In which of the following reactions does a decrease in the volume of the reaction vessel at constant
temperature favor formation of the products?
A) 2H2(g) + O2(g)--> 2H2O(g)
B) NO2(g) + CO(g)--> NO(g) + CO2(g)
C) H2(g) + I2(g)--> 2HI(g)
D) 2O3(g)--> 3O2(g)
E) MgCO3(s)--> MgO(s) + CO2(g)

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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Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] -Where A⁻ is acetate ion and HA is acetic acid-

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

1,7378 =  [A⁻]/[HA] (1)

As concentration of buffer is 0,15M, it is possible to write:

[A⁻] + [HA] = 0,15M (2)

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need acetic acid 0,055M and acetate 0,095M to obtain the buffer you need.

i hope it helps!

To make a 0.15 M buffer solution at a pH of 5.0 using acetic acid (pKa=4.76) and acetate, the concentrations of acetic acid and acetate need to be about 0.064 M and 0.086 M respectively, determined through rearranging the Henderson-Hasselbalch equation.

To prepare a 0.15 M buffer solution at pH 5.0 using acetic acid (pKa = 4.76) and acetate, we first utilize the Henderson-Hasselbalch equation to determine the ratio of acetate (A-) to acetic acid (HA). This rearranged equation is: [A-]/[HA] = 10^(pH - pKa) = 10^(5.0 - 4.76).

From this, the ratio of [A-]/[HA] comes out to be approximately 1.74. Since an ideal buffer has roughly equal concentrations of acid and base, we know that the sum of the acetic acid and acetate concentrations should be 0.15 M. Therefore, to achieve a buffer at pH 5.0, the acetic acid concentration is approximately 0.064 M (0.037 M * (1/(1+1.74))) and the acetate concentration is approximately 0.086 M (0.15 M - 0.064 M).

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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).

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

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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Explanation:

The given data is as follows.

 Weight of solute = 75.8 g,   Molecular weight of solute (toulene) = 92.13 g/mol,    volume = 200 ml

      Molarity = [tex]\frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}[/tex]

                    = [tex]\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}[/tex]

                    = 4.11 M

Hence, molarity of toulene is 4.11 M.

So, we will calculate the molality of toulene as follows.

   Molality = [tex]\frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}[/tex]

             = [tex]\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}[/tex]

             = 8.6 m

Hence, molality of given toulene solution is 8.6 m.

       No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                             = [tex]\frac{75.8 g}{92.13 g/mol}[/tex]

                             = 0.8227 mol

Now, no. of moles of benzene will be as follows.

     No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                             = [tex]\frac{95.6 g}{78.11 g/mol}[/tex]

                             = 1.2239 mol

Hence, the mole fraction of toulene is as follows.

         Mole fraction = [tex]\frac{\text{moles of toulene}}{\text{total moles}}[/tex]

                             = [tex]\frac{0.8227 mol}{(0.8227 + 1.2239) mol}[/tex]

                             = 0.402

Hence, mole fraction of toulene is 0.402.

         Density = [tex]\frac{mass}{volume}[/tex]

     0.857 [tex]g/cm^{3}[/tex] = [tex]\frac{mass}{200 ml}[/tex]      (As 1 [tex]cm^{3}[/tex] = 1 g)

                      mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

      Mass % = [tex]\frac{\text{mass of solute}}{\text{mass of solution}} \times 100[/tex]

                   = [tex]\frac{75.8 g}{171.4 g} \times 100[/tex]

                   = 44.22%

Therefore, mass percent of toulene is 44.22%.

The molarity, molality, mass percent, and mole fraction of toluene in a solution can be calculated using the given information.

To calculate the molarity of toluene in the solution, we first need to calculate the moles of toluene and benzene. The molarity is then the moles of toluene divided by the volume of the solution in liters. The molality is calculated by dividing the moles of toluene by the mass of the solvent in kilograms. The mass percent is calculated by dividing the mass of toluene by the total mass of the solution and multiplying by 100. Finally, the mole fraction of toluene is calculated by dividing the moles of toluene by the total moles of all components in the solution.

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Answer:

c. the soluble compound exists as dissolved ions in a solution while insoluble compound forms as a precipitated.

Explanation:

Let's consider an example of a double displacement reaction that produces a soluble compound and an insoluble one.

BaCl₂(aq) + Na₂SO₄(aq) ⇒ NaCl(aq) + BaSO₄(s)↓

NaCl is a soluble compound so it is dissolved in the solution. It is also an electrolyte so it ionizes in water according to the following equation:

NaCl(aq) ⇒ Na⁺(aq) + Cl⁻(aq)

BaSO₄ is insoluble so it remains as a precipitate.

Answer:

The correct option is: (A) Ketones cannot be oxidized further

Explanation:

Oxidation refers to the gain of oxygen or the formation of carbon-oxygen bond (C-O bond).

Alcohols are organic compounds containing at least one hydroxyl group. These compounds can be further classified into primary (R-CH₂-OH), secondary (R¹R²CH-OH) and tertiary alcohols (R¹R²R³C-OH).

[tex]R-CH_{2}-OH \overset{[O]}{\rightarrow} R-CHO \overset{[O]}{\rightarrow} R-COOH \\R-CH_{2}-OH \overset{[O]}{\rightarrow} R-COOH[/tex]

[tex]R^{1}R^{2}CH-OH \overset{[O]}{\rightarrow} R^{1}R^{2}C=O[/tex]

Therefore, the correct statement regarding oxidation is (A) Ketones cannot be oxidized further.

Answer:

The correct answer is given below

Explanation:

B before the reaction boron is sp2 hybridized and after the reaction boron is it is sp3 hybridized.

    Before the reaction boron is joined to 3 fluorine atoms by 3 covalent bonds between them one is s and rest of the 2 are p.That"s why before the reaction boron is sp2 hybridized.

   Whereas after the reaction boron is joined to 3 fluorine atoms by 3 covalent bonds and one nitrogen atom by one co ordinate bond.That"s why after reaction boron is sp3 hybridized.

B  before the reaction nitrogen is sp3 hybridized and after the reaction it is sp3 hybridized.

  Before the reaction Nitrogen is joined to 3 hydrogen atoms by 3 covalent bonds and contain one lone pair of electron.That"s why before the reaction Nitrogen atom is sp3 hybridized.

 whereas after the reaction Nitrogen is joined to 3 hydrogen atoms by 3 covalent bonds and with one boron atom by a co ordinate bond.That"s why after reaction Nitrogen atom is sp3 hybridized.

Boron atom's hybridization changes from sp2 to sp3 as a result of the reaction while Nitrogen's hybridization remains sp3 both before and after the reaction.

In the given reaction, BF3 + NH3 → F3B―NH3, the hybridization of both Boron (B) and Nitrogen (N) atoms change. For the Boron atom, before the reaction, it is sp2 hybridized. After the reaction, when it bonds with NH3, it becomes sp3 hybridized. Hence, the change in Boron's hybridization as a result of this reaction is from sp2 to sp3.

For the Nitrogen atom, before the reaction, it is sp3 hybridized. After the reaction, it remains sp3 hybridized because it still maintains four regions of electron density. So, there is no change in Nitrogen's hybridization as a result of the reaction. It remains sp3 both before and after the reaction.

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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.

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:

[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 question involves drawing of structures and showing mechanism in which brainly text editor did not support. I made sure I created a pdf file with both the anwsers and explanations in it. The pdf can be found in the attachment below.  

Explanation:

Compound A, C6H12, is most likely cyclohexane, a 6-carbon ring structured alkane. It reacts via a radical addition reaction with HBr/ROOR to produce Compound B, C6H13Br, a bromocyclohexane. Compound C, C6H14, is most likely hexane, an 6-carbon linear alkane. It undergoes a radical reaction with bromine and light to also form bromocyclohexane.

The subject of this question is organic chemistry, specifically dealing with the structures of different compounds and their reactions. Compound A, C6H12, is likely cyclohexane - a 6-carbon cyclic alkane - which reacts with HBr/ROOR (a radical addition reaction) to form compound B, C6H13Br, a bromocyclohexane. Compound C, C6H14, on the other hand, is hexane - a linear alkane. It reacts with bromine and light (again, another radical reaction) to also give bromocyclohexane.

The structure of compound A (cyclohexane) is drawn as a hexagon representing the 6-carbon cyclic structure, with each corner representing one carbon and its associated hydrogen atoms. This is a very simplified version of chemical structures used for ease of drawing. Detailed representation would include all the Hydrogen atoms.

Cyclohexane will readily react with hydrogen bromide in the presence of a radical initiator (ROOR), producing a bromocyclohexane. Hexane, on the other hand, will react with bromine in the presence of light to form the same product. The light provides the energy needed to break the bromine molecule into bromine radicals, which then react with hexane.

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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

Applying the principle of conservation of energy and the heat equation, the final temperature of the ethanol-water mixture is calculated to be 22.7°C

The specific example used in the image is for a mixture of ethanol and water.

Step 1: Understand the principle

The underlying principle is that heat lost by the hot object is equal to the heat gained by the cold object. This principle is known as the law of conservation of energy.

Step 2: Form an equation

We can express this principle mathematically using the following equation:

q₁ = -q₂

where:

q₁ is the heat lost by the hot object

q₂ is the heat gained by the cold object

We can further expand this equation using the following relationship between heat, mass, specific heat capacity, and temperature change:

q = mcΔT

where:

m is the mass of the object

c is the specific heat capacity of the object

ΔT is the change in temperature

Substituting this relationship into the first equation, we get:

m₁c₁(T₁ - T_f) = -m₂c₂(T_f - T₂)

where:

m₁ and c₁ are the mass and specific heat capacity of the hot object, respectively

m₂ and c₂ are the mass and specific heat capacity of the cold object, respectively

T₁ and T₂ are the initial temperatures of the hot and cold objects, respectively

T_f is the final temperature of the mixture

Step 3: Apply the formula to the specific case

In the example given, the hot object is ethanol and the cold object is water. The following values are provided:

m₁ = 45.0 g (mass of ethanol)

c₁ = 2.3 J/g°C (specific heat capacity of ethanol)

T₁ = 9.0°C (initial temperature of ethanol)

m₂ = 45.0 g (mass of water)

c₂ = 4.18 J/g°C (specific heat capacity of water)

T₂ = 28.6°C (initial temperature of water)

T_f = ? (final temperature of the mixture)

Step 4: Solve for the unknown

Substituting the known values into the equation and solving for T_f, we get:

(0.789 g/mL x 45.0 mL) x (2.3 J/g°C) x (T_f - 9.0°C) = -(1.0 g/mL x 45.0 mL) x (4.18 J/g°C) x (T_f - 28.6°C)

Solving this equation, we get:

T_f = 22.7°C

Therefore, the final temperature of the mixture is 22.7°C.

Answer:

The temperature of the nitrogen gas at another section is [tex]148.8^{o}C[/tex]

Explanation:

Energy balance equation for steady state flow of gas under negiligible potential energy.

The negligible heat transfer and no shaft work is as follows.

[tex]\Delta H+\frac{\Delta u^{2}}{2}=Q+W[/tex]

[tex]\Delta H+\frac{\Delta u^{2}}{2}=0..........(1)[/tex]

[tex]\Delta H[/tex] is enthalphy of gas and it is changes with the temperature.

[tex]\Delta H=C_{p}(T_{2}-T_{1}).................(2)[/tex]

[tex]C_{p}[/tex]=  Molar heat capacity of the gas at constant pressure.[tex]T_{1}[/tex]= Initial temperature at section 1

[tex]T_{2}[/tex] = Final temperature at section 2

Substitute the equation (2) in equation (1)

[tex]C_{p}(T_{2}-T_{1})+\frac{u_{2}^2-u_{1}^2}{2}=0[/tex]

Solve the above equation is as follows.

[tex]T_{2}=T_{1}-\frac{u_{2}^{2}-u_{1}^{2}}{2}=0...............(3)[/tex]

From the given,

[tex]T_{1}=150+273=423K[/tex]

[tex]C_{p}=\frac{7R}{2}[/tex]

[tex]u_{1}=2.5\,m/s[/tex]

[tex]u_{2}=50\,m/s[/tex]

Molar mass of nitrogen gas = 0.02802 kg/mol

Substitute the all values in the equation (3)

[tex]T_{2}=423K-\frac{(50m/s)^{2}-(2.5m/s)^{2}}{2\times \frac{7}{2}\times8.314\,J\,mol^{-1}K^{-1}}\times \frac{J/kg}{m^{2}/s^{2}}\times \frac{0.02802\,kg}{mol}[/tex]

=421.8K=148.8^{o}C

Therefore,The temperature of the nitrogen gas at another section is [tex]148.8^{o}C[/tex].

Final answer:

The equation that relates temperature to velocity of an ideal gas flowing through a horizontal tube with changing cross-sectional area is T1/V1 = T2/V2. Using the given values, we can solve for the temperature at the other section where the velocity is given as 50 m/s, which is found to be 300 °C.

Explanation:

Based on the principle of continuity of flow, the equation that relates temperature to velocity of an ideal gas flowing through a horizontal tube with changing cross-sectional area is:

T1/V1 = T2/V2

To solve for the temperature at another section where the velocity is 50 m/s, we can use the given temperature and velocity at the initial section (T1 = 150 °C, V1 = 2.5 m/s) and plug them into the equation:

150 / 2.5 = T2 / 50

By solving for T2, we find that the temperature at the other section is 300 °C.

To find the molarity of the HA solution, calculate the number of moles of NaOH used in the titration and use the ratio of moles of HA to moles of NaOH to determine the number of moles of HA in the solution. Then, calculate the molarity of the HA solution using the moles of HA and the volume of the solution.

To find the molarity of the HA solution, we can first calculate the number of moles of NaOH used in the titration. The balanced chemical equation for the reaction between HA and NaOH is:

HA + NaOH → NaA + H2O

From the equation, we can see that the ratio of moles of HA to moles of NaOH is 1:1. We can use the molarity and volume of NaOH used in the titration to calculate the number of moles of NaOH, and since the ratio is 1:1, this will also be the number of moles of HA in the solution.

First, we calculate the number of moles of NaOH:

Moles of NaOH = Molarity of NaOH * Volume of NaOH

= 0.0906 M * 0.0873 L

= 0.0078958 mol NaOH

Since the ratio of moles of HA to moles of NaOH is 1:1, the number of moles of HA in the solution is also 0.0078958 mol.

Now, we can calculate the molarity of the HA solution:

Molarity of HA = Moles of HA / Volume of HA solution

= 0.0078958 mol / 0.0435 L

= 0.181 M

The molarity of the HA solution is [tex]\boxed{0.1817 \text{ M}}[/tex]

To calculate the molarity of the HA solution, we need to use the concept of a neutralization reaction. The reaction between the monoprotic acid HA and the strong base NaOH is as follows:

[tex]\[ \text{HA} + \text{NaOH} \rightarrow \text{NaA} + \text{H}_2\text{O} \][/tex]

 In this reaction, one mole of HA reacts with one mole of NaOH to produce one mole of NaA (the salt) and one mole of water. The moles of NaOH used in the titration can be calculated using the formula

[tex]\[ \text{moles of NaOH} = \text{volume of NaOH} \times \text{molarity of NaOH} \][/tex]

 Given that the volume of NaOH is 87.3 mL (which we convert to liters by dividing by 1000) and the molarity of NaOH is 0.0906 M, we can calculate the moles of NaOH:

[tex]\[ \text{moles of NaOH} = 0.0873 \text{ L} \times 0.0906 \text{ M} \][/tex]

[tex]\[ \text{moles of NaOH} = 0.00790358 \text{ mol} \][/tex]

Since the stoichiometry of the reaction is 1:1, the moles of HA that reacted with NaOH are equal to the moles of NaOH used. Now we can calculate the molarity of the HA solution using the formula:

[tex]\[ \text{molarity of HA} = \frac{\text{moles of HA}}{\text{volume of HA solution}} \][/tex]

 The volume of the HA solution is given as 43.5 mL, which we convert to liters:

[tex]\[ \text{volume of HA solution} = 0.0435 \text{ L} \][/tex]

Now we can calculate the molarity of HA:

[tex]\[ \text{molarity of HA} = \frac{0.00790358 \text{ mol}}{0.0435 \text{ L}} \][/tex]

[tex]\[ \text{molarity of HA} = 0.18167287 \text{ M} \][/tex]

 Rounding to a reasonable number of significant figures, the molarity of the HA solution is approximately 0.1817 M.

Therefore, the molarity of the HA solution is [tex]\boxed{0.1817 \text{ M}}[/tex]

The answer is: [tex]0.1817 \text{ M}.[/tex]

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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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 new concentration is 2.03M

Explanation:

Step 1: Data given

A 200 mL 3.55 M HBr is diluted with 150 mL

Step 2: The dilution

In a dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution equals the ratio that exists between the volume of the diluted solution and the volume of the stock solution.

Dilution factor = [stock sample]/[diluted sample] = diluted volume / stock volume

In this case, the volume of the stock solution is 200 mL

Adding  150 mL  of water to the stock solution will dilute it to a final volume of 200 + 150 = 350 mL

The dilution factor wll be 350/200 = 1.75

This makes the diluted concentration:

3.55/1.75 = 2.03M

The new concentration is 2.03M

The Gibbs free energy change (ΔG∘) for the reaction can be estimated using the equation ΔG∘ = ΔH∘ - TΔS∘. At a temperature of 718K, if ΔG∘ is negative, the reaction will be spontaneous.

The Gibbs free energy change (ΔG∘) for a reaction can be estimated using the equation ΔG∘ = ΔH∘ - TΔS∘, where ΔH∘ is the change in enthalpy and ΔS∘ is the change in entropy. At a temperature of 718K, you can estimate ΔG∘ by substituting the given values into the equation. If the value of ΔG∘ is negative, the reaction will be spontaneous. If it is positive or zero, the reaction will not be spontaneous.

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The reaction is spontaneous at 298 K and 723 K but non-spontaneous at 842 K as determined by calculating ΔG∘ at each temperature using the Gibbs free energy equation. Negative ΔG∘ values indicate spontaneity while positive values indicate non-spontaneity.

To determine the spontaneity of the reaction 2NO(g) + O₂(g) → 2NO₂(g) at different temperatures (298 K, 723 K, and 842 K), we use the Gibbs free energy equation:

ΔG∘ = ΔH∘ - TΔS∘

Given that ΔH∘ (standard enthalpy change) and ΔS∘ (standard entropy change) are assumed to be constant over the temperature range, we need to calculate ΔG∘ for each temperature:

If ΔG∘ is negative, the reaction is spontaneous at that temperature. Conversely, if ΔG∘ is positive, the reaction is non-spontaneous.

Determining Spontaneity:

Using standard thermodynamic tables or given data for ΔH∘ and ΔS∘:

Suppose: ΔH∘ = -114.1 kJ/mol and ΔS∘ = -146.5 J/(mol·K)

In summary, the reaction is spontaneous at 298 K and 723 K, but non-spontaneous at 842 K

complete question.

Consider the reaction:

2NO(g)+O₂(g)→2NO₂(g)

Estimate ΔG∘ for this reaction at each temperature and predict whether or not the reaction will be spontaneous. (Assume that ΔH∘ and ΔS∘ do not change too much within the given temperature range.)

A) 298 K

B) 723 K

C) 842 K

Answer:

The mass of carbon dioxide is 28.6 grams

Explanation:

Step 1: Data given

Mass of O2 = 41.6 grams

Volume of the O2 container = 2V

Volume of the CO2 container = V

Pressure and temperature are the same

Step 2: Ideal gas law

The ideal gas law = p*V=nRT

For O2: p*2V = n(O2)*R*T

For CO2: p*V = n(CO2)*R*T

n(O2)*R*T / P*2V = n(CO2)*R*T / P*V

Since P, R and T are contstant, we can simplify this to:

n(O2)/2 = n(CO2)

Step 3:  Calculate moles of O2

Moles O2 = mass O2/ Molar mass O2

Moles O2 = 41.6 grams / 32 g/mol

Moles O2 = 1.3 mol O2

The number of moles CO2 = 0.65 mol

Mass of CO2 = Moles CO2 * Molar Mass CO2

Mass of CO2 = 0.65 mol * 44.01 g/mol

Mass of CO2 = 28.6 grams

The mass of carbon dioxide is 28.6 grams

Answer:

A) is true

Explanation:

For the reaction:

O₂(g) + 2F₂(g) ⇄ 2OF₂(g); Kp = 2,3x10⁻¹⁵

kp is defined as:

Kp = 2,3x10⁻¹⁵ = [OF₂]²/[O₂] [F₂]²

A) If the reaction mixture initially contains only OF₂(g), then at equilibrium, the  reaction mixture will consist of essentially only O₂(g) and F₂(g).  TRUE. As the kp is 2,3x10⁻¹⁵ means per 1 of [O₂] [F₂]² you will have just 2,3x10⁻¹⁵ of [OF₂]²

B) For this equilibrium, Kc = Kp.  FALSE. That is true just when moles of reactants are the same than moles of products. Here there are 3 moles of reactants vs 2 moles of products.

C) If the reaction mixture initially contains only OF₂(g), then the total pressure at  equilibrium will be less than the total initial pressure.  FALSE. Because per 2 moles of OF₂(g) you will produce 3 moles of gas increasing pressure.

D) If the reaction mixture initially contains only O₂(g) and F₂(g), then at equilibrium,  the reaction mixture will consist of essentially only OF₂(g).  FALSE. For the same reason of A), the mixture will contains essentially only O₂(g) and F₂(g)

E) If the reaction mixture initially contains only O₂(g) and F₂(g), then the total  pressure at equilibrium will be greater than the total initial pressure. FALSE. If mixture initially contains only O₂(g) and F₂(g), 3 moles will of gas will react to produce 2 moles of gas doing pressure decreases.

I hope it helps!

Answer:

Gases are at 1 bar pressure.

Explanation:

Option a is incorrect . Standard  Pressure in chemistry is always taken in atm and not in bar. The standard condition for electro chemical cell measurements are

Gases must be at 1 atm pressure.

Liquids must be in pure state

Solids must also be in pure state

Solutes are at 1.0 M concentration

And the cell voltages are positive.

In the context of electrochemical cells, the statement 'The cell voltage is always positive' is incorrect as cell voltage can be negative in an electrolytic cell.

The statement that is not true of the standard conditions for electrochemical cell measurements is that 'The cell voltage is always positive.' While the cell voltage can be positive in a galvanic/voltaic cell as it produces an electrical current, it can be negative in an electrolytic cell, where electrical energy is being consumed to bring about a chemical change.

The other statements about gases at 1 bar pressure, liquids and solids being in a pure state, and solutes at a 1.0 M concentration are indeed part of the standard conditions for measuring the properties of an electrochemical cell.

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Answer:

m = 1

n = 1

Explanation:

The rate law is:

[tex]r=k.[A]^{m} .[B]^{n}[/tex]

where,

r is the rate of the reaction

k is the rate constant

m is the order of reaction with respect to A

n is the order of reaction with respect to B

Let's consider trials 1 and 4. We know that [B]₁ = [B]₄ . The rate r₁/r₄ is:

[tex]\frac{r_{1}}{r_{4}} =\frac{k.[A]_{1}^{m}.[B]_{1}^{n}  }{k.[A]_{4}^{m}.[B]_{4}^{n}} \\\frac{r_{1}}{r_{4}} =(\frac{[A]_{1}}{[A]_{4}} )^{m} \\\frac{0.0904M/s}{0.181M/s}=(\frac{0.100M}{0.200M})^{m} \\m=1[/tex]

Let's consider trial 1 and 5. We know that [A]₁ = [A]₅. The rate r₁/r₅ is:

[tex]\frac{r_{1}}{r_{5}} =\frac{k.[A]_{1}^{m}.[B]_{1}^{n}  }{k.[A]_{5}^{m}.[B]_{5}^{n}} \\\frac{r_{1}}{r_{5}} =(\frac{[B]_{1}}{[B]_{5}} )^{n} \\\frac{0.0904M/s}{0.181M/s}=(\frac{0.400M}{0.800M})^{n} \\n=1[/tex]

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:

See figure attached.

Explanation:

Pyridinium chlorochromate (PCC) is a weaker oxidizing agent when compared with oth chromic acid derivatives as dichromate, for example. This results in the fact that PCC wil oxidize a primary alcohol transforming it to a aldehyde.

The entire mechanism of the reaction of 1-butanol with PCC in an anhydrous solvent is shown in the figure attached.

The neutral intermediate and the organic product (1-butanal) are both represented in the figure.

Answer:

Gibbs free-energy of the reaction = (–12.5 kJ/mol)

Explanation:

The Gibbs free-energy of a reaction predicts the spontaneity or feasibility of a given chemical reaction.

Given the standard Gibbs free energy changes:

Phosphocreatine → creatine + Pi,  ∆G° = –43.0 kJ/mol     ...(1)

ATP → ADP + Pi , ∆G° = –30.5 kJ/mol      ....(2)

Now to calculate the Gibbs free-energy of the given chemical reaction: Phosphocreatine + ADP → creatine + ATP; the equation (2) is reversed to give:

ADP + Pi  → ATP, ∆G° = + 30.5 kJ/mol      ...(3)

Now the equation (3) and (1) are added, to give:

Phosphocreatine + ADP + Pi→ creatine + ATP + Pi

⇒ Phosphocreatine + ADP → creatine + ATP  

Therefore, to calculate the Gibbs free-energy of the reaction, the standard Gibbs free energy changes of the equations (1) and (3) are added similarly:

Gibbs free-energy of the reaction: ∆G° = (–43.0 kJ/mol) + ( + 30.5 kJ/mol) = (–12.5 kJ/mol)

Therefore, the Gibbs free-energy of the reaction = (–12.5 kJ/mol)

The overall standard free-energy change for the reaction Phosphocreatine + ADP to form creatine and ATP is the sum of the changes for the individual reactions, resulting in −12.5 kJ/mol.

To determine the overall standard free-energy change (Δ G'°) for the reaction where phosphocreatine and ADP react to form creatine and ATP (“Phosphocreatine + ADP → creatine + ATP”), we need to consider the given free-energy changes for the individual reactions:

By reversing the second reaction, we invert the sign of its standard free-energy change. Therefore, for the reaction ADP + Pi → ATP, the Δ G'° becomes +30.5 kJ/mol.
Now, to find the overall Δ G'° for the reaction Phosphocreatine + ADP → creatine + ATP, we sum up the standard free-energy changes of the two reactions:

Δ G'° (overall) = Δ G'° (Phosphocreatine → creatine + Pi) + Δ G'° (ADP + Pi → ATP)

Δ G'° (overall) = (−43.0 kJ/mol) + (+30.5 kJ/mol) = −12.5 kJ/mol

The result is −12.5 kJ/mol, which indicates that the coupled reaction is also exergonic, releasing energy.

Answer:

1s^22s^22p^63s^23p^63d^24s^2

Explanation:

Titanium with the chemical symbol Ti, is a transition element, and in fact the second transition element after scandium. It has an atomic number of 22. It is the second element in the first transition series.

To write the electronic configuration, we consider the atomic number of titanium. The atomic number of titanium is 22. Hence, since an atom is electrically neutral, the number of electrons in the titanium metal is 22.

It should be noted that while filling,the maximum number of electrons the s subshell is 2. The maximum number of electrons in the p subshell is 6 while the number in the d shell is 10