When CO2(g) reacts with H2(g) to form C2H2(g) and H2O(g), 23.3 kJ of energy are absorbed for each mole of CO2(g) that reacts. Write a balanced thermochemical equation for the reaction with an energy term in kJ as part of the equation. Note that the answer box for the energy term is case sensitive. Use the SMALLEST INTEGER coefficients possible and put the energy term in the last box on the appropriate side of the equation. If a box is not needed, leave it blank. 2CO2 + 5H2 + C2H2 + 4H2O + -46.6kJ

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = [tex]a=31 cm^2[/tex]

Thickness of the gold plating  = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

[tex]V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3[/tex]

Density of the gold = [tex]d=19.3 g/cm^3[/tex]

[tex]m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g[/tex]

Moles of gold = [tex]\frac{29.915 g}{197 g/mol}=0.152 mol[/tex]

[tex]Au^{3+}+3e^-\rightarrow Au[/tex]

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

[tex]\frac{3}{1}\times 0.152 mol=0.456 mol[/tex] of electrons

Number of electrons = N =[tex]0.456\times \times 6.022\times 10^{23}[/tex]

Charge on single electron = [tex]q=1.6\times 10^{-19} C[/tex]

Total charge required = Q

[tex]Q=N\times q[/tex]

Amount of current passes = I = 8 Ampere

Duration of time  = T

[tex]I=\frac{Q}{T}[/tex]

[tex]T=\frac{N\times q}{I}[/tex]

[tex]=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s[/tex]

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

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.

Hope this helped

:)

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)

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:

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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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.

An explosion in a pressure cooker can be explained by Boyle's Law, which states that pressure and volume are inversely related at constant temperature.So,option B is correct.

The explosion of a pressure cooker while making rice can be explained by Boyle's Law, which relates the volume and pressure of a gas under conditions of constant temperature. According to Boyle's Law, if a gas is compressed to a smaller volume without changing the temperature, the pressure of the gas increases. In a pressure cooker, when the steam cannot escape, the pressure continues to rise, and if the volume is constricted, the cooker may not be able to withstand the increased pressure, leading to an explosion. Therefore, the correct answer to the question is B) Boyle’s Law (relationship between pressure and volume).

Answer:

1.20 V

Explanation:

The standard cell potential is calculated from the expression

ε⁰ cell = ε⁰ oxidation + ε⁰  reduction

The species that will be reduced is the one with the higher standard reduction potential and the species that will be oxidized will be the one with the more negative reduction potential.

Thus for our question we will have

oxidation:

Pb(s)  →   Pb2+(aq) + 2 e-       ε⁰ oxidation       =  -   ε⁰  reduction

                                                                          =   - ( - 0.13 V ) = + 0.13 V

reduction    

Br2(l) + 2 e- → 2 Br-(aq)           ε⁰  reduction     = +1.07 V

ε⁰ cell = ε⁰ oxidation + ε⁰  reduction = + 0.13 V + 1.07 V  = 1.20 V

Answer:

A catalyst increases the rate of reaction by decreasing activation energy. ... Enzymes are highly substrate specific and catalyze reactions by providing an alternate pathway of lower activation energy.

Explanation:

I hope this helps :)

To calculate the reaction free energy ΔG for this reaction, we need to use the standard free energy of formation values given in a data tab, the stoichiometry of the reaction, and the specific conditions of the reaction, including the concentrations of Pb2+ and Br−. After a series of calculations, we will get the ΔG value in joules, which can be converted to kilojoules.

The task here is to calculate the reaction free energy ΔG for the Pb2+(aq) + 2Br−(aq) = PbBr2(s) reaction at 25.0°C. From the given information, we can start by calculating the number of moles of PbBr2 from its mass. Then, referring to the thermodynamic data tab of the ALEKS, we find the standard free energy of formation (ΔGf°) values for Pb2+(aq), Br−(aq), and PbBr2(s). Now, we can use these values and the definition of ΔG for a reaction in terms of ΔGf° values and stoichiometry.

ΔG = ΣΔGf°(products) - ΣΔGf°(reactants).

Note that the equation must be balanced so each ΔGf° value is multiplied by the stoichiometric coefficient of that substance in the reaction. It is also important to remember to convert the answer to kilojoules if the ΔGf° values are given in joules/mole. Lastly, the concentrations of Pb2+ and Br− are included in the reaction quotient Q to show the reaction's non-standard conditions.

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The reaction free energy under the given conditions[tex]\Delta G \approx -472 \text{ kJ/mol}[/tex].

The reaction free energy for the given chemical reaction can be calculated using the standard free energy of formation for each substance involved in the reaction. The reaction is as follows:

[tex]\[ \text{Pb}^{2+} (aq) + 2\text{Br}^- (aq) \rightarrow \text{PbBr}_2 (s) \][/tex]

The standard free energy change \( \Delta G^\circ \) for the reaction is given by:

[tex]\[ \Delta G^\circ = \Delta G^\circ_f (\text{PbBr}_2) - \left( \Delta G^\circ_f (\text{Pb}^{2+}) + 2 \cdot \Delta G^\circ_f (\text{Br}^-) \right) \][/tex]

Using the thermodynamic data provided in the ALEKS Data tab, we have:

[tex]- \( \Delta G^\circ_f (\text{PbBr}_2) = -260.8 \text{ kJ/mol} \)\\ - \( \Delta G^\circ_f (\text{Pb}^{2+}) = -24.4 \text{ kJ/mol} \)\\ - \( \Delta G^\circ_f (\text{Br}^-) = -121.5 \text{ kJ/mol} \)[/tex]

Plugging in these values, we get:

[tex]\[ \Delta G^\circ = -260.8 \text{ kJ/mol} - \left( -24.4 \text{ kJ/mol} + 2 \cdot (-121.5 \text{ kJ/mol}) \right) \] \[ \Delta G^\circ = -260.8 \text{ kJ/mol} + 24.4 \text{ kJ/mol} - 2 \cdot 121.5 \text{ kJ/mol} \] \[ \Delta G^\circ = -260.8 \text{ kJ/mol} + 24.4 \text{ kJ/mol} - 243.0 \text{ kJ/mol} \] \[ \Delta G^\circ = -260.8 \text{ kJ/mol} - 218.6 \text{ kJ/mol} \] \[ \Delta G^\circ = -479.4 \text{ kJ/mol} \][/tex]

Relationship:

[tex]\[ \Delta G = \Delta G^\circ + RT \ln(Q) \][/tex]

This reaction is:

[tex]\[ Q = \frac{[\text{PbBr}_2]}{[\text{Pb}^{2+}][\text{Br}^-]^2} \][/tex]

Simplifies to:

[tex]\[ Q = \frac{1}{[\text{Pb}^{2+}][\text{Br}^-]^2} \][/tex]

The initial concentrations are:

[tex]- \( [\text{Pb}^{2+}] = 0.750 \text{ M} \) - \( [\text{Br}^-] = 0.232 \text{ M} \)[/tex]

So:

[tex]\[ Q = \frac{1}{(0.750 \text{ M})(0.232 \text{ M})^2} \] \[ Q = \frac{1}{0.750 \cdot 0.053824 \text{ M}^3} \] \[ Q = \frac{1}{0.040368 \text{ M}^3} \] \[ Q \approx 24.77 \][/tex]

Calculation:

[tex]\[ \Delta G = -479.4 \text{ kJ/mol} + (8.314 \text{ J/(mol·K)} \cdot 298.15 \text{ K}) \cdot \ln(24.77) \] \[ \Delta G = -479.4 \text{ kJ/mol} + (2477.57 \text{ J/mol}) \cdot \ln(24.77) \] \[ \Delta G = -479.4 \text{ kJ/mol} + (2477.57 \text{ J/mol}) \cdot 3.183 \] \[ \Delta G = -479.4 \text{ kJ/mol} + 7896.5 \text{ J/mol} \] \[ \Delta G = -479.4 \text{ kJ/mol} + 7.8965 \text{ kJ/mol} \] \[ \Delta G \approx -471.5 \text{ kJ/mol} \][/tex]

Rounding to the nearest kilojoule:

[tex]\[ \boxed{\Delta G \approx -472 \text{ kJ/mol}} \][/tex]

This is the reaction free energy under the given conditions.

The answer is: [tex]\Delta G \approx -472 \text{ kJ/mol}[/tex].

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:

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.

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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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:

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.

Answer:

c. 9.94 g

Explanation:

From the question,

Using

mt = m₀e⁻kt.................... Equation 1

Where mt = mass of the leaf remaining in the bag, m₀ = original mass of leave that was placed in the bag, k = decay constant, t = time.

Given: m₀ = 33 g, k = 0.04, t = 30 days.

Substitute into equation 1

mt = 33(e⁻(0.04ˣ30))

mt = 33e⁻¹²/¹⁰

mt = 33/e¹²/¹⁰

mt = 33/3.320

mt = 9.94 g.

Hence the right answer is c. 9.94 g

The system equations that could be used to determine the number of tacos sold and the number of burritos sold is 17.5x = 595. The variable x is the number of tacos sold.

Let the number of tacos sold = x

Let the number of burritos sold = y

It has been told the revenue for the selling of both taco and burrito = 595 $.

The price of 1 taco = $3

The price of 1 burrito = $7.25

So, 3x + 7.25y = 595. ......(i)

It has been told that the number of burritos sold is twice the number of taco sold.

Since the number of tacos sold = x

Number of burritos sold = 2x

Number of burritos sold = y

y = 2x.

Substituting the value of y in the equation (i).

3x + 7.25 (2x) = 595

3x + 14.5x = 595

17.5x = 595.

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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:

[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:

Step 1) hydrolysis using NaOH/H2O to form benzylalcohol

Step2) oxidation to Carboxylic acid using KMnO4 followed by decarboxylation to form benzene

3) friedel craft acylation using CH3COCl/AlCl3

Explanation:

The above 3 steps will yield acetophenone from methylbenzoate

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: 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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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.

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.

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:

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.

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Answer:

C2= 0.16M

Explanation:

C1= 2M, V1= 20ml, C2= ?, V2= 250ml

Applying dilution formula

C1V1= C2V2

2×20 =C2×250

C2= 0.16M

Answer:

Aldol

Explanation:

The reaction will be an aldol.