The increase in entropy when the partition is removed is calculated using the entropy change formula ΔS=nRln(Vf/Vi). The total entropy change is the sum of the entropy changes for each gas. For the first case, the total entropy increase is 11.52 J/K; for the 2 moles of gas A case, the increase is 17.29 J/K; and for the case where both compartments contain gas A, the increase is also 11.52 J/K.
Explanation:For this, we will use the formula for the entropy change, ΔS, which is given by ΔS=nRln(Vf/Vi), where n is the number of moles, R is the gas constant, and Vf and Vi are the final and initial volumes respectively. In the first case, the total entropy change when the partition is removed is the sum of the entropy changes for both gas A and gas B, since both volumes double, so ΔS=(1mol)(8.31 J/mol K)ln(2)+(1mol)(8.31 J/mol K)ln(2)=11.52 J/K.
If there are 2 moles of gas A in the first compartment, the entropy change is ΔS=(2mol)(8.31 J/mol K)ln(2)+(1mol)(8.31 J/mol K)ln(2)=17.29 J/K. For the final case where both compartments initially contain gas A, when the partition is removed, the total volume available to the gas doubles, so each gas experiences an entropy change of ΔS=(1mol)(8.31 J/mol K)ln(2), summed over both gases, gives us again 11.52 J/K.
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Case 1 - For Ideal Gas A and B the value of ΔStotal is 11.52 J/K.
Case 2 - For 2 Moles of Ideal Gas A the value of ΔS is 11.52 J/K.
Case 3 - Both Compartments with Ideal Gas A has the value of ΔS as 11.52 J/K.
Case 1: Ideal Gas A and B
For the first case, where there is 1 mole of ideal gas A in one compartment and 1 mole of ideal gas B in the other, both at 1 atm:
The total initial volume of the system is effectively doubled when the partition is removed.The entropy change for each gas can be calculated using the formula:Case 2: 2 Moles of Ideal Gas A
For the second case, with 2 moles of ideal gas A in one compartment:
Similar to the first case, the volume ratio Vf/Vi is 2.The entropy change is calculated for 2 moles:Case 3: Both Compartments with Ideal Gas A
Now, let’s consider the scenarios in which both compartments contain the same type of gas A:
If both compartments had 1 mole each of gas A, using the same logic explained in the first case:The total entropy change is: ΔStotal = 2 * 8.314 * ln(2) ≈ 11.52 J/KIf one compartment had 2 moles of gas A and the other also had 2 moles, the calculation is the same, but for each set of 2 moles:
ΔS = 2 * 8.314 * ln(2) ≈ 11.52 J/K
The key idea is that the type of molecules and their identical properties simplify calculations using the entropy formula for ideal gases.
When can a hypothesis be elevated to the status of a theory?
Answer:
A hypothesis can be described as a tentative statement which can be proved either right or wrong through scientific experiments. If a hypothesis is tested again and again and every time the experiments give the same results, then the hypothesis can take the form of a theory. However, a theory is subjected to change if new researches are made which can annul it. For a theory to be formed, there should be enough explanation behind the phenomenon along with the experiments.
The first-order rate constant for the decomposition of N2O5, 2N2O5(g)→4NO2(g)+O2(g) at 70∘C is 6.82×10−3 s−1. Suppose we start with 2.90×10−2 mol of N2O5(g) in a volume of 1.7 L.
Part A How many moles of N2O5 will remain after 4.0 min ?
Answer:
0.00564 moles
Explanation:
Given that:
The rate constant, k = [tex]6.82\times 10^{-3}[/tex] s⁻¹
Initial concentration [A₀] = [tex]2.90\times 10^{-2}[/tex] mol
Time = 4.0 min = [tex]4.0\times 60[/tex] sec = 240 sec
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
So,
[tex][A_t]=2.90\times 10^{-2}\times e^{-6.82\times 10^{-3}\times 240}=2.9\times \frac{1}{10^2}\times \frac{1}{e^{1.6368}}[/tex]
[tex][A_t]=0.00564\ moles[/tex]
The concentration after four minutes is 3.3 ×10−3.
Let us recall that for a first order reaction;
ln[A] = ln[A]o - kt
Where;
[A] = concentration at time t
[A]o = initial concentration
k = rate constant
t = time
[A]o = 2.90×10−2 mol/1.7 L = 0.0171 M
k = 6.82×10−3 s−1
t = 4 min or 240 s
Substituting values;
ln[A] = ln[0.0171 M] - (6.82×10−3 s−1 × 240 s)
[A] = e^ln[0.0171 M] - (6.82×10−3 s−1 × 240 s)
[A] = 3.3 ×10−3.
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How much energy must be supplied to break a single 21Ne nucleus into separated protons and neutrons if the nucleus has a mass of 20.98846 amu?
What is the nuclear binding energy for 1 {\rm mol} of {\rm ^{21}Ne}?
Answer:
1)There is 2.68 * 10^-11 J of energy needed
2) The nuclear binding energy for 1 mol of Ne is 1.6 *10^13 J/mol
Explanation:
Step 1: Data given
The nucleus of a21Ne atom has a amass of 20.98846 amu.
Step 2: Calculate number of protons and neutrons
The number of electrons and protons in an 21Ne atom = 10
The number of neutrons = 21 -10 =11
Step 3: mass of the atom
Mass of a proton = 1.00727647 u
Mass of a neutron = 1.0086649 u
The mass of the atom = mass of all neutrons + mass of protons
Mass of atom = 11*1.0086649 + 10*1.00727647 = 21.1680786 amu
Step 4: Calculate change of mass
The change in mass = Mass of atom - mass of neon
Δmass = 21.1680786 - 20.98846
Δmass = 0.1796186
Step 5: Calculate mass for a single nucleus
The change of mass for a single nucleus is = Δmass / number of avogadro
Δmass of nucleus = 0.1796186 / 6.022*10^23
Δmass of nucleus =2.98 * 10^-25 grams = 2.98 * 10^-28 kg
Step 6: Calculate energy to break a Ne nucleus
Calculate the amount of energy to break a Ne nucleus
ΔEnucleus = Δmass of nucleus * c²
⇒ with c = 2.9979 *10^8 m/s
ΔEnucleus = 2.98 * 10^-28 kg * (2.9979*10^8)² = 2.68 * 10^-11 J
What is the nuclear binding energy for 1 mol of Ne?
ΔE= ΔEnucleus * number of avogadro
ΔE= 2.68 * 10^-11 J * 6.022*10^23
ΔE= 1.6 *10^13 J/mol
Consider the following reaction at equilibrium. What effect will increasing the temperature have on the system?Fe3O4(s) + CO(g) ↔ 3 FeO(s) + CO2(g) ΔH°= +35.9 kJThe equilibrium constant will decrease.No effect will be observed.The reaction will shift to the right in the direction of products.The equilibrium constant will increase.The reaction will shift to the left in the direction of reactants
Answer:
The reaction will shift to the right in the direction of products.
Explanation:
According to Le Chatelier's Principle, the change in any state of the equilibrium say temperature, volume, pressure, or the concentration, the equilibrium will oppose these changes and will shift in such a way that the effect cause must be nullified.
For an endothermic reaction,
On increasing the temperature, reaction will go in forward direction (towards right), because according to Le Chatelier's principle as we increase the temperature, the equilibrium is will be disturbed , so to again establish the equilibrium, the reaction will go in forward direction as it is endothermic in nature (towards right).
Thus, the [tex]\Delta H^0[/tex] of the given reaction is positive and thus, on increasing temperature, reaction will go in forward direction.
Answer:- The reaction will shift to the right in the direction of products.
The reaction will move to the left in the direction of the reactants as the temperature rises, favouring the endothermic path. The reaction will move left in the direction of the reactants, thus that is the right response.
A chemical process that takes heat from its surroundings is said to be going in an endothermic direction. Because the reactants and products of an endothermic reaction have different energies, heat energy is needed to help the reaction proceed and create the products. An endothermic reaction is one that absorbs heat while it is occurring. This can be felt as a drop in temperature in the immediate environment or shown by thermodynamic calculations. An endothermic process has a positive enthalpy change (H), which means energy is being absorbed. The reaction will move to the left in the direction of the reactants as the temperature rises, favouring the endothermic path. The reaction will move left in the direction of the reactants, thus that is the right response.
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a student dissolves 20.0 g of glucose into 511 mL of water at 25 C the vapor pressure of pure water at 25 C is 3.13
Answer:
3.12 × 10⁻² atm
Explanation:
A student dissolves 20.0 g of glucose into 511 mL of water. At 25 °C, the vapor pressure of pure water at 25 C is 3.13 × 10⁻² atm. I think the question is: "What is the vapor pressure of the solution?"
According to Raoult's law, the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent present.
[tex]P_{solution}=P\°_{solvent}X_{solvent}[/tex]
The molar mass of glucose is 180.16 g/mol. The moles corresponding to 20.0 g of glucose are:
20.0 g × (1 mol/180.16 g) = 0.111 mol
The density of water at 25°C is 0.997 g/mL. The mass corresponding to 511 mL of water is:
511 mL × (0.997 g/mL) = 509 g
The molar mass of water is 18.02 g/mol. The moles corresponding to 509 g of water are:
509 g × (1 mol/18.02 g) = 28.2 mol
The total number of moles is 0.111 mol + 28.2 mol = 28.3 mol
The mole fraction of water is:
[tex]X_{solvent}=\frac{28.2mol}{28.3mol} =0.996[/tex]
The vapor pressure of a solvent above the solution is:
[tex]P_{solution}=3.13 \times 10^{-2} atm \times 0.996 = 3.12 \times 10^{-2} atm[/tex]
In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.0 cm^3. If the sphere were broken down into eight spheres each having a volume of 1.25 cm^3, and the reaction is run a second time, which of the following accurately characterizes the second run?
Choose all that apply.
A. The second run will be faster.
B. The second run will be slower.
C. The second run will have the same rate as the first.
D. The second run has twice the surface area.
E. The second run has eight times the surface area.
F. The second run has 10 times the surface area.
Answer:
D
Explanation:
We know that the
reaction catalyzing power of a catalyst ∝ surface area exposed by it
Given
volume V1= 10 cm^3
⇒[tex]\frac{4}{3} \pi r^3= 10[/tex]
hence r= 1.545 cm
also, surface area S1= [tex]4\pi r^2[/tex]
now when the sphere is broken down into 8 smaller spheres
S2= 8×4πr'^2
now, equating V1 and V2 ( as the volume must remain same )
[tex]\frac{4}{3}\pi r^3=8\times\frac{4}{3} \pi r'^3[/tex]
and solving we get
r'= r/2
therefore, S2=[tex]8\times4\pi\frac{r}{2}^2[/tex]
S2=[tex]2\times4\pi r^2[/tex]
S2= 2S1
hence the correct answer is
. The second run has twice the surface area.
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
(a) 2NH3(g) ⇌ N2(g)+3H2(g) ΔH = 92kJ
(b) N2(g) + O2(g) ⇌ 2NO(g) ΔH =181kJ
(c) 2O3(g) ⇌ 3O2(g) ΔH = − 285kJ
(d) CaO(s) + CO2(g) ⇌ CaCO3(s) ΔH = − 176kJ
An increase in temperature shifts an endothermic reaction to the right and an exothermic reaction to the left. For the reduced volume, it shifts the equilibriums to the side with the fewer moles of gas.
Explanation:The effect of an increase in temperature on the given equilibrium systems depends on the sign of Delta H (ΔH), which represents the heat of reaction. For (a) 2NH3(g) ⇌ N2(g)+3H2(g) with ΔH = 92kJ and (b) N2(g) + O2(g) ⇌ 2NO(g) with ΔH = 181kJ, because these reactions are endothermic (ΔH is positive), an increase in temperature will shift the equilibrium towards the right to absorb the excess heat. On the other hand, reactions (c) 2O3(g) ⇌ 3O2(g) with ΔH = - 285kJ and (d) CaO(s) + CO2(g) ⇌ CaCO3(s) with ΔH = - 176kJ are exothermic (ΔH is negative), an increase in temperature will shift the equilibrium to the left to offset the increase in heat.
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Increasing temperature generally shifts endothermic reactions to the right and exothermic reactions to the left. Decreasing volume typically shifts equilibria towards the side with fewer moles of gas. Each reaction's equilibrium shift depends on its enthalpy change (ΔH) and the moles of gases involved.
(a) 2NH₃(g) ⇌ N₂(g) + 3H₂(g) ΔH = 92kJ
Increase in temperature: The reaction is endothermic (ΔH is positive), so an increase in temperature will shift the equilibrium to the right, favoring the formation of N₂ and H₂.Decrease in volume: Decreasing the volume increases pressure. Since there are more moles of gas on the right side, the equilibrium will shift to the left to reduce pressure, favoring the formation of NH₃.(b) N₂(g) + O₂(g) ⇌ 2NO(g) ΔH = 181kJ
Increase in temperature: Similar to (a), the reaction is endothermic, so increasing temperature shifts the equilibrium to the right, favoring the formation of NO.Decrease in volume: Since there is no change in the number of moles of gas (2 moles reactants, 2 moles products), a decrease in volume has little to no effect on the equilibrium position.(c) 2O₃(g) ⇌ 3O₂(g) ΔH = − 285kJ
Increase in temperature: The reaction is exothermic (ΔH is negative), so an increase in temperature will shift the equilibrium to the left, favoring the formation of O₃.Decrease in volume: Decreasing the volume increases pressure. Since there are more moles of gas on the right side, the equilibrium will shift to the left to decrease the pressure, favoring the formation of O₃.(d) CaO(s) + CO₂(g) ⇌ CaCO₃(s) ΔH = −176kJ
Increase in temperature: This reaction is exothermic, so increasing temperature will shift the equilibrium to the left, favoring the formation of CaO and CO₂.Decrease in volume: Decreasing the volume increases pressure. Since CO₂ is the only gas involved, the equilibrium will shift to the right to decrease the amount of gas, favoring the formation of CaCO₃.A block of aluminum with m = 0.5 kg, T = 20oC is dropped into a reservoir at a temperature of 90oC. Calculate (a) the change in stored energy (ΔE), (b) the amount of heat transfer (Q), (c) the change in entropy (ΔS), (d) the amount of entropy transfer by heat and (e) the entropy generation (Sgen,univ) in the system's universe during the heat transfer process.
Explanation:
The given data is as follows.
m = 0.5 kg, [tex]T = 20^{o}C[/tex], [tex]T_{2} = 90^{o}C[/tex]
It is known that specific heat of aluminium is 0.91 kJ/kg.
As we know that, dQ = dU + dw
where, dQ = heat transfer
dU = change in internal energy
dw = work transfer
For the given system, work transfer "w" is 0.
(a) Hence, change in stored energy will be calculated as follows.
Q = [tex]mC \Delta T[/tex]
= [tex]0.5 \times 0.91 \times (90 - 20)[/tex]
= 31.85 kJ
(b) The amount of heat transferred will be equal to change in stored energy.
So, dQ = Q = 31.85 kJ
(c) Change in entropy will be calculated as follows.
dS = [tex]mC ln \frac{T_{2}}{T_{1}}[/tex]
= [tex]0.5 \times 0.91 \times ln \frac{90}{20}[/tex]
= 0.684 kJ/K
(d) Entropy transfer by heat will be calculated as follows.
[tex]\Delta S = \frac{dQ}{dT}[/tex]
= [tex]\frac{31.85}{(20 + 273)}[/tex]
= 0.1087 kJ/K
(e) Entropy change will be calculated as follows.
Entropy change = entropy transfer + entropy generation
[tex]S_{2} - S_{1} = \frac{dQ}{T} + S^{o}_{gen}[/tex]
0.684 kJ/K = 0.187 + [tex]S^{o}_{gen}[/tex]
[tex]S^{o}_{gen}[/tex] = 0.5752 kJ/K
An FM radio station broadcasts electromagnetic radiation at a frequency of 103.4 MHz (megahertz; MHz =106s–1). Calculate the wavelength of this radiation. The speed of light is 2.998 × 108m/s to four significantdigits.
The FM radio station broadcasting at a frequency of 103.4 MHz produces electromagnetic radiation with a wavelength of approximately 2.9 meters.
Explanation:The wave equation, which is used to calculate the wavelength of electromagnetic radiation, is given as: c = fλ where c = 3.00 × 10^8 m/s is the speed of light in vacuum, f is the frequency of the electromagnetic wave in Hz (s⁻¹) and λ is its wavelength in m.
In this scenario, the radio station is broadcasting at a frequency of 103.4 MHz, which equals 103.4 x 10^6 Hz. From the wave equation, we can rearrange and solve for the wavelength: λ = c/f. Therefore the wavelength of the radio wave is approximately λ = 3.00 ×10^8 m/s / 103.4 ×10^6 s⁻¹ = 2.9 meters. So, the FM radio broadcast at a frequency of 103.4 MHz has a wavelength of approximately 2.9 meters in free space.
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The wavelength of an FM radio station broadcasting at a frequency of 103.4 MHz is calculated by using the formula c = λv, where c is the speed of light, λ is the wavelength and v is the frequency. The frequency is converted from megahertz to hertz, and the formula is rearranged to solve for wavelength. The calculated wavelength is approximately 2.913 m.
Explanation:The question asks for the calculation of the wavelength of the radio waves being broadcasted by an FM station at 103.4 MHz. To find the wavelength, we can use the formula c = λv, where c is the speed of light, λ is the wavelength and v is the frequency.
The frequency (v) needs to be in Hz (hertz), so we must first convert the given frequency from megahertz (MHz) to hertz (Hz) - 1 MHz = 10⁸ Hz, therefore 103.4 MHz = 1.034 × 10⁸ Hz.
Then, using the speed of light, c = 2.998 × 10⁸ m/s and rearranging the formula to solve for λ (wavelength), we get λ = c/v.
Substituting the given values, λ = 2.998 × 10⁸ m/s / 1.03 × 10⁸ Hz = 2.913 m.
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What is the empirical formula for a compound whose molecular formula is P4O10
a. POb. P2O5c. P4O10d. P8O20e. PO2
Answer:
The empirical formula is P2O5 (option B)
Explanation:
An empirical formula does not necessarily represent the actual numbers of atoms present in a molecule of a compound; it represents only the ratio between those numbers.
The actual numbers of atoms of each element that occur in the smallest freely existing unit or molecule of the compound is expressed by the molecular formula of the compound.
The molecular formula of a compound may be the empirical formula, or it may be a multiple of the empirical formula.
If the molecular formula is P4O10, this means for each for P-atoms we have 10-O atoms this is a ratio 4:10 or 1: 2.5
To find the empirical formula we divide the molecular formula by 2 what will give us P2O5
For each 2 P atoms we have 5 O-atoms. This is a ratio 1: 2.5
This is the simpliest form for the compound P4O10.
The empirical formula is P2O5 (option B)
Write Lewis dot structures for the following molecules: NH3 SO2 CH3OH HNO2 N2 CH2O
Answer:
Check it below
Explanation:
1) The dots in Lewis Notation represent the Electronic Valence, in other words, the amount of Valence Electrons. The Lewis structures has the advantage of pictorially displaying the valence electrons around the symbol of the Atom.
2) We can easily find in a Periodic Table the number of bonds. Notice that we must rearrange the dots to observe the rule of the Octets. Each atom tends to have more stability and behave as noble gas, with eight electrons in its outer shell. So, in some examples, we'll have to rearrange the dots in order to have follow the Octet rule.
Check below each Lewis Structure
2.1[tex]NH_{3}[/tex] Ammonia
[tex]Valence:\\ N=5,H=1[/tex] Given by each group number.
2.2[tex]SO_{2}[/tex] Sulfur Dioxide
[tex]Valence:\\ S=6,O=2[/tex]
2.3) [tex]CH_{3}OH[/tex] Methanol
[tex]Valence:\\ C=4,O=2,H=1[/tex]
2.4) [tex]HNO_{2}[/tex]
2.5) [tex]N_{2}[/tex] Nitrogen Gas
2.6) [tex]CH_{2}O[/tex] Formaldehyde
To draw the Lewis dot structures for the given molecules, we need to determine the total number of valence electrons for each molecule and then arrange the atoms and electrons to satisfy the octet rule. The Lewis dot structures for NH3, SO2, CH3OH, HNO2, N2, and CH2O are shown with the appropriate arrangements of atoms and their valence electrons.
Explanation:To draw the Lewis dot structures for the given molecules, we first need to determine the total number of valence electrons for each molecule. For NH3, N has 5 valence electrons and each H has 1 valence electron, giving a total of 5 + 3 = 8 valence electrons. The Lewis dot structure for NH3 shows N as the central atom surrounded by three H atoms, each bonded by a single bond. Each H atom has two dots around it to represent its two valence electrons.
For SO2, S has 6 valence electrons and each O has 6 valence electrons, giving a total of 6 + 2(6) = 18 valence electrons. The Lewis dot structure for SO2 shows S as the central atom bonded to two O atoms by double bonds. Each O atom has six dots around it to represent its six valence electrons.
Continuing with CH3OH, C has 4 valence electrons, H has 1 valence electron, and O has 6 valence electrons, giving a total of 4 + 3(1) + 6 + 1 = 14 valence electrons. The Lewis dot structure for CH3OH shows C as the central atom bonded to three H atoms and one O atom. The O atom is bonded to the C atom by a single bond and has two dots around it to represent its two valence electrons.
For HNO2, H has 1 valence electron, N has 5 valence electrons, and each O has 6 valence electrons, giving a total of 1 + 5 + 2(6) = 18 valence electrons. The Lewis dot structure for HNO2 shows N as the central atom bonded to two O atoms by single bonds. Each O atom has six dots around it to represent its six valence electrons, and the H atom is bonded to one of the O atoms.
For N2, each N atom has 5 valence electrons, giving a total of 2(5) = 10 valence electrons. The Lewis dot structure for N2 shows two N atoms bonded by a triple bond, with each N atom having three dots around it to represent its three valence electrons.
Finally, for CH2O, C has 4 valence electrons, H has 1 valence electron, and O has 6 valence electrons, giving a total of 4 + 2(1) + 6 = 12 valence electrons. The Lewis dot structure for CH2O shows C as the central atom bonded to two H atoms and one O atom. The O atom is bonded to the C atom by a double bond and has four dots around it to represent its four valence electrons.
A galvanic (voltaic) cell consists of an electrode composed of titanium in a 1.0 M titanium(II) ion solution and a second electrode composed of tin in a 1.0 M tin(II) ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25C.
Answer: The standard potential for this cell is +1.49 V at 25C.
Explanation:
[tex]E^0_{[Sn^{2+}/Sn]}=-0.14V[/tex]
[tex]E^0_{[Ti^{2+}/Ti]}=-1.63V[/tex]
As titanium has lower reduction potential, it will act as anode and tin will acts as cathode.
[tex]Ti+Sn^{2+}\rightarrow Ti^{2+}+Sn[/tex]
Using Nernst equation :
[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Ti^{2+}]}{[Sn^{2+}]}[/tex]
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = [tex]25^oC=273+25=298K[/tex]
n = number of electrons in oxidation-reduction reaction = 2
[tex]E^0=E^0_{cathode}- E^0_{anode}=-0.14-(-1.63)=1.49V[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 1.49 V
[tex]E_{cell}[/tex] = emf of the cell = ?
Now put all the given values in the above equation, we get:
[tex]E_{cell}=1.49-\frac{2.303\times (8.314)\times (298)}{1\times 96500}\log \frac{1}{1}[/tex]
[tex]E_{cell}=1.49V[/tex]
The voltage of the voltaic cell is 1.49 V.
What is a voltaic cell?A voltaic cell is a cell in which electrical energy is produced by a spontaneous chemical reaction.
The equation of the cell is; Ti(s) + Si^2+(aq) -----> Ti^2+(aq) + Si(s)
E°cathode = -0.14 V
E°anode = -1.63 V
E°cell = (-0.14 V) - (-1.63 V) = 1.49 V
Using the Nernst equation;
E = E°cell - 0.0592/n log[Ti^2+]/[Si^2+]
E = 1.49 V - 0.0592/2 log(1)/(1)
E = 1.49 V
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For heat transfer purposes, a standing man can be modeled as a 30-cm-diameter, 175-cm-long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of 35 . If the rate of heat loss from this man to the environment at is 336 W, determine the rate of entropy transfer from the body of this person accompanying heat transfer, in W/K.
Answer:
Sgen = 0.0366 W/K
Explanation:
for the body:
∴ Q = - 336 W...rate of heat loss
∴ T surface = 35°C ≅ 308 K
the rate of entropy transfer from the body:
⇒ ΔS = - Q/Ts
for the enviroment:
⇒ ΔS = Q/Te
∴ assuming: T = Tenv = 25°C ≅ 298 K
resulting in a net variation in the universe:
⇒ Sgen = ΔS = Q/Tenviroment - Q/Tsurface = Q(Ts - Te)/Ts*Te
⇒ Sgen = (336( 308-298))/(308×298) = 3360 WK/91784 K² = 0.0366 W/K
The total volume in milliliters of a glucose-water solution is given by the equation below: V = 1001.93 + 111.5282m + 0.64698m^2 where m is the molality of the solution. a. The partial molar volume of glucose, is? b. Find the partial molar volume of glucose in a 0.100m solution of glucose in water.
Answer:
a. Vₐ = 111.5282 + 1.29396m
b. For m = 0.100m; Vₐ = 111.6576
Explanation:
The partial molar volume of compound A in a mixture of A and B is defined as :
[tex]V_a = \frac{dV}{dn_a}[/tex]
Where V is volume and n are moles of a.
a. As molality is proportional to moles of substance, partial molar volume of glucose can be defined as:
Vₐ = dV / dm = d(1001.93 + 111.5282m + 0.64698m²) / dm
Vₐ = 111.5282 + 1.29396m
b. Replacing for m = 0.100m:
Vₐ = 111.5282 + 1.29396×0.100
Vₐ = 111.6576
I hope it helps!
Final answer:
The partial molar volume of glucose is calculated by taking the derivative of the volume-molality equation and then evaluating at a specific molality (0.100m). The result is 111.657596 mL/mol for the partial molar volume of glucose in a 0.100m solution.
Explanation:
The question asks us to calculate the partial molar volume of glucose and then find its value in a 0.100m glucose-water solution. The partial molar volume (denoted as ϕV/ϕm) represents the change in total volume (V) with respect to the change in molality (m) of the solution, holding the amount of solvent constant. This can be computed by taking the derivative of the volume equation with respect to m:
V = 1001.93 + 111.5282m + 0.64698m2
The first derivative will yield the expression for the partial molar volume:
(ϕV/ϕm) = 111.5282 + 2×0.64698m
To calculate the partial molar volume of glucose in a 0.100m solution, simply substitute 0.100 for m in the derived expression:
(ϕV/ϕm) = 111.5282 + 2×0.64698×0.100 = 111.5282 + 0.129396 = 111.657596 mL/mol
Note that to find the mass of the solvent, water, in the solution, a calculation is needed using the molarity and density of the solution. Knowing that the mass of glucose and molality conversions are essential for such calculations, and realizing that glucose's molar mass is needed to find the mass of glucose from its molar amount.
Consider the following isotopic symbol: 137Ba2+
a. Explain how to find the number of protons
b. Explain how to find the number of electrons
c. Explain how to find the number of neutrons
d. The atomic number is not shown in the symbol. Why isn't it needed?
e. How would you write the name of this isotope?
f. Is the 137 the atomic mass or the mass number? Explain your answer.
Answer:
a) 56 protons
b) 54 electrons
c) 81 neutrons
d) The sum of protons and neutrons is shown. The number of protons is always the same. So we can calculate the number of neutrons ( and also the isotopes)
e)137Ba (with 56 protons and 81 neutrons)
f) atomic mass is 136.9 u ; the mass number is the sum of protons and neutrons and is 137
Explanation:
Step 1: Data given
137 Ba2+ is an isotope of barium. The atomic number of barium( and its isotopes) is 56. This shows the number of protons.
For a neutral atom, the number of protons is equal to the number of electrons.
The different isotopes of an element have the same number of protons but a different number of neutrons.
137Ba2+ has 56 protons (this is the same as the atomic number)
137Ba2+ has 54 electrons ( since it's Ba2+, this means it has 2 electrons less than protons, that's why it's charged +2)
137Ba2+ has 81 neutrons ( 137 - 56 = 81)
In the symbol, the atomic number is not shown. The sum of the protons and neutrons is shown. (Since the number of protons is the same for every isotope, we can calculate the number of neutrons that way. By knowing the neutrons, we also know the isotope.
This isotope is 137Ba
Atomic mass is also known as atomic weight. The atomic mass is the weighted average mass of an atom of an element based on the relative natural abundance of that element's isotopes.
The atomic mass of 137Ba2+ is 136.9 u
The mass number is a count of the total number of protons and neutrons in an atom's nucleus.
The mass number of 137Ba2+ is 137
The relative rates of reaction of ethane, toluene, and ethylbenzene with bromine atoms have been measured.
The most reactive hydrocarbon undergoes hydrogen atom abstraction a million times faster than does the least reactive one.
Arrange these hydrocarbons in order of decreasing reactivity.
Answer:
Reactivity: Ethyl benzene > Toluene > Ethane
Explanation:
The order of reactivity depends on how easily the most reactive hydrogen can be abstracted.
The order of reactivity of hydrogen is : 1° < 2° < 3° < Benzyllic based on the stabilizing effects like inductive effect, hyperconjugation , resonance effect.
In ethane, all the hydrogen present are 1°.In toluene, the most reactive hydrogen is benzyllic.In Ethyl Benzene, the most reactive hydrogen is 2° as well as benzyllic.Hence the order of decreasing reactivity : Ethyl Benzene, Toluene , Ethane.
To sum up, ethylbenzene > toluene > ethane is the sequence in which reactivity decreases with bromine, and the previously described components are critical in dictating the rates of reaction.
Hydrocarbons containing bromine atoms exhibit varying degrees of reactivity, which can be attributed mainly to the stability of the free radical generated during hydrogen atom abstraction. When an unpaired electron delocalizes across the aromatic ring, a more stable benzyl radical is produced upon abstraction from a benzylic carbon in toluene and ethylbenzene. Compared to toluene, the benzyl radical generated from ethylbenzene has somewhat more reactivity due to this delocalization. Ethane has the lowest reactivity of the three compounds since it doesn't have an aromatic ring or benzylic hydrogens, which leads to the formation of a primary free radical that is much less stable.
A solid is hard brittle and electrically nonconducting. it's melt ( the liquid form of the substance) and an aqueous solution containing the substance conduct electricity. classify solid.
Answer: ionic solid
Explanation:
In an ionic solid, the ions are bound together by strong electrostatic attraction hence they are immobile and the solid is unable to conduct electricity. If this solid is dissolved in water, the ions move apart due to solvation and become mobile hence the solution conduts electricity. Similarly, when the solid melts, the ions also become free and the melt conduct electricity.
A solid that is hard, brittle, does not conduct electricity in solid form but does in liquid form or when dissolved in water, and has a high melting point is classified as an ionic solid.
Explanation:The characteristics of the solid described are indicative of an ionic solid. These solids are typically hard and brittle, and they have high melting points. As a solid, ionic compounds do not conduct electricity because the ions are locked in place within the crystal lattice and thus cannot move freely. When these compounds melt, however, the ionic lattice breaks down and the ions are free to move, allowing the liquid to conduct electricity.
Similarly, when an ionic compound is dissolved in water, it dissociates into ions, which are free to move in the solution, making the solution an electrical conductor. This is because an electrolyte is present, which is a substance that contains free ions and can behave as an electrical conductor.
The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant.2 SO2(g) + O2(g) ↔ 2 SO3(g) Kc = 1.7 × 106SO3(g) ↔ 1/2 O2(g) + SO2(g) Kc = ?1.3 × 10^31.2 × 10^-68.57.7 × 10^-43.4 × 10^2
Answer:
The value of equilibrium constant for reverse reaction is [tex]7.7\times 10^{-4}[/tex]
Explanation:
The given chemical equation follows:
[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]
The equilibrium constant for the above equation is [tex]1.7\times 10^6[/tex].
We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:
[tex]2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)[/tex]
The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.
If the equation is multiplied by a factor of '1/2', the equilibrium constant of the reverse reaction will be the square root of the equilibrium constant of initial reaction.
So,
[tex]SO_3(g)\rightarrow SO_2(g)+\frac{1}{2} O_2(g)[/tex]
The value of equilibrium constant for half reverse reaction is:
[tex]K_{eq}'=(\frac{1}{1.7\times 10^6})^{\frac{1}{2}}=0.00077=7.7\times 10^{-4}[/tex]
Hence, the value of equilibrium constant for reverse reaction is [tex]7.7\times 10^{-4}[/tex]
Answer:
[tex]7.7\times10^{-4}[/tex]
Explanation:
The equation for which we have to find Kc is obtained by two - step transformation of the equation whose Kc is given.
1st step:
Reversing the reaction:
By reversing the reaction the reactants become products and vice-versa.
The new equilibrium constant will be:
[tex]Kc^{'}=\frac{1}{Kc}[/tex]
2nd step:
Dividing the equation throughout by 2:
New Kc becomes:
[tex]Kc^{''}=\sqrt{Kc^{'}}=\frac{1}{\sqrt{Kc} }[/tex]
[tex]=\frac{1}{\sqrt{1.7\times10^{6} } }=7.7\times10^{-4}[/tex]
Hence the equilibrium constant is [tex]7.7\times10^{-4}[/tex]
If I have a protein with the sequence of amino acids below, what type of structure is being described? Met-Leu-Pro-Ile-His-Ala-Leu
Answer:
Protein's primary structure
Explanation:
The proteins have four different types of structures:
Primary: Describes the sequence of amino acids that is unique for each protein
Secondary: Makes reference to the proteins 3D geometry, it can be alpha-helix (product of the protein coiling) or beta-plated sheet (product of the protein folding).
Tertiary: Refers to the comprehensive 3D structure of the protein. There are many types of tertiary structures, due to hyfrophobic interactions, hydrogen unions, ionic unions.
Quaternary: Resambles the macromolecule formed by many protein molecules. Not all proteins have quaternary structure.
The nuclide As-76 has a half-life of 26.0 hours. If a sample of As-76 weighs 344 g, what mass of As-76 remains after 538 minutes?271 g67.8 g144 g437 g251 g
Answer:
271g
Explanation:
The full explanation is seen in the image attached. See the solution below for details
Choose a depiction of a gas sample containing equal molar amounts of xenon and argon as described by kinetic molecular theory. Red dots are used to represent xenon atoms and blue dots to represent argon atoms. Each atom is drawn with a "tail" that represents its velocity relative to the others in the mixture.
Answer:
On the attached picture.
Explanation:
Hello,
At first, it is important to remember that kinetic molecular theory help us understand how the molecules of a gas behave in terms of motion. In such a way, the relative velocity of a gas molecule has the following relationship with the gas' molar mass:
[tex]V[/tex]∝[tex]\frac{1}{\sqrt{M} }[/tex]
That is, an inversely proportional relationship which allows us to infer that the bigger the molecule the slower it. In this manner, as argon is smaller than xenon, it will move faster.
Now, as the gases are in equal molar amounts and considering that argon moves faster, on the attached picture you will find the suitable depiction of the gas sample, since red dots (argon) have a larger tail than the blue dots (xenon).
Best regards.
The kinetic molecular theory explains gas behavior, showing that at a given temperature, heavier molecules like xenon move slower than lighter molecules like argon, which can be depicted with varying tail lengths in visual models.
Explanation:Understanding Gases through Kinetic Molecular TheoryThe kinetic molecular theory of gases provides an explanation for the properties of gases by modeling them as small, hard spheres with insignificant volume, in constant motion, and undergoing perfectly elastic collisions. According to this theory, the average kinetic energy (KEavg) of gas molecules is the same for all gases at a given temperature, regardless of the molecular mass. However, because the kinetic energy depends only on temperature, lighter molecules will have higher speeds compared to heavier molecules at the same temperature.
Given a gas sample containing equal molar amounts of xenon and argon, depicted by kinetic molecular theory, we would see red dots (xenon) and blue dots (argon) with tails representing their velocities. As the diagrams from the theory suggest, we would expect that, at the same temperature, xenon atoms (being heavier) would have shorter tails (indicating lower speeds) than argon atoms (which are lighter and thus would have longer tails for higher speeds).
This behavior of the molecules can be seen in the average root mean square speed (Urms) trend, where heavier noble gases like xenon show a distribution of speeds peaking at lower values, whereas lighter ones like argon peak at higher speeds. This concept is crucial in the depiction of gas samples in kinetic molecular theory and can be visualized through illustrations that incorporate this difference in molecular speed based on the mass of the gas particles.
Nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. The Swedish chemist Alfred Nobel (1833-1896) founded the Nobel Prizes with a fortune he made by inventing dynamite, a mixture of nitroglycerin and inert ingredients that was safe to handle. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of liquid nitroglycerin () into gaseous dinitrogen, gaseous dioxygen, gaseous water and gaseous carbon dioxide. 2. Suppose of carbon dioxide gas are produced by this reaction, at a temperature of and pressure of exactly . Calculate the mass of nitroglycerin that must have reacted. Be sure your answer has the correct number of significant digits.
Answer:
4 C3H5N3O9 ------> 6N2 + O2 + 10H2O + 12CO2
Explanation:
Nitroglycerin has a chemical formula C3H5N3O9. The balanced chemical equation is as follows:
4 C3H5N3O9 ------> 6N2 + O2 + 10H2O + 12CO2
We suppose that in a reaction, 44g of carbon dioxide is produced. The mass of nitroglycerin that must have reacted will be calculated as under:
Molecular mass of Nitroglycerin = 227g/mol
Molecular mass of Carbon dioxide = 44g/mol
No. of moles of carbon dioxide produced = 44/44 = 1 mole produced.
Now, from balanced chemical equation, we can see that
12 moles of carbon dioxide are produced by = 4 moles of nitroglycerin.
1 mole of carbon dioxide is produced by = 4/12 = 1/3 moles of nitroglycerin.
Mass of nitroglycerin which produced 1 mole of carbon dioxide = 1/3 x 227 = 75.666 grams.
The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H 2 O H2O that can be produced by combining 62.8 g 62.8 g of each reactant? 4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Answer:
42,3g of H₂O
Explanation:
For the reaction:
4NH₃(g) + 5O₂(g) ⟶ 4NO(g) + 6H₂O(g)
62,8 g of NH₃ are:
62,8g×(1mol/17,031g) = 3,69 moles of NH₃
62,8 g of O₂ are:
62,8g×(1mol/32g) = 1,96 moles of O₂
For a complete reaction of 1,96 moles of O₂ you need:
1,96mol O₂×(4mol NH₃ / 5molO₂) = 1,57 moles NH₃. As you have 3,69 moles, limiting reactant is O₂.
Assuming a complete reaction, 1,96mol O₂ produce:
1,96mol O₂×(6mol H₂O / 5molO₂) = 2,35 moles of H₂O. In grams:
2,35 moles of H₂O×(18,01g/1mol) = 42,3g of H₂O
I hope it helps!
The equation below shows the thermite reaction, which has often been used for welding iron.
8 Al (s) + 3 Fe3O4 (s) ----------> 4 Al2O3 (s) + 9 Fe(s)
ΔH = -3350*KJ/mol rxn
1. It is highly exothermic; the reacting mixture can reach temperatures as high as 3000°C. 8 Al(s) + 3 Fe3O4(s) ® 4 Al2O3(s) + 9 Fe(s) DH° = –3350 kJ How much heat would be released by the reaction of 47.6 g of Al with 69.12 g of Fe3O4?
Answer:
335 Joules kJ of heat will be released
Explanation:
Given the balanced equation:
8 Al (s) + 3 Fe3O4 (s) ----------> 4 Al2O3 (s) + 9 Fe(s),
ΔH = -3350*KJ/mol rxn
This is the heat released when 8 moles of Al react with 3 mol Fe3O4.
We then need to calculate the moles of reactants, verify if there is a limiting reagent and proceed to answer the question based on the soichiometry of the reaction.
Atomic weight Al = 26.98 g/mol Molecular Weight Fe3O4 = 231.53 g/mol
mol Al = 47.6 g/26.98 g/mol = 1.76 mol
mol Fe3O4 = 69.12 g/ 231.53 g/mol = 0.30 mol
Limiting reagent calculation:
8 mol Al / 3 mol Fe3O4 x 0.30 mol Fe3O4 = 0.80 mol Al are required and we have 1.76 mol, therefore Fe3O4 is the limiting reagent
Amount of Heat
-3350 kJ/ 3 mol Fe3O4 x 0.30 mol Fe3O4 = -335.00 kJ
In a separate experiment beginning from the same initial conditions, including a temperature Ti of 25.0°C, half the number of moles found in part (a) are withdrawn while the temperature is allowed to vary and the pressure undergoes the same change from 25.7 atm to 4.10 atm. What is the final temperature (in °C) of the gas?
Final answer:
The problem involves physics principles, specifically the ideal gas law. To solve for the final temperature of a gas when pressure changes and the number of moles is halved from an initial condition, the relationship between pressure and temperature must be considered. However, the type of process (isothermal, isobaric, adiabatic) must be known for an accurate calculation.
Explanation:
The student's question involves finding the final temperature of a gas when the pressure changes and the number of moles is reduced by half, starting from an initial temperature (Ti) of 25.0°C. This problem can be solved by applying the ideal gas law and the concept that, for a given amount of gas, the pressure is directly proportional to the temperature (P ∝ T) when volume and the number of moles are constant. Given the initial conditions and the pressure change from 25.7 atm to 4.10 atm, the process is not specified as isothermal, isobaric, or adiabatic; therefore, additional details from the context of the part (a) of the experiment would be required to provide a comprehensive solution.
To find the final temperature based on the given information, one would have to assume the same type of process that occurred in part (a), where volume doubled and pressure got halved. If we assume a similar relationship between temperature and pressure as was demonstrated before, where if the pressure is halved from 2.50 atm, the temperature must also be halved from 303.15 K, we could calculate the final temperature for the new conditions by adjusting for the fact that the number of moles was halved. However, without explicit mention of whether this happens in an isothermal, isobaric, or adiabatic process, a direct calculation cannot be provided here.
Nitrogen and oxygen can react to form various compounds.
Two experiments showed that one compound is formed when 3.62 g of nitrogen and 2.07 g of oxygen react completely, while another compound is formed when 1.82 g of nitrogen reacts completely with 4.13 g of oxygen.
Which of the following are most likely the molecular formulas for the nitrogen oxides obtained in these experiments?(1) NO, N2O(2) NO, NO2(3) N2O, N2O5(4) NO, N2O4(5) N2O, N2O4
Answer:
5- N20 and N2O5
Explanation:
Full working is shown in the image attached. It is important to remember that NO2 dimerizes to N2O4 while N2PO occurs as monomers.
An electrochemical cell is constructed such that on one side a pure nickel electrode is in contact with a solution containing Ni2+ ions at a concentration of 3 × 10−3 M. The other cell half consists of a pure Fe electrode that is immersed in a solution of Fe2+ ions having a concentration of 0.1 M. At what temperature will the potential between the two electrodes be +0.140 V?
Answer: The temperature at which given potential between the two electrodes is attained is 331.13 K
Explanation:
The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction.
The half reaction follows:
Oxidation half reaction: [tex]Fe(s)\rightarrow Fe^{2+}(0.1M)+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V[/tex]
Reduction half reaction: [tex]Ni^{2+}(3\times 10^{-3}M)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.25V[/tex]
Net reaction: [tex]Fe(s)+Ni^{2+}(3\times 10^{-3}M)\rightarrow Fe^{2+}(0.1M)+Ni(s)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Putting values in above equation, we get:
[tex]E^o_{cell}=-0.25-(-0.44)=0.19V[/tex]
To calculate the temperature at which the reaction is taking place, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Fe^{2+}]}{[Ni^{2+}]}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = +0.140 V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.19 V
n = number of electrons exchanged = 2
R = Gas constant = 8.314 J/mol.K
F = Faraday's constant = 96500
T = temperature of the reaction
[tex][Fe^{2+}]=0.1M[/tex]
[tex][Ni^{2+}]=3\times 10^{-3}M[/tex]
Putting values in above equation, we get:
[tex]0.140=0.19-\frac{2.303\times 8.314\times T}{2\times 96500}\times \log(\frac{(0.1)}{(3\times 10^{-3})})\\\\T=331.13K[/tex]
Hence, the temperature at which given potential between the two electrodes is attained is 331.13 K
The temperature at which the potential between the two electrodes will be +0.140 V can be calculated using the Nernst equation. Rearranging the equation allows us to solve for temperature. Substituting the given values allows us to find the temperature.
Explanation:First, we need to find the cell potential using the Nernst equation:
Ecell = E°cell - (0.0592 V / n) * log(Q)
In this case, since both sides of the cell are based on the same half-reaction with different concentrations, the number of electrons transferred (n) is 2. The equilibrium constant (Q) can be calculated using the concentrations of Ni2+ and Fe2+ ions in the half-cells. Rearranging the equation, we can solve for temperature (T):
T = (Ecell - E°cell) / ((0.0592 V / n) * log(Q))
Substituting the given values, we can solve for T.
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Which one of the following solutes is most likely to have low water solubility due to the dissolution process being highly endothermic?
Al2O3
RbF
CaF2
AgCl
FeCl2
Al2O3, or aluminum oxide, is most likely to have low water solubility due to the dissolution process being highly endothermic. It requires more energy to break its intermolecular forces and disperse into water which results in a low solubility.
Explanation:The solute most likely to have low water solubility due to the dissolution process being highly endothermic is Al2O3 (aluminum oxide). In chemical reactions, an endothermic process involves the absorption of heat. Solutes like Al2O3 need more energy to overcome intermolecular forces and disperse into the solvent, thus making the dissolution process highly endothermic and resulting in lower solubility in water.
On the other hand, solutes like RbF, CaF2, AgCl, and FeCl2 generally require less energy to dissolve in water, making their dissolution process less endothermic and more favorable under normal environmental conditions.
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The electron cloud of HF is smaller than that of F2, however, HF has a much higher boiling point than F2 has. Which of the following explains how the dispersion-force model of intermolecular attraction does not account for the unusually high boiling point of HF?
A. F2 is soluble in water, whereas HF is insoluble in water.
B. The F2 molecule has a greater mass than the HF molecule has.
C. Liquid F2 has weak dispersion force attractions between its molecules, whereas liquid HF has strong ionic interactions between H+ and F- ions.
D. Liquid F2 has weak dispersion force attractions between its molecules, whereas liquid HF has both weak dispersion force attractions and hydrogen bonding interactions between its molecules.
Answer:D
Explanation:
The high boiling point of HF is not attributable to the dispersion forces mentioned in the question. In HF, a stronger attraction is in operation, that is hydrogen bonding. This ultimately accounts for the high boiling point and not solely the dispersion model as in F2.
The unusually high boiling point of HF compared to F2 is due to the strong hydrogen bonding interactions between HF molecules.
Explanation:The correct answer is D. Liquid F2 has weak dispersion force attractions between its molecules, whereas liquid HF has both weak dispersion force attractions and hydrogen bonding interactions between its molecules. Dispersion forces are a type of intermolecular force that occurs between all molecules, regardless of polarity.
However, these forces are generally weaker than other types of intermolecular forces such as hydrogen bonding. In HF, the significant electronegativity difference between hydrogen and fluorine leads to the formation of a polar covalent bond, which makes the HF molecules capable of hydrogen bonding, a stronger intermolecular force.
This hydrogen bonding results in a much higher boiling point for HF as compared to F2, which can only interact with other F2 molecules via relatively weaker dispersion forces.
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Rank the following in order of increasing surface tension at a given temperature, and explain your ranking: a. CH3CH2CH2OH b. HOCH2CH(OH)CH2OH c. HOCH2CH2OH.
Explanation:
Surface tension is defined as the attractive forces experienced by the surface molecules of a liquid by the molecules present beneath the surface layer of the liquid.
In a molecule, more is the number of hydroxyl groups present in it more will be the force of attraction faced by surface molecules towards the inner side of the liquid.
Hence, then the surface tension will increase.
Compoud (a) has only one -OH group, compound (b) has 3 -OH groups and compound (c) has 2 -OH groups.
Therefore, the given compounds are arranged on the basis of order of increasing surface tension at a given temperature are as follows.
[tex]CH_3CH_2CH_2OH[/tex] < [tex]HOCH_2CH_2OH[/tex] < [tex]HOCH_2CH(OH)CH_2OH[/tex]
The surface tension in the following molecules in the increasing order is given as, [tex]\rm CH_2CH_2CH_2OH < HOCH_2CH_2OH < HOCH_2CH(OH)CH_2OH[/tex].
What is surface tension?Surface tension can be given as the force applied by the molecules underneath the liquid surface to form the attraction and made the liquid occupy, the least surface area.
The compounds with a more number of attracting groups such as hydroxyl radicals tend to apply more attraction and have more surface tension.
The number of hydroxyl groups in the following compounds are:
[tex]\rm CH_3CH_2CH_2OH=1[/tex][tex]\rm OHCH_2CH(OH)CH_2OH=3[/tex][tex]\rm OHCH_2CH_2OH=2[/tex]The increasing order of the surface tension in the following molecules is, [tex]\rm CH_2CH_2CH_2OH < HOCH_2CH_2OH < HOCH_2CH(OH)CH_2OH[/tex].
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