The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when [ Zn2+] = 3.5 M and [Pb2+] = 2.0⋅10−4 M.
Pb2+ (aq) + Zn (s) → Zn2+ (aq) + Pb (s)
A) 0.84
B) 0.76
C) 0.50
D) 0.63
E) 0.39

Answer:

11.0L of carbon dioxide is produced

Explanation:

Balanced equation: [tex]CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)[/tex]

According to balanced equation, 1 mol of [tex]CaCO_{3}[/tex] produces 1 mol of [tex]CO_{2}[/tex]

So, 0.489 mol of [tex]CaCO_{3}[/tex] produces 0.489 mol of [tex]CO_{2}[/tex]

Let's assume [tex]CO_{2}[/tex] behaves ideally.

So, [tex]P_{CO_{2}}V_{CO_{2}}=n_{CO_{2}}RT[/tex]

where P is pressure, V is volume , n is number of moles, R is gas constant and T is temperature in kelvin

Plug-in all the values in the above equation-

[tex](1atm)\times V_{CO_{2}}=(0.489mol)\times (0.0821L.atm.mol^{-1}.K^{-1})\times (273K)[/tex]

or, [tex]V_{CO_{2}}=11.0L[/tex]

So, 11.0L of carbon dioxide is produced

Answer:

Total pressure = 732.9 torr

Partial pressure of helium =275.7 torr

Partial pressure of  argon = 457.2 torr

Explanation:

Given data:

Volume of flask one = 275 mL

Pressure of helium = 752 torr

Volume of second flask = 475 mL

Pressure of argon = 722 torr

What is partial pressure = ?

What is total pressure = ?

Solution:

V₂ = 275 mL + 475 mL = 750 mL

For helium:

P₁V₁ = P₂V₂

P₂ = P₁V₁ /V₂

P₂ =  752 torr. 275 mL /  750 mL

P₂ = 206800 torr. mL  /750 mL

P₂ = 275.7 torr

For argon:

P₁V₁ = P₂V₂

P₂ = P₁V₁ /V₂

P₂ =  722 torr. 475 mL /  750 mL

P₂ = 342950 torr. mL  /750 mL

P₂ = 457.2  torr

Total pressure = P(helium) + P ( argon)

Total pressure = 275.7 torr +  457.2  torr

Total pressure = 732.9 torr

We have a 275-mL flask with helium at 752 torr and a 475-mL flask with argon at 722 torr. If both flasks are connected, the partial pressure of helium will be 276 torr, the partial pressure of argon will be 457 torr and the total pressure will be 733 torr.

We have 2 flasks:

If the two flasks are connected through a stopcock and the stopcock is opened, the final volume will be the sum of the initial volumes.

[tex]V_2 = 275mL + 475 mL= 750 mL[/tex]

Assuming ideal behavior and constant temperature, we can calculate the final partial pressure of each gas using Boyle's law.

[tex]P_1 \times V_1 = P_2 \times V_2[/tex]

[tex]P_2 = \frac{P_1 \times V_1}{V_2} = \frac{752 torr \times 275mL}{750 mL} = 276torr[/tex]

[tex]P_2 = \frac{P_1 \times V_1}{V_2} = \frac{722 torr \times 475mL}{750 mL} = 457torr[/tex]

The total pressure will be the sum of the partial pressures.

[tex]P = pHe + pAr = 276 torr + 457torr = 733torr[/tex]

We have a 275-mL flask with helium at 752 torr and a 475-mL flask with argon at 722 torr. If both flasks are connected, the partial pressure of helium will be 276 torr, the partial pressure of argon will be 457 torr and the total pressure will be 733 torr.

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Answer:

Option B - reducing; NADH; negative

explanation:

Nicotinamide adenine dinucleotide, or NADH, is a compound, that's more actively used in the transportation of electron chain. It's evident that produced and make use NADH than FADH2 in the process that brings about energy creation.

NADH carries electrons from one reaction to another. Its Cofactor is in two forms in cells: NAD+ accepts electron from other molecule and hence reduced as an oxidizing agent while NADH is formed by this reaction, which can be used to donate electrons because it's a reducing agent. The main function of NAD is to transfer these reactions.

Answer:     E = (13.6 eV) [1/nf² - 1/ni²]  

En = (-13.6 eV)/n²

where n=1,2,3...

Explanation:

According to Bohr's theory each spcified energy value( E1,E2,E3...) is called energy level of the atom and the only allowable values are given by the equation

En = (-13.6 eV)/n²

The energy change (ΔE) that accompaies the leap of an electron from one energy level to another is given by equation

E = (13.6 eV) [1/nf² - 1/ni²]  

The equation for the energy levels of the hydrogen atom is given by the Bohr formula: En = -13.6 eV / n^2. The energy levels are inversely proportional to the square of the principal quantum number.

The question concerns the equation for the energy levels of the hydrogen atom. According to Bohr's model, the energy levels of a hydrogen atom, which consists of a single electron orbiting a single proton, are given by the formula:

En = -13.6 eV / n2

Here, En represents the energy of an electron at a particular level n, which is known as the principal quantum number. This number can be any positive integer (n = 1, 2, 3, ...), and the energy level becomes less negative as n increases. The ground state energy of the hydrogen atom, when n = 1, is

-2.18 x 10-18 joules

Transitions between these energy levels result in the emission or absorption of light at specific wavelengths, giving rise to the hydrogen spectral series. Notably, the Lyman series corresponds to transitions ending at n=1, and the Balmer series ends at n=2.

The difference between stating "The electron is located at a particular point in space" and "There is a high probability that the electron is located at a particular point in space" is in the first statement, we cannot define the exact location of the electron.

The elementary electric charge of the electron is a negative one, making it a subatomic particle. Due to their lack of known components or substructure, electrons, which are members of the first generation of the lepton particle family, are typically regarded to be elementary particles.

Quarks make up protons and neutrons, but not electrons. As far as we can know, quarks and electrons are basic particles that are not composed of lesser particles.

Thus, option D is correct.

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Final Answer:

The enthalpy of vaporization [tex](\( \Delta H_{\text{vap}} \))[/tex]of sulfur dioxide is approximately[tex]\( 28.5 \, \text{kJ/mol} \)[/tex].

Explanation:

To find the enthalpy of vaporization [tex](\( \Delta H_{\text{vap}} \))[/tex], we use the Clausius-Clapeyron equation:

[tex]\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \][/tex]

Given vapor pressures:

[tex]\( P_1 = 462.7 \, \text{mm Hg} \) at \( T_1 = -21.0 \, \text{°C} \),[/tex]

[tex]\( P_2 = 140.5 \, \text{mm Hg} \) at \( T_2 = -44.0 \, \text{°C} \).[/tex]

Convert temperatures to Kelvin:

[tex]\[ T_1 = -21.0 \, \text{°C} + 273.15 = 252.15 \, \text{K} \][/tex]

[tex]\[ T_2 = -44.0 \, \text{°C} + 273.15 = 229.15 \, \text{K} \][/tex]

Now, substitute values into the equation:

[tex]\[ \ln\left(\frac{140.5}{462.7}\right) = -\frac{\Delta H_{\text{vap}}}{8.314}\left(\frac{1}{229.15} - \frac{1}{252.15}\right) \][/tex]

Solving for [tex]\( \Delta H_{\text{vap}} \):[/tex]

[tex]\[ \Delta H_{\text{vap}} = -8.314 \times \ln\left(\frac{140.5}{462.7}\right) \times \frac{1}{229.15 - 252.15} \][/tex]

After calculation, [tex]\( \Delta H_{\text{vap}} \)[/tex]is approximately[tex]\( 28.5 \, \text{kJ/mol} \).[/tex]

In conclusion, the enthalpy of vaporization for sulfur dioxide is determined through the Clausius-Clapeyron equation, with precise calculations yielding an enthalpy value of [tex]\( 28.5 \, \text{kJ/mol} \)[/tex].

Answer: 910.44K

Explanation:

From the question, the parameters given are; ∆H(vaporization)= 48.17 kJ/mol and ΔS(vaporization)= 52.91 J mol-1•K.

Using the formula;

dG = dH - TdS -------------------------(1).

Here, dG=0, dG= change in free energy.

dH(vaporization) = 48,170 J/mol

dS(vaporization) = 52.91 J/mol.K

Boiling point of the substance, T=???

Therefore, we solve for T(temperature, that is the boiling point-boiling temperature) of the compound in Kelvin.

Slotting in the parameters given into equation (1). We have;

0= 48,170 J/mol - T× 52.91 J/mol.K

=> 48,170= 52.91T

T= 910.44 K

The boiling point of a compound can be calculated using the formula T = ΔHvap / ΔSvap. For this compound, with ΔHvap of 48.17 kJ mol-1 and ΔSvap of 52.91 J mol-1•K-1, the estimated boiling point is around 908 K.

Boiling point is the temperature at which a substance changes from a liquid to a gas. To calculate it, you can use the formula ΔGvap = ΔHvap - TΔSvap where ΔGvap is Gibbs free energy of vaporization, ΔHvap is enthalpy of vaporization, T is temperature in Kelvin, and ΔSvap is entropy of vaporization.

To find the boiling point of the compound, rearrange the formula as T = ΔHvap / ΔSvap. Substituting the given values, T = 48.17 kJ mol-1 / (52.91 J mol-1*K-1 / 1000) = approximately 908 K.

Answer:

2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → 2MnSO₄ + 10CO₂ + 8H₂O + K₂SO₄

0,02658g of La in the unknown

Explanation:

The reaction of  La₂(C₂O₄)₃ with acid is:

La₂(C₂O₄)₃ + 6H⁺ → 3H₂C₂O₄ + 2La³⁺

The titration of H₂C₂O₄ with KMnO₄ is:

2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → 2MnSO₄ + 10CO₂ + 8H₂O + K₂SO₄

The moles of KMnO₄ that react are:

0,006363M KMnO₄×0,01804L = 1,148x10⁻⁴moles of KMnO₄

By the titration reaction, 2 moles of KMnO₄ react with 5 moles of H₂C₂O₄, that means:

1,148x10⁻⁴moles of KMnO₄×[tex]\frac{5molesH_{2}C_{2}O_{4}}{2molesKMnO_{4}}[/tex] = 2,870x10⁻⁴ moles of H₂C₂O₄.

In the reaction of La₂(C₂O₄)₃ with acid, 3 moles of H₂C₂O₄ were produced while 2 moles of La³⁺ were produced, that means:

2,870x10⁻⁴ moles H₂C₂O₄× [tex]\frac{2molesLa^{3+}}{3molesH_{2}C_{2}O_{4}}[/tex] = 1,913x10⁻⁴ moles of La³⁺, in grams -Using molar mass of lanthanum-:

1,913x10⁻⁴ moles of La³⁺×[tex]\frac{138,9g}{1mol}[/tex] = 0,02658g of La

There are 0,02658g of La in the unknown

I hope it helps!

Answer:

At 31.88 atm pressure of ethane will have density of 37.2 g/L at 40.0 °C.

Explanation:

To calculate the pressure of gas, we use the equation given by ideal gas equation:

[tex]PV=nRT[/tex]

Number of moles (n)

can be written as: [tex]n=\frac{m}{M}[/tex]

where, m = given mass

M = molar mass

[tex]PV=\frac{m}{M}RT\\\\PM=\frac{m}{V}RT[/tex]

where,

[tex]\frac{m}{V}=d[/tex] which is known as density of the gas

The relation becomes:

[tex]PM=dRT[/tex]

Where :

P = pressure of the gas

R = universal gas constant

T = temperature of the gas

We are given with:

Density of the gas = d = 37.2 g/L

Molar mass of the ethane gas = M = 30 g/mol

Temperature of the ethane gas = T = 40.0°C= 313.15 K

Pressure of the ethane gas = P

[tex]P=\frac{dRT}{M}=\frac{37.2 g/L\times 0.0821 atm L/mol K\times 313.15 K}{30 g/mol}[/tex]

P = 31.88 atm

At 31.88 atm pressure of ethane will have density of 37.2 g/L at 40.0 °C.

The pressure of an ethane gas with a density of 37.2 g/L at 40.0 °C is equal to 31.87 atm.

Given the following data:

Molar mass ([tex]C_2H_6[/tex]) = [tex](12 \times 2 + 1 \times 6) = (24+6) =30 \;g/mol[/tex]

Ideal gas constant, R = 0.0821L⋅atm/mol⋅K

Conversion:

Temperature = 40.0 °C to K = [tex]273 +40=[/tex] 313K

To find the pressure of ethane gas, we would use the ideal gas law equation;

[tex]PV = nRT[/tex]

Where;

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

In terms of density, the ideal gas law equation becomes:

[tex]Density = \frac{M_MP}{RT}[/tex]

Where;

Making P the subject of formula, we have;

[tex]P = \frac{DRT}{M_M}[/tex]

Substituting the given parameters into the formula, we have;

[tex]P = \frac{37.2 \times 0.0821 \times 313}{30}\\\\P = \frac{955.94}{30}[/tex]

Pressure, P = 31.87 atm.

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Answer:

The type of chemical mutagen to choose depends on the intended effect. In this case, the best ones are acridines and nitrous acid.

Explanation:

Brenner et al. proposed that acridines induce mutations by causing deletions or additions of single base pairs during replication. Acridines bind to DNA by intercalation between adjacent base pairs. Acridines inactivate extracellular phage  by photodynamic action but the necessary conditions for this killing

are avoided in the procedure for acridine-induced mutation of reproducing phage. The lack of reported acridine-induced mutation in organisms other than phage raises some questions as to the generality of its

mutagenesis, thus making it a good type of compounds to induce specific mutations.

In the other hand,  nitrous acid deaminates the amino bases adenine, cytosine  (and hydroxymethylcytosine) , and guanine in nucleic acids.

Analysis of the effect of differences of pH during nitrous acid treatment  

of phage DNA showed that the rate of killing was affected similarly to

the rate of guanine deamination, and that the rates of induced r mutation was affected similarly to the rates of adenine and hydroxymethylcytosine deamination. Ascribing the induced mutations to deamination of adenine and cytosine is reasonable in terms of the hydrogen  bonding of their products and the Watson-Crick base pairing schemes. Since this inorganic acid is molecule-specific, it would also be used to induce certain mutations in bacteria without causing transition mutations.

Explanation:

It is given that mass is 11 kg. Convert mass into grams as follows.

                [tex]11 kg \times \frac{1000 g}{1 kg}[/tex]

               = 11000 g               (as 1 kg = 1000 g)

Now, calculate the number of moles as follows.

    No. of moles = [tex]\frac{\text{Mass of Cr}}{\text{Molar mass of Cr}}[/tex]

                          = [tex]\frac{11000 g}{52 g/mol}[/tex]

                          = 211.54 mol

For [tex]Cr^{3+}[/tex], 3 moles of electrons are required

Hence,       [tex]3 \times 211.54 mol[/tex]

                  = 634.62 mol

As 1 mol of electrons contain 96500 C of charge. Therefore, charge carried by 634.62 mol of electrons will be calculated as follows.

             Q = [tex]634.62 mol \times 96500 C/mol [/tex]

                  = [tex]61.24 \times 10^{6} C[/tex] of charge

We know that relation between charge, current and time is as follows.

                    Q = [tex]I \times t[/tex]

Current is given as 41.5 A and charge is calculated as [tex]61.24 \times 10^{6} C[/tex]. Therefore, calculate the time as follows.

                Q = [tex]I \times t[/tex]

      [tex]61.24 \times 10^{6} C = 41.5 A \times t[/tex]        

                   t = [tex]1.47 \times 10^{6} sec[/tex]

As there are 3600 seconds in one 1 hour. Therefore, converting [tex]1.47 \times 10^{6} sec[/tex] into hours as follows.

            [tex]\frac{1.47 \times 10^{6} sec}{3600 sec/hr}[/tex]

               = 4.09 hr

Thus, we can conclude that it takes 4.09 hours to plate 11.0 kg of chromium onto the cathode if the current passed through the cell is held constant at 41.5 A.

The process of electrorefining chromium involves oxidation of Chromium III ions at anode and reduction at the cathode. The total charge required for this process can be calculated using Faraday's law, and the time required to pass this charge can be calculated using the formula Q/I. Therefore, it will take approximately 408 hours to plate 11kg of chromium onto the cathode with 41.5 A current.

The process of electrorefining chromium involves two half-reactions at anode and cathode in which chromium is oxidized at the anode and is then reduced at the cathode. Specifically, Chromium III ion (Cr³+) is oxidized at the anode with the reaction Cr³+ (aq) + 3e-, and is then reduced at the cathode. The stoichiometry of this process requires three moles of electrons for each mole of chromium(0) produced. Given this stoichiometry and the current passed through the cell, we can calculate the time required to plate 11kg of chromium.

First, since the molar mass of chromium is approximately 52 g/mol, 11kg is equal to about 211.5 mol. This number of moles of chromium requires [tex]211.5 * 3 = 634.5[/tex] mol of electrons. One mol of electrons carries a charge of approximately 96485 Coulombs (C), so the total charge is approximately [tex]6.115 * 10^7 C.[/tex]

If this charge is being passed at 41.5 C/s, the time required to pass the total amount of charge can be calculated by the formula Q/I, where Q represents the total charge and I represents the current. Therefore, the time required to plate 11kg of chromium onto the cathode with 41.5 A current is approximately [tex]1.47 * 10^6[/tex] seconds which is approximately 408 hours.

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Answer:

Explained

Explanation:

the various ways of input of CO_2 into the atmosphere.

1.Dissolved  CO 2 in the ocean is released back into the atmosphere by heating ocean surface water

2.Plant and animal respiration, an exothermic reaction involving the breakdown into CO 2 and water of organic molecules.

3. Degradation of fungi and bacteria which are responsible for breaking down carbon compounds in dead animals and plants(fossils) and convert carbon into CO2 when oxygen or methane is present.

4.Combustion of organic matter (that includes deforestation and combustion of fossil fuels) oxidizing to produce CO2;

5. Cement production when calcium carbonate (limestone) is heated to produce calcium oxide(lime), cement component, and CO2 are released;

Answer:

The answer is acceleration.

Explanation:

Acceleration- When it comes to science, Force is defined as a "push or pull" of an object with mass that causes it to change velocity.

Balanced Force- This occurs when two forces of the same mass or size are acting on a particular object in opposite direction. The object being acted upon could either stay in its place or move in a uniform speed and direction.

Unbalanced Force- This occurs when the forces applied are not equal and are of different direction. For example, a 55-kg person pulling a 30-kg cart. The cart will move to the direction of the pull.

Thus, when forces acting on an object are unbalanced, it causes the object to accelerate. This acceleration is directly proportional to the "net force", but is inversely proportional to the mass or size.

A net force is the total amount of force that is acting on the object.

Answer:

Mass of AgCl = 2.87 g

[Ca²⁺] = 0.075 M

[Cl⁻] = 0.05 M

[NO₃⁻] = 0.10 M

Explanation:

The reaction between silver nitrate (AgNO₃) and calcium chloride (CaCl₂) is:

2AgNO₃(aq) + CaCl₂(aq) → 2AgCl(s) + Ca(NO₃)₂(aq)

The number of moles of the reactants are:

AgNO₃ = Volume (L)*concentration (M) = 0.1*0.2 = 0.02 mol

CaCl₂ = 0.1*0.15 = 0.015 mol

First, let's find which reactant is limiting. Testing for AgNO₃, the stoichiometry is:

2 moles of AgNO₃ ------------------------- 1 mol of CaCl₂

0.02 mol ------------------------- x

By a simple direct three rule:

2x = 0.02

x = 0.01 mol of CaCl₂, which is higher than was used, so CaCl₂ is in excess and AgNO₃ is the limiting reactant.

The stoichiometry between AgNO₃ and AgCl is:

2 moles of AgNO₃ ---------------- 2 moles of AgCl

0.02 mol ---------------- x

By a simple direct three rule

x = 0.02 mol of AgCl

The molar mass of AgCl is 143.32 g/mol, so the mass formed is:

m = 143.32*0.02 = 2.87 g

All the silver nitrate had reacted, and by the stoichiometry, only 0.01 mol of CaCl₂ reacted, so the number of moles remaining is:

nCaCl₂ = 0.015 - 0.01 = 0.005 mol

And the number of moles of Ca(NO₃)₂ formed is the same as CaCl₂ that reacted: 0.01 mol.

Both substances are in aqueous form, so, they produce ions:

CaCl₂ → Ca²⁺ + 2Cl⁻

By the stoichiometry: nCa²⁺ = 0.005 mol; nCl⁻ = 0.01 mol

Ca(NO₃)₂ → Ca²⁺ + 2NO₃⁻

By the stoichiometry: nCa²⁺ = 0.01 mol; nNO₃⁻ = 0.02 mol

The total volume is 0.2 L, so the ions concentrantions are:

[Ca²⁺] = (0.005 + 0.01)/0.2 = 0.075 M

[Cl⁻] = 0.01/0.2 = 0.05 M

[NO₃⁻] = 0.02/0.2 = 0.10 M

The concentrations of each ion remaining in solution after precipitation is complete is Mass of AgCl = 2.87 g, [Ca²⁺] = 0.075 M, [Cl⁻] = 0.05 m, [NO₃⁻] = 0.10 M.

Silver chloride is an ionic compound in which the silver is cation and the chloride is anion.

The reaction is

[tex]2AgNO_3(aq) + CaCl_2(aq) = 2AgCl(s) + Ca(NO_3)_2(aq)[/tex]

The number of moles of the reactants are:

[tex]AgNO_3 = Volume (L) \times concentration (M) \\\\= 0.1 \times 0.2 = 0.02 mol[/tex]

[tex]CaCl_2 = 0.1 \times 0.15 = 0.015 mol[/tex]

Now, finding the limiting reactant.

Testing for AgNO₃,

The stoichiometry is:

2 moles of AgNO₃  = 1 mol of CaCl₂

0.02 mol = x

By direct three rule:

2x = 0.02

[tex]AgNO_3[/tex] is the limiting reactant.

The stoichiometry between [tex]AgNO_3[/tex] and AgCl is:

2 moles of [tex]AgNO_3[/tex] =  2 moles of AgCl

0.02 mol = x

By a simple direct three rule

x = 0.02 mol of AgCl

The mass formed is

If the molar mass of AgCl is 143.32 g/mol

[tex]m = 143.32\times0.02 = 2.87 g[/tex]

Remaining number of moles are

nCaCl₂ = 0.015 - 0.01 = 0.005 mol

The substances will produce ions, because they are aqueous form.

[tex]CaCl_2 = Ca^2^++ 2Cl^-[/tex]

By the stoichiometry

nCa²⁺ = 0.005 mol

nCl⁻ = 0.01 mol

[tex]Ca(NO_3)_2 = Ca^2^+ + 2NO_3^-[/tex]⁻

By the rule of stoichiometry

nCa²⁺ = 0.01 mol

nNO₃⁻ = 0.02 mol

The volume is 0.2 L,

The concentration of ions are

[Ca²⁺] =[tex]\dfrac{(0.005 + 0.01)}{0.2} = 0.075 M[/tex]

[Cl⁻] = [tex]\dfrac{0.01}{0.2} = 0.05 M[/tex]

[NO₃⁻] = [tex]\dfrac{0.02}{0.2} = 0.10\; M[/tex]

Thus, the concentrations of each ion remaining in solution after precipitation is complete is Mass of AgCl = 2.87 g, [Ca²⁺] = 0.075 M, [Cl⁻] = 0.05 m, [NO₃⁻] = 0.10 M.

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Answer:

The specific heat capacity of a metal is 1.31 J/g°C = C

Explanation:

A classical excersise of calorimetry to apply this formula:

Q = m . C . ΔT

177.5 J = 15 g . C (34°C - 25°C)

177.5 J = 15g . 9°C . C

177.5 J /15g . 9°C = C

1.31 J/g°C = C

Final answer:

To find the specific heat capacity of the metal, the heat energy absorbed is divided by the product of mass and temperature change, which yields a specific heat capacity of 1.315 J/g\u00b0C.

Explanation:

To calculate the specific heat capacity of a metal with the given data, we can use the formula:

q = m \\cdot c \\cdot \\Delta T,

where q is the heat energy absorbed (in joules), m is the mass of the substance (in grams), c is the specific heat capacity (in J/g\u00b0C), and \\Delta T is the change in temperature (in degrees Celsius).

Rearranging the formula to solve for c gives:

c = q / (m \\cdot \\Delta T)

Substituting the given values:

q = 177.5 J,

m = 15.0 g,

\\Delta T = 34\u00b0C - 25\u00b0C = 9\u00b0C.

Therefore:

c = 177.5 J / (15.0 g \\cdot 9\u00b0C)

c = 177.5 J / 135 g\u00b0C

c = 1.315 J/g\u00b0C

The specific heat capacity of the metal is 1.315 J/g\u00b0C.

Answer:

ΔS > 0 only for choice E: CH4(g) + H2O (g) → CO(g) + 3 H2(g)

Explanation:

Our strategy in this question is to use the trend in entropies :

S (solids)  less than S (liquids) less than S (gases)

Also we have to look for the  molar quanties involved of each state and their change to answer the question:

A. N2(g) + 3 H2(g) → 2 NH3(g)

Here we have 4 moles gases going to 2 moles of products, so the change in entropy is negative.

B. Na2CO3(s) + H2O(g) + CO2(g) → 2 NaHCO3(s)

The change in entropy is negative since we have 2 mol gases in the reactants and zero in the products.

C. CH3OH(l) → CH3OH(s)

A liquid has a higher entropy than a solid so ΔS is negative

D. False see A,B,C

E. The change in moles of gases is 4 - 2= 2, therefore  ΔS is greater than O.

The reaction CH4(g) + H2O (g) → CO(g) + 3 H2(g) will have ΔS > 0.

The term entropy refers to the degree of disorder of a system. Hence, the change in entropy is positive (greater than zero) when there is an increase in the degree of disorderliness of the system.

As such, the reaction;

will experience an increase in entropy since there is an increase in the number of molecules of  gaseous species from left to right.

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Answer:

Increase; remain constant, remain constant; No; by adjusting the acid:base ratio of the buffer.

Explanation:

On a 10-fold dilution of a weak acid, the pH will INCREASE and that is because strong acids are not found in buffers, therefore, diluting a strong acid reduces the amount of hydrogen ions,H^+ present, consequently increasing the pH.

On a 10-fold dilution of a buffered solution, the pH will REMAIN CONSTANT because diluting a buffer does not affect the pH of the buffer.

If one adds a small amount of strong base to a buffered solution, the pH will REMAIN CONSTANT because the buffer will equalize the strong base, however, the pH may increase just a little.

One can not make a buffer solution using a strong acid because  Buffers are composed of a weak acid and its conjugate base or a weak base and its conjugate acid, and weak acids or bases only dissociate partially. Strong acids dissociate completely, they overpower the reaction and move the reaction to completion.

pH of a buffer solution can be adjusted using by adjusting the acid:base ratio of the buffer.

Final answer:

Diluting a weak acid increases its pH, while a buffered solution maintains a consistent pH upon dilution or the addition of a small amount of strong base. Buffers cannot be made with strong acids, and adjusting a buffer's pH involves altering the weak acid to conjugate base ratio.

Explanation:

Changes in pH with Dilution and Additions:

One cannot make a buffer using a strong acid because strong acids fully dissociate in water, and a buffer relies on the equilibrium between a weak acid and its conjugate base to maintain pH. To adjust the pH of a buffer solution, one can vary the ratio of the weak acid to its conjugate base. The pH of a buffer is ideally maintained within ±1 unit of the pKa of the weak acid.

Answer:

P flask = 860.966 torr

Explanation:

⇒ P open tube = Patm + (δHg)(g)(h) = P flask

∴ δHg = 13.6 g/cm³

∴ g = 980 cm/s²

∴ h = 10.5 cm

⇒ (δHg)(g)(h) = (13.6 g/cm³)(980 cm/s²)(10.5 cm) = 139944 g/cm.s²

⇒ (δHg)(g)(h) = (13994.4 Kg/m.s²(Pa))×( 0.00750062 torr/Pa)

⇒ (δHg)(g)(h) = 104.966 torr

⇒ P flask = 756 torr + 104.966 torr

⇒ P flask = 860.966 torr

Using the manometer equation, the pressure in the flask is calculated as 898.8 torr. The height difference of the mercury columns, 10.5 cm, is converted to torr using the density of mercury.

- Atmospheric pressure [tex](\(P_{\text{atm}}\))[/tex] = 756 torr

- Height difference = 10.5 cm

Convert the height difference to torr using the density of mercury [tex](\(13.6 \, \text{g/cm}^3\)):\[ \text{height difference in torr} = 10.5 \, \text{cm} \times \left(\frac{13.6 \, \text{g/cm}^3}{1 \, \text{cm}}\right) = 142.8 \, \text{torr} \][/tex]

Now, apply the manometer equation:

[tex]P_{flask} - P_{atm}[/tex] = height difference in torr

P_flask - 756 torr = 142.8 torr

Solve for P_flask:

[tex]\[ P_{\text{flask}} = 756 \, \text{torr} + 142.8 \, \text{torr} = 898.8 \, \text{torr} \][/tex]

Therefore, the pressure in the flask is 898.8 torr.

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Answer:true

Explanation:

In covalent bond, lone pair of electrons are shared by the reacting species inorder to achieve a stable duplet or octet condition.as a result, molecules and not ions are formed

Answer:

pH = 4.8

Explanation:

A buffer is formed by a weak acid (0.145 M HC₂H₃O₂) and its conjugate base (0.202 M C₂H₃O₂⁻ coming from 0.202 M KC₂H₃O₂). The pH of a buffer system can be calculated using Henderson-Hasselbalch's equation.

[tex]pH = pKa + log\frac{[base]}{[acid]} \\pH = -log(1.8 \times 10^{-5} )+log(\frac{0.202M}{0.145M} )\\pH=4.8[/tex]

Answer:

the height of a iodododecane barometer column =66.46mm

Explanation:

we will the density of liquid 1-iodododecane and the density of mecury (compare with the given value of mercury)

when the atmospheric pressure is 751 torr ?

the compound 1-iodododecane has a density of 1.2g/mL

the density of mercury is 13.56g/mL

for mercury at

1 torr corresponds to 1 mm of a liquid mercury column= 13.56g/ml

for 1-iodododecane

751toor = ? at 1.2g/mL

1 torr = 1mm = 13.56

751 torr of mercury = 751mm = 13.56

751 toor of iodo = ? = 1.2

751mm = 13.56

?  ====1.2

1.2 * 751)/13.56 =66.46mm

the height of a iodododecane barometer column =66.46mm

The height of a barometer column with 1-iodododecane at 751 torr would depend on the density of 1-iodododecane relative to mercury. Using the hydrostatic pressure formula, the height can be determined with the known density of 1-iodododecane.

When considering the height of a barometer column based on 1-iodododecane at an atmospheric pressure of 751 torr, it's important to account for the density of 1-iodododecane relative to that of mercury. Since the density of the 1-iodododecane will be different from mercury (13.6 g/cm³), the height of the barometer column for the same atmospheric pressure would also be different. To find the height, one would need to use the formula for hydrostatic pressure, p = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the column. With the density of 1-iodododecane known, one could rearrange the formula to solve for h to find the expected height of the barometer column at 751 torr.

Answer:

2.76 atm

Explanation:

Boyle's law states that, for an isothermic process (temperature remaining the same), the product of the pressure and volume is constant.

Dalton's law states that in a gas mixture, the total pressure is the sum of the partial pressure of the components.

So:

P1*V1 + P2*V2 = P*V

Where P1 is the pressure in the bulb 1, V1 is the volume in the bulb 1, P2 is the pressure in the bulb 2, V2 is the volume in the bulb 2, P is the pressure at the mixture after the valve was opened, and V is the final volume (5.00 L).

1.80*2.00 + 3.40*3.00 = P*5.00

5P = 13.80

P = 2.76 atm

We have that for the Question "The valve between a 2.00-L bulb, in which the gas pressure is 1.80 atm, and a 3.00-L bulb, in which the gas pressure is 3.40 atm, is opened. What is the final pressure in the two bulbs, the temperature remaining constant?" it can be said that the final pressure in the two bulbs, the temperature remaining constant

P_f=2.44atm

From the question we are told

The valve between a 2.00-L bulb, in which the gas pressure is 1.80 atm, and a 3.00-L bulb, in which the gas pressure is 3.40 atm, is opened. What is the final pressure in the two bulbs, the temperature remaining constant?

Generally the equation for the ideal gas  is mathematically given as

[tex]Pv=nRT[/tex]

Where

[tex]n_1=\frac{PV}{RT}\\\\n_1=\frac{1.80*3}{RT}\\\\n_1=\frac{5.4}{RT}\\\\[/tex]

[tex]n_2=\frac{PV}{RT}\\\\n_2=\frac{3.4*2}{RT}\\\\n_2=\frac{5.4}{RT}+\frac{6.8}{RT}\\\\[/tex]

Where

[tex]the total moles =\frac{5.4}{RT}+\frac{6.8}{RT}\\\\the total moles =\frac{12.2}{RT}[/tex]

Therefore

[tex]P_f*5=\frac{12.2}{RT}*RT[/tex]

P_f=2.44atm

Hence, the final pressure in the two bulbs, the temperature remaining constant

P_f=2.44atm

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Answer:

0.925 atm

Explanation:

By Dalton's Law, the total pressure of a gas mixture is the sum of the partial pressure of its components. The vapor pressure of the water is the pressure that some molecules that evaporated do under the liquid surface. The gas and the liquid are at equilibrium. So, the gas mixture is water vapor and hydrogen gas.

Ptotal = Pwater + PH₂

745 = 42 + PH₂

PH₂ = 703 torr

Transforming to atm:

1 atm ------------------760 torr

x ------------------ 703 torr

By a simple direct three rule

760x = 703

x = 0.925 atm

C

When the alpha particle hits the beryllium atoms at high speeds, it splits the atomic nuclei hence causing the nuclei particles flying. When exposed to an electric field, the path of the proton is curved towards the negative pole while neutrons are unaffected.  

Explanation:

Neutrons are found in the dense part of atoms (the nucleus) along with protons. Unlike protons, however, that are positively charged, neutrons are uncharged particles. Neutrons are important in the stability of the atomic nuclei because they ensure that the positively charged particles (protons), which are cramped together in a tight space, do not repel each other because like-charges repel.

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Answer:

B.

Explanation:

1g of Boron has the most number of atoms. This is simply because it has the highest number of moles.

Since 1 mole contain 6.22 × 10^23 atoms, the atom that has most moles closer to 1 will contain most atoms.

This in fact can be calculated from the fact that the number of moles equal mass divided by the atomic mass.

The mass here is equal I.e 1g and thus the dividing factor will be the atomic mass. The atom with the highest atomic mass here us thorium and thus will give the lowest number of moles. Zinc follows suit in that order with Boron at the top of the other and thus will contain the highest number of atoms.

Answer:

The answer to your question is 92.7%

Explanation:

Balanced Chemical reaction

                             3 Zn  + Fe₂(SO₄)₃   ⇒   2Fe   +   3ZnSO₄

Molecular weight

Zinc = 65.4 x 3 = 196.2g

Iron (III) = 56 x 2 = 112 g

Proportions  

                           196.2 g of Zinc ------------------ 112 g of Iron

                            20.4 g of Zinc  -----------------   x

                            x = (20.4 x 112) / 196.2

                            x = 2284.8/196.2

                            x = 11.65 g of Iron

% yield = [tex]\frac{10.8}{11.65}  x 100[/tex]

% yield = 0.927 x 100

% yield = 92.7

Answer: NH3 and NH3

Explanation:

If hydrogen is bonded to a highly electronegative element, a strong dipole-dipole attraction is set up with strength ten times greater than normal dipole-dipole attractions, and it is crystal clear that electronegativity increase from left to right across the period and it decreases down the group. Nitrogen being electronegative more than Oxygen will exhibit a greater dipole-dipole attraction. As such, NH3 and NH3 is  the right answer.

The pair of molecules with the strongest dipole-dipole interactions among the given options is NH3 and NH3 due to the molecule's polar nature and resultant net dipole.

Among the options listed, the pair of molecules with the strongest dipole-dipole interactions is NH3 and NH3. Dipole-dipole interactions are attractive forces between the positive end of one polar molecule and the negative end of another polar molecule. They occur when the molecule is polar, meaning there is an unequal distribution of electrons and hence a net dipole. NH3, or ammonia, is a polar molecule because it has a pyramidal shape with a net dipole. This results in strong dipole-dipole interactions between NH3 molecules, stronger than CH4 (methane) with CH4, CO2 (carbon dioxide) with CO2, or NH3 with CH4 as these molecules are nonpolar and thus exhibit weaker London dispersion forces and not dipole-dipole interactions.

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Answer:

b. The transmittance of the cuvette must be measured in the same place each time.

Explanation:

When using a spectrophotometer, light passes not only through the liquid sample, it also passes through the cuvette. This means that each time a reading is made, you not only measure the transmittance/absorbance of the sample, but of the cuvette as well.

For this reason it's important that the reading of the cuvette's absorbance remains the same through all the process, so the answer is b), because different faces of the cuvette may have different absorbances.

Answer:

Expected energy change is -162kJ

Explanation:

For the reaction:

NBr₃(g) + 3H₂O(g) → 3HOBr(g) + NH₃(g) ΔH = +81kJ

For the reverse reaction, ΔH changes sign, thus:

3HOBr(g) + NH₃(g) → NBr₃(g) + 3H₂O(g) ΔH = -81kJ

If 2 moles of NH₃ react, the ΔH must be multiplied twice:

6HOBr(g) + 2NH₃(g) → 2NBr₃(g) + 6H₂O(g) ΔH = -162kJ

As you have 9 moles of HOBr, the limitng reactant is NH₃. Thus, expected energy change is -162kJ

I hope it helps!

Answer:

Can either be a solid, a liquid or a gas

Explanation:

A solvent can either be a solid, liquid or a gas. It is the carrier medium in a solution. It is the one in which the solute is dissolved.

Although quite unusual, a solvent might also be a solid. An important application of this can be seen in the production of alloys. Alloys are mixture of metals. To produce let’s say an alloy containing just two metals, the use of a solid solvent is needed. Here, one of the two metals is known as the base metal. It is this base metal that will serve as the carrier medium for the other metal