Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the copper(II) acetate is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer:

[tex]CS_{2}[/tex] > [tex]CF_{4}[/tex] > [tex]SCl_{2}[/tex]

Explanation:

The X-A-X bond angle means the angle between the surrounding 'X' atoms and the central 'A' atom. The compound [tex]CS_{2}[/tex] has two bond pairs and it is linear in shape. Its bond angle is 180 degrees. The compound [tex]CF_{4}[/tex] has four bond pairs and it is tetrahedral in shape. Its bond angle is 109.5 degrees. The compound [tex]SCl_{2}[/tex] has a bond angle of approximately 109.5 degrees. Therefore the decreasing order of bond angle is:

[tex]CS_{2}[/tex] > [tex]CF_{4}[/tex] > [tex]SCl_{2}[/tex]

The correct order of decreasing X-A-X bond angle is CS2 > CF4 >SCl2.

The term bond angle refers to the dihedral angle that exists between two atoms that are bonded to the same central atom. Usually, the central atom is the least electronegative atom of the three.

Looking at the compounds involved, we will see that the correct order of decreasing X-A-X bond angle, where A represents the central atom and X represents the outer atoms in each molecule is CS2 > CF4 >SCl2.

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Answer:

1)There is 2.68 * 10^-11 J of energy needed

2) The nuclear binding energy for 1 mol of Ne is 1.6 *10^13 J/mol

Explanation:

Step 1: Data given

The nucleus of a21Ne atom has a amass of 20.98846 amu.

Step 2: Calculate number of protons and neutrons

The number of electrons and protons in an 21Ne atom = 10

The number of neutrons = 21 -10 =11

Step 3: mass of the atom

Mass of a proton = 1.00727647 u

Mass of a neutron = 1.0086649 u

The mass of the atom = mass of all neutrons + mass of protons

Mass of atom = 11*1.0086649 + 10*1.00727647  = 21.1680786 amu

Step 4: Calculate change of mass

The change in mass = Mass of atom - mass of neon

Δmass = 21.1680786 - 20.98846

Δmass = 0.1796186

Step 5: Calculate mass for a single nucleus

The change of mass for a single nucleus is = Δmass / number of avogadro

Δmass of nucleus = 0.1796186 / 6.022*10^23

Δmass of nucleus =2.98 * 10^-25 grams = 2.98 * 10^-28 kg

Step 6: Calculate energy to break a Ne nucleus

Calculate the amount of energy to break a Ne nucleus

ΔEnucleus = Δmass of nucleus * c²

⇒ with c = 2.9979 *10^8 m/s

ΔEnucleus = 2.98 * 10^-28 kg * (2.9979*10^8)² = 2.68 * 10^-11 J

What is the nuclear binding energy for 1 mol of Ne?

ΔE= ΔEnucleus * number of avogadro

ΔE= 2.68 * 10^-11 J * 6.022*10^23

ΔE= 1.6 *10^13 J/mol

Al2O3, or aluminum oxide, is most likely to have low water solubility due to the dissolution process being highly endothermic. It requires more energy to break its intermolecular forces and disperse into water which results in a low solubility.

The solute most likely to have low water solubility due to the dissolution process being highly endothermic is Al2O3 (aluminum oxide). In chemical reactions, an endothermic process involves the absorption of heat. Solutes like Al2O3 need more energy to overcome intermolecular forces and disperse into the solvent, thus making the dissolution process highly endothermic and resulting in lower solubility in water.

On the other hand, solutes like RbF, CaF2, AgCl, and FeCl2 generally require less energy to dissolve in water, making their dissolution process less endothermic and more favorable under normal environmental conditions.

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Answer: The temperature at which given potential between the two electrodes is attained is 331.13 K

Explanation:

The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction.

The half reaction follows:

Oxidation half reaction:  [tex]Fe(s)\rightarrow Fe^{2+}(0.1M)+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V[/tex]

Reduction half reaction:  [tex]Ni^{2+}(3\times 10^{-3}M)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.25V[/tex]

Net reaction:  [tex]Fe(s)+Ni^{2+}(3\times 10^{-3}M)\rightarrow Fe^{2+}(0.1M)+Ni(s)[/tex]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

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

Putting values in above equation, we get:

[tex]E^o_{cell}=-0.25-(-0.44)=0.19V[/tex]

To calculate the temperature at which the reaction is taking place, we use the Nernst equation, which is:

[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Fe^{2+}]}{[Ni^{2+}]}[/tex]

where,

[tex]E_{cell}[/tex] = electrode potential of the cell = +0.140 V

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.19 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol.K

F = Faraday's constant = 96500

T = temperature of the reaction

[tex][Fe^{2+}]=0.1M[/tex]

[tex][Ni^{2+}]=3\times 10^{-3}M[/tex]

Putting values in above equation, we get:

[tex]0.140=0.19-\frac{2.303\times 8.314\times T}{2\times 96500}\times \log(\frac{(0.1)}{(3\times 10^{-3})})\\\\T=331.13K[/tex]

Hence, the temperature at which given potential between the two electrodes is attained is 331.13 K

The temperature at which the potential between the two electrodes will be +0.140 V can be calculated using the Nernst equation. Rearranging the equation allows us to solve for temperature. Substituting the given values allows us to find the temperature.

First, we need to find the cell potential using the Nernst equation:

Ecell = E°cell - (0.0592 V / n) * log(Q)

In this case, since both sides of the cell are based on the same half-reaction with different concentrations, the number of electrons transferred (n) is 2. The equilibrium constant (Q) can be calculated using the concentrations of Ni2+ and Fe2+ ions in the half-cells. Rearranging the equation, we can solve for temperature (T):

T = (Ecell - E°cell) / ((0.0592 V / n) * log(Q))

Substituting the given values, we can solve for T.

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Explanation:

The given data is as follows.

     m = 0.5 kg,     [tex]T = 20^{o}C[/tex],     [tex]T_{2} = 90^{o}C[/tex]

It is known that specific heat of aluminium is 0.91 kJ/kg.

As we know that,   dQ = dU + dw

where,     dQ = heat transfer

                dU = change in internal energy

                dw = work transfer

For the given system, work transfer "w" is 0.

(a)    Hence, change in stored energy will be calculated as follows.

               Q = [tex]mC \Delta T[/tex]

                   = [tex]0.5 \times 0.91 \times (90 - 20)[/tex]

                   = 31.85 kJ

(b)    The amount of heat transferred will be equal to change in stored energy.

So,              dQ = Q = 31.85 kJ

(c)     Change in entropy will be calculated as follows.

                dS = [tex]mC ln \frac{T_{2}}{T_{1}}[/tex]          

                      = [tex]0.5 \times 0.91 \times ln \frac{90}{20}[/tex]

                      = 0.684 kJ/K

(d)     Entropy transfer by heat will be calculated as follows.

             [tex]\Delta S = \frac{dQ}{dT}[/tex]

                          = [tex]\frac{31.85}{(20 + 273)}[/tex]

                          = 0.1087 kJ/K

(e)    Entropy change will be calculated as follows.

              Entropy change = entropy transfer + entropy generation

           [tex]S_{2} - S_{1} = \frac{dQ}{T} + S^{o}_{gen}[/tex]

            0.684 kJ/K = 0.187 + [tex]S^{o}_{gen}[/tex]

                  [tex]S^{o}_{gen}[/tex] = 0.5752 kJ/K

Answer:

On the attached picture.

Explanation:

Hello,

At first, it is important to remember that kinetic molecular theory help us understand how the molecules of a gas behave in terms of motion. In such a way, the relative velocity of a gas molecule has the following relationship with the gas' molar mass:

[tex]V[/tex]∝[tex]\frac{1}{\sqrt{M} }[/tex]

That is, an inversely proportional relationship which allows us to infer that the bigger the molecule the slower it. In this manner, as argon is smaller than xenon, it will move faster.

Now, as the gases are in equal molar amounts and considering that argon moves faster, on the attached picture you will find the suitable depiction of the gas sample, since red dots (argon) have a larger tail than the blue dots (xenon).

Best regards.

The kinetic molecular theory explains gas behavior, showing that at a given temperature, heavier molecules like xenon move slower than lighter molecules like argon, which can be depicted with varying tail lengths in visual models.

The kinetic molecular theory of gases provides an explanation for the properties of gases by modeling them as small, hard spheres with insignificant volume, in constant motion, and undergoing perfectly elastic collisions. According to this theory, the average kinetic energy (KEavg) of gas molecules is the same for all gases at a given temperature, regardless of the molecular mass. However, because the kinetic energy depends only on temperature, lighter molecules will have higher speeds compared to heavier molecules at the same temperature.

Given a gas sample containing equal molar amounts of xenon and argon, depicted by kinetic molecular theory, we would see red dots (xenon) and blue dots (argon) with tails representing their velocities. As the diagrams from the theory suggest, we would expect that, at the same temperature, xenon atoms (being heavier) would have shorter tails (indicating lower speeds) than argon atoms (which are lighter and thus would have longer tails for higher speeds).

This behavior of the molecules can be seen in the average root mean square speed (Urms) trend, where heavier noble gases like xenon show a distribution of speeds peaking at lower values, whereas lighter ones like argon peak at higher speeds. This concept is crucial in the depiction of gas samples in kinetic molecular theory and can be visualized through illustrations that incorporate this difference in molecular speed based on the mass of the gas particles.

Answer:

4 C3H5N3O9 ------> 6N2 + O2 + 10H2O + 12CO2

Explanation:

Nitroglycerin has a chemical formula C3H5N3O9. The balanced chemical equation is as follows:

4 C3H5N3O9 ------> 6N2 + O2 + 10H2O + 12CO2

We suppose that in a reaction, 44g of carbon dioxide is produced. The mass of nitroglycerin that must have reacted will be calculated as under:

Molecular mass of Nitroglycerin = 227g/mol

Molecular mass of Carbon dioxide = 44g/mol

No. of moles of carbon dioxide produced = 44/44 = 1 mole produced.

Now, from balanced chemical equation, we can see that

12 moles of carbon dioxide are produced by = 4 moles of nitroglycerin.

1 mole of carbon dioxide is produced by = 4/12 = 1/3 moles of nitroglycerin.

Mass of nitroglycerin which produced 1 mole of carbon dioxide =  1/3 x 227 = 75.666 grams.

Answer:

D

Explanation:

We know that the

reaction catalyzing power of a catalyst ∝ surface area exposed by it

Given

volume V1= 10 cm^3

⇒[tex]\frac{4}{3} \pi r^3= 10[/tex]

hence r= 1.545 cm

also, surface area S1= [tex]4\pi r^2[/tex]

now when the sphere is broken down into 8 smaller spheres

S2= 8×4πr'^2

now, equating V1 and V2 ( as the volume must remain same )

[tex]\frac{4}{3}\pi r^3=8\times\frac{4}{3} \pi r'^3[/tex]

and solving we get

r'= r/2

therefore, S2=[tex]8\times4\pi\frac{r}{2}^2[/tex]

S2=[tex]2\times4\pi r^2[/tex]

S2= 2S1

hence the correct answer is

. The second run has twice the surface area.

An increase in temperature shifts an endothermic reaction to the right and an exothermic reaction to the left. For the reduced volume, it shifts the equilibriums to the side with the fewer moles of gas.

The effect of an increase in temperature on the given equilibrium systems depends on the sign of Delta H (ΔH), which represents the heat of reaction. For (a) 2NH3(g) ⇌ N2(g)+3H2(g) with ΔH = 92kJ and (b) N2(g) + O2(g) ⇌ 2NO(g) with ΔH = 181kJ, because these reactions are endothermic (ΔH is positive), an increase in temperature will shift the equilibrium towards the right to absorb the excess heat. On the other hand, reactions (c) 2O3(g) ⇌ 3O2(g) with ΔH = - 285kJ and (d) CaO(s) + CO2(g) ⇌ CaCO3(s) with ΔH = - 176kJ are exothermic (ΔH is negative), an increase in temperature will shift the equilibrium to the left to offset the increase in heat.

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Increasing temperature generally shifts endothermic reactions to the right and exothermic reactions to the left. Decreasing volume typically shifts equilibria towards the side with fewer moles of gas. Each reaction's equilibrium shift depends on its enthalpy change (ΔH) and the moles of gases involved.

(a) 2NH₃(g) ⇌ N₂(g) + 3H₂(g) ΔH = 92kJ

(b) N₂(g) + O₂(g) ⇌ 2NO(g) ΔH = 181kJ

(c) 2O₃(g) ⇌ 3O₂(g) ΔH = − 285kJ

(d) CaO(s) + CO₂(g) ⇌ CaCO₃(s) ΔH = −176kJ

Answer:

a. Vₐ = 111.5282 + 1.29396m

b. For m = 0.100m; Vₐ = 111.6576

Explanation:

The partial molar volume of compound A in a mixture of A and B is defined as :

[tex]V_a = \frac{dV}{dn_a}[/tex]

Where V is volume and n are moles of a.

a. As molality is proportional to moles of substance, partial molar volume of glucose can be defined as:

Vₐ =  dV / dm =  d(1001.93 + 111.5282m + 0.64698m²) / dm

Vₐ = 111.5282 + 1.29396m

b. Replacing for m = 0.100m:

Vₐ = 111.5282 + 1.29396×0.100

Vₐ = 111.6576

I hope it helps!

Final answer:

The partial molar volume of glucose is calculated by taking the derivative of the volume-molality equation and then evaluating at a specific molality (0.100m). The result is 111.657596 mL/mol for the partial molar volume of glucose in a 0.100m solution.

Explanation:

The question asks us to calculate the partial molar volume of glucose and then find its value in a 0.100m glucose-water solution. The partial molar volume (denoted as ϕV/ϕm) represents the change in total volume (V) with respect to the change in molality (m) of the solution, holding the amount of solvent constant. This can be computed by taking the derivative of the volume equation with respect to m:

V = 1001.93 + 111.5282m + 0.64698m2

The first derivative will yield the expression for the partial molar volume:

(ϕV/ϕm) = 111.5282 + 2×0.64698m

To calculate the partial molar volume of glucose in a 0.100m solution, simply substitute 0.100 for m in the derived expression:

(ϕV/ϕm) = 111.5282 + 2×0.64698×0.100 = 111.5282 + 0.129396 = 111.657596 mL/mol

Note that to find the mass of the solvent, water, in the solution, a calculation is needed using the molarity and density of the solution. Knowing that the mass of glucose and molality conversions are essential for such calculations, and realizing that glucose's molar mass is needed to find the mass of glucose from its molar amount.

Answer:

42,3g of H₂O

Explanation:

For the reaction:

4NH₃(g) + 5O₂(g) ⟶ 4NO(g) + 6H₂O(g)

62,8 g of NH₃ are:

62,8g×(1mol/17,031g) = 3,69 moles of NH₃

62,8 g of O₂ are:

62,8g×(1mol/32g) = 1,96 moles of O₂

For a complete reaction of 1,96 moles of O₂ you need:

1,96mol O₂×(4mol NH₃ / 5molO₂) = 1,57 moles NH₃. As you have 3,69 moles, limiting reactant is O₂.

Assuming a complete reaction, 1,96mol O₂ produce:

1,96mol O₂×(6mol H₂O / 5molO₂) = 2,35 moles of H₂O. In grams:

2,35 moles of H₂O×(18,01g/1mol) = 42,3g of H₂O

I hope it helps!

Answer: ionic solid

Explanation:

In an ionic solid, the ions are bound together by strong electrostatic attraction hence they are immobile and the solid is unable to conduct electricity. If this solid is dissolved in water, the ions move apart due to solvation and become mobile hence the solution conduts electricity. Similarly, when the solid melts, the ions also become free and the melt conduct electricity.

A solid that is hard, brittle, does not conduct electricity in solid form but does in liquid form or when dissolved in water, and has a high melting point is classified as an ionic solid.

The characteristics of the solid described are indicative of an ionic solid. These solids are typically hard and brittle, and they have high melting points. As a solid, ionic compounds do not conduct electricity because the ions are locked in place within the crystal lattice and thus cannot move freely. When these compounds melt, however, the ionic lattice breaks down and the ions are free to move, allowing the liquid to conduct electricity.

Similarly, when an ionic compound is dissolved in water, it dissociates into ions, which are free to move in the solution, making the solution an electrical conductor. This is because an electrolyte is present, which is a substance that contains free ions and can behave as an electrical conductor.

Answer:

The reaction will shift to the right in the direction of products.

Explanation:

According to Le Chatelier's Principle, the change in any state of the equilibrium say temperature, volume, pressure, or the concentration, the equilibrium will oppose these changes and will shift in such a way that the effect cause must be nullified.

For an endothermic reaction,

On increasing the temperature, reaction will go in forward direction (towards right), because according to Le Chatelier's principle as we increase the temperature,  the equilibrium is will be disturbed , so to again establish the equilibrium, the reaction will go in forward direction as it is endothermic in nature (towards right).

Thus, the [tex]\Delta H^0[/tex] of the given reaction is positive and thus, on increasing temperature, reaction will go in forward direction.

Answer:- The reaction will shift to the right in the direction of products.

The reaction will move to the left in the direction of the reactants as the temperature rises, favouring the endothermic path. The reaction will move left in the direction of the reactants, thus that is the right response.

A chemical process that takes heat from its surroundings is said to be going in an endothermic direction. Because the reactants and products of an endothermic reaction have different energies, heat energy is needed to help the reaction proceed and create the products. An endothermic reaction is one that absorbs heat while it is occurring. This can be felt as a drop in temperature in the immediate environment or shown by thermodynamic calculations. An endothermic process has a positive enthalpy change (H), which means energy is being absorbed. The reaction will move to the left in the direction of the reactants as the temperature rises, favouring the endothermic path. The reaction will move left in the direction of the reactants, thus that is the right response.

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Answer:

5- N20 and N2O5

Explanation:

Full working is shown in the image attached. It is important to remember that NO2 dimerizes to N2O4 while N2PO occurs as monomers.

Answer:

The value of equilibrium constant for reverse reaction is [tex]7.7\times 10^{-4}[/tex]

Explanation:

The given chemical equation follows:

[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]

The equilibrium constant for the above equation is [tex]1.7\times 10^6[/tex].

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

[tex]2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)[/tex]

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '1/2', the equilibrium constant of the reverse reaction will be the square root of the equilibrium constant  of initial reaction.

So,

[tex]SO_3(g)\rightarrow SO_2(g)+\frac{1}{2} O_2(g)[/tex]

The value of equilibrium constant for half reverse reaction is:

[tex]K_{eq}'=(\frac{1}{1.7\times 10^6})^{\frac{1}{2}}=0.00077=7.7\times 10^{-4}[/tex]

Hence, the value of equilibrium constant for reverse reaction is [tex]7.7\times 10^{-4}[/tex]

Answer:

[tex]7.7\times10^{-4}[/tex]

Explanation:

The equation for which we have to find Kc is obtained by two - step transformation of the equation whose Kc is given.

1st step:

Reversing the reaction:

By reversing the reaction the reactants become products and vice-versa.

The new equilibrium constant will be:

[tex]Kc^{'}=\frac{1}{Kc}[/tex]

2nd step:

Dividing the equation throughout by 2:

New Kc becomes:

[tex]Kc^{''}=\sqrt{Kc^{'}}=\frac{1}{\sqrt{Kc} }[/tex]

[tex]=\frac{1}{\sqrt{1.7\times10^{6} } }=7.7\times10^{-4}[/tex]

Hence the equilibrium constant is [tex]7.7\times10^{-4}[/tex]

Answer:

39g

Explanation:

Details of the solution is shown below. From the information provided regarding the N2 produced, we could calculate the amount of N2 produced and use that to find the mass of sodium azide reacted.

The balanced chemical equation for the decomposition of solid sodium azide (NaN3) is 2 NaN3(s) = 2 Na(s) + 3 N2(g). Using the ideal gas law, we can calculate the number of moles of dinitrogen gas produced, which is 1.07 mol. From the balanced equation, we find that 1 mole of NaN3 decomposes to produce 3 moles of N2. Therefore, the mass of sodium azide that reacted is 23.40 g.

The balanced chemical equation for the decomposition of solid sodium azide (NaN3) into solid sodium and gaseous dinitrogen is:

2 NaN3(s) → 2 Na(s) + 3 N2(g)

Given that 22.0 L of dinitrogen gas are produced by this reaction at a temperature of 11.0°C and a pressure of exactly 1 atm, we can use the ideal gas law to calculate the number of moles of dinitrogen gas produced:

n = PV / RT = (1 atm)(22.0 L) / (0.0821 atm·L/mol·K)(11.0°C + 273.15 K) = 1.07 mol

From the balanced chemical equation, we can see that 1 mole of NaN3 decomposes to produce 3 moles of N2. Therefore, the number of moles of NaN3 that reacted is 1.07 / 3 = 0.36 mol.

To calculate the mass of sodium azide that must have reacted, we can use the molar mass of NaN3 which is 65.01 g/mol:

mass = moles × molar mass = 0.36 mol × 65.01 g/mol = 23.40 g

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Answer:

Dispersion forces.

Explanation:

CO2 contains dispersion forces, and covalent bonds. It is a linear molecule, and the bond angle of O-C-O is 180 degree. O is more electronegative than C, the C-O contains polar bond with the having negative end pointing towards the O.

CO contains two C-O bonds. They cancel each other out because of the dipoles point in opposite directions. Although, CO2 contains polar bonds, it is known as a nonpolar molecule. So, the only intramolecular forces which CO2 having are London dispersion forces.

Answer:

271g

Explanation:

The full explanation is seen in the image attached. See the solution below for details

Final answer:

The problem involves physics principles, specifically the ideal gas law. To solve for the final temperature of a gas when pressure changes and the number of moles is halved from an initial condition, the relationship between pressure and temperature must be considered. However, the type of process (isothermal, isobaric, adiabatic) must be known for an accurate calculation.

Explanation:

The student's question involves finding the final temperature of a gas when the pressure changes and the number of moles is reduced by half, starting from an initial temperature (Ti) of 25.0°C. This problem can be solved by applying the ideal gas law and the concept that, for a given amount of gas, the pressure is directly proportional to the temperature (P ∝ T) when volume and the number of moles are constant. Given the initial conditions and the pressure change from 25.7 atm to 4.10 atm, the process is not specified as isothermal, isobaric, or adiabatic; therefore, additional details from the context of the part (a) of the experiment would be required to provide a comprehensive solution.

To find the final temperature based on the given information, one would have to assume the same type of process that occurred in part (a), where volume doubled and pressure got halved. If we assume a similar relationship between temperature and pressure as was demonstrated before, where if the pressure is halved from 2.50 atm, the temperature must also be halved from 303.15 K, we could calculate the final temperature for the new conditions by adjusting for the fact that the number of moles was halved. However, without explicit mention of whether this happens in an isothermal, isobaric, or adiabatic process, a direct calculation cannot be provided here.

Explanation:

Surface tension is defined as the attractive forces experienced by the surface molecules of a liquid by the molecules present beneath the surface layer of the liquid.

In a molecule, more is the number of hydroxyl groups present in it more will be the force of attraction faced by surface molecules towards the inner side of the liquid.

Hence, then the surface tension will increase.

Compoud (a) has only one -OH group, compound (b) has 3 -OH groups and compound (c) has 2 -OH groups.

Therefore, the given compounds are arranged on the basis of order of increasing surface tension at a given temperature are as follows.

    [tex]CH_3CH_2CH_2OH[/tex] < [tex]HOCH_2CH_2OH[/tex] < [tex]HOCH_2CH(OH)CH_2OH[/tex]

The surface tension in the following molecules in the increasing order is given as, [tex]\rm CH_2CH_2CH_2OH < HOCH_2CH_2OH < HOCH_2CH(OH)CH_2OH[/tex].

Surface tension can be given as the force applied by the molecules underneath the liquid surface to form the attraction and made the liquid occupy, the least surface area.

The compounds with a more number of attracting groups such as hydroxyl radicals tend to apply more attraction and have more surface tension.

The number of hydroxyl groups in the following compounds are:

The increasing order of the surface tension in the following molecules is, [tex]\rm CH_2CH_2CH_2OH < HOCH_2CH_2OH < HOCH_2CH(OH)CH_2OH[/tex].

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Answer:

Reactivity: Ethyl benzene > Toluene > Ethane

Explanation:

The order of reactivity depends on how easily the most reactive hydrogen can be abstracted.

The order of reactivity of hydrogen is : 1° < 2° < 3° < Benzyllic based on the stabilizing effects like inductive effect, hyperconjugation , resonance effect.

Hence the order of decreasing reactivity : Ethyl Benzene, Toluene , Ethane.

To sum up, ethylbenzene > toluene > ethane is the sequence in which reactivity decreases with bromine, and the previously described components are critical in dictating the rates of reaction.

Hydrocarbons containing bromine atoms exhibit varying degrees of reactivity, which can be attributed mainly to the stability of the free radical generated during hydrogen atom abstraction. When an unpaired electron delocalizes across the aromatic ring, a more stable benzyl radical is produced upon abstraction from a benzylic carbon in toluene and ethylbenzene. Compared to toluene, the benzyl radical generated from ethylbenzene has somewhat more reactivity due to this delocalization. Ethane has the lowest reactivity of the three compounds since it doesn't have an aromatic ring or benzylic hydrogens, which leads to the formation of a primary free radical that is much less stable.

Answer:

K= 0.06611

Explanation: The rate of reaction is defined as the change in concentration of any of reactant or products per unit time. From the given reaction, the rate of reaction may be equal to the rate of disappearance of reactant which is equal to the rate of appearance of products.

The average rate of disappearance of H2O2 over the time period from t=0 min at 8.92×10^-2 to t=9.63min at 4.72×10^-2 is given as -4.36×10^-3Mmin-1.

We can say:

•The initial concentration [H2O2]o is 8.92×10^-2M

•The concentration at time t. [H2O2]t is 4.72×10^-2

•The time (t) is 9.63 min

The expression of rate constant for a first order reaction is shown as

K=2.303/t log[H2O2]o/ [H2O2]t

Substitute the values of t, [H2O2]o and [H2O2]t in the equation of rate constant.

K=2.303/9.63 log [8.92×10^-2]/ [4.72×10^-2]

K= 0.2391 (log 8.92×10^-2 - log 4.72×10^-2)

K= 0.2391 [-1.0496-(-1.3261)]

K= 0.2391 (-1.0496+1.3261)

K= 0.2391 (0.2765)

K= 0.06611

Since the value of k is almost constant, the decomposition of H2O2 is a first order reaction.

Answer:

A hypothesis can be described as a tentative statement which can be proved either right or wrong through scientific experiments. If a hypothesis is tested again and again and every time the experiments give the same results, then the hypothesis can take the form of a theory. However, a theory is subjected to change if new researches are made which can annul it.  For a theory to be formed, there should be enough explanation behind the phenomenon along with the experiments.

Answer:

a) 56 protons

b) 54 electrons

c) 81 neutrons

d) The sum of protons and neutrons is shown. The number of protons is always the same. So we can calculate the number of neutrons ( and also the isotopes)

e)137Ba (with 56 protons and 81 neutrons)

f) atomic mass is 136.9 u ; the mass number is the sum of protons and neutrons and is 137

Explanation:

Step 1: Data given

137 Ba2+ is an isotope of barium. The atomic number of barium( and its isotopes) is 56. This shows the number of protons.

For a neutral atom, the number of protons is equal to the number of electrons.

The different isotopes of an element have the same number of protons but a different number of neutrons.

137Ba2+ has 56 protons (this is the same as the atomic number)

137Ba2+ has 54 electrons ( since it's Ba2+, this means it has 2 electrons less than protons, that's why it's charged +2)

137Ba2+ has 81 neutrons ( 137 - 56 = 81)

In the symbol, the atomic number is not shown. The sum of the protons and neutrons is shown. (Since the number of protons is the same for every isotope, we can calculate the number of neutrons that way. By knowing the neutrons, we also know the isotope.

This isotope is 137Ba

Atomic mass is also known as atomic weight. The atomic mass is the weighted average mass of an atom of an element based on the relative natural abundance of that element's isotopes.

The atomic mass of 137Ba2+ is 136.9 u

The mass number is a count of the total number of protons and neutrons in an atom's nucleus.

The mass number of 137Ba2+ is 137

Answer:

0.00564 moles

Explanation:

Given that:

The rate constant, k = [tex]6.82\times 10^{-3}[/tex] s⁻¹

Initial concentration [A₀] = [tex]2.90\times 10^{-2}[/tex] mol

Time = 4.0 min = [tex]4.0\times 60[/tex] sec = 240 sec

Using integrated rate law for first order kinetics as:

[tex][A_t]=[A_0]e^{-kt}[/tex]

Where,  

[tex][A_t][/tex] is the concentration at time t

So,  

[tex][A_t]=2.90\times 10^{-2}\times e^{-6.82\times 10^{-3}\times 240}=2.9\times \frac{1}{10^2}\times \frac{1}{e^{1.6368}}[/tex]

[tex][A_t]=0.00564\ moles[/tex]

The concentration after four minutes is 3.3 ×10−3.

Let us recall that for a first order reaction;

ln[A] = ln[A]o - kt

Where;

[A] = concentration at time t

[A]o = initial concentration

k = rate constant

t = time

[A]o = 2.90×10−2 mol/1.7 L = 0.0171 M

k = 6.82×10−3 s−1

t = 4 min or 240 s

Substituting values;

ln[A] = ln[0.0171 M] - (6.82×10−3 s−1 × 240 s)

[A]  = e^ln[0.0171 M] - (6.82×10−3 s−1 × 240 s)

[A]  = 3.3 ×10−3.

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Answer:

The empirical formula is P2O5 (option B)

Explanation:

An empirical formula does not necessarily represent the actual numbers of atoms present in a molecule of a compound; it represents only the ratio between those numbers.

The actual numbers of atoms of each element that occur in the smallest freely existing unit or molecule of the compound is expressed by the molecular formula of the compound.

The molecular formula of a compound may be the empirical formula, or it may be a multiple of the empirical formula.

If the molecular formula is P4O10, this means for each for P-atoms we have 10-O atoms this is a ratio 4:10 or 1: 2.5

To find the empirical formula we divide the molecular formula by 2 what will give us P2O5

For each 2 P atoms we have 5 O-atoms. This is a ratio 1: 2.5

This is the simpliest form for the compound P4O10.

The empirical formula is P2O5 (option B)

Answer:

FCC: r = 0.414R

BCC: r = 0.291R

Explanation:

For an FCC unit cell, the interstitial site is located at the middle of the edge. An atom that can occupy the interstitial site will have a diameter of 2*r. And we know that:

2*r = a - 2*R        equation (1.0)

a = [tex]2*\sqrt{2}*R[/tex]

Therefore, substituting the expression for 'a' in equation (1.0)

2*r =  [tex]2*\sqrt{2}*R[/tex] - 2*R

r = R*([tex]2\sqrt{2} - 2[/tex])/2 = 0.414R

For a BCC unit cell, there is a right-angle triangle formed by 3 arrows. Using the triangle, we have:

[tex]\frac{a^{2} }{2} +\frac{a^{2} }{4} = (R+r)^{2}[/tex]        equation (2.0)

a = [tex]\frac{4R}{\sqrt{3} }[/tex]

replacing the expression for a in equation (2.0), we have:

[tex]\frac{4R^{2} }{2\sqrt{3} } + \frac{4R^{2} }{4\sqrt{3} } = R^{2} + 2Rr + r^{2}[/tex]

Further simplification and rearrangement, the expression above is simplified to:

[tex]r^{2} + 2Rr - 0.667R^{2} = 0[/tex]

Solving the above quadratic equation, we have:

[tex]r = \frac{-2R - 2.582R}{2}or\frac{-2R + 2.582R}{2}[/tex]

r = - 2.291R or 0.291 R

Since the value of r can only be positive, the correct answer is r = 0.291R

To find the radius of an impurity atom in a FCC octahedral site, use the length of the face diagonal and the atomic radius of the host atom. For a BCC tetrahedral site, consider the relationship between the cation and anion radii.

In an FCC structure, the radius of an impurity atom that will just fit into an octahedral site can be calculated using the length of the face diagonal and the atomic radius of the host atom. The length of the diagonal is equal to four times the host atom radius, so we can use this information to find the radius of the impurity atom.

For a BCC structure, the radius of an impurity atom that will just fit into a tetrahedral site can be calculated by considering the relationship between the cation and anion radii. The cation radius is typically a certain percentage of the anion radius, and this information can be used to determine the radius of the impurity atom.

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Answer:

B.red

Explanation:

Electromagnetic spectrum is range of the frequencies and their respective wavelengths of the various type of the electromagnetic radiation.

In order of the decreasing wavelength the spectrum are:  

Red , Orange, Yellow, Green, Blue, Indigo, Violet

Increasing wavelength is the opposite trend. Thus, The longest visible wavelength is red and the shortest is violet.

Also, Violet light gets scattered the most while the red light gets scattered the least.

During the time of the sunset, the Earth is rotating away from the Sun. Thus, most of the light colors scatters in the ways and the color that least scatter which is red reaches the Earth.

That's why, at the time of sunrise and sunset, the sky looks red.

Answer: The standard potential for this cell is +1.49 V at 25C.

Explanation:

[tex]E^0_{[Sn^{2+}/Sn]}=-0.14V[/tex]

[tex]E^0_{[Ti^{2+}/Ti]}=-1.63V[/tex]

As titanium has lower reduction potential, it will act as anode and tin will acts as cathode.

[tex]Ti+Sn^{2+}\rightarrow Ti^{2+}+Sn[/tex]

Using Nernst equation :

[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Ti^{2+}]}{[Sn^{2+}]}[/tex]

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = [tex]25^oC=273+25=298K[/tex]

n = number of electrons in oxidation-reduction reaction = 2

[tex]E^0=E^0_{cathode}- E^0_{anode}=-0.14-(-1.63)=1.49V[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 1.49 V

[tex]E_{cell}[/tex] = emf of the cell = ?

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

[tex]E_{cell}=1.49-\frac{2.303\times (8.314)\times (298)}{1\times 96500}\log \frac{1}{1}[/tex]

[tex]E_{cell}=1.49V[/tex]

The voltage of the voltaic cell is 1.49 V.

A voltaic cell is a cell in which electrical energy is produced by a spontaneous chemical reaction.

The equation of the cell is; Ti(s) + Si^2+(aq) -----> Ti^2+(aq) + Si(s)

E°cathode = -0.14 V

E°anode = -1.63 V

E°cell = (-0.14 V) - (-1.63 V) = 1.49 V

Using the Nernst equation;

E = E°cell - 0.0592/n log[Ti^2+]/[Si^2+]

E = 1.49 V - 0.0592/2 log(1)/(1)

E = 1.49 V

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Answer:D

Explanation:

The high boiling point of HF is not attributable to the dispersion forces mentioned in the question. In HF, a stronger attraction is in operation, that is hydrogen bonding. This ultimately accounts for the high boiling point and not solely the dispersion model as in F2.

The unusually high boiling point of HF compared to F2 is due to the strong hydrogen bonding interactions between HF molecules.

The correct answer is D. Liquid F2 has weak dispersion force attractions between its molecules, whereas liquid HF has both weak dispersion force attractions and hydrogen bonding interactions between its molecules. Dispersion forces are a type of intermolecular force that occurs between all molecules, regardless of polarity.

However, these forces are generally weaker than other types of intermolecular forces such as hydrogen bonding. In HF, the significant electronegativity difference between hydrogen and fluorine leads to the formation of a polar covalent bond, which makes the HF molecules capable of hydrogen bonding, a stronger intermolecular force.

This hydrogen bonding results in a much higher boiling point for HF as compared to F2, which can only interact with other F2 molecules via relatively weaker dispersion forces.

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