1. Choose the alkyl halide(s) from the following list of C6H13Br isomers that meet each criterion below. 1) 1-bromohexane 2) 3-bromo-3-methylpentane 3) 1-bromo-2,2-dimethylbutane 4) 3-bromo-2-methylpentane 5) 2-bromo-3-methylpentane a) the compound(s) that can exist as enantiomers b) the compound(s) that can exist as diastereomers c) the compound that gives the fastest SN2 reaction with sodium methoxide d) the compound that is least reactive to sodium methoxide in methanol e) the compound(s) that undergo an SN1 reaction to give rearranged products f) the compound that gives the fastest SN1 reaction

Answer:

ΔS = -114.296 J/K

Explanation:

Step 1: Data given

Temperature = 298 K

Number of moles N2 = 2.19 moles

S°(N2) = 191.6 J/K*mol

S°(O2)= 161.1 J/k*mol

S°(N2O) = 219.96 J/K*mol

Step 2: The balanced equation

2N2(g) + O2(g) ⇆ 2N2O(g)

Step 3: Calculate ΔSrxn

ΔSrxn = ∑S°products -  ∑S°reactants

ΔSrxn = (2*219.96) - (161.1 + 2*191.6) J/K*mol

ΔSrxn = -104.38 J/K

Step 4: Calculate ΔS for 2.19 moles

The reaction is for 2 moles N2

ΔS = -114.296 J/K

Final answer:

The entropy change for a chemical reaction can be determined using standard absolute entropies. In this case, the given reaction of N2(g) and O2(g) to form N2O(g) can be analyzed using these principles.

Explanation:

The entropy change for the given reaction can be calculated using the standard absolute entropies at 298K. The formula to calculate the entropy change is ΔS = ΣnS(products) - ΣnS(reactants). In this case, for the reaction 2N2(g) + O2(g) → 2N2O(g), the entropy change can be calculated using the standard absolute entropies of each component.

Step 1: Data given

Temperature = 298 K

Number of moles N2 = 2.19 moles

S°(N2) = 191.6 J/K*mol

S°(O2)= 161.1 J/k*mol

S°(N2O) = 219.96 J/K*mol

Step 2: The balanced equation

2N2(g) + O2(g) ⇆ 2N2O(g)

Step 3: Calculate ΔSrxn

ΔSrxn = ∑S°products -  ∑S°reactants

ΔSrxn = (2*219.96) - (161.1 + 2*191.6) J/K*mol

ΔSrxn = -104.38 J/K

Step 4: Calculate ΔS for 2.19 moles

The reaction is for 2 moles N2

ΔS = -114.296 J/K

Answer:

The heat is  [tex]H= -8.044KJ[/tex]

Explanation:

From the question we are told that

   The pressure is  [tex]P = 1 \ atm[/tex]

    The boiling point is  [tex]B_P = 34^oC[/tex]

     The heat of vaporization at 34°C is  = [tex]26.5 kJ/mol[/tex]

       The specific heat of the liquid is [tex]c_p = 2.32 J/g^oC[/tex]

      The  mass is  [tex]m = 22.5g[/tex]

  The no of moles of the sample of [tex]C_2 H_5OC_2H_5[/tex]  is given as

               [tex]No \ mole (n) = \frac{Mass \ of \ sample }{Molar \ mass }[/tex]

    The  Molar mass for [tex]C_2 H_5OC_2H_5[/tex]  is a value = [tex]= 74.12 g/mol[/tex]

Substituting the value into the above equation

          [tex]n = \frac{22.5}{74.12}[/tex]

                           [tex]= 0.30356 \ mol[/tex]

      The heat H is mathematically as

                [tex]H =- nH_{vap}[/tex]

The negative sign show that the heat is for condensing

            [tex]H = 0.30356 * 26.5 *10^{3} J/mol[/tex]

                [tex]H= -8.044KJ[/tex]

The enthalpy change for condensing a 22.5 g sample of gaseous ether at its normal boiling point of 34.6 °C is approximately -8.045 kJ.

The question asks about the enthalpy change (H) during the condensation of a sample of ether. To calculate the enthalpy change for condensing 22.5 g of diethyl ether at its boiling point, we'll use the given enthalpy of vaporization ([tex]H_{vap}[/tex]) and the mass of the ether. Considering that the enthalpy of vaporization ([tex]H_{vap}[/tex]) is the amount of energy required to vaporize 1 mole of a substance at its boiling point, the enthalpy of condensation will be the negative of this value since condensation is the reverse process.

First, we convert the mass of ether to moles:

22.5 g * (1 mol / 74.12 g/mol) ≈ 0.3036 mol

Next, multiply the moles by the enthalpy of vaporization (Hvap) but with a negative sign for condensation:

-0.3036 mol * 26.5 kJ/mol ≈ -8.045 kJ

This is the enthalpy change for the condensation process of the ether sample.

Answer:

Here's what I get  

Explanation:

1. Requirements for a liquid/liquid extraction solvent

(a) Distribution Coefficient

The solute must be more soluble in the solvent (the extract phase) than in the raffinate phase (the other liquid).

(b) Insolubility

The solvent and the other liquid must be mutually immiscible. This means they will form two layers in the separatory funnel.

(c) Removability

It must be easy to separate the solvent from the solutes. This usually means the solvent must have a much lower boiling point than the solutes.

2. Extract layer

The density of ethyl acetate is 0.920 g/mL.

The density of water is 1.00 g g/mL.

The ethyl acetate layer will be on the top.

3. Separability of mixture

Two miscible liquids must have a boiling point difference of at least 40 °C to be separable by simple distillation.

4. Type of distillation  

The boiling point difference is only 30°C, so you must use fractional distillation.

5. Apparatus

Your microscale kit must include an air condenser or a fractionating column.  

Perhaps it looked like the fractionating head in the kit in Fig. 1.

The alternate distillation (simple distillation) would use a distillation head like that in Fig. 2.

Final answer:

The three main criteria for the liquid/liquid extraction solvent are solvent properties, low boiling point, and chemical inertness. Ethyl acetate will be in the upper layer. Water can be removed from the organic layer using drying agents or azeotropic distillation. The mixture of Q and R should have a significant difference in boiling points and the desired composition to be separated by distillation. Fractional distillation would be carried out in this case, using the distilling flask, condenser, and receiving flask from the microscale kit.

Explanation:

The three main criteria for the liquid/liquid extraction organic solvent are:

Ethyl acetate will be in the upper layer as it has a lower boiling point than both the major product Q and the side product R.

To remove water from the organic layer, two methods can be used:

The mixture of liquids Q and R needs to satisfy the following criteria to be separated by simple or fractional distillation:

In this case, a fractional distillation would be carried out because the boiling points of Q and R are relatively close, and fractional distillation provides better separation of two volatile liquids with similar boiling points.

The parts of the microscale kit that would be used for the chosen fractional distillation would include the distilling flask, condenser, and receiving flask. These are essential components for the distillation process. The parts that would not be used for the alternate distillation method would depend on the specific method chosen but could include components such as the fractionating column or any additional specialized equipment.

CO2 is a chemical formula that is representing a molecule.

Explanation:

All the  given chemical formulas represent molecules as they are made up of two or more than two atoms .

Chemical  formula is a way of representing the number of atoms present in a compound or molecule.It is written with the help of symbols  of elements. It also makes use of brackets and subscripts.

Subscripts are used to denote number of atoms of each element and brackets indicate presence of group of atoms. Chemical formula does not contain words. Chemical formula in the simplest form  is called empirical formula.

It is not the same as structural formula and does not have any information regarding structure.It does not provide any information regarding structure of molecule as obtained in structural formula.

There are four types of chemical formula:

1)empirical formula

2) structural formula

3)condensed formula

4)molecular formula

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Answer:

a. Copper (II) oxide is the limiting reactant.

b. [tex]m_{N_2}=5.30g[/tex]

c. [tex]Y=87\%[/tex]

d. [tex]m_{NH_3}=2.60gNH_3[/tex]

Explanation:

Hello,

In this case, for the given reaction:

[tex]2 NH_3(g) + 3 CuO(s) \rightarrow N_2(g) + 3 Cu(s) + 3 H_2O(l)[/tex]

a. The limiting reactant is identified by computing the available moles of ammonia and the moles of ammonia that react with 45.2 g of copper (I) oxide as shown below:

[tex]n_{NH_3}^{Available}=9.05gNH_3*\frac{1molNH_3}{17gNH_3}=0.532molNH_3\\n_{NH_3}^{Reacted}=45.2gCuO*\frac{1molCuO}{79.545gCuO}*\frac{2molCuO}{3molCuO} =0.379molNH_3[/tex]

In such a way, as there more ammonia available than that is reacted, we say it is in excess and the copper (II) oxide the limiting reactant.

b. Here, with the reacting moles of ammonia, we compute the yielded grams of nitrogen:

[tex]m_{N_2}=0.379molNH_3*\frac{1molN_2}{2molNH_3}*\frac{28gN_2}{1molN_2}\\m_{N_2}=5.30g[/tex]

c. Now, since the 5.30 g of nitrogen are the expected grams of it, the percent yield of nitrogen is compute by dividing the real obtained mass over the theoretical previously computed mass:

[tex]Y=\frac{4.61g}{5.30g} *100\%\\Y=87\%[/tex]

d. Finally, as 0.532 moles of ammonia are available, but just 0.379 moles react, the unreacted moles are:

[tex]n_{NH_3}=0.532molNH_3-0.379molNH_3=0.153molNH_3[/tex]

That in grams are:

[tex]m_{NH_3}=0.153molNH_3*\frac{17gNH_3}{1molNH_3}=2.60gNH_3[/tex]

Best regards.

The limiting reactant is CuO, the percent yield of nitrogen gas is 29%.

The equation of the reaction is;

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)

Number of moles of NH3 =  9.05 g/17 g/mol = 0.53 moles

Number of moles of CuO = 45.2 g/80 g/mol = 0.565 moles

Since 2 moles of NH3 reacts with 3 moles of CuO

0.53 moles of NH3 reacts with 0.53 moles × 3 moles/ 2 moles

= 0.795 moles

We can see that there is not enough CuO in the system hence it is the limiting reactant.

Number of moles of N2 produced = 0.565 moles × 28 g/mol =15.82 g of N2

Percent yield = Actual yield/Theoretical yield × 100/1

Percent yield = 4.61 g/15.82 g  × 100/1

Percent yield = 29%

If 2 moles of NH3 reacts with 3 moles of CuO

x moles of NH3 reacts with 0.565 moles of CuO

x =  2 moles × 0.565 moles/3 moles

x = 0.38 moles

Number of moles of excess reactant left over = 0.53 moles - 0.38 moles

= 0.15 moles

Mass of excess reactant left over = 0.15 moles × 17 g/mol = 2.55 g

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Answer:

First structure is sophorose, second structure is turanose, third structure is trehalose, fourth structure is none of the above.

Trehalose is a non-reducing disaccharide, similar to sucrose, with two glucose units linked at their anomeric carbons. Sophorose is a reducing sugar yielding two glucose anomers upon hydrolysis. Turanose is also a reducing sugar but yields one molecule each of glucose (aldose) and fructose (ketose) upon hydrolysis.

Identifying trehalose, sophorose, and turanose among the disaccharides requires understanding their structural characteristics and behavior upon hydrolysis. Trehalose is a non-reducing sugar, which means it doesn't have a free anomeric carbon that can form an open-chain aldehyde or ketone upon hydrolysis. From the given information, we know that trehalose upon hydrolysis gives two molecules of aldoses, which means it must have two glucose units. As trehalose is a non-reducing sugar, it is similar to sucrose in having a glycosidic bond between the two anomeric carbons.

Sophorose is a reducing sugar that upon hydrolysis yields two molecules of an aldose that are anomers of each other, implying the presence of two glucose molecules. Given that it is reducing, at least one of the anomeric carbons must not be involved in the glycosidic linkage, allowing it to open and reduce other compounds.

Turanose, like sophorose, is a reducing sugar, but it gives one molecule of an aldose and one molecule of a ketose upon hydrolysis. This indicates that turanose is composed of one glucose and one fructose molecule. The glucose component would be linked through its anomeric carbon, leaving the fructose part with a potential ketose structure.

Answer:

The peak at mass 100 with a 8% relative abundance is the molecular ion peak

Explanation:

Molecular ion peak has the highest charge to mass ratio,

Mass of 100 is same as mass to charge ratio =100

Answer: Third stream’s temperature is [tex]33.2^{o}C[/tex]  and its relative humidity is 50%.

Explanation:

According to the Psychrometric chart.

For the first stream, we have the following.

  [tex]T_{1} = 40^{o}C [/tex],       R.H = 40%

      V = 3L/s ,         W = 19 gm of moisture/kg of dry air

    [tex]h_{1}[/tex] = 89 kj/kg ,     [tex]v_{1} = 0.913 m^{3}/kg[/tex]

For the second stream, we have the following.

   [tex]T_{2} = 15^{o}C[/tex]

       R.H = 80%

        V = 1 L/s

     W = 8.5 gm of moisture/kg of dry air

  [tex]h_{2}[/tex] = 36.5 kj/kg

  [tex]v_{2} = 0.828 m^{3}/kg[/tex]

Now,

     [tex]ma_{1} = \frac{V}{v_{1}}[/tex]

              = [tex]\frac{3}{0.913}[/tex]

              = 3.286 kg/s

     [tex]ma_{2} = \frac{V}{v_{2}}[/tex]

              = [tex]\frac{1}{0.828}[/tex]

              = 1.2077 kg/s

We will calculate the value of [tex]ma_{3}[/tex] as follows.

  [tex]ma_{3} = ma_{1} + ma_{2}[/tex]

            = 3.286 kg/s  + 1.2077 kg/s

             = 4.5 kg/s

Now, heat balance will be as follows.

     [tex]ma_{1}h_{1} + ma_{2}h_{2} = ma_{3}h_{3}[/tex]

        [tex]h_{3}[/tex] = 74.7855 kj/kg

Hence, the moisture balance will be calculated as follows.

        [tex]ma_{1}W_{1} + ma_{2}W_{2} = ma_{3}W_{3}[/tex]

       [tex]W_{3}[/tex] = 16.155

Now, again using the psychrometric chart we will find the value of  temperature and relative humidity as follows.

Values of [tex]T_{3}[/tex] and [tex]R.H_{3}[/tex] against [tex]W_{3}[/tex] and [tex]h_{3}[/tex] are:

     [tex]T_{3} = 33.2^{o}C[/tex]

           R.H = 50%

Thus, we can conclude that third stream’s temperature is [tex]33.2^{o}C[/tex]  and its relative humidity is 50%.

To calculate the temperature and relative humidity of the third stream formed by mixing two humid streams, we can use the concept of partial pressure and the definition of relative humidity. Using the given data, we can determine that the third stream's temperature is approximately 27.4°C and its relative humidity is 59.4%.

To calculate the temperature and relative humidity of the third stream formed by adiabatically mixing two humid streams, we can use the concept of partial pressure and the definition of relative humidity.

First, we need to calculate the mole fraction of water vapor in each stream using the given relative humidity values.

Then, we can calculate the partial pressures of water vapor in each stream using the mole fraction and the saturation vapor pressure at the respective temperatures.

Finally, the total pressure of the third stream can be calculated using the ideal gas law, and the mole fraction of water vapor in the third stream can be used to calculate its relative humidity.

Using the given data, we can determine that the third stream's temperature is approximately 27.4°C and its relative humidity is 59.4%.

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Answer:

We need 113000 J of heat (option 2 is correct)

Explanation:

Step 1: Data given

Mass of liquid water = 50.0 grams

ΔHVap = 2260 J/g

Temperature = 100 °C

ΔHVap = The amount of heat released to change phase of a liquid water to steam = 2260 J/g

Step 2: Calculate the heat needed

Q =m* ΔHVap

⇒with Q = the amount of heat needed = TO BE DETERMINED

⇒with m = the mass of water = 50.0 grams

⇒with ΔHVap = 2260 J/g

Q = 50.0 grams * 2260 J

Q = 113000 J

We need 113000 J of heat (option 2 is correct)

Here is the correct question

You mix 125 mL of 0.170 M CsOH with 50.0 mL of 0.425 M HF in a coffee-cup calorimeter, and the temperature of both solutions rises from 20.20 °C before mixing to 22.17 °C after the reaction. What is the enthalpy of reaction per mole of ? Assume the densities of the solutions are all 1.00 g/mL, and the specific heat capacities of the solutions are 4.2 J/g · K. Enthalpy of reaction = kJ/mol

Answer:

75.059 kJ/mol

Explanation:

The formula for calculating density  is:

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

Making mass the subject of the formula; we have :

mass = density × volume

which can be rewritten as:

mass of the solution =  density × volume of the solution

= 1.00 g/mL × (125+ 50 ) mL

= 175 g

Specific heat capacity = 4.2 J/g.K

∴ the energy absorbed is = mcΔT

= 175 × 4.2 × (22.17 - 20.00) ° C

= 1594.95 J

= 1.595 J

number of moles of CsOH =  [tex]\frac{125}{1000} *100[/tex]

= 0.2125 mole

Therefore; the enthalpy of the reaction = [tex]\frac{Energy \ absorbed }{number \ of \ moles}[/tex]

= [tex]\frac{1.595}{0.02125}[/tex]

= 75.059 kJ/mol

Final answer:

The enthalpy of reaction is calculated by first determining the total heat absorbed or released (∆Q) using the mass of the solutions, specific heat capacity, and temperature change from the calorimetry experiment. Then, by adjusting ∆Q for the amount of reactant, the ∆H per mole is found.

Explanation:

The question concerns the calculation of the enthalpy of reaction from a mixing experiment in a coffee-cup calorimeter. Given are volumes and molarities of two solutions mixed, along with the temperature change upon mixing. The enthalpy of reaction (∆H) is calculated using the concept that the heat absorbed or released by the solution (∆Q) during the reaction, adjusted for the amount of reactant, is equivalent to the enthalpy change.

To find ∆H, first, the mass of the solution needs to be calculated assuming the density is 1.00 g/mL. Then ∆Q can be determined using the specific heat capacity (4.2 J/g·K), the mass of the solution, and the temperature change. Finally, ∆H per mole of reactant can be calculated by dividing ∆Q by the moles of the limiting reactant. This approach illustrates how calorimetry experiments can provide valuable insights into the thermochemical properties of reactions.

The enolate ion is formed from a carbonyl function via a strong base, leading to resonance structures between keto and enolate forms. The resonance involves shifting of electron pairs, with the enolate form being more prevalent under basic conditions.

An enolate ion is formed when a strong base, B, abstracts a proton α to a carbonyl function. The negatively charged oxygen donates a pair of electrons, pushing the electrons from the double bond into the alpha carbon to form the enolate form. The curved arrow represents the movement of a pair of electrons.

The resonance of the keto form to the enolate form involves shifting of electron pairs, with the electron pair on the carbon moving to form a C=O bond and the π bond between carbon and oxygen moving to the oxygen, forming a p-orbital with a lone pair of electrons and making the oxygen negatively charged.

By this mechanism, the keto and enolate forms are interconvertible with the enolate form being more prevalent under basic conditions. The full resonance structures for both the keto and enolate forms would consist of all their bonds, charges, and nonbonding electron pairs.

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Answer:

No, the copper metal weighs more

Explanation:

When the electrochemical cell is set up, zinc will function as the oxidation half cell (anode) while copper will function as the reduction half cell(cathode).

The cathode increases in size due to deposition of the anode metal on the cathode rod. Hence copper (the cathode) will be heavier in size while the anode zinc will be lighter and smaller in size than the cathode.

Answer: pH of the solution after the addition of 600.0 ml of LiOH is 7.

Explanation:

The given data is as follows.

    Volume of [tex]HClO_{4}[/tex] = 900.0 ml = 0.9 L,

   Molarity of [tex]HClO_{4}[/tex] = 0.18 M,

Hence, we will calculate the number of moles of [tex]HClO_{4}[/tex] as follows.

       No. of moles = Molarity × Volume

                             = 0.18 M × 0.9 L

                             = 0.162 moles

Volume of NaOH = 600.0 ml = 0.6 L

Molarity of NaOH = 0.27 M

No. of moles of NaOH = Molarity × Volume

                                     = 0.27 M × 0.6 L

                                     = 0.162 moles

This shows that the number of moles of [tex]HClO_{4}[/tex] is equal to the number of moles of NaOH.

Also we know that,

         [tex]HClO_{4} + NaOH \rightarrow NaClO_{4} + H_{2}O[/tex]

As 1 mole of [tex]HClO_{4}[/tex] reacts with 1 mole of NaOH then all the hydrogen ions and hydroxide ions will be consumed.

This means that pH = 7.

Thus, we can conclude that pH of the solution after the addition of 600.0 ml of LiOH is 7.

Answer:

The correct answer is the first option. No, the structures above cannot undergo a Fischer esterification reaction to form an ester.

Explanation:

The reaction that will take place can be found in the attached file. The reaction does not require any catalyst and it cannot undergo a Fischer esterification reaction to form an ester.

The accurate answer is the first option which is No, the structures above cannot undergo a Fischer esterification response to form an ester.

The reaction that will take the spot can be found in the connected file. The reaction does not require any catalyst and it cannot experience a Fischer esterification reaction to create an ester.

Therefore the correct option is No, the structures above cannot experience a Fischer esterification reaction to form an ester.

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Answer:

35.6 g

Explanation:

Step 1: Calculate the [H⁺]

We use the following expression

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -13.90 = 1.259 × 10⁻¹⁴ M

Step 2: Calculate the [OH⁻]

We use the ionic product of water (Kw).

Kw = 10⁻¹⁴ = [H⁺] × [OH⁻]

[OH⁻] = 0.7943 M

Step 3: Calculate the moles of OH⁻

[tex]\frac{0.7943mol}{L} \times 0.800 L = 0.635 mol[/tex]

Step 4: Calculate the required moles of KOH

KOH is a strong base that dissociates according to the following equation.

KOH → K⁺ + OH⁻

The molar ratio of KOH to OH⁻ is 1:1. Then, the required moles of KOH are 0.635 moles.

Step 5: Calculate the mass of KOH

The molar mass of KOH is 56.11 g/mol.

[tex]0.635 mol \times \frac{56.11g}{mol} =35.6 g[/tex]

All angiosperms have flowers at some stage in their life.

Angiosperms have small pollen grains that spread genetic information from flower to flower.

All angiosperms have stamens.

Angiosperms have much smaller female reproductive parts than non-flowering plants, allowing them to produce seeds more quickly.

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Answer: Co lose two electrons and thus gets oxidized and acts as reducing agent.

[tex]F_2[/tex] gain two electrons and thus gets reduced and acts as oxidizing agent.

Explanation:

Oxidation reaction : When there is an increase in oxidation state number by loss of electrons

Reduction reaction : when there is a decrease in oxidation state number by gain of electrons.

[tex]Co(s)+F_2(g)\rightarrow Co^{2+}(aq)+2F^-(aq)[/tex]

Cobalt metal has undergone oxidation, as its oxidation state is changing from 0 to 2+.

Florine gas has undergone reduction, as its oxidation state is changing from 0 to -1.

The chemical agent which itself get oxidized and reduce others is called reducing agent. Thus Co is a reducing agent.

The chemical agent which itself get reduced and oxidize others is called oxidizing agent. [tex]F_2[/tex] is an oxidizing agent.

Find the figure in the attachment

Answer:

copper

Explanation:

These are usually copper wires with no insulation. They make the path through which the electricity flows. One piece of the wire connects the current from the power source (cell) to the load

Answer:

1. The balanced equation is given below:

CaCO3 —> CaO + CO2

2. The coefficients are: 1, 1, 1

Explanation:

Step 1:

The word equation is given below:

calcium carbonate decomposes to produce calcium oxide and carbon dioxide.

Step 2:

The elemental equation. This is illustrated below:

Calcium carbonate => CaCO3

calcium oxide => CaO

carbon dioxide => CO2

CaCO3 —> CaO + CO2

A careful observation of the above equation shows that the equation is balanced.

The coefficients are: 1, 1, 1

Final answer:

The balanced equation for the decomposition reaction of calcium carbonate is CaCO3(s) → CaO(s) + CO2(g), representing the formation of calcium oxide and carbon dioxide gas from calcium carbonate.

Explanation:

The decomposition reaction of calcium carbonate forming calcium oxide and carbon dioxide can be represented by the following balanced chemical equation:

CaCO3(s) → CaO(s) + CO2(g)

This equation indicates that one mole of calcium carbonate (CaCO3) decomposes to form one mole of calcium oxide (CaO) and one mole of carbon dioxide (CO2) gas upon heating. The reactants and products are represented using their smallest possible integer coefficients, reflecting the principle of conservation of mass and the stoichiometry of the decomposition reaction.

Answer:

B. CH3COOH pH > 4.7 (4.8)

Explanation:

∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol

⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol

⇒ C CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)

⇒ C CH3COOH = 0.0905 M

∴ mol CH3COONa = (0.05 L )*(0.20 mol/L) = 0.01 mol

⇒ C CH3COONa =  (0.01 mol + 5 E-4 mol) / (0.105 L )

⇒ C CH3COONa = 0.1 M

∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5

⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])

⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]

⇒ [H3O+]² + 0.1000175[H3O+] - 1.584 E-6 = 0

⇒ [H3O+] = 1.5835 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = - Log (1.5835 E-5)

⇒ pH = 4.8004 > 4.7

Final answer:

After adding NaOH to the acetate buffer, the acetate ion concentration increases and the pH of the solution will be greater than the pKa of acetic acid (4.7). Option D, with C2H3O2- and pH > 4.7, is correct.

Explanation:

The question involves calculating the pH of an acetate buffer solution after the addition of sodium hydroxide, NaOH. The buffer consists of acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2). The pKa of acetic acid is 4.7, and equal concentrations of the buffer components are typically ideal. When NaOH is added, it neutralizes some of the acetic acid, converting it to acetate and increasing the pH slightly.

Before the addition of NaOH, the buffer has equal concentrations of acetic acid and acetate, so its pH is close to the pKa of acetic acid, which is 4.7. After adding NaOH, more acetic acid is converted to acetate, but the pH will still be maintained close to the pKa since the buffer action resists changes in pH. Therefore, the correct pairing is that the acetate ion (C2H3O2-) will be present in greater concentration and the pH of the solution will be greater than 4.7 after NaOH is added, making option D correct.

Question: The question is incomplete. Below is the complete question and the answer;

While ethanol (CH3CH2OH is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene CH2CH2) with water vapor at elevated temperatures. A chemical engineer studying this reaction fills a 50.0 L tank at 22. °C with 24. mol of ethylene gas and 24. mol of water vapor. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 15.4 mol of ethylene gas and 15.4 mol of water vapor The engineer then adds another 12. mol of water, and allows the mixture to come to equilibrium again. Calculate the moles of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

Answer:

Number of moles of ethanol = 11 mol

Explanation:

SEE THE ATTACHED FILE FOR THE CALCULATION

Answer:

Fe²⁺(aq) →  Fe³⁺(aq) + 1e⁻

Ce⁴⁺(aq) + 1e⁻ →  Ce³⁺(aq)

Explanation:

The reaction that takes place is:

The oxidation half-reaction is:

It is an oxidation because the oxidation state of Fe increases from 2+ to 3+.

The reduction half-reaction is:

It is a reduction because the oxidation state of Ce decreases from 4+ to 3+.

Answer:

Problem 1 => 8778 joules

Problem 2 => 14,630 joules

Explanation:

Reference the Heating Curge of Water Problem posted earlier. These are temperature change fragments of that type problem. As you read the problem note the 'temperature change' phrase in the problem. This should signal you to use the q = m·c·ΔT expression as opposed to the q = m·ΔH expression where no temperature change is noted; i.e., melting or evaporation/boiling.

For the listed problems...

Problem 1: Amount of heat needed to heat 150 grams water from 21.0 to 35.0 Celcius.

Note the temperature change in the problem context => use q = m·c·ΔT ...

q = (150g)(4.18j/g·°C)(35.00°C - 21.0°C) = 8778 joules (4 sig. figs. based on the 150.0g value of water)

Problem 2: Amount of heat needed to heat 250.0 grams water from 31.0 to 45.0 Celcius.

Same type problem. Note temperature change in text of problem => use q = m·c·ΔT ...

q = (250.0g)(4.18j/g·°C)(45.0°C - 31.0°C) = 14,630 joules

Hope this helps. Doc

Final answer:

The balanced thermochemical equation for the endothermic reaction of CO2(g) and H2(g) forming C2H2(g) and H2O(g), with the energy absorption of 23.3 kJ per mole of CO2, is 2CO2(g) + 5H2(g) + 46.6 kJ → 2C2H2(g) + 4H2O(g). The energy term is positive and on the reactant side due to the endothermic nature of the reaction.

Explanation:

The question involves writing a balanced thermochemical equation for the reaction where carbon dioxide (CO2) and hydrogen gas (H2) react to form acetylene (C2H2) and water (H2O), with an energy absorption of 23.3 kJ per mole of CO2 that reacts. Since energy is absorbed, it is an endothermic reaction, and the energy term will appear on the reactant side of the equation with a positive sign.

The balanced thermochemical equation is:

2CO2(g) + 5H2(g) + 46.6 kJ → 2C2H2(g) + 4H2O(g)

Note that we multiply the energy absorbed per mole by the number of moles of CO2, which is 2, resulting in a total energy absorption of 46.6 kJ for the reaction as written.

Answer:

[H3O+]= 2.88×10^-12M

pH= 11.54

Explanation:

pOH= -log[OH] = -log[3.5*10^-3]= 2.46

pH= 14- pOH= 14-2.46= 11.54

High pH means that a solution is basic while high pOH means that a solution is acidic,

Hence the solution is basic

Answer: The Molar mass of Cro2 (or Chromium (IV) oxide) is 83.9949 g/mol.

Explanation: The atomic mass of Chromium is 51.9961 and the atomic mass of Oxygen is 15.9994. You then add both numbers and multiply by 2. Put it in the calculator like this: 51.9961 + 15.9994 x 2 and you will get 83.9949 g/mol.

Answer:

The structure in the first image file attached below shows the arrangement of atoms in hexagonal close packing

we need to show that the ratio between the height of the unit cell divided by its edge length is 1.633

In the structure, the two atoms are shown apart. But in fact the two atoms are touching. Therefore, the edge length is the sum of the radius of two atoms. If we assume r as radius then the expression for edge length will be as follows.

a = 2r    ..............(1)

other attached files show additional solutions in steps

The ideal c/a ratio in an HCP unit cell is 1.633, depicting the most compact arrangement of spheres. However, real HCP metals, influenced by bonding characteristics and atomic sizes, display varying c/a ratios. For Mg with known atomic radius in HCP arrangement, the lattice constants, c/a ratio, and theoretical density can be calculated.

The HCP unit cell (hexagonal-close-packed) is one of the simplest lattice types in crystal structures. It includes two types of atoms, one at the corners (and center) of a hexagonal prism and the other at the top and bottom of the prism. The ideal ratio of c/a (height of the unit cell c divided by edge length a) corresponds to the most compact possible arrangement of spheres, and is theoretically calculated to be √(8/3), or approximately 1.633.

In real-world applications, various HCP metals display c/a ratios varying slightly, such as 1.58 for Be and 1.89 for Cd. This depends on bonding characteristics and varying atomic sizes in the metal crystal structure. For example, for Mg (magnesium) which has a known atomic radius of 0.1605 nm in HCP arrangement, we can calculate the lattice constants, c and a, and thus the c/a ratio. Theoretical density can be determined using the mass of atoms in a unit cell, Avogadro's number, and the volume of unit cell.

Learn more about HCP Unit Cell Geometry here:

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Answer: The value of [tex]E^{o}[/tex] for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          [tex]AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)[/tex]

Its solubility product will be as follows.

       [tex]K_{sp} = [Ag^{+}][Cl^{-}][/tex]

Cell reaction for this equation is as follows.

     [tex]Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)[/tex]

Reduction half-reaction: [tex]Ag^{+} + 1e^{-} \rightarrow Ag(s)[/tex],  [tex]E^{o}_{Ag^{+}/Ag} = 0.799 V[/tex]

Oxidation half-reaction: [tex]Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}[/tex],   [tex]E^{o}_{AgCl/Ag}[/tex] = ?

Cell reaction: [tex]Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)[/tex]

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, [tex]K_{sp} = [Ag^{+}][Cl^{-}][/tex]

                                               = [tex]1.77 \times 10^{-10}[/tex]

Therefore, according to the Nernst equation

           [tex]E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}[/tex]

At equilibrium, [tex]E_{cell}[/tex] = 0.00 V

Putting the given values into the above formula as follows.

         [tex]E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}[/tex]

        [tex]0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}[/tex]    

       [tex]E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}[/tex]

                  = [tex]0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}[/tex]

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           [tex]E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}[/tex]

       [tex]0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}[/tex]

       [tex]0.577 V = 0.799 V - E^{o}_{AgCl/Ag}[/tex]

       [tex]E^{o}_{AgCl/Ag}[/tex] = 0.222 V

Thus, we can conclude that value of [tex]E^{o}[/tex] for the half-cell reaction is 0.222 V.

The E° for the half-reaction can be calculated by using the Nernst equation, solubility product constant (Ksp) and the standard reduction potential for Ag+(aq) + e− ⇀ ↽ − Ag(s) which is +0.799 V.

The E° for the half-reaction, AgCl (s) + e− ⇀ ↽ − Ag (s) + Cl− (aq) can be calculated using the Nernst equation, considering the relationship between standard reduction potential and solubility product constant.

First, we can rewrite the half-reaction for the dissolution of AgCl: AgCl(s) ⇀ ↽ − Ag+(aq) + Cl−(aq). The solubility product constant (Ksp) for this reaction is 1.77 × 10^−10.

We also know that the standard reduction potential of Ag+(aq) + e− ⇀ ↽ − Ag(s) is +0.799 V.

We can relate this data through the Nernst equation: E=E°−(RT/nF)lnQ. Here, we can set Q as Ksp and adjust the equation to solve for our wanted E.

Using these values in Nernst equation would yield the E for the half-reaction.

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The following are the definition for the following terms:

Explanation:

1.  Kinetic molecular theory:

(a) The particles that the compose a gas are so small compared to the distances between them that the volume of the individual particles can be assumed to be neglisible.

2.  The definition of a Diffusion:

(a) The movement of a gas particle from one area to another.

3.  The definition of Effusion:

(c) The escape of a gas thorugh a small opening in a barrier in to a region of lower pressure.

Diffusion is the process of gaseous molecules moving from regions of high concentration to low concentration until a uniform concentration is achieved. Effusion, similar to diffusion, involves gas escaping from a container to a vacuum through a small opening, often from high pressure to low pressure.

In the realm of physics, Diffusion can be defined as the process whereby gaseous atoms and molecules move from regions of relatively higher concentration to regions of lower concentration. The molecules, moving freely and randomly, are transferred due to their inherent kinetic energy. This movement continues until there is a uniform concentration of the molecules throughout, achieving a state of dynamic equilibrium.

On the other hand, Effusion is a similar process, with the key difference being that the gaseous species escape from a container to a vacuum through very small openings or orifices. This escape often occurs from regions of high pressure to regions of low pressure. Just like in diffusion, the rates of effusion are influenced by factors such as the molar mass of the gas involved; however, the rates are not equal.

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