What is the % by volume of 50mL of ethylene glycol dissolved in 950mL of H2O?

The bond angles for ozone (O₃), boron trifluoride (BF₃), oxygen difluoride (OF₂), beryllium dichloride (BeCl₂), phosphorus trifluoride (PF₃), and silicon tetrachloride (SiCl₄) are <120°, 120°, <109.5°, 180°, <109.5°, and 109.5° respectively, based on their molecular geometries.

The correct value for the indicated bond angle in each of the compounds is based on the molecular geometry and electron-pair repulsion theory. Here are the answers:

Ozone (O₃): The O-O-O angle in O₃ is expected to be <120° due to its bent molecular structure.

Boron trifluoride (BF₃): The F-B-F angle in BF3 is 120° because of its trigonal planar molecular geometry.

Oxygen difluoride (OF₂): The F-O-F angle in OF2 is expected to be <109.5° because fluoro substituents cause greater repulsion than hydrogens, narrowing the bond angle from the tetrahedral

Beryllium dichloride (BeCl₂): The Cl-Be-Cl angle in BeCl₂ is 180° due to its linear molecular structure.

Phosphorus trifluoride (PF₃): The F-P-F angle in PF₃ is expected to be <109.5° due to its trigonal pyramidal molecular structure.

Silicon tetrachloride (SiCl₄): The Cl-Si-Cl angle in SiCl₄ is 109.5° since it has a tetrahedral molecular geometry.

Answer:Yes it affects the conditions of the hair.

Explanation:most shampoos have pH values higher than the pH of the hair shaft of 3.6 and also higher than the pH of the hair scalp of 5.5. Therefore the usage of hair shampoo with pH values above 5.5 may likely increase the rate of friction which may cause hair breakage and facilitate hair tangling.

Shampoos with various pH ratings have different effects on hair. The pH level of your shampoo influences the health of your hair. Shampoos that match the hair's natural pH of 5 to 7 are typically best for maintaining hair health.

The pH measurement, which ranges from 0 to 14, indicates whether a substance is acidic, neutral, or alkaline. Most shampoos usually have a pH level between 5 and 7, similar to hair's natural pH. When a shampoo claims to 'balance' the pH of your hair, it means it tries to keep this pH level, which is slightly acidic for a healthy scalp and hair. Meanwhile, neutral shampoos have a pH of 7, and alkaline shampoos have a pH higher than 7, which could potentially cause hair to become dry and brittle over time, especially with overuse. So, the pH of your shampoo does indeed affect your hair condition. Using a shampoo that aligns with the natural pH of your hair is generally considered optimal for maintaining hair health.

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Answer: Time required for 100 molecules of valinomycin to transport enough [tex]K^{+}[/tex] to change the concentration inside an erythrocyte is 10 minutes.

Explanation:

The given data is as follows.

 Ions to be transported = [tex](10 mM) \times (100 \mu^{3})(N)[/tex]

            = [tex]0.010 mol \times 10^{-13} L \times 6.02 \times 10^{23} ions/mol[/tex]

             = [tex]6.02 \times 10^{6}[/tex] ions

Here, there are 100 ionophores are present and each one of ionophores are required to transport a value of  ions.

Therefore, the time required for 100 molecules of valinomycin to transport enough [tex]K^{+}[/tex] to change the concentration inside an erythrocyte is as follows.

   [tex]6.02 \times 10^{6} ions \times \frac{1 sec}{10^{4} ions}[/tex]

              = 602 sec

or,          = 10 minutes        (as 1 min = 60 sec)

Thus, we can conclude that time required for 100 molecules of valinomycin to transport enough [tex]K^{+}[/tex] to change the concentration inside an erythrocyte is 10 minutes.

When 32.40 mL of 0.258 M NaOH is used to titrate a 1.00 mL sample, the concentration of acetic acid in the mixture is 8.36 M.

Let's consider the neutralization reaction between sodium hydroxide and acetic acid.

NaOH + CH₃COOH ⇒ CH₃COONa + H₂O

32.40 mL of 0.258 M NaOH react. The reacting moles of NaOH are:

[tex]0.03240 L \times \frac{0.258mol}{L} = 8.36 \times 10^{-3} mol[/tex]

The molar ratio of NaOH to CH₃COOH is 1:1. The moles of CH₃COOH required to react with 8.36 × 10⁻³ moles of NaOH are:

[tex]8.36 \times 10^{-3} mol NaOH \times \frac{1molCH_3COOH}{1molNaOH} = 8.36 \times 10^{-3} mol CH_3COOH[/tex]

8.36 × 10⁻³ moles of CH₃COOH are in 1.00 mL of solution. The concentration of CH₃COOH is:

[tex]M = \frac{8.36 \times 10^{-3} mol}{1.00 \times 10^{-3} L} = 8.36 M[/tex]

When 32.40 mL of 0.258 M NaOH is used to titrate a 1.00 mL sample, the concentration of acetic acid in the mixture is 8.36 M.

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The concentration of acetic acid in the mixture can be found by titrating with NaOH, using a stoichiometric reaction to equate the moles of NaOH to moles of acetic acid, and then calculating the concentration of the acid.

This problem is about using acid-base reactions to determine the concentration of acetic acid in a solution. The reaction between the acetic acid and sodium hydroxide (NaOH) is a stoichiometric reaction where one mole of acetic acid reacts with one mole of NaOH. We use that information to calculate the number of moles of acetic acid.

First, we calculate the number of moles of NaOH used in the titration, which is the concentration of NaOH multiplied by the volume used (in liters): moles NaOH = (0.258 mol/L) * (32.40 mL * 1L/1000 mL) = 0.00835 mol NaOH.

Since the reaction is stoichiometric, the number of moles of NaOH is equivalent to the number of moles of acetic acid.

Finally, the concentration of acetic acid is the number of moles divided by the volume of the sample: concentration = moles / volume = 0.00835 mol / 0.001 L = 8.35 M.

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Explanation:

The only flaw I can find is you squared 3 instead of cubing it and it will be 27X^4 instead of 9x^4.

This reduces the amount slightly, but the number is still incredibly high (about 10 ^ 5 L is what I've calculated). Your professor might want to point out that this will not be a effective experiment due to the large volume of saturated

The Ksp value of Ca(OH)2 on the site (I used 5.5E-6 [a far more soluble compound than Al(OH)3]) and estimated how much of it will be needed. My calculation was approximately 30 ml. If you were using that much in the experiment, it implies so our estimates for Al(OH)3 are right, that the high amount is unreasonably big and that Al(OH)3 will not be a suitable replacement unless the procedure was modified slightly.

Answer : The specific heat capacity of lead is, [tex]0.128J/g^oC[/tex]

Explanation :

Formula used:

[tex]q=m\times c\times (T_2-T_1)[/tex]

where,

q = heat produced = 3840 J

m = mass of lead = 3 kg = 3000 g

c = specific heat capacity of lead = ?

[tex]T_1[/tex] = initial temperature = [tex]40^oC[/tex]

[tex]T_2[/tex] = final temperature = [tex]50^oC[/tex]

Now put all the given values in the above formula, we get:

[tex]3840J=3000g\times c\times (50-40)^oC[/tex]

[tex]c=0.128J/g^oC[/tex]

Therefore, the specific heat capacity of lead is, [tex]0.128J/g^oC[/tex]

Answer:

Answer: Kp = 4.5

Explanation:

The balanced equation for the production of synthetic gas is

    CH4 +  H20 ⇒ CO + 3H2

Let x be the change in the concentration of each species at equilibrium

 CH4  +   H20   ⇔ CO  +  3H2

Initial                  0.60       2.6        0          0

Change               -x           -x          +x       +3x

Equilibrium      0.60-x      2.6-x       x        3x

Given that the equilibrium partial pressure of H2= 1.4 atm

then, 3x= 1.40

x= 1.4/3 = 0.466667

The equilibrium concentrations are

{CH4} = 0.60- x = 0.60 - 0.466667 = 0.133333atm

{H2O} = 2.60- x = 2.60 - 0.466667 = 2.133333atm

{CO} = x = 0.466667atm

{H2} = 1.4atm (given)

Kp  = {CO}{H2}³

        {CH4}{H2O}

Kp = (0.466667)(1.4)³

         (0.133333)(2.133333)

=  4.501875

Kp = 4.5

pH before KOH: 4.23. After 25.0 mL KOH: 4.23. After 35.0 mL: 9.98. After 50.0 mL: 11.26. After 60.0 mL: 11.85.

To solve this problem, we'll use the stoichiometry of the reaction between HClO and KOH to find the amount of each reactant remaining at each point of the titration. Then, we'll use the relevant equilibrium equations to calculate the pH.

Step 1: Calculate the initial pH before adding any KOH.

[tex]\[ \text{HClO} \rightleftharpoons \text{H}^+ + \text{ClO}^- \][/tex]

[tex]\[ K_a = \frac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]} \][/tex]

[tex]\[ [\text{H}^+] = \sqrt{K_a [\text{HClO}]} \][/tex]

[tex]\[ [\text{H}^+] = \sqrt{(3.5 \times 10^{-8})(0.110)} \][/tex]

[tex]\[ [\text{H}^+] \approx 5.92 \times 10^{-5} \, \text{M} \][/tex]

[tex]\[ \text{pH} = -\log{[\text{H}^+]} = -\log{(5.92 \times 10^{-5})} \][/tex]

[tex]\[ \text{pH} \approx 4.23 \][/tex]

Step 2: Calculate the pH after adding 25.0 mL of KOH.

Moles of HClO initially = [tex]\( 0.110 \, \text{M} \times 0.0500 \, \text{L} = 0.00550 \, \text{mol} \)[/tex]

Moles of KOH added = [tex]\( 0.110 \, \text{M} \times 0.0250 \, \text{L} = 0.00275 \, \text{mol} \)[/tex]

Remaining moles of HClO = [tex]\( 0.00550 \, \text{mol} - 0.00275 \, \text{mol} = 0.00275 \, \text{mol} \)[/tex]

Calculate the new concentration of HClO:

[tex]\[ \text{New volume of HClO} = 50.0 \, \text{mL} - 25.0 \, \text{mL} = 25.0 \, \text{mL} = 0.0250 \, \text{L} \][/tex]

[tex]\[ \text{New} \, [\text{HClO}] = \frac{0.00275 \, \text{mol}}{0.0250 \, \text{L}} \][/tex]

[tex]\[ \text{New} \, [\text{HClO}] = 0.110 \, \text{M} \][/tex]

Now, we use the equilibrium equation for the dissociation of HClO to find the concentration of [tex]\( \text{H}^+ \):[/tex]

[tex]\[ \text{HClO} \rightleftharpoons \text{H}^+ + \text{ClO}^- \][/tex]

[tex]\[ K_a = \frac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]} \][/tex]

[tex]\[ [\text{H}^+] = \sqrt{K_a [\text{HClO}]} \][/tex]

[tex]\[ [\text{H}^+] = \sqrt{(3.5 \times 10^{-8})(0.110)} \][/tex]

[tex]\[ [\text{H}^+] \approx 5.92 \times 10^{-5} \, \text{M} \][/tex]

[tex]\[ \text{pH} = -\log{[\text{H}^+]} = -\log{(5.92 \times 10^{-5})} \][/tex]

[tex]\[ \text{pH} \approx 4.23 \][/tex]

The pH remains the same after adding 25.0 mL of KOH.

Let's now calculate the pH after adding 35.0 mL, 50.0 mL, and 60.0 mL of KOH.

Final answer:

Evidence of a chemical reaction can be determined by changes such as temperature fluctuations, formation of a precipitate, gas production, and color changes. Temperatures changing, solid products forming, gas releasing, and explosions are all indications of chemical reactions.

Explanation:

Evidence indicating that a chemical reaction has occurred includes:

Based on these observations, the correct answers from the choices provided are: (a) The temperature decreased, (b) A solid brown product formed, (c) The temperature increased, (d) A gas formed, and potentially (e) An explosion occurred, as an explosion indicates a rapid energy transfer and creation of new substances.

Explanation:

Water Content of Epidermal Cells

Temperature: Increase in the temperature causes stomata to open

Answer:

Temperature: Increase in the temperature causes stomata to open

Explanation:

Answer: The final temperature of the mixture is  49°C

Explanation:

To calculate the mass of water, we use the equation:

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

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

[tex]1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g[/tex]

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

[tex]1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g[/tex]

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]

The equation used to calculate heat released or absorbed follows:

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]      ......(1)

where,

q = heat absorbed or released

[tex]m_1[/tex] = mass of hot water = 120 g

[tex]m_2[/tex] = mass of cold water = 230 g

[tex]T_{final}[/tex] = final temperature = ?°C

[tex]T_1[/tex] = initial temperature of hot water = 95°C

[tex]T_2[/tex] = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

[tex]120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)][/tex]

[tex]T_{final}=49^oC[/tex]

Hence, the final temperature of the mixture is  49°C

Answer:

The wavelength the student should use is 700 nm.

Explanation:

Attached below you can find the diagram I found for this question elsewhere.

Because the idea is to minimize the interference of the Co⁺²(aq) species, we should choose a wavelength in which its absorbance is minimum.

At 400 nm Co⁺²(aq) shows no absorbance, however neither does Cu⁺²(aq). While at 700 nm Co⁺²(aq) shows no absorbance and Cu⁺²(aq) does.

Answer:

Spectrophotometric analysis is a quantitative measure of the interaction of spectrum rays and certain molecules.  The wavelength used by student is 700 nanometers.

Explanation:

Spectrophotometric analysis is a method used to measure the amount of absorbance or transmittance done by a substance.

From the knowledge of spectrometric analysis:

In the given question, to minimize the interference by cobalt ion, the student should use the wavelength of 700 nanometers or more. The copper is absorbed at wavelength of 700 nm or more, whereas the cobalt ion is absorbed between the range of 400 to 500 nanometres.

Therefore, the student will use wavelength of 700 nanometers because copper is absorbed at this wavelength. The cobalt is absorbed at 400 to 500 nanometers, which will minimize its interference.

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Answer:

O3 + NO. = O2. + NO2

2O. +NO2 = NO. + 2O2

Explanation:

NO. = Radical

O. & O2. = Radical

Final answer:

Nitric oxide (NO) catalytically destroys ozone (O3) in the stratosphere via a series of propagation steps, contributing to ozone layer depletion and increasing exposure to harmful UV rays.

Explanation:

Nitric oxide (NO) plays a significant role in the destruction of ozone (O3) through a series of reactions in the stratosphere. The presence of NO contributes to the depletion of the ozone layer, which can increase the risk of harmful UV exposure to humans and the environment. The relevant ozone depletion mechanism proceeds with nitric oxide acting as a catalyst through the following propagation steps:

These reactions show how NO continues to regenerate and is not consumed, which is why it acts as a catalyst, increasing the rate of ozone decomposition. The nitric oxide is primarily produced by high-flying aircraft and motor vehicles, raising concerns about its environmental impact.

Answer:

Motorcycle

Explanation:

Final answer:

To calculate the percent dissociation of trimethylacetic acid in an aqueous solution, use the equilibrium constant expression and the initial concentration of the acid.

Explanation:

To calculate the percent dissociation of trimethylacetic acid in an aqueous solution, we need to use the equilibrium constant expression and the initial concentration of the acid. The equilibrium expression for the dissociation of trimethylacetic acid is written as:

Ka = [CH₃COOH]/[CH₃C₃H₇OH]

where [CH₃COOH] represents the concentration of trimethylacetic acid and [CH₃C₃H₇OH] represents the concentration of the dissociated form of the acid.

By rearranging the equation, we can solve for the percent dissociation:

Percent Dissociation = ([CH₃C₃H₇OH]/[CH₃COOH]) x 100

Using the given information or data from the ALEKS Data resource, we can substitute the values into the equation and calculate the percent dissociation.

Answer:

Yes

Explanation:

A reaction normally takes place between metals and non metals. The metals acts as the electron donors and the non metals acts as the electron acceptors. This exchange of electrons form bonds such as ionic or covalent.

A good example of a reaction between a metal and non metal is Sodium metal and Chlorine(non metal). They form an ionic bond and the product is Sodium chloride.

Answer:

11.6 L will be the number of liters of carbon dioxide measured at STP.

Explanation:

The balanced equation for this combustion reaction is:

C₃H₈  +  5O₂ →  3CO₂  + 4H₂O

where 1 mol of propane reacts to 5 moles of oxygen in order to produce 3 moles of carbon dioxide and 4 moles of water.

We assume the oxygen in excess, so the limiting reagent is the propane. Now, we determine the moles: 7.65 g . 1 mol/ 44 g = 0.174 moles

Ratio is 1:3. 1 mol of propane can produce 3 moles of CO₂

Therefore, 0.174 moles will produce (0.174 . 3) / 1 = 0.521 moles of CO₂

As 1 mol of gas is contained in 22.4L at STP conditions, we propose

22.4L / 1 mol = V₂ / 0.521 mol

22.4 L / 1 mol . 0.521 mol = V₂ → 11.6 L

Answer:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (g)

11.6 L of CO2 will be produced

Explanation:

Step 1: Data given

Mass of propane = 7.65 grams

Molar mass propane = 44.1 g/mol

Burning = combustion reation = adding O2. The products will be CO2 and H2O

Step 2: The balanced equation

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (g)

Step 3: Calculate moles propane

Moles propane = 7.65 grams / 44.1 g/mol

Moles propane = 0.173 moles

Step 4: Calculate moles CO2

For 1 mol propane we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O

For 0.173 moles propane we'll have 3*0.173 = 0.519 moles CO2

Step 5: Calculate volume of CO2

1 mol = 22.4 L

0.519 moles = 22.4 L * 0.519 = 11.6 L

11.6 L of CO2 will be produced

Answer:

The half life is     [tex]H_{1/2}= 7333.3sec[/tex]

Explanation:

   The half life of a first order  reaction is mathematically represented as

                          [tex]H_{1/2} = \frac{0.693}{Rate Constant }[/tex]

 Substituting   [tex]9.45 * 10^{-5}s^{-1}[/tex]  for the rate constant

                         [tex]H_{1/2} = \frac{0.693}{9.45*10^{-5}}[/tex]

                                 [tex]H_{1/2}= 7333.3sec[/tex]

Final answer:

The half-life of the isomerization of methylisonitrile to acetonitrile, which is a first-order reaction with a rate constant of 9.45 x 10⁻⁵ s⁻¹, is approximately 7330 seconds.

Explanation:

The student's question asks for the calculation of the half-life of the reaction where methylisonitrile isomerizes to acetonitrile (CH₃NC to CH₃CN). This reaction is first order with respect to methylisonitrile, and the rate constant k is given as 9.45 x 10⁻⁵ s⁻¹ at 478 K. To find the half-life ([tex]t^{1/2[/tex]), we can use the first-order half-life equation, [tex]t^{1/2[/tex] = (ln 2) / k. Substituting the given rate constant into this equation

[tex]t^{1/2[/tex] = (ln 2) / (9.45 x 10⁻⁵ s⁻¹)

[tex]t^{1/2[/tex] = (0.693) / (9.45 x 10⁻⁵ s⁻¹)

[tex]t^{1/2[/tex] ≈ 7330 seconds

The half-life of the reaction under the given conditions is approximately 7330 seconds.

The nucleus of an atom, comprised of protons and neutrons, contains most of the atom's mass. Despite its relatively minuscule size, it is much heavier than the electron cloud surrounding it, which occupies most of the atom's volume.

The statement that correctly describes the nucleus of an atom is: D. the nucleus of the atom contains most of the mass of an atom.

Modern atomic theory has shown us that the nucleus of an atom is made up of positively charged protons and uncharged neutrons, and overall, carries a positive charge. The nucleus is incredibly tiny, with its diameter being about 100,000 times smaller than the diameter of the atom. However, it contains almost all of the atom's mass as protons and neutrons are significantly heavier than electrons, thus making statement D accurate.

Electrons, on the other hand, occupy the majority of an atom's volume, constituting an 'electron cloud' around the nucleus. Because they are very lightweight, their specific locations can't be definitively pinpointed and are often represented as probabilities within this cloud. Hence, options A, B, and C are incorrect.

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Answer:

Yes is possible

Explanation:

MnSO4*7H2O(s) → MnSO4(s) + 7H2O(g)

This should be an endothermic reaction. The water molecules are interacting with the Mn2+ and SO42- ions within the crystal lattice of the compound. You must overcome those interactions, and so that would require the input of energy.

Answer:

One mole of a substance is equal to 6.022 × 10²³ units of that substance (atoms, molecules, or ions)

Explanation:

This number is Avogadro's number. The concept of the mole can be used to convert between mass and number of particles. its used to compare very large numbers.

Answer:

A. In the formation of water, H2 is the limiting reactant and O2 is the excess reactant.

B. In the Combustion of methane (CH4), O2 is the limiting reactant and CH4 is the excess reactant.

C. In the formation of ammonia (NH3), H2 is the limiting reactant and N2 is the excess reactant.

Explanation:

The question suggests that each reactant has 5 moles.

A. Formation of water. Water is produced when H2 and O2 combine together according to the balanced equation below:

2H2 + O2 —> 2H2O

From the balanced equation above,

2 moles of H2 reacted with 1 mole of O2.

Therefore, 5 moles of H2 will react with = 5/2 = 2.5 moles of O2.

From the above calculations, we can see that there are left over for O2 as only 2.5 moles reacted out of the 5 moles that was given.

Therefore, H2 is the limiting reactant and O2 is the excess reactant.

B. Combustion of methane. Combustion is simply a reaction in the presence of oxygen. The balance equation for the Combustion of methane (CH4) is given below:

CH4 + 2O2 —> CO2 + 2H2O

From the balanced equation above,

1 mole of CH4 reacted with 2 moles of O2.

Therefore, 5 moles of CH4 will react with = 5 x 2 = 10 moles of O2.

From the calculations made above, we can see that it requires higher amount of O2 to react with 5 moles CH4. Therefore, O2 is the limiting reactant and CH4 is the excess reactant.

C. Formation of ammonia.

Ammonia is obtained when N2 and H2 combine according to the balanced equation below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 5 moles of N2 will react with = 5 x 3 = 15 moles of H2.

From the calculations made above, a higher amount of H2 is required to react with 5 moles of N2. Therefore, H2 is the limiting reactant and N2 is the excess reactant.

Final answer:

To determine limiting or excess reactants, compare the stoichiometry of the balanced equations to the amounts provided. For reactions involving the formation of water or ammonia, and the combustion of methane, the limiting reactant is based on the stoichiometric ratio required for each reactant. Hence, O2 typically ends up as the limiting reactant for the formation of water and combustion of methane, while N2 is the limiting reactant in the formation of ammonia.

Explanation:

When determining whether a reactant is a limiting reactant or an excess reactant, you must look at the balanced chemical equations for the reactions. Consider the stoichiometric coefficients, which tell you the proportions in which reactants combine to form products. If starting with five moles of each reactant:

To identify the limiting reactant, compare the molar ratios of the reactants to those in the balanced equation. The reactant that is in a smaller proportion relative to its stoichiometric coefficient in the equation is the limiting reactant. The one in the larger proportion is the excess reactant.

At equilibrium, the composition of the system is determined by the stoichiometry of the reaction. Doubling the pressure will not have a significant effect on the composition of the system at equilibrium.

At equilibrium, the composition of the system can be determined using the stoichiometry of the reaction. In this reaction, one N2 molecule reacts with one C2H2 molecule to produce two HCN molecules. Therefore, at equilibrium, the concentration of N2 and C2H2 will both decrease by the same amount, while the concentration of HCN will increase by twice that amount.

If the pressure is doubled, the system will try to relieve the stress by shifting the equilibrium to the side with fewer moles of gas. In this case, the total number of moles of gas will remain the same, so doubling the pressure will not have a significant effect on the composition of the system at equilibrium.

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Final answer:

The formation of xenon compounds such as xenon difluoride is made possible due to the displacement of xenon's outer electrons, which can occur under certain conditions like high pressure and temperature.

Explanation:

The formation of compounds containing Xe is made possible by the fact that the outer electrons of the larger noble gases like Xe (xenon) are far enough away from the nucleus that they can be displaced under certain conditions. The noble gases were long thought to be entirely unreactive due to their filled outer electron shells, which provide a stable electronic configuration. Despite this, experiments in the early 1960s confirmed that noble gas compounds can indeed be synthesized, with xenon forming stable compounds with fluorine. For instance, xenon difluoride (XeF2), xenon tetrafluoride (XeF4), and xenon hexafluoride (XeF6) are all compounds that form when xenon reacts with varying amounts of fluorine, producing stable crystals that are inert at room temperature. These reactions typically require the noble gas to be exposed to high pressure and temperature conditions.

Answer:

First of all, exterior paint is more resistant to extreme temperatures, humidity, pressure differences, the presence of droughts and this is due to its chemical composition, which is more resistant and because it penetrates the surface to be painted better (than in This case is the external wall of a home), by better penetrating the surface to which it is applied, there is less risk of detachment, of accumulating bubbles or air at the wall-paint interface, it adheres much better ... as explained this? Well, this phenomenon is due to the fact that it presents a solvent that helps to impregnate the wall (primer) and transports the paint as a vehicle to the internal pores of the wall.

On the other hand, in interior environments, the paints are less resistant to the aforementioned adverse factors and less impregnate the surfaces because the solvent is in less quantity or does not have the same impregnation capacity or essential primer as in the case of exterior paints. .

Explanation:

There are some specialists in the field who nowadays advise even interior paints with insulating or waterproof lacquers since the interior ones bear so little factors such as humidity that they no longer meet the basic requirements, that is why some interior paints they show fungus or wet stains against walls with wet content.

Answer:

Water on the outside of the paint evaporate faster than the inside. To test this hypothesis this student might place water in containers with different sizes and shapes in locations with different temperatures. This way it can be measured how fast water in different places evaporate.

Explanation:

A change in temperature can affect the  physical properties of water.  Water circulates through a continuous  cycle.  The sun’s energy drives weather patterns, and enables water to change through different stages in the water cycle : evaporation, condensations,  precipitation, ground water,  transpiration.

By placing water in containers with different sizes and shapes in locations with different temperatures, the student will see that the sun’s energy increases the amount of water that evaporates. More water evaporates when the temperature is warmer and when there is a large surface area or mouth/opening.

Answer:

Check Explanation

Explanation:

The Mathematical expression of the first law is given as

ΔU = Q - W

Note that depending on convention, the mathematical expression for the first law can be written as

ΔU = Q + W

a) Yes, the gas does work on its surroundings. Since it expands; indicating that there is a change in volume.

The work done is given as -PΔV if we are using the convention of (ΔU = Q + W). And this work is negative work done, since the system expands and does work on the surroundings.

But if we are using the convention of (ΔU = Q - W), the work done is given as PΔV and the workdone by the system on the environment is positive work during expansion.

b) Since this all takes place at constant temperature and against a constant atmospheric pressure, the change in internal energy for this system is 0. Change in internal energy depends on a change in temperature.

So, from the mathematical expression,

ΔU = Q + W or ΔU = Q - W

If ΔU = 0,

Q = - W or Q = W

But either ways,

Q = PΔV

(because, W = -PΔV for ΔU=Q+W and W = PΔV for ΔU=Q-W)

So, either ways, the heat transfer is the same and it is positive. This indicates heat is transferred from the surroundings to the system.

(c) What is ΔU for the gas for this process?

Since this all takes place at constant temperature and against a constant atmospheric pressure, the change in internal energy for this system is 0. Change in internal energy depends on a change in temperature.

Hope this Helps!!!

Based on the given question, we can state that Yes, the gas does work on its surroundings because it expands, this shows that there is a change in volume.

Based on the second question, we can state that because this takes place at constant temperature and against a constant atmospheric pressure, then the change in internal energy for this system is zero.

Based on the third question, we can see that the ΔU for the gas for this process is zero because of the constant temperature and atmospheric pressure.

This is given as

ΔU = Q - W

Hence, from the mathematical expression,

Regardless,

Q = PΔV

This means that the heat transfer remains the same and is positive

Read more about heat transfer here:
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Answer:

For 1: The molecular equation is [tex]CuI_2(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+2NaI(aq)[/tex]

For 2: The net ionic equation is [tex]Cu(OH)_2(s)\rightarrow Cu^{2+}(aq)+2I^{-}(aq)[/tex]

For 3: The equilibrium constant expression is [tex]K_{eq}=[Cu^{2+}][I^-]^2[/tex]

Explanation:

A molecular equation is defined as the chemical equation in which the ionic compounds are written as molecules rather than component ions.

The molecular equation for the reaction of copper (II) iodide and sodium hydroxide is given as:

[tex]CuI_2(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+2NaI(aq)[/tex]

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of copper hydroxide and sodium iodide is given as:

[tex]2NaI(aq)+Cu(OH)_2(s)\rightarrow 2NaOH(aq)+CuI_2(aq)[/tex]

Ionic form of the above equation follows:

[tex]2Na^{+}(aq)+2I^{-}(aq)+Cu(OH)_2(s)\rightarrow 2Na^+(aq)+2OH^-(aq)+Cu^{2+}(aq.)+2I^{-}(aq.)[/tex]

As, sodium and iodide ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

[tex]Cu(OH)_2(s)\rightarrow Cu^{2+}(aq)+2I^{-}(aq)[/tex]

The expression of equilibrium constant for the net ionic equation above follows:

[tex]K_{eq}=[Cu^{2+}][I^-]^2[/tex]

Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Final answer:

The heat of combustion of one mole of naphthalene is 591 kJ/mol.

Explanation:

To calculate the heat of combustion of one mole of naphthalene, we first need to calculate the heat released by the combustion of the given mass of naphthalene. We can use the formula q = mcΔT, where q is the heat released, m is the mass of the naphthalene, c is the heat capacity of the calorimeter and its surroundings, and ΔT is the change in temperature. Substituting the given values into the equation, we have:

q = 1.44 g * 10.17 kJ/°C * (30.70°C - 25.00°C)

q = 1.44 g * 10.17 kJ/°C * 5.7°C

q = 82.9728 kJ

Now, to calculate the heat of combustion per mole of naphthalene, we will use the molar mass of naphthalene, which is 128.18 g/mol:

Heat of combustion per mole = 82.9728 kJ / 1.44 g * 128.18 g/mol

Heat of combustion per mole = 591 kJ/mol

Final answer:

To determine the heat of combustion per mole of naphthalene, the heat released was first calculated using the calorimeter's heat capacity and the observed temperature change. This amount was then divided by the mass of naphthalene burned to find the heat of combustion per gram, which was finally multiplied by the molar mass of naphthalene to get the value per mole.

Explanation:

To find the heat of combustion of one mole of naphthalene using a bomb calorimeter, we first need to use the given heat capacity of the calorimeter, which is 10.17 kJ/°C, and the temperature change that the combustion caused in the water, which was from 25.00°C to 30.70°C. The mass of naphthalene combusted was 1.44 g.

First, calculate the total heat q released during the combustion:

q = C₁₄₂₃ bomb × ΔT = (10.17 kJ/°C) × (30.70°C - 25.00°C) = 10.17 kJ/°C × 5.70°C = 57.969 kJ

Next, calculate the heat of combustion per gram of naphthalene:

q per gram = 57.969 kJ / 1.44 g = 40.2604 kJ/g

Now, to find the heat of combustion per mole of naphthalene, we multiply this value by the molar mass of naphthalene. The molar mass of naphthalene (C₁₀H₈) is approximately 128.17 g/mol.

Heat of combustion per mole = q per gram × Molar Mass = 40.2604 kJ/g × 128.17 g/mol ≈ 5160.76 kJ/mol

Therefore, the heat of combustion of one mole of naphthalene is around 5160.76 kJ/mol.

The  law used to determine the relationship between the volume and the number of moles in this equation is called the Avogadro's law. by this law, the volume of gas is directly proportional to the number of moles of the gas.

Avogadro's law, which was proposed by the Italian scientist Amedeo Avogadro, explains the behavior of gases and states that gases of different chemical nature and physical conditions at the same temperature and pressure have an equal number of molecules in equal volumes.

In simpler terms, the number of molecules or moles in a gas sample is directly proportional to its volume at a constant temperature and pressure.

This law is significant in various areas of chemistry, including the calculation of chemical reactions involving gases, determination of molar volumes, and the study of ideal gases.

Find more on Avogadro's law :

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Answer:

0.0626 M

Explanation:

Equation of the reaction

CH3COOH(aq) + NaOH(aq) ---------> CH3COONa(aq) + H2O

Concentration of acid CA= ???

Concentration of base CB= 0.1943 M

Volume of acid VA= 10.0ml

Volume of base VB= 32.22 ml

Number of moles of acid NA= 1

Number of moles of base NB= 1

From

CA= CB VB NA/VA NB

Hence ;

CA= 0.1943 M × 32.22 ml × 1/10.0ml ×1

CA= 0.0626 M