The activation energy for the reaction NO2 (g )+ CO (g) ⟶ NO (g) + CO2 (g) is Ea = 217 kJ/mol and the change in enthalpy for the reaction is ΔH = -293 kJ/mol .
What is the activation energy for the reverse reaction?

Enter your answer numerically and in terms of kJ/mol.

Explanation:

The chemical reaction is as follows.

            [tex]CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O[/tex]

It is given that 2094 [tex]ft^{3}/hr[/tex]. And, it is known that 1 [tex]m^{3}/s[/tex] = 127133 [tex]ft^{3}/hr[/tex]

Hence, convert 2094 [tex]ft^{3}/hr[/tex] into [tex]m^{3}/s[/tex] as follows.

                [tex]\frac{2094 ft^{3}/hr}{127133 ft^{3}/hr} \times 1 m^{3}/s[/tex]

                  = [tex]0.0165 m^{3}/s[/tex]

As ideal gas equation is PV = nRT. So, calculate the number of moles as follows.

                   n = [tex]\frac{PV}{RT}[/tex]

                      = [tex]\frac{1 atm \times 0.0165 m^{3}}{0.0821 \times 298 K}[/tex]

                       = 0.673 mol/sec

According to the stoichiometry of the given reaction, 1 mol of methane reacts with 2 mol of oxygen.

So, 1 mol [tex]CH_{4}[/tex] = 2 mol [tex]O_{2}[\tex]

Hence, [tex]O_{2}[\tex] required theoretically = [tex]2 \times 0.673 mol/s[/tex] = 1.346 mol/s

Hence, air required theoretically = [tex]\frac{1.346}{0.21}[/tex] = 6.4095 mol/s.

Since, 6% of excess air is being supplied. Therefore, total air supplied will be calculated as follows.

                Total air supplied = [tex]6.4095 mol/s [1 + \frac{6}{100}][/tex]

                                              = 6.794 mol/s

Now, calculate the volume using ideal gas law equation as follows.

                            PV = nRT

           [tex]1 atm \times V = 6.794 mol/s \times 8.21 \times 10^{-5} Latm/K mol \6 times 298 K[/tex]

                           V = 0.166229 [tex]m^{3}/s[/tex]

Converting calculated volume into [tex]ft^{3}/hr[/tex] as follows.

                  1 [tex]m^{3}/s[/tex] = 127133 [tex]ft^{3}/hr[/tex]

So,        0.166229 [tex]m^{3}/s[/tex] = [tex]0.166229 m^{3}/s \times 127133 \frac{ft^{3}/hr}{1 m^{3}/s}[/tex]  

                                        = 21133.191 [tex]ft^{3}/hr[/tex]

Thus, we can conclude that 21133.191 [tex]ft^{3}/hr[/tex] of air are drawn from outside  per hour by the fan that supplies the air.

Answer:

The pump work is 3451 kJ/s

Explanation:

Pump work (W) is calculated as

[tex] W = (h_f - h_i) \times \dot{m}[/tex]

where

[tex] h_f [/tex] is the enthalpy of water at its final state

[tex] h_i [/tex] is the enthalpy of water at its initial state

[tex]\dot{m} = 10 kg/s [/tex] is water mass flow

For liquids, properties are evaluated as saturated liquid. From the figure attached, it can be seen that

[tex] h_f = 762.81 kJ/kg [/tex]

[tex] h_i = 417.46 kJ/kg [/tex]

Replacing

[tex] W = (762.81 kJ/kg - 417.46 kJ/kg) \times 10 kg/s[/tex]

[tex] W = 3451 kJ/s [/tex]

Answer:

a. T and P remain the same (T=298 K and P=1 bar)

b. 11.23J/K

Explanation:

a. Since the mixing process of an idea gas doesn't present a change in the enthalpy, we could state that no change in neither temperature and pressure are given.

b. It is not necessary to know enthalpy data, the following formula is enough to compute the entropy change:

Δ[tex]S_{mix}=-n_{N_2}R ln(x_{N_2})-n_{O_2}R ln(x_{O_2})[/tex]

Thus, the molar fractions are equal to 0.5, and the result yields:

Δ[tex]S_{mix}=-(1mol)[(8.314J/(mol*K)]ln(0.5)-(1mol)[(8.314J/(mol*K)]ln(0.5)[/tex]

Δ[tex]S_{mix}=11.23J/K[/tex]

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Answer: The mass of water produced in the reaction is 47.25 grams.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of ammonia = 29.7 g

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of ammonia}=\frac{29.7g}{17g/mol}=1.75mol[/tex]

The given chemical reaction follows:

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]

By stoichiometry of the reaction:

4 moles of ammonia produces 6 moles of water.

So, 1.75 moles of ammonia will produce = [tex]\frac{6}{4}\times 1.75=2.625mol[/tex] of water.

Now, calculating the mass of water by using equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 2.625 moles

Putting values in equation 1, we get:

[tex]2.625mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=47.25g[/tex]

Hence, the mass of water produced in the reaction is 47.25 grams.

The complete reaction of 29.7 grams of NH3 would produce 47.099 grams of water by converting the given mass of NH3 to moles, using the balanced chemical equation to find the ratio of NH3 to H2O, and then converting the moles of H2O to grams.

To find out how many grams of water are produced in the complete reaction of 29.7 grams of ammonia (NH3), we need to use stoichiometry, which involves several steps:

Therefore, the complete reaction of 29.7 grams of NH3 would produce 47.099 grams of water (H2O).

Answer:

Ka = [tex]2.32 \times 10^{-4}[/tex]

pH = 2.12

Explanation:

Calculation of Ka:

% Dissociation = 3% = 0.03

Concentration of solution = 0.25 M

HA dissociated as:

       [tex]HA \rightarrow H^+ + A^{-}[/tex]

      C(1 - 0.03)    C×0.03   C×0.03

HA] after dissociation = 0.25×0.97 = 0.2425 M

[tex][H^+]= 0.25\times0.03 = 0.0075 M[/tex]

[tex][A^{-}]= 0.25 \times 0.03 = 0.0075 M[/tex]

[tex]Ka= \frac{[H^+][A^{-}]}{[HA]}[/tex]

[tex]Ka = \frac{(0.0075)^2}{0.2475} =2.32 \times 10^{-4}[/tex]

Calculation of pH of the solution

[tex]pH = -log [H^+][/tex]

[tex]H^+[\tex] = 0.0075 M[/tex]

[tex]pH = -log 0.0075 = 2.12[/tex]

The Ka of the weak acid is calculated using the concentrations of the dissociated and non-dissociated parts of the weak acid and the pH is derived from the concentration of H+ ions.

The subject in concern relates to the dissociation of a weak acid and the calculation of the Acid dissociation constant (Ka) and the pH of the solution.

We are given that 3% of the weak acid, HA, is dissociated. Meaning, 3% of 0.25 M HA dissociates into H+ and A-. This will be 0.03 * 0.25 M = 0.0075 M.

The equilibrium for the reaction is HA <--> H+ + A-. Due to water autoionization, the concentration of water is approximately taken as constant in the denominator of Ka calculation. Hence, Ka for HA can be approximated as [H+][A-] / [HA]. We know that [H+] = [A-] = x, and [HA] = 0.25 - x. We approximate x as 0.0075 since only 3% of HA dissociates. Hence Ka=~(0.0075)^2/(0.25-0.0075).

The pH of the solution is -log[H+] = -log(0.0075).

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To calculate the mass of iodine solid that participates in a chemical reaction, multiply the number of moles of iodine by its molar mass.

To calculate the mass of iodine solid that participates in a chemical reaction, we need to use the concept of moles and the molar mass of iodine. Given that 0.0200 moles of iodine solid participate in the reaction, we can convert moles to grams by multiplying the number of moles by the molar mass of iodine. The molar mass of iodine (I2) is 253.8 grams per mole. Multiplying 0.0200 moles by 253.8 grams/mole gives us a mass of 5.08 grams of iodine solid that participates in the chemical reaction.

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Answer: The molarity of sulfuric acid solution is 2.3 M

Explanation:

The relationship between specific gravity and density of a substance is given as:

[tex]\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}[/tex]

Specific gravity of sulfuric acid solution = 1.13

Density of water = 1.00 g/mL

Putting values in above equation we get:

[tex]1.13=\frac{\text{Density of sulfuric acid solution}}{1.00g/mL}\\\\\text{Density of sulfuric acid solution}=(1.13\times 1.00g/mL)=1.13g/mL[/tex]

We are given:

20% (m/m) sulfuric acid solution. This means that 20 g of sulfuric acid is present in 100 g of solution

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of solution = 1.13 g/mL

Mass of Solution = 100 g

Putting values in above equation, we get:

[tex]1.13g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.13g/mL}=88.5mL[/tex]

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

We are given:

Mass of solute (sulfuric acid) = 20 g

Molar mass of sulfuric acid = 98 g/mol

Volume of solution = 88.5 mL

Putting values in above equation, we get:

[tex]\text{Molarity of solution}=\frac{20g\times 1000}{98g/mol\times 88.5mL}\\\\\text{Molarity of solution}=2.3M[/tex]

Hence, the molarity of sulfuric acid solution is 2.3 M

Explanation:

The given data is as follows.

                 [tex]mass_{1}[/tex] = 5 g,        [tex]Volume_{1}[/tex] = 1 L

                 [tex]mass_{2}[/tex] = 1 g,         [tex]Volume_{1}[/tex] = ?

No. of moles of helium present in 5 g helium gas are as follows.

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

                                         = [tex]\frac{5 g}{4 g/mol}[/tex]  

                                         = 1.25 mol

No. of moles of helium present in 1 g helium gas are as follows.

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

                                         = [tex]\frac{1 g}{4 g/mol}[/tex]  

                                         = 0.25 mol

According to the ideal gas equation, PV = nRT. And, since temperature and pressure are held constant. Therefore,

                      [tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]

                      [tex]\frac{1 L}{1.25 mol} = \frac{V_{2}}{0.25 mol}[/tex]

                             [tex]V_{2}[/tex] = 0.2 L

Thus, we can conclude that the new volume of the balloon is 0.2 L.

The final volume of the balloon after adding 5.00 g of helium to the already present 1.00 g of helium, under constant temperature and pressure, is 6.00 L.

The subject of your question pertains to the ideal gas law which is represented by the formula PV = nRT, where P is pressure, V is volume, n is the number of moles, R refers to the gas constant, and T is temperature. Since the question states that there is no change in temperature or pressure, you can use Avogadro's law which states that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules.

In your problem, you are adding more helium to the balloon. Therefore, the volume of the balloon will increase proportionally. Since the initial volume of 1.00 L corresponds to 1.00 g of Helium, 5.00 g of Helium will correspond to 5.00 L. This is because the amount of Helium has increased by a factor of 5, and thus, the volume will also increase by a factor of 5.

Therefore, the final volume of the balloon will be the sum of the initial volume and the increase, which equals 1.00 L + 5.00 L = 6.00 L.

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Answer:

Head loss in turbulent flow is varying as square of velocity.

Explanation:

As we know that head loss in turbulent flow given as

[tex]h_F=\dfrac{FLV^2}{2gD}[/tex]

Where

F is the friction factor.

L is the length of pipe

V is the flow velocity

D is the diameter of pipe.

So from above equation we can say that

[tex]h_F\alpha V^2[/tex]

It means that head loss in turbulent flow is varying as square of velocity.

We know that loss in flow are of two types

1.Major loss :Due to surface property of pipe

2.Minor loss :Due to change in momentum of fluid.

Answer:

The correct answer is head varies directly with square of velocity of flow

Explanation:

The head loss in a pipe as given by Darcy Weisebach equation is

[tex]h_L=\frac{flv^2}{2gD}[/tex]

where

'f' is friction factor whose value depends on the nature of flow (Laminar/turbulent)

'L' is the length of the section in which the head loss is calculated

'v' is the velocity of the flow

'D' is the diameter of the duct

Thus we can see that the head loss varies with square of velocity of the fluid.

Answer:

[tex]\frac{N_C}{N} = 1.42\times 10^{-4}[/tex]

Explanation:

given data:

temperature  = 552 degree celcius  = 825 Kelvin

energy for vacancy formation is given as 0.63 eV/atom

fraction of atom can be obtained from following formula

[tex]N_C = N e^{\frac{-Q_V}{kT}}[/tex]

Where, K IS BOLTZMAN CONSTANT [tex]= 8.62\times 10^{-5}eV/K[/tex]

[tex]\frac{N_C}{N} =e^{\frac{-0.63}{8.62\times 10^{-5} \times 825}[/tex]

[tex]\frac{N_C}{N} = e^{-8.8858}[/tex]

[tex]\frac{N_C}{N} = 1.42\times 10^{-4}[/tex]

The fraction of atom sites that are vacant for silver at 552°C is approximately [tex]\( 7.347 \times 10^{-5} \).[/tex]

The fraction of atom sites that are vacant for silver at a given temperature can be calculated using the formula derived from the statistics of a canonical ensemble:

[tex]\[ f = \exp\left(-\frac{\Delta G_{\text{form}}}{k_B T}\right) \][/tex]

Now, we can plug in the values:

[tex]\[ f = \exp\left(-\frac{0.63 \text{ eV/atom}}{(8.617 \times 10^{-5} \text{ eV/K}) \times (825.15 \text{ K})}\right) \] \[ f = \exp\left(-\frac{0.63}{8.617 \times 10^{-5} \times 825.15}\right) \] \[ f = \exp\left(-\frac{0.63}{0.0658}\right) \] \[ f = \exp\left(-9.574\right) \] \[ f \approx \exp\left(-9.574\right) \] \[ f \approx 7.347 \times 10^{-5} \][/tex]

Answer : The molecular weight of a substance is 157.3 g/mol

Explanation :

As we are given that 7 % by weight that means 7 grams of solute present in 100 grams of solution.

Mass of solute = 7 g

Mass of solution = 100 g

Mass of solvent = 100 - 7 = 93 g

Formula used :  

[tex]\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=k_f\times\frac{\text{Mass of substance(solute)}\times 1000}{\text{Molar mass of substance(solute)}\times \text{Mass of water(solvent)}}[/tex]

where,

[tex]\Delta T_f[/tex] = change in freezing point

[tex]T_f^o[/tex] = temperature of pure water = [tex]0^oC[/tex]

[tex]T_f[/tex] = temperature of solution = [tex]-0.89^oC[/tex]

[tex]K_f[/tex] = freezing point constant of water = [tex]1.86^oC/m[/tex]

m = molality

Now put all the given values in this formula, we get

[tex](0-(-0.89))^oC=1.86^oC/m\times \frac{7g\times 1000}{\text{Molar mass of substance(solute)}\times 93g}[/tex]

[tex]\text{Molar mass of substance(solute)}=157.3g/mol[/tex]

Therefore, the molecular weight of a substance is 157.3 g/mol

Answer:

12 mL

Explanation:

Density is expressed as mass divided by volume, so the expression can be rearranged to solve for volume:

D = m/V ⇒ V = m/D

V = m/D = (8.410 g)/(0.71g/mL) = 12 mL

Final answer:

The volume of a substance with a mass of 8,410 g and a density of 0.71 g/mL is 12,000 mL, rounded to two significant figures to match the precision of the given density.

Explanation:

To calculate the volume of a substance, you can use the density formula: Density = Mass / Volume. Given the density of the substance is 0.71 g/mL and the mass is 8,410 g, the volume can be found by rearranging the formula to Volume = Mass / Density.

Volume = 8,410 g / 0.71 g/mL = 11,845.07042 mL. This raw calculation has more significant figures than is justified by the precision of the given data. Since the density value has two significant figures, the volume should also be reported with two significant figures, resulting in 12,000 mL.

Answer:

Biuret test does not work for all kinds of proteins. It only works with proteins and peptides that have 2 or more peptide bonds.

Explanation:

The Biuret test is used to determine proteins and polypeptides that contain 2 or more peptide bonds.

It is made of potassium hydroxide and cupric sulfate, along with sodium and potassium tartrate.

It consists of treating a protein with Cu ++ in alkaline medium, producing a violet coloration by forming a coordination complex between the Cu ++ and the free electron pairs of the nitrogens of the amino groups of the peptide junction. At least two peptide bonds are necessary for the reaction to take place.

Therefore, this test does not serve to determine the amount of amino acids and peptides with a single peptide bond such as: tyrosine (amino acid), tryptophan (amino acid), alanine (amino acid), aspartame (peptide)

Final answer:

One photon of microwave radiation with a wavelength of 0.122m has an energy of approximately 1.63 x 10⁻²⁴ joules, calculated using Planck's equation (E=hc/λ).

Explanation:

The energy of a photon can be calculated using the equation E = hc/λ, where E is the energy of the photon in joules, h is Planck's constant (6.626 x 10⁻³⁴ J·s), c is the speed of light in a vacuum (3.00 x 10⁸ m/s), and λ is the wavelength of the photon in meters. Using the given wavelength of 0.122 meters for the microwave radiation, we can calculate the energy of one photon by substituting the values into the equation:

E = (6.626 x 10⁻³⁴ J·s)(3.00 x 10⁸ m/s) / 0.122 m

E = 1.63 x 10⁻²⁴ Joules per photon. Therefore, one photon of microwave radiation with a wavelength of 0.122m has an energy of approximately 1.63 x 10⁻²⁴ joules.

Answer:

Amount of Carbon dioxide equals 4.49625 grams.

Explanation:

From the basic stichometric reaction between carbon and oxygen we know that 1 mole of carbon combines with 1 mole of oxygen to form 1 mole of carbon dioxide.

Thus we can say that 12 grams of carbon combines with 32 grams of oxygen to form 44 grams of carbon dioxide.

In the given question assuming that there is no limited supply of carbon we can find the find the amount of carbon dioxide formed from 3.27 grams of Oxygen using ratio and proportion method.

As we can see that 32 grams of oxygen form 44 grams of carbon dioxide thus we can say 1 gram of oxygen yields [tex]\frac{44}{32}grams[/tex] Carbon Dioxide

Thus the carbon dioxide formed by 3.27 grams of Oxygen equals

[tex]3.27\times \frac{44}{32}=4.49625grams[/tex]

Answer : The correct option is, (A) -x kJ/mol

Explanation :

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

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

The given chemical reaction is,

[tex]A\rightarrow P[/tex] [tex]\Delta H=x\text{ kJ/mole}[/tex]

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

[tex]P\rightarrow A[/tex] [tex]\Delta H'=?[/tex]

According to the Hess’s law, if we reverse the reaction then the sign of [tex]\Delta H[/tex] change.

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

[tex]\Delta H'=-(x\text{ kJ/mole})[/tex]

[tex]\Delta H'=-x\text{ kJ/mole}[/tex]

Hence, the value of [tex]\Delta H[/tex] for the reaction is -x kJ/mole.

Answer:

The number of moles of calcium phosphate is 0.0485 mol. The mass is 15.1 g.

Explanation:

The balanced equation is:

[tex]3 Ca(OH)_{2} + 2 H_{3}PO_{4} => Ca_{3}(PO_{4})_{2} + 6H_{2}O[/tex]

As we can see, 3 moles of calcium hidroxide produced 1 mol of calcium phosphate. So the quantity of moles of Calcium posphate is:

[tex]10.8g Ca(OH)_{2}*\frac{1 molCa(OH)_{2}}{74.093gCa(OH)_{2}} *\frac{1mol Ca_{3}(PO_{4})_{2}}{3molCa(OH)_{2}} =0.0485 molCa_{3}(PO_{4})_{2} [/tex]

The mass of calcium phosphate in grams is:

[tex]0.0485 molCa_{3}(PO_{4})_{2}*\frac{310.176g}{mol} =15.1 g[/tex]

Answer:

FALSE

Explanation:

Grunge refers to the genre of rock music and the fashion inspired by it. It originated in the mid-1980s in Seattle, Washington State.

Grunge was described as the fusion of punk rock and heavy metal.

This genre of music became popular in the early mid-1990s and included lyrics based on the theme of emotional and social alienation, betrayal, abuse, trauma etc.    

The statement that grunge is a rock style from Detroit is false; grunge originated in Seattle, Washington. This music genre is characterized by a stripped-down aesthetic, distorted guitars, and socially conscious lyrics, gaining popularity in the late 1980s and early 1990s.

Grunge is indeed a style of rock music, but it is not from Detroit; that statement is false. Grunge, often referred to as the Seattle sound, originated in Seattle, Washington and became widely popular in the late 1980s and early 1990s. It represented a departure from the extravagant stage productions and fashion of contemporary rock bands, favoring a more stripped-down sound with an emphasis on distorted electric guitars and direct, often socially conscious lyrics.

The grunge movement produced several iconic bands, including Nirvana, Pearl Jam, and Soundgarden, which significantly shaped the alternative rock landscape. Seattle's geographic isolation played a pivotal role in the development of grunge music. Local bands evolved their unique sound and aesthetic away from the influence of mainstream music industry expectations, which at the time were centered around Los Angeles. This isolation contributed to grunge's authenticity and appeal when it eventually burst onto the national and international stages.

Contrary to the glamorous rock scene in Los Angeles, grunge musicians typically wore simple, unremarkable clothing, often sourced from second hand stores, and their lyrics addressed themes like political issues, mental health, and substance abuse, resonating with the disaffected Generation X.

Answer:

You can use a thermal analysis called Differential Scanning Calorimetry (DSC)

Explanation:

The glass transition temperature (Tg) and melting temperature (Tm) are thermal properties of substances, such as polymers. You can determine them with a very low sample quantity by differential scanning calorimetry (DSC).

In DSC, there are two aluminium pans. In one pan, you put the sample, and the other pan is empty because is the reference pan. The pans are placed in a chamber of a DSC calorimeter and they are heated at a temperature range you select at a given speed (usually in ºC/min). The system register the difference of heat between the pans (sample pan and reference pan) and it shows you a plot of heat flow (in J/s) vs temperature.                                                                      

Tg is commonly detected as a step in the plot

Tm is usually an endothermic peak (upwards in the plot), because is a process in which heat is absorbed.

Differential scanning calorimetry (DSC) under quasi-isothermal conditions is utilized to accurately determine the glass transition temperature and melting temperature of a polymer, reflecting its chain interactions and molecular weight.

Determining Glass Transition and Melting Temperature

To easily and accurately determine the glass transition temperature (Tg) and melting temperature (Tm) of a polymer, differential scanning calorimetry (DSC) is the most commonly used technique. The glass transition is a measure of the change in the polymer's heat capacity (Cp) as it crosses certain heat energy thresholds, and it can be more accurately determined under quasi-isothermal conditions as these conditions produce precise Cp measurements as a function of time. When a polymer is heated to its Tg, it enters a molten state, and upon cooling, becomes brittle. If the heating continues beyond Tg, the polymer will soften, allowing it to be molded or deformed.

The glass transition temperature can also provide insight into a polymer's chain length and structure, as it reflects the number of van der Waals or entangling chain interactions. Additionally, the molecular weight of a polymer influences Tg, typically following the Flory-Fox equation. The melting temperature (Tm) is the temperature at which a polymer transitions from a crystalline to a molten state and is detectable as a peak on a DSC trace above Tg.

Overall, by using DSC and applying quasi-isothermal steps across the apparent glass transition range, we can observe changing Cp rates that indicate the exact Tg range. Similarly, for Tm, we can detect the transition between crystalline and molten states as the temperature rises above Tg.

Answer:

The pressure in the vessel is 13,3 atm.

Explanation:

The reaction that occurs in vessel (where limestone is 96% of CaCO₃) is:

2 HCl (aq)+ CaCO₃ (s) → CaCl₂(aq)+ H₂O(l)+ CO₂(g)

The increase in the pressure of the vessel after the reaction is by formation of a gas (CO₂). So we have to find the produced moles of this gas and apply the gas ideal law to find the pressure.

We have to find the limit reactant, to do so, we have to calculate the moles of each reactant in the reaction, the one that have the less moles will be the limit reactant:

HCl:

1,0L × (35/100) × (1000 cm³/1L) × (1,28 g/ 1cm³) × (1mol HCl/ 36,46 g) ÷ 2mol

(Concentration)      (L to cm³)         (cm³ to g)      (g to mol)  (moles of reaction)

moles of HCl= 6,14 mol

CaCo₃:

   1,0 kg     ×       (96/100)                ×   (1000 g/1kg) × (1 mol/100,09g)

(Limestone) (CaCo₃ in limestone)          (kg to g)            (g to mol)

moles of CaCo₃= 9,59 mol

So, reactant limit is HCl

This reaction have a yield of 97%. So, the CO₂ moles are:

6,14 mol × 97÷ = 5,96 mol CO₂

The ideal gas formula to obtain pressure is:

P = nRT/V

Where: n = 5,96mol; R= 0,082 atm×L/mol×K; T = 273,15 (until STP conditions) and V= 10,0 L

Replacing this values in the equation the pressure is

P = 13,3 atm

I hope it helps!

Answer:

The strong acid reacts with the weak base in the buffer to form a weak acid, which produces few H+ ions in solution and therefore only a little change in pH.

Explanation:

When a strong acid is added to the buffer, the acid dissociates and furnish hydrogen ions which combine with the conjugate of the weak acid, forming weak acid. The weak acid dissociates to only some extent and can furnish only some protons and there is no significant change in the pH.

Hence, option B is correct.

In a buffer, a strong acid reacts with the weak base to form a weak acid, yielding fewer H+ ions, so only a slight change in pH is noted. Buffer solutions, including weak conjugate acid-base pairs, resist pH changes. However, the buffering ability may be disrupted if large amounts of acid or base are added, a concept known as buffer capacity.

A buffer essentially resists changes in pH when an acid or base is added to the solution. This is due to the presence of a weak conjugate acid-base pair. In the buffer, a strong acid will react with the weak base to form a weak acid, thus producing fewer H+ ions. This, in essence, leads to only a slight change in pH.

A typical example of a buffer might include a solution of acetic acid and sodium acetate. This buffer consists of a weak acid and its salt. Similarly, a solution of ammonia and ammonium chloride could serve as a buffer consisting of a weak base and its salt.

It's crucial to note, however, that buffers do not have infinite capacity to resist pH changes. This is known as the buffer capacity. If a large amount of acid or base is added to the buffer solution, potentially lowering the concentration of the conjugate acid or base pair, the ability of the buffer to regulate pH may be disrupted.

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Answer:

The energy transferred between samples of matter because of a difference in their temperatures is called a. heat.

Explanation:

The first law of thermodynamics establishes that when two bodies with different temperatures are put in contact they will find thermic equilibrium to a final temperature by transferring heat. Thus the correct answer is (a).

Thermochemistry is the study of the transformations of heat energy on the chemical reactions. Chemical kinetics is the study of the rate of chemical reactions. And temperature is the measure of the heat.

Answer: The volume of oxygen gas at STP is 446 mL

Explanation:

STP conditions are:

Pressure of the gas = 1 atm

Temperature of the gas = 273 K

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1,V_1\text{ and }T_1[/tex] are the initial pressure, volume and temperature of the gas

[tex]P_2,V_2\text{ and }T_2[/tex] are the final pressure, volume and temperature of the gas

We are given:

[tex]P_1=1.50atm\\V_1=346mL\\T_1=45^oC=(45+273)K=318K\\P_2=1atm\\V_2=?\\T_2=273K[/tex]

Putting values in above equation, we get:

[tex]\frac{1.50atm\times 346mL}{318K}=\frac{1atm\times V_2}{273K}\\\\V_2=446mL[/tex]

Hence, the volume of oxygen gas at STP is 446 mL

Explanation:

The given data is as follows.

     [tex]V_{1}[/tex] = 346 mL,     [tex]T_{1}[/tex] = [tex]45.0^{o}C[/tex] = (45 + 273) K = 318 K,

      [tex]P_{1}[/tex] = 1.50 atm,   [tex]V_{2}[/tex] = ?,   [tex]T_{2}[/tex] = 273 K,

      [tex]P_{2}[/tex] = 1 atm

And, according to ideal gas equation,

               [tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]

Now, putting the given values into the above formula and we will calculate the final volume as follows.

          [tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]

           [tex]\frac{1.50 atm \times 346 mL}{318 K} = \frac{1 atm \times V_{2}}{273 K}[/tex]

            [tex]V_{2}[/tex] = 445.56 mL

Thus, we can conclude that the volume of this [tex]O_{2}[/tex] gas sample at STP is 445.56 mL.

Answer:

Raoult's law is not a good approximation when the compound is very dilute in the liquid phase.

Explanation:

Thermodynamics defines ideal mixture as a mixture in which all the molecules of the different species involved are so similar between each other that intermolecular forces between different molecules are the same to those of similar molecules. Thus, Raoult's law takes this definition and stablish that when a compound is highly concentrated in the solution behaves as if it is in a pure solution.

Also, the Raoult's law stablish that the partial pressure ([tex]P_{i}[/tex]) of every component in an ideal mixture of liquids is equal to the pressure of the pure component ([tex]P_{i}^{*}[/tex]) multiplicated by the molar fraction in the solution of that component ([tex]x_{i}[/tex])

[tex]P_{i} = P_{i}^{*}. x_{i}[/tex]

The answer "When the compound has a high mole fraction in the liquid phase" is wrong because this is the case that follows the Raoult's law. And the answer "When the compound has a high mole fraction in the vapor phase." talks about the fraction in the vapor, and Raoult's law use the molar fraction in the solution.

Answer:

Ethyl alcohol and isopropyl alcohol are the most commonly used agents.

Explanation:

Isopropyl alcohol is used in a 10% concentration.

- not true, to be useful as a chemical control agent its is the most potent  with a (water) solution containing 70% isopropyl alcohol.

Ethyl alcohol and isopropyl alcohol are the most commonly used agents.

-true

Dipping small instruments in ethyl alcohol for a few seconds will result in disinfection.

- ethyl alcohol is denaturing proteins, inhibiting metabolic processes so it has bactericidal and fungicidal properties. However the proces is not so fast so the instruments are not disinfected in few seconds.

Alcohols will eliminate bacterial spores.

-alcohols are not sporicidal, however they do inhibit the processes of sporulation and germination but they do not eliminate the spores.

Answer:

a) mass flow = 56940 Kg/h

b) mass flow = 202.5 mol/s

Explanation:

∴ δ C6H6 = 876 Kg/m³,,,,,wwwcarlroth.com

⇒ 65m³/h * 876 Kg/m³ = 56940 Kg/h

⇒ 56940 Kg/h * ( 1000 g/Kg ) * ( mol/ 78.11 g) * ( h/3600s )= 202.5 mol/s

The mass flow rate of benzene is calculated using its density of 0.876 kg/m³, resulting in 56.94 kg/h. The molar flow rate is determined using the molar mass of benzene, 78.11 g/mol, which yields 20.34 mol/s.

Mass Flow and Molar Flow Calculations

To calculate the mass flow fed in kg/h for benzene, the density of benzene is required. Benzene has a density of approximately 0.876 kg/m³ at room temperature. Using this density, the mass flow can be calculated as follows:

Mass flow (kg/h) = Volumetric flow (m³/h) × Density (kg/m³)

Mass flow = 65 m³/h ×  0.876 kg/m³ = 56.94 kg/h

To calculate the molar flow in mol/s, we use the molar mass of benzene which is approximately 78.11 g/mol:

Molar flow (mol/s) = Mass flow (kg/h) × (1000 g/kg) / (Molar mass (g/mol) × (3600 s/h))

Molar flow = (56.94 kg/h × 1000 g/kg) / (78.11 g/mol ×  3600 s/h) = 20.34 mol/s

Note: The molecular formula of benzene is C₆H₆, not CH as mentioned in the information provided. Therefore, the combustion analysis would yield different amounts of CO₂ and H₂O than suggested by the empirical formula CH. It's important to use the correct molecular formula for accurate calculations.

The answer is unknown

Answer:

The energy of the photon generated in the transition is 3.14*10⁻¹⁹ J

Explanation:

There are two equation that we need to use in order to solve this problem:

The first one is Planck's equation, which describes the relationship between energy and frequency:

E = h*v    eq. 1)

Where E is energy, h is Planck's constant (6.626 * 10⁻³⁴ J*s) and v is the radiation frequency.

In order to know the frequency, we use the second equation, which is the wave equation:

c = λ*v    eq. 2)

Where c is the speed of light in vacuum (aprx 3 * 10⁸ m/s), and λ is the wavelength. If we solve that equation for v we're left with

v=c/λ    eq. 3)

We replace v in eq. 1):

E= c*h/λ

Lastly we put the data we know and solve the equation, keeping in mind the correct use of units (converting 652 nm into m):

[tex]E=\frac{3*10^{8}ms^{-1} *6.626*10^{-34}Js}{633*10^{-9}m }=3.14*10^{-19}J[/tex]

Final answer:

The energy of the photon generated in the transition using the given formula. Calculating gives approximately 3.04 x 10^-19 joules.

Explanation:

The energy of the photon generated in the transition can be calculated using the formula:

E = hc / λ

Where E is the energy of the photon, h is the Planck's constant, c is the speed of light, and λ is the wavelength of the light. Substituting the given values:

E = (6.63 x 10^-34 J s x 3 x 10^8 m/s) / (652 x 10^-9 m)

Calculating this gives the energy of the photon as approximately 3.04 x 10^-19 joules.

Answer:

100.223°C is the boiling point of an aqueous solution.

Explanation:

Osmotic pressure of the solution = π = 10.50 atm

Temperature of the solution =T= 25 °C = 298 .15 K

Concentration of the solution = c

van'y Hoff factor = i = 1 (non electrolyte)

[tex]\pi =icRT[/tex]

[tex]c=\frac{\pi }{RT}=\frac{10.50 atm}{0.0821 atm L/mol K\times 298.15 K}[/tex]

c = 0.429 mol/L = 0.429 mol/kg = m

(density of solution is the same as  pure water)

m = molality of the solution

Elevation in boiling point = [tex]\Delta T_b[/tex]

[tex]\Delta T_b=iK_b\times m[/tex]

[tex]\Delta T_b=T_b-T[/tex]

T = Boiling point of the pure solvent

[tex]T_b[/tex] = boiling point of the solution

[tex]K_b[/tex] = Molal elevation constant

We have :

[tex]K_b=0.52^oC/m[/tex] (given)

m = 0.429 mol/kg

T = 100° C (water)

[tex]\Delta T_b=1\times 0.52^oC/m\times 0.429 mol/kg[/tex]

[tex]\Delta T_b=0.223^oC[/tex]

[tex]\Delta T_b=T_b-T[/tex]

[tex]T_b=\Delta T_b+T=0.223^oC+100^oC=100.223^oC[/tex]

100.223°C is the boiling point of an aqueous solution.

Answer:

8.83 moles of Hydrogen

Explanation:

We are told that 1.92 moles of H2 occupies 22.2 L at a certain pressure and temperature, and we are asked to calculate the quantity of moles of H2 are necessary to fill a container of 102.1 L. So, if the pressure and temperature are constant, then the equality expressed before 1.92 H2 moles = 22.2 L maintains. Therefore, we can calculate what is being asked as follows:

22.2 L ----- 1.92 H2 moles

102.1 L ---- x = (102.1 L × 1.92 H2 moles)/22.2 L = 8.83 H2 moles

This means that 8.83 H2 moles are necessary to fill a 102.1 L container at a certain pressure and temperature.