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.
ΔG is zero, ΔH is positive, and ΔS is negative.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.
Purification of chromium can be achieved by electrorefining chromium from an impure chromium anode onto a pure chromium cathode in an electrolytic cell. How many hours will it take to plate 11.0 kg of chromium onto the cathode if the current passed through the cell is held constant at 41.5 A? Assume the chromium in the electrolytic solution is present as Cr₃. ______ h.
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.
Explanation: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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A 100 W light bulb is placed in a cylinder equipped with a moveable piston. The light bulb is turned on for 2.0×10−2 hour, and the assembly expands from an initial volume of 0.85 L to a final volume of 5.88 L against an external pressure of 1.0 atm.a)Use the wattage of the light bulb and the time it is on to calculate ΔU in joules (assume that the cylinder and light bulb assembly is the system and assume two significant figures). Express your answer using two significant figures.b) Calculate w. Express your answer using two significant figures.c) Calculate q. Express your answer using two significant figures.
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.
Explanation: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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What is the equation for the energy levels of the hydrogen atom
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.
Explanation: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.
An experiment reacts 20.4 g of zinc metal with a solution containing an excess of iron (III) sulfate. After the reaction, 10.8 grams of iron metal are recovered. What is the percent yield of the experiment?
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
Solution: The ΔHvap of a certain compound is 48.17 kJ mol-1 and its ΔSvap is 52.91 J mol-1•K-1. What is the boiling point of this compound?
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.
One liter of a gas is in a sealed chamber containing a moveable piston. If the piston is moved so that the volume of the gas is compressed to a volume of one-half liter, what will happen to the pressure on the gas? (Assume the temperature is constant and no gas particles are lost.)
a)The pressure will be twice the original value.
b)The pressure will be half of the original value.
c)It would be impossible to move the piston since gases are not compressible.
d)The pressure will remain the same.
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]
Which pair of molecules has the strongest dipole-dipole interactions?
CH4 and CH4 NH3 and CH4 CO2 and CH4 NH3 and NH3 CO2 and CO2
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.
Explanation: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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A 50.00-mL sample containing La31 was treated with sodium oxalate to precipitate La2(C2O4)3, which was washed, dissolved in acid, and titrated with 18.04 mL of 0.006 363 M KMnO4. Write the titration reaction and find [La31] in the unknown.
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!
What is the specific heat capacity of a metal if it requires 177.5 J to change the temperatures of 15.0g of the metal from 25.00 degree celcius to 34 degree celcius
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.
Sulfur dioxide has a vapor pressure of 462.7 mm Hg at –21.0 °C and a vapor pressure of 140.5 mm Hg at –44.0 °C. What is the enthalpy of vaporization of sulfur dioxide? (R = 8.314 J/K⋅mol)
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].
Without doing any calculations, determine which sample contains the most atoms.
a. a 1-g sample of thorium
b. a 1-g sample of boron
c. a 1-g sample of zinc
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.
You would like to induce a transversion mutation into a sequence of dna. Which type of chemical mutagen would give you the best chance of inducing the correct mutation without causing transition mutations as well?
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.
Given the coefficients of the balanced equation, the enthalpy of the reaction can be expressed in terms of the amount of ammonia as follows: ΔHrxn=−906 kJ4 mol NH3 Next, find the number of moles of NH3 that react to determine the heat associated with 345 g of NH3.
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
Classify the solid state of the following substances as ionic crystals, covalent crystals, molecular crystals, or metallic crystals:
(a) CO2
(b) B12
(c) S8
(d) KBr
(e) Mg
(f) SiO2
(g) LiCl
(h) Cr
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
A 1.00-L gas sample at 100.°C and 500. torr contains 52.0% helium and 48.0% xenon by mass. What are the partial pressures of the individual gases?
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.
A Sample of 61.8g of hyroboric acid(H3BO3), a weak acid, is dissolved in 1,000g of water to make a 1.0 molal solution. Which would be the best procedure to determine the molarity if the solution?
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
How is carbon moved from the hydrosphere to the atmosphere?
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;
Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single σ bond or around a double σ plus π bond, or would they be the same?
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.
Explanation: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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Calculate the ph of a buffer that is 0.145 m hc2h3o2 and 0.202 m kc2h3o2. The ka for hc2h3o2 is 1.8 × 10-5.
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]
The energy change, ∆H, associated with the following reaction is +81 kJ.
NBr3(g) + 3 H2O(g) → 3 HOBr(g) + NH3(g)
What is the expected energy change for the reverse reaction of nine moles of HOBr and two moles of NH3?
1.) -81 kJ
2.) +162 kJ
3.) +243 kJ
4.) -365 kJ
5.) -243 kJ
6.) +365 kJ
7.) +81 kJ
8.) -162 kJ
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!
Name the following structure:
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.
covalent bonds are formed from the sharing of valence electrons
true or false give a brief simple explanation
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
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?
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 the following questions: On a 10-fold dilution of a weak acid, the pH will ___________________. On a 10-fold dilution of a buffered solution, the pH will ______________. If one adds a small amount of strong base to a buffered solution, the pH will _______________. Can you make a buffer using a strong acid? Explain. How can you adjust the pH of a buffer solution? Explain.
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:
On a 10-fold dilution of a weak acid, the pH will increase, because the concentration of hydrogen ions decreases, making the solution less acidic.On a 10-fold dilution of a buffered solution, the pH will remain relatively constant. This stability is due to the presence of both a weak acid and its conjugate base which can absorb added acids or bases without significantly changing the pH.If one adds a small amount of strong base to a buffered solution, the pH will increase slightly. However, due to the buffering action, this change is much less than it would be in an unbuffered solution.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.
The wavelength of light associated with the n = 2 to n = 1 electron transition in the hydrogen spectrum is 1.216 × 10–7 m. By what coefficient should this wavelength be multiplied to obtain the wavelength associated with the same electron transition in the Li2+ ion?
Go through each step, and explain in detail.
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.
Explanation: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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The value of ΔH for the reaction below is -72 kJ. __________ kJ of heat are released when 1.0 mol of HBr is formed in this reaction. H 2 (g) + Br 2 (g) → 2HBr (g)
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.
What is Hess's law?The heat of any reaction ΔH for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction:[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.
Explanation: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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At what pressure does ethane (C₂H₆) have a density of 37.2 g/L at 40.0 °C?
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:
Density = 37.2 g/LTemperature = 40.0 °CMolar 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;
P is the pressure. V is the volume. n is the number of moles of substance. R is the ideal gas constant. T is the temperature.[tex]Density = \frac{Mass}{Volume}[/tex]
In terms of density, the ideal gas law equation becomes:
[tex]Density = \frac{M_MP}{RT}[/tex]
Where;
[tex]M_M[/tex] is the molar mass.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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A bomb calorimetric experiment was run to determine the enthalpy of combustion of methanol. The reaction is CH3OH(l)+3/2O2(g)→CO2(g)+2H2O(l) The bomb calorimeter has a heat capacity of 250.0 J/K. Burning 0.028 g of methanol resulted in a rise in temperature from 21.50 ∘C to 23.41 ∘C. Calculate the change in internal energy for the combustion of methanol in kJ/mol.
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.
An open-tube manometer is used to measure the pressure in flask. The atmospheric pressure is 756 torr and the Hg column is 10.5 cm higher on the open end. What is the pressure in the flask?
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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An example of a strong _____ agent important in biochemistry is _____. The redox potential for this molecule is _____. Please choose the correct answer from the following choices
a. oxidizing; O2;
b. negative reducing; NADH;
c. positive oxidizing; NADH;
d. positive reducing; O2;
e. negative reducing; NADH;
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.