‘A’ is an element which belongs to period 3, having 6 electrons in its valence shell. Below is a list of successive ionization energies (in kJ/mol) for period 3. IE2 = 2250 IE3 = 3360 IE4 = 4560 IE5 = 7010 IE6 = 8500 IE7 = 27,100 Identify the element ‘A’

In spite of ethylene and benzene has double bonded carbon-carbon, Benzene is a cyclic molecule with a special property called aromaticity. Aromaticity gives to the ring electronic properties which makes harder for benzene being reactive as ethylene.

Benzene has the formula C6H6 being cyclic and aromatic because have conjugated double bonds and those double bonds has a property called resonance unusual elevated making the structure very stable. That can be seen under the conditions of reaction of alkene compounds in which benzene doesn't react, like bromination with Br2 in CCl4.

On the other hand, ethylene is an alkene which has the formula C2H4, the double bond carbon carbon is available to suffer the different reactions of alkenes being more reactive.

Ethylene is more reactive than benzene primarily due to ethylene's discrete, highly reactive double bond, compared to benzene's delocalized and stabilized double bonds within its aromatic ring. The stability of the delocalized electrons in the benzene ring reduces its reactivity, while ethylene's electron-rich double bond invites addition reactions.

The difference in reactivity between ethylene (C2H4) and benzene (C6H6) toward molecular bromine is primarily due to the structural differences in their carbon-carbon double bonds. Ethylene contains a discrete C=C double bond that is highly reactive because it can open up to allow additional atoms to attach to the carbon atoms (an addition reaction). In contrast, the C=C bonds in benzene are part of a delocalized electron system over the entire aromatic ring, giving it extraordinary stability. This delocalization makes benzene less reactive towards addition reactions, such as with bromine, because the addition would disrupt the stable aromatic system.

Benzene's unique stability is due to the alternating single and double C-C bonds, which form a resonance-stabilized structure. This stability is a concept commonly understood as aromaticity, which significantly reduces the chemical reactivity of benzene's double bonds compared to those in alkenes like ethylene. Although the C=C bond is nonpolar, the reactivity of ethylene is also influenced by the electron-rich region around the σ bond, making it more prone to reactions with nucleophiles.

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Statement a is true because of Avogadro's Law. Statement b is also true since molar mass of O2 is greater than H2, so 5.0g of O2 contains fewer molecules than H2. Statement c is false as the number of molecules in given volume and temperature is constant.

The subject of this question is chemistry, specifically the concept of Avogadro's law. Avogadro's law states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.

Statement a is TRUE. Given equal volumes and temperatures, the flasks do contain the same number of molecules. This principle is referred to as Avogadro's Law.

Statement b is TRUE. The molar mass of O2 is indeed greater than H2, meaning that 5.0 g of O2 has fewer molecules than 5.0 g of H2, contrary to what statement b suggests.

Statement c is FALSE. Equal volume flasks at the same temperature will have the same number of molecules, regardless of pressure. Which specific gas is involved does not change this fact.

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The correct options are as follows:

a) False. Because the molar mass of O2 is greater than the molar mass of H2, 5.0 g of O2 will contain fewer molecules than 5.0 g of H2.

b) True. Because the molar mass of O2 is greater than the molar mass of H2, 5.0 g of O2 will contain fewer molecules than 5.0 g of H2.

c) False. Regardless of the pressure, if the volumes and temperatures are the same, the number of molecules will be the same due to Avogadro's Law.

Let's analyze each statement:

a) This statement is false. The volume of a gas is directly proportional to the number of moles of gas at a constant temperature and pressure (Avogadro's Law). Since the molar mass of O2 (approximately 32 g/mol) is much greater than that of H2 (approximately 2 g/mol), 5.0 g of O2 will represent fewer moles than 5.0 g of H2. Consequently, the flask containing O2 will have fewer molecules than the flask containing H2, despite the volumes being equal.

b) This statement is true. As explained above, because O2 has a higher molar mass than H2, 5.0 g of O2 will contain fewer moles and hence fewer molecules than 5.0 g of H2.

c) This statement is false. According to Avogadro's Law, equal volumes of gases at the same temperature and pressure contain the same number of molecules. Therefore, the pressure does not affect the number of molecules if the volume and temperature are constant. The number of molecules in each flask will be the same because both flasks are at the same temperature and have the same volume. The pressure within each flask will adjust according to the number of moles of gas present (as per the ideal gas law, PV = nRT), but this does not change the number of molecules.

To calculate the number of moles for each gas:

 For O2:

Number of moles of O2 = mass of O2 / molar mass of O2

Number of moles of O2 = 5.0 g / 32 g/mol ≈ 0.156 mol

For H2:

Number of moles of H2 = mass of H2 / molar mass of H2

Number of moles of H2 = 5.0 g / 2 g/mol = 2.5 mol

Clearly, the number of moles of H2 is greater than the number of moles of O2 for the same mass, which confirms that the number of molecules in the H2 flask is greater than in the O2 flask.

Answer:

The equilibrium concentration of [PCl₅] = 0.24M

Explanation:

This is the reaction.

PCl₅(g) → PCl₃(g) +Cl₂(g)

Now let's make, the expression for Kc

Remember that concentrations must be in M

Kc = ( [PCl₃] . [Cl₂] ) / [PCl₅]

4.2x10⁻² = [0.1] . [0.1] / [PCl₅]

[PCl₅] =  [0.1] . [0.1] / 4.2x10⁻²

[PCl₅] = 0.24M

Using the given equilibrium constant (Kc) and the concentrations of PCl3 and Cl2 at equilibrium, the equilibrium concentration of PCl5 is determined to be 0.24 M.

This question refers to the concept of equilibrium in chemistry, specifically the application of the equilibrium constant (Kc) in determining the concentrations of reactants and products in chemical reactions.

The given chemical reaction is: PCl₅(g)  PCl(g) + Cl₂(g). The equilibrium constant expression for the reaction is Kc = [PCl₃][Cl₂] / [PCl₅]. Here, [PCl₃], [Cl₂], and [PCl₅] represent molar concentrations of PCl₃, Cl₂, and PCl₅, respectively, at equilibrium.

We are given that Kc = 4.2 x 10-2,, [PCl₃]= 0.10M, [Cl₂]= 0.10M. Substituting these values into the equilibrium constant expression, we can solve for [PCl₅], giving:

[PCl₅]= [PCl₃][Cl₂] / Kc = (0.10)(0.10) / (4.2 x 10-2) = 0.24 M.

So, the equilibrium concentration of PCl₅ in this reaction is 0.24 M, thus the correct option is (e) 0.24 M.

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Answer:

0.008

Explanation:

From the balanced equation, 3 moles of oxygen gas produced 2 moles of the product. Then, 0.012 moles of oxygen gas will produces 0.012 x 2/3 = 0.008

When 0.012 mol of O2 completely reacts with excess aluminum according to the balanced chemical equation, 0.008 mol of Al2O3 is produced, rounded to two significant figures.

To calculate the number of moles of Al2O3 produced from the complete reaction of 0.012 mol of O2, we use stoichiometry based on the balanced chemical equation:

4Al(s) + 3O2(g) → 2Al2O3(s).

From the equation, 3 moles of oxygen gas (O2) produce 2 moles of aluminum oxide (Al2O3). To find the moles of Al2O3 produced, use the mole ratio:

(mol Al2O3) = (mol O2) × (2 mol Al2O3 / 3 mol O2)

Substituting the given value:

(mol Al2O3) = (0.012 mol O2) × (2 mol Al2O3 / 3 mol O2) = 0.008 mol Al2O3

So, when 0.012 mol of O2 completely reacts, 0.008 mol of Al2O3 is produced, assuming excess aluminum is present. This answer is rounded to two significant figures.

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Answer:

1.877 x 10⁷ kg

Explanation:

To solve this problem we first use the equation:

Where Q is the heat needed to increase the temperature of a substance, m is the mass, c is the specific heat and ΔT is the difference in temperature.

For this problem:

We put the data in the equation and solve for Q:

So that's the energy required to heat 1000 kg of water, now we calculate the mass of wood using the equation:

Where b is the heating value of wood and m its mass:

Answer:

The substance has a specific heat of 1.176 J/g°C

Explanation:

Step 1: Data given

Temperature change = 34 °C

Mass of the substance = 20 kg = 20000 grams

The substance gained 800 kJ of heat during this temperature change

Step 2: Calculate the specific heat

q = m*c*ΔT

⇒ with q = heat gained = 800 kJ = 800000 J

⇒ with m = the mass of the substance = 20 kg = 20000 grams

⇒ with c = the specific heat of the substance = TO BE DETERMINED

⇒ with ΔT = the change of temperature = T2 -T1 = 48° - 14 ° = 34°

c = q/(m*ΔT)

c = 800000 / (20000 * 34)

c = 1.176 J/g°C

The substance has a specific heat of 1.176 J/g°C

The volume of NO₂ gas collected over water at 25.0 °C is 1.67 L

We'll begin by calculating the number of mole of Cu. This can be obtained as follow:

Mass of Cu = 2.01 g

Molar mass of Cu = 63.55 g/mol

Mole = mass / molar mass

Mole of Cu = 2.01 / 63.55

Next, we shall determine the number of mole of NO₂ produced by 2.01 g (i.e 0.0316 mole) of Cu. This can be obtained as follow:

Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2H₂O(l) + 2NO₂(g)

From the balanced equation above,

1 mole of Cu reacted to produce 2 moles of NO₂.

Therefore,

0.0316 mole of Cu will react to produce = 0.0316 × 2 = 0.0632 mole of NO₂.

Finally, we shall determine the volume of NO₂ gas obtained from the reaction. This can be obtained as follow:

Number of mole of NO₂ (n) = 0.0632 mole

Temperature (T) = 25 °C = 25 + 273 = 298 K

Total pressure = 726 mm Hg

Vapor pressure of water = 23.8 mm Hg

Pressure of NO₂ (P) = 726 – 23.8 = 702.2 / 760 = 0.924 atm

Gas constant (R) = 0.0821 atm.L/Kmol

0.924 × V = 0.0632 × 0.0821 × 298

0.924 × V = 1.54623856

Divide both side by 0.924

[tex]V = \frac{1.54623856}{0.924} \\\\[/tex]

Therefore, the volume of NO₂ collected over water at 25.0 °C is 1.67 L

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The chemical reaction is defined as the reaction of reactants to form products. The reaction between copper and nitric acid will result in the formation of copper(II)nitrate, water, and nitrogen dioxide. The volume of NO[tex]_2[/tex] collected will be 1.67 L.

Give that,

The number of moles produced by nitrogen dioxide (2.01 g) will be:

Now, from the equation,

Now, from the ideal gas equation,

Therefore, the volume of nitrogen dioxide collected over water at 25-degree celcius is 1.67 L.

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Answer:

1. C has been oxidized in the reaction.

2. Mn has been reduced in the reaction.

3/4 The reducing agent is the C and the oxidizing agent is the Mn.

The reaction in an acidic solution is:

16H⁺  +  2MnO₄⁻  + 5C₂O₄⁻²   →  10CO₂  +  2Mn²⁺  +  8H₂O

10 moles of electrons are been transferred.

Explanation:

MnO₄⁻ (aq) + H₂C₂O₄ (aq)  →  Mn²⁺ (aq) + CO₂ (g)

In the permanganate, Mn acts with +7 as oxidation number, and in product side, we have Mn2+. It has decrease the oxidation number, so it has been reduced. This is the oxidizing agent.

In the oxalic acid, carbon has +3 as oxidation number, and in CO2, we see that acts with +4 so, it has increase it. This element is the reducing agent and has been oxidated.

8H⁺  +  MnO₄⁻  +  5e⁻  →  Mn²⁺  +  4H₂O

It is completed on the opposite side where there are oxygen, with as many water as there are oxygen, and on the opposite side complete with protons to balance the H

C₂O₄⁻²   →  2CO₂  +  2e⁻  

In oxalate, the carbon that acted with +3 gained an electron to oxidize to +4, but since there are 2 carbons, it gained 2 electrons. The oxygen is balanced by adding a 2 in stoichiometry

The halfs reaction have to be multipplied .2 (reduction) and .5 (oxidation) to balance the electrons in the main equation.

(8H⁺  +  MnO₄⁻  +  5e⁻  →  Mn²⁺  +  4H₂O) . 2

(C₂O₄⁻²   →  2CO₂  +  2e⁻ ) . 5

16H⁺  +  2MnO₄⁻  +  10e⁻  + 5C₂O₄⁻²   →  10CO₂  +  10e⁻  +  2Mn²⁺  +  8H₂O

16H⁺  +  2MnO₄⁻  + 5C₂O₄⁻²   →  10CO₂  +  2Mn²⁺  +  8H₂O

Answer:

1 mol of sucarose has 7.22 x10²⁴ atoms of C, 1.32 x10²⁵ atoms of H and 6.62 x10²⁴ atoms of O

Explanation:

C12H22O11 is the molecular formular for sucarose

So in 1 mol of sacarose, we have 12 mols of carbon, 22 moles of hydrogen, and 11 mols of oxygen.

As you know, 1 mol has 6.02x10²³ atoms so these are the rule of three to calculate the number of atoms of each type.

1 mol ______ 6.02x10²³

12 moles ______ 12 . 6.02x10²³ = 7.22 x10²⁴

22 moles ______ 22 . 6.02x10²³ = 1.32 x10²⁵

11 moles _______ 11 . 6.02x10²³ = 6.62 x10²⁴

Answer:

2. ΔG is zero, ΔH is positive, and ΔS is positive

Explanation:

When the ice is being converted to water ate 0ºC and 1 atm, there is an equilibrium between the solid and the liquid. At the equilibrium point, ΔG (the free energy) is zero. It is negative for spontaneous reactions and positive for nonspontaneous reactions.

For the phase change happens, the ice must absorb heat from the surroundings, so it's an endothermic reaction, and because of that ΔH (the enthalpy) must be positive. It is negative for exothermic reactions.

In the liquid state, the molecules have more energy and the randomness is higher than the solid-state. The entropy (S) is the measure of the randomness, so if it's increasing, ΔS must be positive.

Final answer:

The conversion of ice to water at 0°C and 1 atm has a negative ΔG, negative ΔH, and positive ΔS.

Explanation:

For the conversion of ice to water at 0°C and 1 atm:

ΔG is negative, ΔH is negative, and ΔS is positive.

ΔG is zero, ΔH is positive, and ΔS is positive.

ΔG is positive, ΔH is negative, and ΔS is positive.

Option 1 is the correct answer. When ice converts to water at 0°C and 1 atm, the Gibbs free energy change (ΔG) is negative, indicating that the process is spontaneous. The enthalpy change (ΔH) is negative, as heat is absorbed from the surroundings, and the entropy change (ΔS) is positive, as there is an increase in disorder.

SiO₂(s) is not an example of a molecular solid

A molecular solid is a of solid in which its molecules are held together by weak intermolecular forces such as van der Waals forces.

Molecular solids are soft, often volatile, have low density, have low melting temperatures, and are electrical insulators.

Example of molecular solid include CO₂(s) (dry ice), iodine (I₂(s)), C₂₅H₅₂(s) (paraffin wax), H₂O(s)

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Answer:

5 pumpkin pies could be made with 7.5 cups of sugar.

Explanation:

The conversion factor to solve the problem is:

[tex]7.5 cups of sugar\times\frac{1 pie}{1.5 cups of sugar} = 5 pies[/tex]

Answer:

a) ΔU = 71 kJ

b) the pressure will be higher

Explanation:

Step 1: Data given

A gas does 171 kJ of work on its surroundings

At the same time there is 242 kJ of heat added to the gas

Step 2: Calculate change of internal energy

Change of internal energy ΔU  by the gas due to the 171 kJ work done by the system and addition of 242 kJ heat to the system.

heat energy (ΔH) is the summation of heat capacity (ΔU) and work done by a system

ΔU = Q + W

The work is done by the system on its surroundings, what means energy is lost. W will have a negative value: -171 kJ

The heat is added to the system, this means we gain energy. Q will have a positive value: 242 kJ

Total change of intern energy will be:

ΔU = Q - W

ΔU = 242 kJ - 171 kJ

ΔU = 71 kJ

ΔU = nR*ΔT

For an ideal gas  n and R are constant. . Now consider PV = nRT.

P and T are directly related. Therefore, an increase in temperature will result in a higher pressure.

Final answer:

The change in internal energy of the gas during the process is 71 kJ, and the pressure of the gas will generally be lower at the end of the process.

Explanation:

The question addresses the concept of the first law of thermodynamics as it applies to heat transfer, work done by a gas, and changes in internal energy in thermodynamic processes. According to the first law of thermodynamics, the change in internal energy of a system (ΔU) is equal to the heat (Q) added to the system minus the work (W) done by the system on its surroundings. In this case:

ΔU = Q - W

The given data states that the gas sample does 171 kJ of work on its surroundings (W = 171 kJ) and 242 kJ of heat is added to the gas (Q = 242 kJ). Therefore:

ΔU = 242 kJ - 171 kJ = 71 kJ

(a) The change in internal energy of the gas during the process is 71 kJ.

(b) As the gas does work on the surroundings and expands, the pressure of the gas will generally be lower at the end of the process, as described by the ideal gas law and assuming other factors such as temperature remain constant.

Answer : The solubility of [tex]Al(OH)_3[/tex] is [tex]1.0\times 10^{-27}M[/tex]

Explanation :

First we have to calculate the pOH.

[tex]pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-12\\\\pOH=2[/tex]

Now we have to calculate the concentration of [tex]OH^-[/tex].

[tex]pOH=-\log [OH^-][/tex]

[tex]2=-\log [OH^-][/tex]

[tex][OH^-]=0.01[/tex]

The solubility equilibrium reaction will be:

[tex]Al(OH)_3\rightleftharpoons Al^{3+}+3OH^{-}[/tex]

The expression for solubility constant for this reaction will be,

[tex]K_{sp}=[Al^{3+}][OH^{-}]^3[/tex]

Now put all the given values in this expression, we get:

[tex]1.0\times 10^{-33}=[Al^{3+}]\times (0.01)^3[/tex]

[tex][Al^{3+}]=1.0\times 10^{-27}M[/tex]

As, the solubility of [tex]Al(OH)_3[/tex] = [tex][Al^{3+}][/tex] = [tex]1.0\times 10^{-27}M[/tex]

Thus, the solubility of [tex]Al(OH)_3[/tex] is [tex]1.0\times 10^{-27}M[/tex]

The question seeks to determine the amount of heat released during the decomposition of 57.0 g of Fe2O3. After converting the given weight to moles, we find that 57.0 g corresponds to about 147 kJ. Therefore, the answer is E. the release of 147 kJ of heat.

The question pertains to the heat change associated with the decomposition of Iron (III) oxide (Fe2O3). Given that Fe2O3 molecular weight is ~159.69 g/mol and the fact that every 2 mol of Fe2O3 releases 824.2 kJ of heat, we can calculate the energy associated with 57 grams of Fe2O3 as follows:

First, deduce the number of moles of Fe2O3 in 57 grams: 57g / 159.69 g/mol = 0.357 mol of Fe2O3

Since 2 mol of Fe2O3 correspond to 824.2 kJ, we can conclude that 0.357 mol corresponds to: (824.2 kJ / 2) x 0.357 = 147 kJ

Considering the question described the decomposition process as 'results in the release of', it suggests the heat change is exothermic or heat-releasing. Hence, the correct answer should be: E. the release of 147 kJ of heat.

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The correct electron configurations are:

A) O2−: [He]2s²2p⁶.

D) Si: [Ne]3s²3p².

E) Ni: [Ar]4s²3d⁸.

The correct electron configurations from the provided list are as follows:

A) The electron configuration of O2− (the oxygen ion with a charge of -2) is [He]2s²2p⁶. This configuration accounts for the addition of two electrons to the neutral oxygen atom's electron configuration of 1s²2s²2p⁴, resulting in a total of eight electrons.

D) The electron configuration of Si (silicon) is [Ne]3s²3p². Silicon, with an atomic number of 14, has 14 electrons. The electron configuration follows the Aufbau principle, filling the 1s, 2s, 2p, and 3s subshells.

E) The electron configuration of Ni (nickel) is [Ar]4s²3d⁸. Nickel, with an atomic number of 28, has 28 electrons. This configuration represents the filling of the 1s, 2s, 2p, 3s, 3p, 4s, and 3d subshells.

Option B and C are not correct:

B) The electron configuration of Cs (cesium) is [Xe]6s¹. Cesium, with an atomic number of 55, has only one valence electron in the 6s orbital.

C) The electron configuration of Ag+ (the silver ion with a charge of +1) is [Kr]5s²4d⁹. This configuration arises from the removal of one electron from the neutral silver atom's electron configuration, [Kr]5s²4d¹⁰, leaving it with 47 electrons.

Understanding electron configurations is essential in chemistry, as they determine the chemical properties and reactivity of elements. These configurations are based on the filling of electron orbitals following specific rules and principles, such as the Aufbau principle, Pauli exclusion principle, and Hund's rule.

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Answer:

[tex]CCl_{2}FCF_{3}[/tex]

Explanation:

To solve this equation we can use Graham's Law of effusion to calculate the molar mass of the chloroflourocarbon

[tex]\sqrt{\frac{M_{1} }{M_{2} } } = \frac{t_{1} }{t_{2} } \\\\\sqrt{\frac{M_{1} }{39.948 } } = \frac{157 }{76}\\\\\\M_{1}  = 170.477\\[/tex]

Which chloroflorocarbon has this molar mass?

C2Cl2F4

Using Graham's Law of Effusion and given effusion times, we calculated the molar mass of the chlorofluorocarbon to be approximately 170.55 g/mol. This formula suggests the gas is likely Cl₂F₂ (chlorodifluoromethane).

The formula of the chlorofluorocarbon, we can use Graham's Law of Effusion. This law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The formula for Graham's Law is: (Rate of effusion of Gas 1) / (Rate of effusion of Gas 2) = √(Molar mass of Gas 2 / Molar mass of Gas 1)

Given:

First, find the rate of effusion for each gas:

Using Graham's Law:

(Rate of effusion of chlorofluorocarbon) / (Rate of effusion of argon) = sqrt(Molar mass of argon / Molar mass of chlorofluorocarbon)

Substitute the values:

(50 mL / 157 s) / (50 mL / 76 s) = √(39.948 g/mol / Molar mass of chlorofluorocarbon)

Simplify:

76 / 157 = √(39.948 / Molar mass of chlorofluorocarbon)

(76 / 157[tex])^2[/tex] = 39.948 / Molar mass of chlorofluorocarbon

Molar mass of chlorofluorocarbon = 39.948 / (76 / 157[tex])^2[/tex] = 39.948 / (0.484[tex])^2[/tex] =170.55 g/mol

An example of a chlorofluorocarbon with a molar mass close to 170.55 g/mol is Cl₂F₂ (chlorodifluoromethane), with a molar mass of approximately 120.91 + 37.996 = 169.91 g/mol.

Answer: Mineral A changed because Molecular energy transferred is equal or greater than than its Activation energy

Mineral B didn't change because Molecular energy transferred is less than its Activation energy.

Explanation:

The molecules of Mineral A has been disturbed by the addition of energy causing a change and the entropy is increased. The Molecules of Mineral B has not been disturbed because the residual energy has not been overcome and therefore yielding no visible change

The plausible reason for the change in mineral A will be the lower activation energy than the transferred energy.

The addition of energy will result in the change in the minerals if the activation energy is exceeded.

The minerals found in mining A and B have transferred the energy. The minimum amount of energy required to do the transition in the atoms in the activation energy.

The possible reason for the change in the mineral A will be the lower activation energy for the mineral A. The mineral B does not change because the activation energy of mineral B is higher as compared to mineral A.

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Answer:

The strong acids are fully ionized in aqueous solution, and they contains higher concentration of hydrogen ions. Strong acids are lower pH in nature. Some examples of strong acids are:

1) Hydrochloric acid.

2) Nitric acid.

3) Sulfuric acid.

The weak acids are not fully ionized, means they are partially ionized in aqueous solution, and they contains lower concentration of hydrogen ions. Weak acids are higher pH in nature than strong acid. Some examples of weak acids are:

1) Ethanoic acid.

2) Acetic acid.

3) Nitrous acid.

Answer :  The partial pressure of [tex]H_2[/tex] is, 726.2 mmHg

Solution :

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

[tex]P_T=p_{H_2}+p_{H_2O}[/tex]

where,

[tex]P_T[/tex] = total partial pressure = barometric pressure = 746 mmHg

[tex]P_{H_2}[/tex] = partial pressure of hydrogen gas = ?

[tex]P_{H_2O}[/tex] = partial pressure of water vapor = 19.8 mmHg  (assume)

Now put all the given values is expression, we get the partial pressure of the hydrogen gas.

[tex]746mmHg=p_{H_2}+19.8mmHg[/tex]

[tex]p_{H_2}=726.2mmHg[/tex]

Therefore, the partial pressure of [tex]H_2[/tex] is, 726.2 mmHg

Answer: B (Items with densities lower than water will sink due to surface tension)

Explanation:

Surface tension is an intermolecular force exerted on the surface of water making it like a stretch elastic skin. Surface tension enables items with lesser densities than water, to float and slide on a water surface. Examples insects, leaves, paper etc

According to Archimedes principle an object denser than the fluid will sink. While objects less dense than the fluid will float.

Answer is (B) is false because items with densities higher then water will sink due to surface tension

Items with densities lower than water will sink due to surface tension. Therefore, the correct option is option B.

A intriguing phenomena called surface tension occurs when a liquid meets a gas or another liquid. It results from the cohesive forces between the molecules that are present at a liquid's surface. Surface tension is a result of the thin layer that is formed by these cohesive forces that holds the molecules closely together. A liquid droplet placed on a surface can be used to illustrate the idea of surface tension. The droplet takes on a spherical shape as a result of the cohesive forces between the water molecules. The molecules at the surface are drawn inside to reduce their exposure to the surroundings.

Therefore, the correct option is option B.

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Answer:

The temperature will change and become 2/3 of its original.

Explanation:

Using Ideal gas equation for same mole of gas as

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]

Given ,  

The volume of the sample gets reduced 1/3 of the original. So,

V₂ = 1/3V₁

The pressure of the sample is doubled of the original. So,

P₂ = 2P₁

Using above equation as:

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{2\times P_1}\times {\frac{1}{3}\times V_1}}{T_2}[/tex]

[tex]T_2=\frac{2}{3}\times T_1[/tex]

The temperature will change and become 2/3 of its original.

Answer: A. There are more reactants than products at equilibrium.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{eq}[/tex]

For a general chemical reaction:

[tex]aA+bB\rightarrow cC+dD[/tex]

The expression for [tex]K_{eq}[/tex] is written as:

[tex]K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

There are 3 conditions:

When [tex]K_{eq}>1[/tex]; the reaction is product favored.

When [tex]K_{eq}<1[/tex]; the reaction is reactant favored.

When [tex]K_{eq}=1[/tex]; the reaction is in equilibrium.

From the above expression, the equilirbium constant is directly dependent on product concentration. Thus, more is the concentration of product, more will be the equilibrium constant.

The values of [tex]K_{eq}[/tex] less than 1 , means the reactant are more than the products.

The mole fraction of nitrogen in the gas mixture can be calculated using the ideal gas law and the given pressure, volume, and temperature. The calculation suggests a mole fraction of 1.509, which might indicate that we've made an erroneous assumption about the mixture's composition.

First, in order to find the mole fraction of nitrogen in the mixture, we should first find the number of moles of the total mixture. To do this, we can use the ideal gas law, which states that the volume, pressure, and temperature of a gas can be related to the number of moles of the gas. This law can be written as: PV=nRT.

To find the pressure in the correct units, we convert mmHg to atm by dividing by 760 (since 1 atm = 760 mmHg). The pressure is therefore 870.2 / 760 = 1.145 atm. Similarly, we should convert the temperature to Kelvin by adding 273 to the Celsius temperature: 31.2 + 273 = 304.2 K.

So, our equation becomes: 1.145 atm * 15.1 L = n * 0.0821 atm*L/mol/K * 304.2 K. Solving for n, we find that the total number of moles in the mixture is n = 0.570 moles.

Given that molecular nitrogen (N₂) has a molar mass of 28.01 g/mol, we can calculate the number of moles of nitrogen in the 24.1-g mixture. The mass of nitrogen divided by its molar mass gives the number of moles, which is 24.1 g / 28.01 g/mol = 0.860 moles. Therefore, the mole fraction of nitrogen (X_N2) in the mixture can be calculated as the number of moles of nitrogen divided by the total number of moles, or 0.860 / 0.570 = 1.509.

Therefore, the mole fraction of nitrogen in the mixture is 1.509. This value may seem unusual as mole fractions are typically less than 1, but this discrepancy could be due to the initial assumption that the entire 24.1 g consisted of nitrogen. In a real situation, you would need to know the mass of each component in the mixture to accurately calculate the mole fractions.

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The mole fraction of nitrogen in the gas mixture is calculated to be approximately 43.8 percent using the Ideal Gas Law and the given molar masses of nitrogen and carbon dioxide.

To find the mole fraction of nitrogen in this mixture, we first need to use the Ideal Gas Law (PV=nRT) to find the total moles (n) of the gases in the mixture. Given the volume (V) of the gas mixture is 15.1 L, the pressure (P) is 870.2 mmHg or 1.145 atm, and the temperature (T) is 31.2oC or 304.35 K, with R being the ideal gas constant which equals 0.0821 L·atm/K·mol, the total number of moles is found to be 0.573 mol.

Then, we need to know that the molar masses of nitrogen (N₂) and carbon dioxide (CO₂) are 28.01 g/mol and 44.01 g/mol respectively. So, we can set up the equation x(28.01) + (1-x)(44.01) = 24.1 (where x is the mole fraction of nitrogen in the mixture) to solve for x. We find that x equals 0.438.

So, the mole fraction of nitrogen in this mixture is approximately 0.438 or 43.8 percent.

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Answer:

-41,9kJ/mol NaOH

Explanation:

For the solution process:

NaOH(s) →Na⁺(aq) + OH⁻(aq)

The released heat is:

Q = -C×m×ΔT

Where Q is the released heat, C is specific heat of the solution (4,18J/g°C), m is the mass of water (100,0g) and ΔT is (32,0°C-23,9°C)

Replacing:

Q = -3385,8J

This heat is released per 3,23g of NaOH. Now, the heat released (ΔH) per mole of NaOH is:

[tex]\frac{-3385,8J}{3,23gNaOH} *\frac{40g}{1mol}[/tex]= -41929J/molNaOH ≡

-41,9kJ/mol NaOH

I hope it helps!

Answer:

158.0 pm

Explanation:

In this case, the bond length of the HI molecule is equal to the sum of the atomic radii of its components. Which is to say:

Bond length HI = Atomic Radius H + Atomic Radius I

Bond Length = 25.0 pm + 133 pm

Bond Length = 158.0 pm

Final answer:

The bond length of the HI molecule can be predicted by summing the atomic radii of hydrogen (25.0 pm) and iodine (133 pm), resulting in an estimated bond length of 158 pm.

Explanation:

The question seeks to predict the bond length of the HI molecule based on the atomic radii of hydrogen (H) and iodine (I). Bond length is crucial for understanding molecular structure and is reflective of the optimal distance between two bonded atoms where the potential energy of the system is at its lowest. Given that the atomic radius of H is 25.0 pm and that of I is 133 pm, the bond length of the HI molecule can be estimated by summing these radii.

To predict the bond length of HI, we simply add the atomic radii of H and I:

Atomic radius of H = 25.0 pm

Atomic radius of I = 133 pm

Estimated bond length of HI = (25.0 pm + 133 pm) = 158 pm

This approach assumes that the bond length is approximately the sum of the individual atomic radii, which is often a reasonable approximation for such predictions.

Answer:

1.20 V

Explanation:

[tex]Pb(s) + Br_2(l)\rightarrow Pb^{2+}(aq) + 2Br^-(aq)[/tex]

Here Pb undergoes oxidation by loss of electrons, thus act as anode. Bromine undergoes reduction by gain of electrons and thus act as cathode.

[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]

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

Given,

[tex]Pb^{2+}(aq) + 2 e^-\rightarrow Pb(s)[/tex]

[tex]E^0_{[Pb^{2+}/Pb]}= -0.13\ V[/tex]

[tex]Br_2(l) + 2 e^-\rightarrow 2 Br(aq)[/tex]

[tex]E^0_{[Br_2/Br^{-}]}=+1.07\ V[/tex]

[tex]E^0=E^0_{[Br_2/Br^{-}]}- E^0_{[Pb^{2+}/Pb]}[/tex]

[tex]E^0=+1.07- (-0.13)\ V=1.20\ V[/tex]

The cell potential for the given reaction is calculated by subtracting the anode's potential from the cathode's potential since the reaction at the anode is an oxidation and the cathode involves a reduction. The standard cell potential for this reaction is 1.20 V, and this positive value indicates the reaction is spontaneous under standard conditions.

The standard cell potential is calculated by adding the standard reduction potential of the cathode to the standard reduction potential of the anode. According to the reaction, Pb(s) is being oxidized to Pb2+(aq), thus acting as the anode: E° = -0.13 V. Meanwhile, Br2(l) is being reduced to 2Br-(aq), acting as the cathode: E° = +1.07 V. When calculating the overall cell potential, we subtract the anode's potential from the cathode's potential, because the reaction at the anode is oxidation (loss of electrons), while the cathode involves a reduction (gain of electrons).

Therefore, the standard cell potential, E°cell, is calculated as E°cathode - E°anode = 1.07 V - (-0.13 V) = 1.20 V. This value is positive, indicating that the reaction is spontaneous under standard conditions as per the cells' relative oxidizing strength.

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Answer:

69%

Explanation:

Say the hydrocarbon is C2H4, the equation of reaction would be;

C2H4 + H2 ----------> C2H6

The hydrocarbon, C2H4 is the limiting reagent.

From the question, the hydrocarbon C2H4 flows into a catalytic reactor at 26.2 atm and 250°C with a flow rate of 1100 L/min

Using, PV=nRT--------------------(1).

n= PV/RT to find the number of moles.

n= 26.2 atm × 1100 L/ 0.0821 L. atm/mol per kelvin × 523 K.

=28,820 atm.L/ 42.9383 L.atm/mol.

= 671.2 mole.

One mole of C2H4 produced one mole of C2H6.

Mass of C2H6 = 30 × 671.2

= 20,136 g = 20.136 kg.

Percent yield = actual yield/ theoretical yield × 100% -------(2)

Actual yield= 13.9 kg, theoretical yield = 20.136 kg

Substitute the parameters into equation (2). We have;

Percent yield = 13.9 kg/ 20.136× 100

Percent yield= 0.69 × 100

Percent yield= 69%

Note: P= pressure, V= volume, T= temperature and n = number of moles

Answer:

ii) They emit various tyoes of radiation

Explanation:

They fact that they emit radiation makes them ideal to use as a tracer because that radiation can be detected and followed. This means that you can know where the isotope is going throught.

Cause of that, radioisotopes are very helpful in medical applications such as tracing blood veins.

Final answer:

A positive ion, or cation, has more protons than electrons, while a negative ion, or anion, has more electrons than protons, resulting in their net electrical charges.

Explanation:

The difference between a positive and negative ion lies in the balance between protons and electrons within an atom or molecule. When an atom has more protons than electrons, it becomes a positive ion or, as it's scientifically named, a cation. Conversely, an atom becomes a negative ion or anion if it gains extra electrons, surpassing the number of protons. This difference in the number of electrons leads to a net electrical charge, which determines whether the ion is positive or negative.

DID YOU KNOW? The names for these ions are pronounced "CAT-eye-ons" (cations) and "ANN-eye-ons" (anions), reflecting their respective charges.