A mixture of H2 and water vapor is present in a closed vessel at 20°C. The total pressure of the system is 755.0 mmHg. Partial pressure of water vapor at 20°C equals 17.5 mmHg. What is the partial pressure of H2? mmHg

Answer:

a) The pressure wiil be twice the original value

Explanation:

Pressure in inversely proportional to volume, that means; when one value decreases at the same rate that the other increases, if the temperature and amount of gas remain unchanged within a closed system. The above derived from the same definition of pressure.  The pressure is defined as the crashes and frequency of the same agains surface of objects, if certain amount of gas is  contained in a recipent, It will crash more times if the volume is minor. The Boyle´s law  express this mathematically.

[tex]P1V1= P2V2\\\\P2=\frac{P1V1}{V2} \\\\P2=\frac{P1*1L}{0.5L} \\\\P2=2P1\\\\\\where\\\\V1=  1L\\V2= 0.5L[/tex]

Explanation:

There are four type of crystals possible in solids state , given as follows -

1. Ionic crystals -

In this type of crystals ionic bond is involved for the bonding between two atoms , which connects two oppositely charged atoms , i.e. , cation and anion together .

The ionic crystal from the given options are -

(d) KBr and (g) LiCl .

2. Covalent crystals -

The bond formed in these type of crystals are from share the electrons , i.e. making a covalent bond .

Hence ,

From the given options , the covalent crystals are -

(b) B₁₂  and  (f) SiO₂ .

3. Molecular crystals -

weak intermolecular forces , like the dispersion forces , holds the atoms together , generating a molecular crystal .

Hence ,

From the given options , the molecular crystals are -

(a) CO₂ and (c) S₈

4. Metallic crystal -

The type of bond , which make up a metallic crystal is , the metal cation with the delocalized electrons .

Hence ,

From the given options , the metallic crystals are -

(e) Mg and (h) Cr

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:

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:

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:

The heat capacity of the bomb calorimeter is 7.58 J/°C.

Explanation:

[tex]C_6H_6(I) + \frac{15}{2} O_2 (g) \rightarrow 6CO_2 (g) + 3H_2O (g) ,\Delta H^o = -3267.5 kJ[/tex]

First, we will calculate energy released on combustion:

[tex]\Delta H[/tex] = enthalpy change =  -3267.5 kJ/mol

q = heat energy released

n = number of moles benzene= [tex]\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{2.15 g g}{78 g /mol}=0.02756 mol[/tex]

[tex]\Delta H=-\frac{q}{n}[/tex]

[tex]q=\Delta H\times n =-3267.5 kJ/mol\times 0.02756 mol=-90.0657 kJ[/tex]

q = -90.0657 kJ = -90,065.7 J

Now we  calculate the heat gained by the calorimeter let it be Q.

Q = -q= -(-90,065.7 J) = 90,065.7 J (conservation of energy)

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

where,

Q = heat gained by calorimeter

c = specific heat capacity of calorimeter =?

[tex]T_{final}[/tex] = final temperature = [tex]34.34^oC[/tex]

[tex]T_{initial}[/tex] = initial temperature = [tex]22.46^oC[/tex]

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

[tex]90,065.7 J=c\times (34.34-22.46)^oC[/tex]

[tex]c=\frac{90,065.7 J}{(34.34-22.46)^oC}=7.58 J/^oC[/tex]

The heat capacity of the bomb calorimeter is 7.58 J/°C.

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:

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: 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:

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:

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: Ethyl ethanoate

Explanation:

An ester composed of the alkyl section (Ethyl) derived from ethanol and the alkanoate section (ethanoate) derived from ethanoic acid.

Answer:

The correct answer is d) ethyl ethanoate.

Explanation:

This organic compound corresponds to an ester, also called ethyl acetate. It is synthesized by the esterification of Fischer using acetic acid and ethanol in the presence of a catalyst.

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:

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;

The work done by a gas on its surrounds is positive when the volume of the gas increases because the gas is doing work to push against the external pressure.

The work done by a gas on its surroundings is defined as positive when the gas expands, meaning when the volume of the gas increases. This is because the gas is doing work to push against the external pressure and expand its volume. So, the correct answer to your question 'When the volume of a gas increases, is the work done by the gas on its surroundings positive, negative, or zero?' would be (c) positive.

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When the volume of a gas increases, the work done by the gas on its surroundings is positive. This occurs because the gas is doing work on its surroundings, resulting in energy loss for the gas. The key equation is w = -Pext ΔV.

So the answer is C)Positive

To determine this, we need to understand the concept of work in thermodynamics. Specifically, the work done by a gas during expansion is given by the equation:

w = -[tex]P_{ext[/tex] ΔV,

where:

When the volume of the gas increases, ΔV is positive because the final volume ([tex]V_{final[/tex]) is greater than the initial volume ([tex]V_{initial[/tex]). According to the equation, since ΔV is positive and there is a negative sign in front, the work (w) done by the gas is negative.

This negative sign indicates that the system (the gas) is doing work on its surroundings, thus losing energy. Therefore, when the volume of a gas increases, the work done by the gas on its surroundings is positive.

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:Reliability

Explanation:

Reliability of a test refers to how consistently a test measures a characteristic under the same conditions.

Reliability can be defined as the degree of consistency of which a chemical test gives a similar result. measure. A test is said to be highly reliable when it gives the same repeated result under the same conditions of measure.

But when a test gives different results under the same condition of measure it has a low reliability.

Hence, If a test yields consistent results every time it is used, it has a high degree of reliability.

Answer:

4C3H5N3O9 (l)  ---------> 12CO2 (g)  +  H20 (g)  + 6N2 (g) + 6O2 (g)

Explanation:

Nitroglcerin is a drug basically used to treat chest pain. It is a dense, colourless and explo9sive liquid. Its molecular formula is C3H5N3O9.

It decomposes to gaseous dinitrogen, gaseous dioxygen, gaseous water and gaseous carbon di oxide

The equation for its decomposition is shown below;

4C3H5N3O9 (l)  ---------> 12CO2 (g)  +  H20 (g)  + 6N2 (g) + 6O2 (g)

Answer:

The change in internal energy during the combustion reaction is- 545.71 kJ/mol.

Explanation:

First we have to calculate the heat gained by the calorimeter.

[tex]q=c\times (T_{final}-T_{initial})[/tex]

where,

q = heat gained = ?

c = specific heat = [tex]250.0 J/^oC[/tex]

[tex]T_{i}[/tex] = Initial temperature = [tex]21.50^oC=294.65 K[/tex]

[tex]T_{f}[/tex] = Final temperature = [tex]23.41^oC=296.56 K[/tex]

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

[tex]q=250.0 J/K\times (296.56 -294.65 )K[/tex]

[tex]q=477.5 J [/tex]

Now we have to calculate the enthalpy change during the reaction.

[tex]\Delta H=-\frac{q}{n}[/tex]

where,

[tex]\Delta H[/tex] = enthalpy change = ?

q = heat gained = -477.5 J

n = number of moles methanol = [tex]\frac{\text{Mass of methanol}}{\text{Molar mass of methanol}}=\frac{0.028 g}{32 g/mol}=0.000875 mol[/tex]

[tex]\Delta H=-\frac{477.5 }{0.000875 mol}=-545,714.28 J/mol=-545.71 kJ/mol[/tex]

Therefore, the change in internal energy during the combustion reaction is- 545.71 kJ/mol.

Answer:

Partial pressure of He=486 torr, partial pressure of Xe= 14 torr

Explanation:

Using the equation, PV=nRT----------------------------------------(1)

Making n the subject of the formula;

n= PV/RT---------------------(2)

Where n= number of moles, v= volume, T= temperature, P= volume.

n= (500 torr/760 torr × 1 atm)× 1L ÷ 373K ×0.082 L atmK^-1. Mol^-1

n= 0.6579 atm.L/ 30.6233

n= 0.0215 mol.

Let the total mass of the gas= 2b(in g).

Mass of helium gas = b (in g) = mass of Xenon gas

Mole of helium gas= b(in g) / 4 gmol^-1

=b/4 mol

Mole of Xenon= b g/131.3 gmol^-1

= b/131.3 mol.

Solving for b, we have;

b/4+b/131.3 = 0.0215 mole

(131.3+4)b/525.2= 0.0215

Multiply both sides by 1/135.3.

b= 11.2918/135.3

b= 0.0835 g

Mole of He gas= 0.0835/4= 0.0209

Mole of Xe gas= 0.0215- 0.0209

= 0.0006 mol

Mole fraction of He = 0.0209/0.0215

= 0.972

Mole fraction of Xe= 0.0006/0.0215

= 0.028

Partial pressure of He gas= 0.972× 500 torr= 486 torr

Partial pressure of Xe gas= 0.028 ×500 torr = 14 torr

Final answer:

Dalton's Law is used to find the partial pressures of He (388.7 torr) and Xe (111.3 torr) in a 1.00 L, 500 torr mixture by calculating their mole fractions based on mass percentages and multiplying by the total pressure.

Explanation:

To determine the partial pressures of helium and xenon in a 1.00-L gas sample at 100.°C and 500. torr, we can use Dalton's Law of Partial Pressures. First, we calculate the mass of each gas based on the given percentage by mass. Then, we convert the mass of each gas to moles using their respective molar masses. Finally, using Dalton's Law, the partial pressure of each gas is calculated as the mole fraction of that gas multiplied by the total pressure.

First, let's convert the total mass percentage to actual mass assuming we have 100 grams of the mixture:

Mass of helium (He) = 52.0% of 100 g = 52.0 g

Mass of xenon (Xe) = 48.0% of 100 g = 48.0 g

Next, we'll convert mass to moles using the molar mass of helium (4.00 g/mol) and xenon (131.29 g/mol):

Moles of He = 52.0 g / 4.00 g/mol = 13.0 moles

Moles of Xe = 48.0 g / 131.29 g/mol = 0.365 moles

The total moles of gas is the sum of the moles of He and Xe:

Total moles of gas = Moles of He + Moles of Xe

Total moles of gas = 13.0 moles + 0.365 moles = 13.365 moles

Next, we calculate the mole fraction for each gas:

Mole fraction of He = Moles of He / Total moles of gas = 13.0 / 13.365

Mole fraction of Xe = Moles of Xe / Total moles of gas = 0.365 / 13.365

Lastly, we determine each gas's partial pressure:

Partial pressure of He = Mole fraction of He × Total pressure

Partial pressure of Xe = Mole fraction of Xe × Total pressure

Substituting the values we get:

Partial pressure of He = (13.0 / 13.365) × 500 torr

Partial pressure of Xe = (0.365 / 13.365) × 500 torr

Partial pressures calculated are for He and Xe respectively.

We can calculate the change in internal energy of the system by multiplying the power of the light bulb by the time it's on, resulting in 7200 Joules. The work done can be calculated using the formula for work done against pressure, resulting in -508.64 Joules. The heat energy is determined using the first law of thermodynamics, resulting in 6691.36 Joules.

The student is asked to calculate the change in internal energy (ΔU), work done (w), and the heat transferred (q) in a system involving a moving piston and a light bulb. Let's address each part individually:

(a) The power (P) of the light bulb is given as 100 W. Power is energy per unit time, so we can find the energy by multiplying the power by the time it's used. Time is given as 2.0×10−2 hour, which needs to be converted to seconds for consistency: 2.0×10−2 hour × 3600 s/hour = 72 seconds. Thus, the change in internal energy (ΔU) is P × t = 100 W × 72 s = 7200 Joules.

(b) Work done by the system (w) equals -P*ΔV where P is the external pressure and ΔV is change in volume. The system expands, so work is done against the external pressure. Here, ΔV = final volume - initial volume = 5.88 L - 0.85 L = 5.03 L. We must convert volume to m³ (cubic meters), 1 L = 1.0e-3 m³, so, ΔV = 5.03e-3 m³. Also convert pressure to Pa from atm, 1 atm = 1.013e5 Pa, so P = 1.013e5 Pa. Inserting these values into the equation gives w = -(1.013e5 Pa)(5.03e-3 m³) = -508.64 Joules.

(c) Finally, to determine the heat energy (q) we utilize the first law of thermodynamics that states ΔU = q - w. Solving for q, we get, q = ΔU + w = 7200 Joules + (-508.64 Joules) = 6691.36 Joules.

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Answer:

3.6

Explanation:

The pH of a solution can be defined as the negative logarithm to base 10 of the concentration of hydrogen ions in the solution. In the equation form, it can be written as :

pH = -Log [H+]

In the question, we have been given the concentration of the hydrogen ions but we do not know the value of the pH. We then insert the value of the concentration into the equation.

This means [H+] = 2.5 * 10^-4

pH = -Log [ 2.5 * 10^-4] = 3.6

Answer:

Molarity = 1.0 M

Explanation:

Step 1:  Data given

Mass of H3BO3 = 61.8 grams

Mass of water = 1,000 grams = 1kg

Molality = 1.0 molal

Step 2: Calculate number of moles

1.0 molal = 1 mol of solute/ 1kg of solvent

Number of moles = molality * mass of the solvent = 1.0 mol/kg * 1kg = 1 mol

This means 61.8 grams of H3BO3 should be 1 mol

We can control this:

Number of moles = mass / molar mass

Number of moles H3BO3 = 61.8/ 61.83 g/mol = 1 mol

Step 3: Calculate the volume of water

Let's suppose the density of water is 1g/mL

Volume of water = mass / density

Volume = 1,000 grams / 1g/mL

Volume = 1,000 mL = 1.00 L

Step 4: Calculate molarity

Molarity = number of moles / volume

Molarity = 1 mol / 1.00 L

Molarity = 1.0 M

For dilute AQUEOUS solutions molality  ≅  molarity

Answer:

q = -4588.89 kJ

Explanation:

First, we need to write the equation taht is taking place. In this case, is the combustion of NH3 so the reaction is as follow:

4NH3 + 5O2 --------> 4NO + 6H2O      ΔHrxn = -906 kJ

Now, in order to get the number of moles that react here and determine the heat, we first calculate the experimental moles of NH3 within the 345 g of ammonia:

moles = m/MM

The molar mass of NH3 is 17 g/mol so:

moles = 345 / 17 = 20.29 moles

Now, we will stablish a relation between the theorical moles and the experimental moles:

q = 20.29 moles * (-906) kJ / 4 moles

q = -4,588.89 kJ

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:

it is easier to rotate and single bond rather than a double bond made of a sigma bond and pi bond

Explanation:

Rotation around a single bond happens easily but it is very limited around a double bond because of the overlapping electron cloud above and below the imaginary axis between the two atoms.

It is easier to rotate around a single sigma (σ) bond than around a double bond that includes both a sigma and a pi bond. This is because single bonds are not dependent on the orientation of the atom's orbitals, unlike double bonds. In the latter, any rotation can break the bond.

The matter at hand relates to the ease of rotation around different types of bonds - a single sigma (σ) bond versus a double bond made of one sigma (σ) and one pi (π) bond. Simplifying the science, it is easier to rotate around a single sigma (σ) bond than it is around a double bond that includes both a sigma and a pi bond. This is because the orbital overlap, which forms these bonds among atoms, doesn't rely on the relative orientation of the orbitals on each atom for a single bond. So, twisting or rotating doesn't affect the bonding. Whereas, with a double bond, the σ and π bonds form from different types of orbital overlaps. With this type of bond, the stable configuration is planar (flat) as seen in ethene molecules. Any rotation would misalign their overlapping, unstable, and effectively break the pi bond.

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The study of chemicals and bonds is called chemistry. There are different types of elements are there and these are metal and nonmetal.

The correct answer is 36 KJ of heat is released when 1.0 mole of HBr is formed.

[tex]H_2 (g) + Br_2 (g) ---> 2HBr (g) \ \ \ \ H = -72 KJ[/tex]

This is the energy released when 2 moles of HBr is formed from one mole each of H2 and Br2. Therefore, Heat released for the formation of 1 mol HBr would behalf on this.

ΔHreq = -36 kJ

36 KJ of heat is released when 1.0 mole of HBr is formed.

Hence, the correct answer is 36KJ.

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The energy change ΔH for the reaction of H2 with Br2 to form 2 moles of HBr is -72 kJ. Therefore, when 1 mole of HBr is formed, half of this energy, which is -36 kJ, is released.

In your question, the chemical equation shows that 2 moles of HBr are formed from the reaction of H2 with Br2, with the release of -72 kJ of heat. This energy change, ΔH, represents the amount of heat released during the reaction, indicating it's an exothermic reaction.  Given that energy change corresponds to the formation of 2 moles of HBr, the heat released when 1 mole of HBr is formed would be -72 kJ divided by 2.

This results in -36 kJ released per mole of HBr formed. Thus, when 1.0 mol of HBr is formed in this reaction, -36 kJ of heat is released.

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9 moles of reactants chemically change into 11 moles of product.

Explanation:

In the reaction above, 9 moles of reactants chemically change into 11 mole of products. The coefficients in a reaction is the number of moles of the reacting atoms .

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Answer:

A) 9 moles of reactants chemically change into 11 moles of product.

Explanation:

C5H12(l) + 8O2(g) -> 5CO2(g) + 6H2O(g)

The first step is to check if the equation is indeed a balanced one before proceeding to compare the moles.

This is done by comparing the number of atoms on the reactant side with that in the product side, if equal the reaction is indeed a balanced one.

Number of moles of Carbon atoms

Reactant = 5

Product = 5

Number of moles of Hydrogen atoms

Reactant = 12

Product = 12

Number of moles of Oxygen atoms

Reactant = 16

Product = 10 + 6 = 16

Since the equation is balanced, we can now compare the moles.

From the reactant side we have a total number of 9 (1 + 8 ) moles reacting to yield a total of 11 ( 5 + 6 ) moles of product.

This alone leads us  to take option A as our answer.

Let us see how other options are wrong however;

Option B) It cannot be grams as what chemical equations shows us the molar (number of moles) relationship between elememts/molecules/compounds/ions etc. The gram relationship can be calculated however if the molar mass of the compounds is taken into consideration.

Option C) This is wrong again due to same reasons mentioned for option B.

Option C) In the compound C5H12 alone, there are 17 (5 C and 12 H) atoms. Hence this option is also wrong.

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:

The coefficient should be 0.1111.

Explanation:

Using Rydberg's Equation:

[tex]\frac{1}{\lambda}=R_{H}\times Z^2\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]

Where,

[tex]\lambda[/tex] = Wavelength of radiation

[tex]R_H[/tex] = Rydberg's Constant

[tex]n_f[/tex] = Higher energy level

[tex]n_i[/tex]= Lower energy level

Z = Atomic number

1) For n = 2 to n = 1 in hydrogen atom:

Z = 1

[tex]\frac{1}{\lambda}=R\times 1^2\left(\frac{1}{2^2}-\frac{1}{1^2} \right )[/tex]

[tex]\frac{1}{\lambda}=-R\times \frac{3}{4}[/tex]..[1]

2) For n = 2 to n = 1 in lithium ion i.e. [tex]Li^{2+}[/tex] :

Z = 3

[tex]\frac{1}{\lambda}=R\times 3^2\left(\frac{1}{2^2}-\frac{1}{1^2} \right )[/tex]

[tex]\frac{1}{\lambda '}=-9R\times \frac{3}{4}[/tex]..[2]

[2] ÷ [1]

[tex]\frac{\frac{1}{\lambda '}}{\frac{1}{\lambda }}=\frac{-9R\times \frac{3}{4}}{-R\times \frac{3}{4}}[/tex]

[tex]\frac{\lambda }{\lambda '}=9[/tex]

[tex]\lambda '=\lambda \times \frac{1}{9}[/tex]

[tex]\lambda '=\lambda \times 0.1111[/tex]

The coefficient should be 0.1111.

To find the wavelength associated with the same electron transition in the Li2+ ion, use the formula wavelength_Li2+ = (wavelength_Hydrogen) * (Z^2) / (n^2), where Z is the atomic number of the Li2+ ion and n is the principal quantum number.

To find the wavelength associated with the same electron transition in the Li2+ ion, we need to use the formula:

wavelength_Li2+ = (wavelength_Hydrogen) * (Z^2) / (n^2)

where Z is the atomic number of the Li2+ ion (which is 3) and n is the principal quantum number (which is 2 in this case).

Plugging in the values, we have:

wavelength_Li2+ = (1.216 × 10–7 m) * (3^2) / (2^2)

Simplifying, we get the wavelength associated with the n = 2 to n = 1 electron transition in the Li2+ ion to be approximately 1.824 × 10–7 m.

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