ (aq) + 2e-

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Answer: The half reactions are written below.

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

[tex]X\rightarrow X^{n+}+ne^-[/tex]

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

[tex]X^{n+}+ne^-\rightarrow X[/tex]

We are given a chemical cell which is Fe-Sn cell. The half reaction follows:

Oxidation half reaction:  [tex]Fe\rightarrow Fe^{2+}+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V[/tex]

Reduction half reaction:  [tex]Sn^{2+}+2e^-\rightarrow Sn;E^o_{Sn^{2+}/Sn}=-0.14V[/tex]

The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced.

Total cell reaction: [tex]Fe+Sn^{2+}\rightarrow Fe^{2+}+Sn[/tex]

Hence, the half reactions are written above.

The overall balanced reaction for a Fe-Sn galvanic cell is: Fe + Sn^2+ -> Fe^2+ + Sn. This reaction is obtained by combining the respective half-reactions for the oxidation of iron and the reduction of tin.

To determine the overall balanced reaction for a Fe-Sn galvanic cell, we need to consider the half-reactions for the oxidation of iron (Fe) and the reduction of tin (Sn).

In this case, the half-reaction showing the oxidation of iron is: Fe -> Fe^2+ + 2e^-

And the half-reaction showing the reduction of tin is: Sn^2+ + 2e^- -> Sn

When these two reactions are combined to give the overall reaction for the galvanic cell, we get: Fe + Sn^2+ -> Fe^2+ + Sn

This is the overall balanced reaction for the Fe-Sn galvanic cell. Please note that the reduction potentials of Sn^2+/Sn and Cr^3+/Cr are different which explains why you would expect different cell potentials if iron was coupled with either tin or chromium.

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Answer: The pressure of NO and [tex]NO_2[/tex] in the mixture is 0.58 atm and 0.024 atm respectively.

Explanation:

We are given:

Equilibrium partial pressure of [tex]O_2[/tex] = 0.29 atm

For the given chemical equation:

                   [tex]2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)[/tex]

Initial:              a

At eqllm:        a-2x          2x          x

Calculating for the value of 'x'

[tex]\Rightarrow x=0.29[/tex]

Equilibrium partial pressure of NO = 2x = 2(0.29) = 0.58 atm

Equilibrium partial pressure of [tex]NO_2[/tex] = a - 2x = a - 2(0.29) = a - 0.58

The expression of [tex]K_p[/tex] for above equation follows:

[tex]K_p=\frac{p_{O_2}\times (p_{NO})^2}{(p_{NO_2})^2}[/tex]

We are given:

[tex]K_p=158[/tex]

Putting values in above expression, we get:

[tex]158=\frac{0.29\times (0.58)^2}{(a-0.58)^2}\\\\a=0.555,0.604[/tex]

Neglecting the value of a = 0.555 because it cannot be less than the equilibrium concentration.

So, [tex]a=0.604[/tex]

Equilibrium partial pressure of [tex]NO_2[/tex] = (a - 0.58) = (0.604 - 0.58) = 0.024 atm

Hence, the pressure of NO and [tex]NO_2[/tex] in the mixture is 0.58 atm and 0.024 atm respectively.

Answer:

Methacrylaldehyde

Explanation:

The first step is the calculation of the IHD (index hydrogen deficiency):

[tex]IHD=~\frac{2C+2+N-H-X}{2}[/tex]

[tex]IHD=~\frac{2(2)+2-6}{2}=2[/tex]

This value indicates that we have 2 double bonds. Now, if we check the IR info we can conclude that we have an oxo group (C=O) due to the signal in 1705 cm^-1 . So, the options that we can have are aldehyde or ketone.

If we analyze the NMR info we have a signal in 194.7 with only 1 hydrogen. This indicates that necessary we have an aldehyde due to the hydrogen. Also, for the signal in 14 we will have a [tex]CH_3[/tex], for the signal at 134.2 we will have a [tex]CH_2[/tex] and for the signal at 146.0 we will have a quaternary carbon (no hydrogens present).

So, we will have a [tex]CH_3[/tex], [tex]CH_2[/tex], C (without hydrogens), an aldehyde group and a double bond.

When we put all this together we will obtain the Methacrylaldehyde (see figure).

N-terminal Arg-Lys-Phe C-terminal

The universal codon code is attached for reference. The central dogma (DNA --> RNA--> Proteins) concept is used to translate the information in DNA into proteins.

Explanation:

We shall begin with the transcription of the DNA strand to form an mRNA. Remember that RNA has no Thiamine base but instead in its place is Uracil;

3' GCATTCAAG 5'       DNA strand

5’ CGUAAGUUC 3’  RNA strand

We can then begin translating the mRNA into protein using the universal codon code (attached). Remember a codon is a 3 nucleotide sequence that codes for a specific amino acid;

5’ CGUAAGUUC 3’     RNA strand

N-terminal Arg-Lys-Phe C-terminal   Peptide strand

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The DNA template strand 3' GCATTCAAG 5' can be transcribed into complementary DNA sequence 5' CGTAAGTTC 3'. This is then transcribed into mRNA sequence 5' CGUAAGUUC 3'. Using the genetic code, this is translated into the amino acid sequence: Arg‑Lys‑Phe.

To determine the amino acid sequence, we first need to establish the complementary DNA sequence for the provided DNA template strand 3' GCATTCAAG 5'. DNA complementary sequence pairs adenine (A) with thymine (T) and cytosine (C) with guanine (G) so we get the complementary DNA sequence (5' to 3' direction) as 5' CGTAAGTTC 3'.

Next, this DNA sequence is transcribed into mRNA using the similar pairing rules but thymine is replaced by uracil (U) in RNA. Thus, the mRNA sequence is 5' CGUAAGUUC 3'. Then we translate this mRNA into amino acids. The genetic code translates each mRNA codon into an amino acid. Therefore, we get the following amino acid sequence: Arg‑Lys‑Phe. Arg stands for Arginine, Lys for Lysine and Phe for Phenylalanine. These abbreviations are internationally accepted standard to represent the corresponding amino acids.

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Answer:

[HI] = 0.097 M

Explanation:

Let's consider the following reaction.

2 HI(g) ⇄ H₂(g) + I₂(g)

The order of reaction for HI is 1. Thus, we can calculate the concentration of HI ([HI]) at certain time using the following expression:

ln [HI] = ln [HI]₀ - k. t

where,

[HI]₀ is the initial concentration of HI

k is the rate constant

t is the time elapsed

When [HI]₀ = 0.440 M and t = 0.210 s, the concentration of HI is

ln [HI] = ln (0.440) - 7.21 s⁻¹ × 0.210 s

ln [HI] = -2.33

[HI] = 0.097 M

Answer:

The molecular shape and ideal bond angle of the [tex]PH_{3}[/tex] is trigonalbipyramidal and [tex]109.5^{o}[/tex] respectively.

Explanation:

The structure of  [tex]PH_{3}[/tex]  is as follows.(in attachment)

From the structure,

Phosphor atom has one lone pair and three hydrogens are bonded by six electrons.

Therefore, total electrons invovled in the formation [tex]PH_{3}[/tex]  is eight.

Hence, four electron groups which indicate the tetrahedral shape. But one pair is lone pair i.e, present on the phosphor atom.

Therefore, ideal geometry of the [tex]PH_{3}[/tex]  molecule is Trigonalbipyramidal.

The ideal angle of trigonalbipyrmidal is [tex]109.5^{o}[/tex].

All three bonds of P-H has [tex]109.5^{o}[/tex].

Therefore, ideal bond angle is [tex]109.5^{o}[/tex].

The electron-group arrangement of PH3 is tetrahedral, the molecular shape is trigonal pyramidal, and the ideal bond angle is 104.5°.

In the case of PH3, the electron-group arrangement is tetrahedral and the molecular shape is trigonal pyramidal. The ideal bond angle for PH3 is 104.5°.

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To calculate the pKa for the acid-base indicator using the given absorbance and molar absorptivity values, apply Beer's Law to determine the concentrations of HIn and In-, then use the Henderson-Hasselbalch equation.

To calculate the pKa for the acid-base indicator HIn, we must use the absorbance data provided and apply Beer's Law, which relates the absorbance to the concentrations of the protonated form (HIn) and deprotonated form (In-) of the indicator.

We are given that at pH 6.12, the absorbance (A) is 0.818, the molar absorptivity (ε) of HIn is 2929 M-1 cm-1 and that of In- is 20060 M-1 cm-1, and the total concentration of the indicator (C) is 0.000127 M.

From Beer's Law we know that:

A = εHInb[HIn] + εnb[In-]

where b is the path length of the cuvette used, which is 1 cm. The concentration of HIn and In- sum up to the total concentration of the indicator:

[HIn] + [In-] = C

We must also consider the Henderson-Hasselbalch equation, which relates the pKa to the pH and the ratio of deprotonated to protonated forms:

pH = pKa + log([In-]/[HIn])

Using the absorbance at pH 6.12 and the total indicator concentration, we can calculate the fractions of HIn and In- and then use the Henderson-Hasselbalch equation to solve for the pKa.

Explanation:

1). The given data is as follows.

       [tex]T_{i} = 25.87^{o}C[/tex],      [tex]T_{f} = 38.13^{o}C[/tex]

               C = 5.73 [tex]kJ/^{o}C[/tex]

Hence, calculate the change in enthalpy of the reaction as follows.

      [tex]\Delta E_{rxn} = -C \times \Delta T[/tex]

                     = [tex]-5.73 \times (38.13 - 25.87)^{o}C[/tex]

                     = -70.25 KJ

As,  number of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                                    = [tex]\frac{1.55}{(6 \times 12 + 14 \times 1)}[/tex]

                                    = 0.018 mol

Therefore, enthalpy of reaction in kJ/mol hexane is as follows.

            [tex]\Delta E_{rxn} = \frac{-70.25 KJ}{0.018 mol}[/tex]

                          = [tex]-3.90 \times 10^{3}[/tex] kJ/mol

Thus, we can conclude that [tex]\Delta E_{rxn}[/tex] for the reaction in kJ/mol hexane is [tex]-3.90 \times 10^{3}[/tex] kJ/mol .

2).  As we know that,

     Number of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                                  = [tex]\frac{1.55}{(7 \times 12 + 8 \times 1)}[/tex]

                                   = 0.017 mol

       [tex]\Delta E_{rxn} = \E_{rxn} per mol \times \text{number of moles}[/tex]

                      = [tex]-3.91 \times 10^{3} \times 0.017[/tex]

                      = -65.875 kJ

As,    [tex]\Delta E_{rxn} = C \times \Delta T [/tex]

                    -65.875 kJ = [tex]-C \times (37.57 - 23.12)^{o}C[/tex]

                           C = 4.56 [tex]kJ/^{o}C[/tex]

Hence, heat capacity of the bomb calorimeter is 4.56 [tex]kJ/^{o}C[/tex].

When 1.550 g of liquid hexane (C₆H₁₄) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 °C to 38.13 °C. The change in the internal energy of the reaction is -3.90 × 10³ kJ/mol.

When 1.55 g of toluene (C₇H₈) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12 °C to 37.57 °C. The heat capacity of the bomb calorimeter is 4.55 kJ/°C.

First, we will calculate the moles corresponding to 1.550 g of hexane using its molar mass (86.18 g/mol).

[tex]n = 1.550 g \times \frac{1mol}{86.18 g} = 0.01799mol[/tex]

Then, we will calculate the heat (Q) absorbed by the bomb calorimeter using the following expression.

[tex]Q = C \times \Delta T = \frac{5.73kJ}{\° C} \times (38.13\° C - 25.87\° C) = 70.2 kJ[/tex]

According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qb) and the heat released by the combustion (Qc) is zero.

[tex]Qb + Qc = 0\\\\Qc = -Qb = -70.2 kJ[/tex]

Finally, we will calculate the change in the internal energy of the reaction (ΔErxn) using the following expression.

[tex]\Delta Erxn = \frac{Qc}{n} = \frac{-70.2kJ}{0.01799mol} = -3.90 \times 10^{3} kJ/mol[/tex]

First, we will calculate the moles corresponding to 1.55 g of toluene using its molar mass (92.14 g/mol).

[tex]1.55 g \times \frac{1mol}{92.14g} = 0.0168 mol[/tex]

The combustion of toluene has a ΔErxn of –3.91 × 10³ kJ/mol. The heat released by the combustion of 0.0168 moles of toluene is:

[tex]0.0168 mol \times \frac{-3.91\times 10^{3}kJ }{mol} = -65.7 kJ[/tex]

According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qb) and the heat released by the combustion (Qc) is zero.

[tex]Qb + Qc = 0\\\\Qb = -Qc = 65.7 kJ[/tex]

We can calculate the heat capacity of the bomb calorimeter (C) using the following expression.

[tex]C = \frac{Qb}{\Delta T } = \frac{65.7 kJ}{37.57 \° C - 23.12 \° C } = 4.55 kJ/ \° C[/tex]

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Answer: - 521.6kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

[tex]H_2(g)+F_2(g)\rightarrow 2HF(g)[/tex]    [tex]\Delta H^0_1=-546.6kJ[/tex]   (1)

[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(l)[/tex] [tex]\Delta H^0_2=-571.6kJ[/tex]  (2)

The final reaction is:

[tex]2F_2(g)+2H_2O(l)\rightarrow 4HF(g)+O_2(g)[/tex]  [tex]\Delta H^0_3=?[/tex]   (3)

By multipling (1) by 2

[tex]2H_2(g)+2F_2(g)\rightarrow 4HF(g)[/tex]    [tex]\Delta H^0_1'=-2\times 546.6kJ=-1093.2kJ[/tex]  (1')

Subtracting (2) from (1')

[tex]2F_2(g)+2H_2O(l)\rightarrow 4HF(g)+O_2(g)[/tex]  [tex]\Delta H^0_3=?[/tex]   (3)

Hence [tex]\Delta H^0_3=\Delta H^0_1'-\Delta H^0_2=-1093.2-(-571.6)kJ=-521.6kJ[/tex].

The enthalpy for the reaction is -521.6kJ

Answer:

Molar mass of silane = 32.15 g/mol

Explanation:

Given:  

Pressure = 2.35 atm

Temperature = 22 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (22 + 273.15) K = 295.15 K  

Volume = 42.9 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

2.35 atm × 4.81 L = n × 0.0821 L.atm/K.mol × 295.15 K  

⇒n = 0.4665 moles

Given, Mass = 15.0 g

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]0.4665\ mole= \frac{15.0\ g}{Molar\ mass}[/tex]

Molar mass of silane = 32.15 g/mol

Answer:

a) non spontaneous

b) spontaneous

c) spontaneous

d) non spontaneous

Answer:

Explanation:

a. Ni(s) + Zn2 + (aq) ¡ Ni2 + (aq) + Zn(s)   non spontaneous

b. Ni(s) + Pb2 + (aq) ¡ Ni2 + (aq) + Pb(s)      spontaneous

c. Al(s) + 3 Ag+ (aq) ¡ Al3 + (aq) + 3 Ag(s)    Reduction spontaneous

d. Pb(s) + Mn2 + (aq) ¡ Pb2 + (aq) + Mn(s)     non spontaneous

Answer:  - 436.5 kJ.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The given chemical reaction is,

[tex]2KCl(s)\rightarrow 2K(s)+Cl_2(g)[/tex]  [tex]\Delta H_1=873kJ[/tex]

Now we have to determine the value of [tex]\Delta H[/tex] for the following reaction i.e,

[tex]K(s)+\frac{1}{2}Cl_2(g)\rightarrow KCl(s)[/tex] [tex]\Delta H_2=?[/tex]

According to the Hess’s law, if we divide the reaction by half then the [tex]\Delta H[/tex] will also get halved and on reversing the reaction , the sign of enthlapy changes.

So, the value [tex]\Delta H_2[/tex] for the reaction will be:

[tex]\Delta H_2=\frac{1}{2}\times (-873kJ)[/tex]

[tex]\Delta H_2=-436.5kJ[/tex]

Hence, the value of [tex]\Delta H_2[/tex] for the reaction is -436.5 kJ.

The standard enthalpy change for the formation of KCl from K(s) and Cl2(g) is -436.5 kJ, calculated by reversing and halving the given reaction's enthalpy change of 873 kJ for 2 moles of KCl.

The student asks about the standard enthalpy change for the reaction where potassium (K) reacts with chlorine gas (Cl2) to form potassium chloride (KCl).

To find the standard enthalpy change for the formation of KCl from K(s) and Cl2(g), we use the given reaction 2 KCl(s) → 2 K(s) + Cl2(g) with a ΔH° of 873 kJ. Since this is the reverse reaction of the formation of KCl, and because enthalpy is a state function, the enthalpy change of the forward reaction is the negative of the reverse reaction. Furthermore, the given reaction involves 2 moles of KCl, so we need to divide the enthalpy by 2 to find the enthalpy change for the formation of 1 mole of KCl.

Therefore, the standard enthalpy change for the reaction K(s) + 0.5 Cl2(g) → KCl(s) is -873 kJ / 2 = -436.5 kJ.

Answer: The number of moles of gas remaining in the lungs is 0.063 moles

Explanation:

The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

where,

[tex]V_1\text{ and }n_1[/tex] are the initial volume and number of moles

[tex]V_2\text{ and }n_2[/tex] are the final volume and number of moles

We are given:

[tex]V-1=1.9L\\n_1=0.080mol\\V_2=1.5L\\n_2=?[/tex]

Putting values in above equation, we get:

[tex]\frac{1.9}{0.080}=\frac{1.5}{n_2}\\\\n_2=\frac{1.5\times 0.080}{1.9}=0.063[/tex]

Hence, the number of moles of gas remaining in the lungs is 0.063 moles

Answer:

A.) Rate=k[A]

Explanation:

The rate law has the following general form:

Rate = k . [A]ᵃ . [B]ᵇ

where,

k is the rate constant

[A] and [B] are the molar concentrations of the reactants

a and b are the reaction orders with respect to A and B

a + b is the overall order of reaction

The rate expression for a first order reaction could be

A.) Rate=k[A]          YES. The reaction order is 1.

B.) Rate=k[A]²[B]     NO. The reaction order is 2 + 1 = 3.

C.) Rate=k[A][B]      NO. The reaction order is 1 + 1 = 2.

D.) Rate=k[A]²[B]²  NO. The reaction order is 2 + 2 = 4

Answer:

Lewis structure is shown in the image below.

Explanation:

Acetate ion (CH₃COO⁻)

Valence electrons of carbon = 4

Valence electrons of oxygen = 6

Valence electrons of hydrogen = 1

Charge = 1 (Negative which means that the electrons are being added)

The total number of the valence electrons  = 2(4) + 2(6) + 3(1) + 1 = 24

The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete. So,  

The Lewis structure is shown in the image below.

The Lewis structure of the acetate ion consists of a carbon atom bonded to two oxygen atoms. There are also resonance structures with the double bond and a negative charge on different atoms. The Lewis dot structure includes lone pairs of electrons for each atom.

The Lewis structure of the acetate ion (CH3COO-) can be represented by drawing the individual Lewis structures for each resonance form. The main structure consists of a carbon atom single-bonded to two oxygen atoms, with one oxygen atom also bonded to a hydrogen atom. The remaining carbon atom is double-bonded to one of the oxygen atoms and also bonded to three hydrogen atoms. Each O atom has three lone pairs of electrons.

The resonance structures can be depicted by moving the double bond between the carbon atoms and by moving the lone pairs of electrons around. One of the resonance structures places a negative charge on the single-bonded oxygen, while the other resonance structure places the negative charge on the double-bonded oxygen.

The Lewis dot structure for the acetate ion includes lone pairs of electrons for all the atoms, as well as the bonding pairs of electrons representing the chemical bonds between the atoms. The oxygen atoms each have three lone pairs of electrons, the carbon atoms each have one lone pair of electrons, and the hydrogen atoms each have no lone pairs of electrons.

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Answer:

60.0mL

Explanation:

For the process of titration , the formula we use is -

M₁V₁ = M₂V₂

Where ,

M₁ = initial concentration

V₁ = initial volume

M₂ = final concentration

V₂ = final volume .

Hence , from the question ,

The given data is ,

M₁ = 0.050 M

V₁ = 30.0mL

M₂ = 0.025 M

V₂ = ?

Now, to determine the unknown quantity , the formula can be applied ,

Hence ,

M₁V₁ = M₂V₂

Putting the respective values ,

0.050 M * 30.0mL = 0.025 M * ?

solving the above equation ,

V₂ = ? = 60.0mL

This titration reaction occurs between a weak base (NH3) and a strong acid (HCl). The distinct pH stages refer to the volume of HCl added. Initially, the pH is determined by the NH3. However, as more HCl is added and reacts with NH3, the pH is determined by the balance of NH3 and NH4+ (from NH3 + HCl -> NH4Cl). If there’s an excess of HCl, the pH drops becoming more acidic.

This question relates to the titration process in chemistry. Let's start by identifying the reactants: NH3 (a weak base) and HCl (a strong acid). The chemical reaction for this titration can be written as NH3 + HCl -> NH4Cl.

To calculate after specific volumes of titrant have been added, we must first understand the principle of titration. In titration, the reaction proceeds until one of the reactants is completely used up. The point at which this happens is known as the equivalence point.

a) At 0 mL HCl, no reaction has occurred so the pH is determined by the NH3 present.

b) At 20.0 mL of HCl, a partial reaction has occurred. The moles of NH3 originally present were 30.0 mL * 0.050 M = 1.5 mmol. After 20 mL of HCl added (0.5 mmol), we have 1.0 mmol NH3 left.

c) At 59.2 mL of HCl added, all the NH3 has reacted because the moles of HCl added are equal to the moles of NH3 initially present.

d) At 60.0 mL of HCl, an excess of HCl has been added and we have a solution of NH4Cl plus excess HCl.

e) At 71.3 mL and f) at 73.3 mL, we still have an excess of HCl, leading to a more acidic solution.

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Answer:

The correct answer is:

The pH of pure water decreases as the temperature is increased.

Explanation:

Ionization constant of water :

[tex]2H_2O(l)\rightleftharpoons H_3O^+(aq) + OH^-(aq)[/tex]

Water will dissociate into equal amount of hydronium and hydroxide ions

[tex]K_w=[H_3O^+]\times [OH^-][/tex]

As we can see that from the given information that ionic product of water increase with increase in temperature which means concentration of hydronium ion concentration increases.

And according to definition of pH of the solution is a negative logarithm of hydronium ions concentration.

[tex]pH=-\log [H_3O^+][/tex]

The pH of the solution of solution is inversely proportional to the concentration of hydronium ions.

As the hydronium ion concentration of water increases with increase in temperature , the pH of the water will decrease.

Answer: The value of [tex]K_c[/tex] for the reaction at 550.3 K is 247.83

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{c}[/tex]

For a general chemical reaction:

[tex]aA+bB\rightarrow cC+dD[/tex]

The expression for [tex]K_{c}[/tex] is written as:

[tex]K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

The chemical equation for the production of methanol follows:

[tex]CO+2H_2\rightleftharpoons CH_3OH[/tex]

The expression of [tex]K_c[/tex] for above equation follows:

[tex]K_c=\frac{[CH_3OH]}{[CO][H_2]^2}[/tex]

We are given:

[tex][CH_3OH]=0.0401mol/L[/tex]

[tex][CO]=0.02722mol/L[/tex]

[tex][H_2]=0.07710mol/L[/tex]

Putting values in above equation, we get:

[tex]K_c=\frac{0.0401}{0.02722\times (0.07710)^2}\\\\K_c=247.83[/tex]

Hence, the value of [tex]K_c[/tex] for the reaction at 550.3 K is 247.83

Final answer:

The equilibrium constant (Kc) for the synthesis of methanol from CO and H₂ at 550.3 K is 30.36, calculated using the given equilibrium concentrations.

Explanation:

The question concerns calculating the equilibrium constant (Kc) at 550.3 K for the synthesis of methanol from CO and H₂. The balanced chemical equation for this process is CO(g) + 2H₂(g) => CH₃OH(g). Given equilibrium concentrations: [H₂] = 0.07710 mol/L, [CO] = 0.02722 mol/L, and [CH₃OH] = 0.0401 mol/L, the equilibrium constant (Kc) can be calculated using the expression Kc = [CH₃OH]/([CO][H₂]²). Plugging in the given values yields Kc = (0.0401)/((0.02722)(0.07710)²) = 30.36.

41Ca (Calcium 41) undergoes electron capture to become 41K (Potassium 41). This 41K then undergoes alpha decay to become 37Cl (Chlorine 37).

The starting nuclide is 41Ca (Calcium 41), which undergoes electron capture (EC). This process involves the nucleus of an atom drawing in a nearby electron and combining it with a proton to form a neutron. The result is a nuclide of an element with atomic number one less than the original. Therefore, 41Ca becomes 41K (Potassium 41).

Next, 41K undergoes alpha decay, a type of radioactive decay where an atomic nucleus emits an alpha particle (consisting of two protons and two neutrons - essentially a helium nucleus). This results in a decrease of both atomic and mass number (by 2 and 4 respectively). Applying this to 41K, we get 37Cl (Chlorine 37) as the product of the alpha decay.

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Answer:

ΔH° = 206.1 kJ

ΔG° = 142.1 kJ

Explanation:

Let's consider the first step in the synthesis of methanol.

Step 1: CH₄(g) + H₂O(g) ⟶ CO(g) + 3 H₂(g) ΔS° = 214.7 J / K

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

ni are the moles of reactants and products

ΔH°f(p) are the standard enthalpies of formation of reactants and products

ΔH° = [1 mol × ΔH°f(CO(g)) + 3 mol × ΔH°f(H₂(g))] - [1 mol × ΔH°f(CH₄(g)) + 1 mol × ΔH°f(H₂O(g))]

ΔH° = [1 mol × (-110.5 kJ/mol) + 3 mol × (0 kJ/mol)] - [1 mol × (-74.81 kJ/mol) + 1 mol × (-241.8 kJ/mol)]

ΔH° = 206.1 kJ

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

ΔG° = 206.1 kJ - 298 K × (214.7 × 10⁻³ kJ/K)

ΔG° = 142.1 kJ

The values of ΔH∘ and ΔG∘ for the reaction can be calculated using their respective formulas and knowing the standard enthalpy and free energy of formation of all the reactants and products. The data is usually found in Chemistry textbooks, specifically in Appendix G.

To calculate ΔH∘ and ΔG∘ at 298K for step 1, we use the Gibb's Helmholtz equation, ΔG = ΔH - TΔS. However, the problem doesn't provide enough information to calculate these directly. Let's presume, however, that we know the values of ΔH∘ for methane and water. Using Appendix G data (usually found in Chemistry textbooks), we could apply the formula ΔH = [ΔH products] - [ΔH reactants] to get ΔH∘.

Similarly, for ΔG∘, if the standard free energy of formation of the compounds is known, we could apply the formula ΔG = [ΔG products] - [ΔG reactants].

Since these values are usually tabulated, this could give us the values of ΔH∘ and ΔG∘ for step 1 of the reaction.

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You may find the structure of compounds as well the oxidation number of sulfur in each compound written with blue color. The nonbonded electrons are represented by black dots.

Explanation:

In the methyl sulfenic acid sulfur have a oxidation number of 0, because we have +1 from carbon and -1 from hydroxil group.

In the methyl sulfinic acid sulfur have a oxidation number of +2 , because we have +1 from carbon, -1 from hydroxil group and -2 from oxygen.

In the  methyl sulfonic acid sulfur have a oxidation number of +4 , because we have +1 from carbon, -1 from hydroxil group, -2 from one oxygen and -2 form the other oxygen.

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Final answer:

The structures and oxidation states of the three sulfur containing acids are: I. Methyl sulfenic acid with sulfur's oxidation state of +2, II. Methyl sulfinic acid with sulfur's oxidation state of +4, and III. Methyl sulfonic acid with sulfur's oxidation state of +6.

Explanation:

I. Methyl sulfenic acid:
Structure: CH3-S(=O)(=O)H
Oxidation state of sulfur: +2

II. Methyl sulfinic acid:
Structure: CH3-S(=O)(=O)-H
Oxidation state of sulfur: +4

III. Methyl sulfonic acid:
Structure: CH3-S(=O)(=O)-OH
Oxidation state of sulfur: +6

Answer: High P and Low T

Where P is the Pressure and T is the Temperature.

Explanation:

1. High Pressure : The above chemical equation has the reactant and product sides.

The product side has total moles of 3 and reactant side has total moles of 4.

To obtain maximum yield in the production of CH4 , pressure must be high because it will favour the side the less number of moles(volume) which is the product side.

2.Low Temperature : The enthaply change indicated as negative shows it is an exothermic reaction. And for an exothermic reaction, the temperature must be lower so as to favor the forward reaction and to make up for the heat/energy loss.

Answer: Correct answer is D.) Every option is true and correct for a galvanic cell.

All the given statements are true: a galvanic cell consists of two half-cells connected by a salt bridge, electrons flow from the anode to the cathode, and reduction occurs at the cathode.

The statement 'D.) All of the above are true' is the correct selection regarding galvanic cells. A.) In a galvanic cell, the two half-cells are indeed connected by a salt bridge. The salt bridge serves to balance the charge, allowing ions to migrate and maintain a neutral charge in the cell. B.) Electrons flow from the anode to the cathode. This is the basic concept of electricity – the flow of electrons. The anode is the electrode where oxidation (loss of electrons) happens, and the electrons released from the oxidation reaction at the anode flow through a wire to the cathode. C.) Reduction occurs at the cathode, meaning that it gains electrons. The cathode is the electrode where reduction (gain of electrons) happens in redox (oxidation-reduction) reactions.

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The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

[tex]rate=k[N_2]^2[H_2]^2[/tex]

At a certain concentration of [tex]N_2[/tex] and [tex]H_2[/tex], the initial rate of reaction is 0.770 M/s. What would the initial rate of the reaction be if the concentration of [tex]N_2[/tex] were halved? Be sure your answer has the correct number of significant digits.

Answer : The initial rate of the reaction will be 0.192 M/s

Explanation :

Rate law expression for the reaction:

[tex]rate=k[N_2]^2[H_2]^2[/tex]

As we are given that:

Initial rate = 0.770 M/s

Expression for rate law for first observation:

[tex]0.770=k[N_2]^2[H_2]^2[/tex] ....(1)

Expression for rate law for second observation:

[tex]R=k(\frac{[N_2]}{2})^2[H_2]^2[/tex] ....(2)

Dividing 1 by 2, we get:

[tex]\frac{R}{0.770}=\frac{k(\frac{[N_2]}{2})^2[H_2]^2}{k[N_2]^2[H_2]^2}[/tex]

[tex]\frac{R}{0.770}=\frac{1}{4}[/tex]

[tex]R=0.192M/s[/tex]

Therefore, the initial rate of the reaction will be 0.192 M/s

If assuming first order, halving the concentration of N2 would halve the initial rate of the reaction, thus the reaction rate would be 0.385 M / s. But the exact rate would depend on the order of the reaction.

Assuming that the reaction rate is directly proportional to the reactant concentration (a first order reaction), if you halve the concentration of N2, the initial rate of the reaction would also be halved. As such, the initial rate of the reaction would be 0.770 M / s divided by 2, which equals to 0.385 M / s.

However, if the reaction is second order with respect to N2 (meaning the rate of reaction is proportional to the square of the concentration of N2), halving the concentration would quarter the reaction rate, if it's zero order (rate independent of concentration), halving the concentration wouldn't change the reaction rate. Hence, additional information such as the order of the reaction with respect to N2 is needed for a more accurate answer.

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Answer:

-105 kJ

Explanation:

The enthalpy change of a reaction is the sum of the energy of the bonds of the reactants and the products. The bonds at the reactants are being broken, so it's an endothermic reaction, so the bond energy must be positive.

The bonds at the products are being formed, so the process is exothermic, and the bond energy must be negative. There are being broken 1 N≡N bond and 3 H-H bonds, and are being formed 6 N-H bonds:

Reactants: 945 + 3*432 = 2241 kJ

Products: 6*(-391) = -2346 kJ

ΔH = 2241 - 2346

ΔH = -105 kJ

The standard enthalpy of formation, which has been established for a huge variety of compounds, is enthalpy change. The reactants undergo chemical changes and combine to produce products in any general chemical reaction. Here the enthalpy change is -297 kJ.

The reaction enthalpy is the change in enthalpy, denoted by the symbol ΔrH.  By deducting the total enthalpies of all the reactants from the total enthalpies of the products, the reaction enthalpy is determined.

Here we will multiply the equation (1) with "2" and then will subtract it from equation "2".

N₂(g) + 2O₂ (g)→ 2NO₂(g) ...... ΔH° = 66.4 kJ  (1) × (2)

2N₂O(g) → 2N₂(g) + O₂(g)  .......ΔH° = -164.2 kJ (2)

2N₂(g) + 4O₂ (g)→ 4NO₂(g) .......  ΔH° = 132.8 kJ

2N₂O(g) + 3O₂(g) → 4NO2(g)

ΔH° = ΔH°(eq 2) - 2ΔH°(eq 1) = -164.2 -(2X66.4) = -297 kJ

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Final answer:

Using Hess's Law and manipulating the given reactions to achieve the target reaction, the standard enthalpy change for the reaction 2N2O(g) + 3O2(g) -> 4NO2(g) is found to be -31.4 kJ.

Explanation:

To find the standard enthalpy change ΔH° for the reaction 2N2O(g) + 3O2(g) → 4NO2(g), we need to apply Hess's Law, which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of individual steps that lead to the overall reaction.

The reactions given are:

(1) N2(g) + 2O2(g) → 2NO2(g)...... ΔH° = 66.4 kJ

(2) 2N2O(g) → 2N2(g) + O2(g)......ΔH° = -164.2 kJ

To use these reactions to find the enthalpy change for (3), we need to manipulate them so that when added, they result in the target reaction. We notice that reaction (1) must be doubled to provide the 4NO2(g) as in reaction (3). This also doubles its enthalpy change. Then, we can simply add the first and second reactions together to get the desired reaction.

So, we have:

(1) x 2: 2N2(g) + 4O2(g) → 4NO2(g)...... ΔH° = 2 x 66.4 kJ = 132.8 kJ

(2):  2N2O(g) → 2N2(g) + O2(g)......ΔH° = -164.2 kJ

To find the ΔH° for (3), we add the ΔH° from the doubled reaction (1) to the ΔH° from reaction (2):

132.8 kJ + (-164.2 kJ) = -31.4 kJ

Hence, the standard enthalpy change for the reaction 2N2O(g) + 3O2(g) → 4NO2(g) is -31.4 kJ.

Answer:

1132 MeV/atom

Explanation:

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Also, atomic number of I = 5

3

Thus, the number of protons = 53

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Mass number = Number of protons + Number of neutrons

135 =  53 + Number of neutrons

Number of neutrons = 82

Mass of neutron = 1.008665 amu

Mass of proton = 1.007825 amu

Calculated mass = Number of protons*Mass of proton + Number of neutrons*Mass of neutron

Thus,

Calculated mass = (53*1.007825 + 82*1.008665) amu = 136.125255 amu

Mass defect = Δm = |136.125255 - 134.910023| amu = 1.215232 amu

The conversion of amu to MeV is shown below as:-

1 amu = 931.5 MeV

So, Energy = 1.215232*931.5 MeV/atom = 1132 MeV/atom

Answer:

The correct answer is B the tertiary halides reacts faster than primary halides.

Explanation:

During SN2 reaction the nucleophile attack the alkyl halide from the opposite side resulting in the formation of transition state in which a bond is not completely broken or a new bond is not completely formed.

   After a certain period of time the nucleophile attach with the substrate by substituting the existing nuclophile.

  An increase in the bulkiness in the alkyl halide the SN2 reaction rate of that alkyl halide decreases.This phenomenon is called steric hindrance.

  So from that point of view the that statement tertiary halides reacts faster that secondary halide is not correct.

The correct answer is C) The transition state species has a pentavalent carbon atom.

The correct answer is C) The transition state species has a pentavalent carbon atom.

In an Sn2 reaction, the mechanism consists of a single step with no intermediates, which eliminates option A as a correct description. Option B is also incorrect because tertiary halides react slower than secondary halides due to steric hindrance. Option D is true, as the rate of the Sn2 reaction depends on the concentrations of both the alkyl halide and the nucleophile.

Thus, option C does not correctly describe Sn2 reactions of alkyl halides.

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