Answer:
-767,2kJ
Explanation:
It is possible to sum enthalpies of half-reactions to obtain the enthalpy of a global reaction using Hess's law. For the reactions:
1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁= −241.8 kJ
2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂= +361.7 kJ
3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl (g) ΔH₃= −92.3 kJ
4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄= − 607.9 kJ
5) H₂O(g) ⟶ H₂O(l) ΔH₅= − 44.0 kJ
The sum of (4) - (2) produce:
6) XCl₄(s) + O₂(g) ⟶ XO₂(s) + 2Cl₂(g) ΔH₆ = ΔH₄ - ΔH₂ = -969,6 kJ
(6) + 4×(3):
7) XCl₄(s) + 2H₂(g) + O₂(g) ⟶ XO₂(s) + 4HCl(g) ΔH₇ = ΔH₆ + 4ΔH₃= -1338,8 kJ
(7) - 2×(1):
8) XCl₄(s) + 2H₂O(g) ⟶ XO₂(s) + 4HCl(g) ΔH₈ = ΔH₇ - 2ΔH₁= -855,2kJ
(8) - 2×(5):
9) XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g) ΔH₉ = ΔH₈ - 2ΔH₅= -767,2kJ
I hope it helps!
To calculate the enthalpy change for the reaction XCl4 (s) + 2 H2O (l) ⟶ XO2 (s) + 4 HCl (g), we can use Hess's law and the enthalpy values of the given reactions.
Explanation:The enthalpy change for the reaction XCl4 (s) + 2 H2O (l) ⟶ XO2 (s) + 4 HCl (g) can be calculated using Hess's law and the enthalpy values of the given reactions.
To represent the desired reaction, we can combine reactions 2, 3, 4, and 5 as follows:
X (s) + 2 Cl2 (g) ⟶ XCl4 (s) [multiply this reaction by 2] 1/2 H2 (g) + 1/2 Cl2 (g) ⟶ HCl (g) [multiply this reaction by 4] X (s) + O2 (g) ⟶ XO2 (s) [multiply this reaction by 2] H2 (g) + 1/2 O2 (g) ⟶ H2O (l)By summing these equations, we get the desired equation:
XCl4 (s) + 2 H2O (l) ⟶ XO2 (s) + 4 HCl (g)
--
The enthalpy change for the reaction is ΔH = ΔH2 + ΔH3 + ΔH4 + ΔH5.
Equilibrium Position -- A + B = C -- Effect of Dilution
Iodine is sparingly soluble in pure water. However, it does `dissolve' in solutions containing excess iodide ion because of the following reaction: I-(aq) + I2(aq)= I3-(aq) K = 710. For each of the following cases calculate the equilibrium ratio of [I3-] to [I2]. 6.00×10-2 mol of I2 is added to 1.00 L of 6.00×10-1 M KI solution.
The solution above is diluted to __.
To calculate the equilibrium ratio of [I3-] to [I2], first calculate the moles of I3- formed from the moles of I2 added. Then, calculate the concentrations of I3- and I2 in the diluted solution. Finally, divide the concentration of I3- by the concentration of I2 to get the equilibrium ratio.
Explanation:In this problem, we are given the concentration of a solution of KI and the amount of I2 added to the solution. We are asked to calculate the equilibrium ratio of [I3-] to [I2]. To solve this problem, we need to use the K value and the stoichiometry of the reaction.
Calculate the moles of I3- formed from the moles of I2 added.Calculate the concentrations of I3- and I2 in the diluted solution.Finally, divide the concentration of I3- by the concentration of I2 to get the equilibrium ratio.Learn more about equilibrium position here:https://brainly.com/question/31576470
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Final answer:
Calculating the equilibrium ratio of [I3-] to [I2] involves applying the equilibrium constant and concentration changes due to reaction, although the exact effect of dilution cannot be determined without the specified dilution factor.
Explanation:
To calculate the equilibrium ratio of [I3-] to [I2] when 6.00×[tex]10^{-2}[/tex] mol of I2 is added to 1.00 L of 6.00×[tex]10^{-1}[/tex] M KI solution, we first recognize the reaction I-(aq) + I2(aq) ⇒ I3-(aq) with a given equilibrium constant (K) of 710. Initially, [I-] is 0.6 M, and [I2] is 0.06 M. Let x be the change in concentration of I2 and I3- at equilibrium. Hence, [I2] at equilibrium becomes (0.06 - x) M, and [I3-] is x M.
The equilibrium expression for the reaction is K = [I3-]/([I-][I2]). Substituting the known values and solving for x will provide the equilibrium concentrations, allowing us to calculate the desired ratio of [I3-] to [I2]. Unfortunately, the exact dilution factor is not specified in the question, which would be necessary to calculate the effect of dilution accurately.
Picture (a) Picture (a) Dipole-dipole forces London dispersion forces Ion-dipole forces Hydrogen bonding Request Answer Part B Picture (b) Picture (b) Hydrogen bonding Dipole-dipole forces London dispersion forces Ion-dipole forces Request Answer Part C Picture (c) Picture (c) Ion-dipole forces Dipole-dipole forces Hydrogen bonding London dispersion forces Request Answer Part D Picture (d) Picture (d) Dipole-dipole forces Hydrogen bonding Ion-dipole forces London dispersion forces Request Answer Part E Predict which of the four interactions is the weakest Predict which of the four interactions is the weakest Forces on the picture (a). Forces on the picture (b). Forces on the picture (c). Forces on the picture (d).
Answer:
(a) Hydrogen bonding
(b) Dispersion forces
(c) Ion-dipole forces
(d) Dipole-dipole forces
Ion-dipole forces (c) are the strongest of the 4 interactions while dispersion forces are the weakest (b).
Explanation:
The picture is missing but I think is the one that I'm uploading.
Picture (a)
HF is a polar molecule with a high difference in electronegativity between H and F. As a consequence, the force between HF molecules is Hydrogen bonding.
Picture (b)
In picture (b) we have F₂ molecules, which are nonpolar due to their atoms having the same electronegativity. The forces between nonpolar molecules are dispersion forces.
Picture (c)
Na⁺ is an ion and H₂O a dipole. Therefore, they experience ion-dipole forces.
Picture (d)
SO₂ molecules are polar, that is, they form dipoles and experience dipole-dipole forces.
Ion-dipole forces (c) are the strongest of the 4 interactions while dispersion forces are the weakest (b).
Final answer:
The strength of intermolecular forces varies greatly, affecting physical properties of substances. London dispersion forces are the weakest followed by dipole-dipole interactions, hydrogen bonding, and ion-dipole forces being the strongest.
Explanation:
Understanding the nature and strength of intermolecular forces is essential to explain many physical properties of substances, such as boiling points, melting points, and solubilities. Intermolecular forces can be categorized into several types, including London dispersion forces, dipole-dipole interactions, hydrogen bonding, and ion-dipole forces.
London Dispersion Forces
London dispersion forces are the weakest intermolecular force and occur between all molecules, polar or nonpolar. They result from the instantaneous distribution of electrons in one molecule inducing a dipole in a neighboring molecule. These forces are significant in molecules with a high number of electrons and become stronger as molecular size increases.
Dipole-Dipole Interactions
Dipole-dipole interactions occur between polar molecules where positive ends of molecules are attracted to negative ends of other molecules. While stronger than London dispersion forces, they are not as strong as hydrogen bonds.
Hydrogen Bonding
Hydrogen bonding is a special type of dipole-dipole interaction that occurs when a hydrogen atom bonded to a highly electronegative atom, like oxygen, nitrogen, or fluorine, is attracted to an electronegative atom in another molecule. Hydrogen bonding is significantly stronger than both London dispersion forces and general dipole-dipole interactions, contributing to the unique properties of water and other substances.
Ion-Dipole Forces
Ion-dipole forces occur between an ion and a polar molecule and are significant in mixtures of ionic substances with polar solvents. These are stronger than hydrogen bonds and are crucial for the solubility of ionic compounds in water.
To rank these interactions from weakest to strongest: London dispersion forces, dipole-dipole interactions, hydrogen bonding, and ion-dipole forces.
For a galvanic cell that uses the following two half-reactions, Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l) Pb(s) → Pb2+(aq) + 2 ehow many moles of Pb(s) are oxidized by three mol es of Cr2O72-?
A) 3
B) 6
C) 9
D) 18
Answer:
C
Explanation:
The detailed solution is found in the image attached. It is necessary to note that the oxidation half equation is multiplied by three to balance electron gain and loss. This is adequately shown in the image below. Inferences are only drawn from balanced redox reaction equation hence the first step is to balance the redox reaction equation.
Every night when the sun sets, the air temperature decreases. How does the relative humidity change (increase, decrease, or stay the same) when the air temperature decreases?
As the air temperature decreases, the relative humidity generally increases because the air's capacity to hold water vapor also reduces. On evenings when the temperature reduces enough to reach the dew point, fog might form as a result. This relationship between temperature and humidity also affects the likelihood of condensation and the range of temperature fluctuations in different regions.
Explanation:The term humidity often refers to relative humidity, which indicates how much water vapor is present in the air compared to the maximum possible. This maximum relies on the air's temperature: as the air temperature decreases, the amount of water vapor the air can hold also decreases. Consequently, if the quantity of water vapor stays the same, the relative humidity will increase.
In the evening, when the air temperature declines, the relative humidity usually rises, sometimes to the point of reaching the dew point, the temperature at which the relative humidity is 100% and fog can form due to the condensation of small water droplets that stay in suspension. When the dew point is below 0°C, a greater possibility of freezing temperatures exists, which is a concern for farmers. In arid regions, the low humidity means low dew-point temperatures, hence, condensation is unlikely, resulting in a larger range of temperature fluctuations compared to regions with higher humidity.
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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.
A Cu-30% Zn alloy tensile bar has a strain hardening coefficient of 0.50. The bar, which has an initial diameter of 1 cm and an initial gage length of 3 cm, fails at an engineering stress of 120 MPa. At the moment of fracture, the gage length is 3.5 cm and the diameter is 0.926 cm. No necking occurred. Calculate the true stress when the true strain is 0.05 cm/cm.
Answer:
79.74*10^6 Pa
Explanation:
Based on the parameters provided, we have:
ε[tex]_{t}[/tex] = ln([tex]l_{f}/l_{i}[/tex])
Where initial gauge length = 3 cm and the final gauge length is 3.5 cm. Therefore:
ε[tex]_{t}[/tex] = ln(3.5/3) = ln(1.167) = 0.154
Similarly,
σ[tex]_{E}[/tex] = F/[3.142*(di^2)/4]
Where σ[tex]_{E}[/tex] = 120*10^6 Pa and di = 1 cm = 0.01 m
Therefore,
F = 120*10^6 * 3.142*(0.01^2)/4 = 9426 N
σ[tex]_{t}[/tex] = F/[3.142*(df^2)/4 = 9426/[3.142*(0.00926^2)/4 = 9426/6.74*10^-5 = 139.95*10^6 Pa
σ[tex]_{t}[/tex] = k*ε[tex]_{t} ^{0.5}[/tex] = 139.95*10^6
k = 139.95*10^6/(0.154)^0.5 = 356.63*10^6 Pa
Therefore, when ε[tex]_{t}[/tex] = 0.05 cm/cm
σ[tex]_{t}[/tex] = 356.63*10^6 (0.05)^0.5 = 79.74*10^6 Pa
The true stress when the true strain is 0.05 cm/cm is approximately 122.97 MPa.
The true stress at a true strain of 0.05 cm/cm for the Cu-30% Zn alloy tensile bar can be calculated using the relationship between true stress and true strain, which is given by the equation:
[tex]\[ \sigma_{true} = \sigma_{eng}(1 + \epsilon_{true}) \][/tex]
where [tex]\( \sigma_{true} \)[/tex] is the true stress, [tex]\( \sigma_{eng} \)[/tex] is the engineering stress at the moment of fracture, and [tex]\( \epsilon_{true} \)[/tex] is the true strain.
Given that the engineering stress at failure is 120 MPa, we can calculate the true stress at the given true strain of 0.05 cm/cm as follows:
Now, we can calculate the true stress using the given strain hardening coefficient [tex]\( n = 0.50 \)[/tex] and the true strain [tex]\( \epsilon_{true} = 0.05 \)[/tex]:
[tex]\[ \sigma_{true} = \sigma_{eng}(1 + \epsilon_{true})^n \][/tex]
[tex]\[ \sigma_{true} = 120 \text{ MPa} \times (1 + 0.05)^{0.50} \][/tex]
[tex]\[ \sigma_{true} = 120 \text{ MPa} \times (1.05)^{0.50} \][/tex]
[tex]\[ \sigma_{true} = 120 \text{ MPa} \times 1.0247 \][/tex]
[tex]\[ \sigma_{true} = 122.97 \text{ MPa} \][/tex]
Part D 2ClO2(g)+2I−(aq)→2ClO−2(aq)+I2(s) Drag the appropriate labels to their respective targets. ResetHelp e−→e Superscript- rightarrow ←e−leftarrow e Superscript- Cathode Cathode Anode Anode II Superscript- ClO−2C l O Subscript 2 Superscript- Request Answer Part E Indicate the half-reaction occurring at Anode. Express your answer as a chemical equation. Identify all of the phases in your answer. nothing
Answer: The half reaction occurring at anode is [tex]2I^-(aq.)\rightarrow I_2(s)+2e^-[/tex]
Explanation:
The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction.
For the given chemical equation:
[tex]2ClO_2(g)+2I^-(aq)\rightarrow 2ClO^{-2}(aq.)+I_2(s)[/tex]
The half reaction follows:
Oxidation half reaction: [tex]2I^-(aq.)\rightarrow I_2(s)+2e^-;E^o_{I_2/I^-}=0.53V[/tex]
Reduction half reaction: [tex]ClO_2(g)+e^-\rightarrow ClO_2^-(aq.);E^o_{ClO_2/ClO_2^-}=+0.954V[/tex] ( × 2 )
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
Hence, the half reaction occurring at anode is [tex]2I^-(aq.)\rightarrow I_2(s)+2e^-[/tex]
The half reaction occurring at anode is:
[tex]2I^-(aq)---- > I_2(s)+2e^-[/tex]
Half reaction for the cell:The substance having highest positive potential will always get reduced and will undergo reduction reaction.
Balanced chemical equation:
[tex]2ClO_2(g)+2I^-(aq)----- > 2ClO^{2-}(aq)+I_2(s)[/tex]
The half reaction follows:
Oxidation half reaction: [tex]2I^-(aq)---- > I_2(s)+2e^-[/tex] , Reduction potential is 0.53V
Reduction half reaction: [tex]ClO_2(g)+e^----- > ClO_2^-[/tex] ( × 2 ), Oxidation potential is +0.954 V
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
Hence, the half reaction occurring at anode is :
[tex]2I^-(aq)---- > I_2(s)+2e^-[/tex]
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2) Compound A is neutral and Compound B is acidic. Both are water-insoluble solids. A and B are dissolved in dichloromethane (DCM) and extracted with aqueous base. The layers are then separated. What must be done to obtain the compound in the aqueous layer?
(A) Acidify and evaporate
(B) Acidify and vacuum filter
(C) Add Sodium Sulfate and Evaporate
(D) Add Sodium Sulfate and vacuum filter
Answer:
answer is a
Explanation:
Acidify and evaporate
A certain substance, X, has a triple-point temperature of 20°C at a pressure of 2.0 atm. Which one of the statements cannot possibly be true?
A. X can exist as a liquid above 20°C.
B. X can exist as a solid above 20°C.
C. Liquid X can exist as a stable phase at 25°C, 1 atm.
D. Both liquid and solid X have the same vapor pressure at 20°C.
E. All of the above are true.
Answer:
Option B is incorrect.
Explanation:
Observe the graph that has been attached along with the answer, that is the general representation of a substance approaching and during triple point phase.
Now, checking the options one by one
A. We verify that a substance can be in liquid form above its triple point, which in this case is 20°C. So this is a true statement.
B.This is a false statement because all the solid state is exhibited before the triple point temperature ie. 20°C. So, false.
C. By stable phase, we mean that a substance exhibits only one state. On increasing the temperature and lowering the pressure, we see the only state possible is the vapor phase. So this statement has to be true.
D. This is a true statement and it is a fact.
E. All the statements are not true so this is not the correct option.
Therefore, only option B is an incorrect option.
The statement that 'X can exist as a solid above 20°C' cannot be true. Given that the substance has a triple-point temperature of 20°C, it implies that the substance cannot exist as a solid above this temperature if it is a typical substance that expands upon heating.
Explanation:The triple point of a substance is a specific temperature and pressure at which the three phases (solid, liquid, and gas) of the substance coexist in thermodynamic equilibrium. Given that substance X has a triple-point temperature of 20°C, it means that at this temperature and at a pressure of 2.0 atm, solid, liquid, and gaseous X can exist simultaneously.
From the choices given, statement B: 'X can exist as a solid above 20°C.' This cannot be true. If we are above the triple-point temperature and X is a typical substance that expands upon heating, it will not exist as a solid at temperatures above its triple point.
The other statements can be true depending on the actual phase diagram of X. The phase a substance is in, is determined both by its temperature and pressure. Therefore, without more information on substance X, it is difficult to conclusively verify or disprove those statements.
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For the galvanic cell reaction, expressed below using shorthand notation, what half-reaction occurs at the cathode?
Zn(s) Zn2+(aq) Ni2+(aq) Ni(s)
A) Zn(s) → Zn2+(aq) + 2 e
B) Zn2+(aq) + 2 e- → Zn(s)
C) Ni(s) → Ni2+(aq) + 2 e
D) Ni2+(aq) + 2 e- → Ni(s)
Answer:
D
Explanation:
In considering the half cell reactions in electrochemical cells, we consider the standard electrode potential of the two half cells. The more negative electrode potential will be the anode and the less negative electrode potential will be the cathode. The electrode potentials of Ni2+(aq)/Ni(s) is -0.25V while that of Zn2+(aq)/Zn(s) is -0.76.
Hence the selected option is the cathodic half reaction equation
In the given galvanic cell reaction, the reduction half-reaction Ni2+(aq) + 2 e- → Ni(s) occurs at the cathode. Nickel ions are gaining electrons to become solid Nickel.
Explanation:In a galvanic cell, the half-reaction that occurs at the cathode is the reduction reaction. Reduction is a chemical process where a species gains electrons. This stands as the acronym OIL RIG, which stands for Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). Given the provided shorthand notation for the galvanic cell, Option D: Ni2+(aq) + 2 e- → Ni(s) would be the half-reaction occurring at the cathode. This is because Nickel ions (Ni2+) are gaining electrons, thus going through a reduction process, to form solid Nickel (Ni).
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Calculate the number of moles in the 2.00-L volume of air in the lungs of the average person. Note that the air is at 37.0°C (body temperature) and that the total volume in the lungs is several times the amount inhaled in a typical breath as given in
Answer:
The number of mole = 0.079 mole.
Explanation:
According to ideal gas equation,
PV = nRT.................... equation 1
Where P = pressure of the air, V = volume of the air, n = number of moles, R = molar gas constant, T = Temperature in (Kelvin), R = molar gas constant
Making n the subject of equation 1
n = PV/RT................... equation 2.
Where P = 101325 pa = 1 atm
V = 2.00 L = 2.00 dm³
T = 37 = 37 + 273 = 310 K, R= 0.082 atm/(dm³Kmol)
Applying these values in equation 1,
n = (2×1)/(310×0.082)
n = 2/25.42
n = 0.079 mole
Therefore the number of mole = 0.079 mole.
Final answer:
To find the number of moles in a 2.00-L volume of air at body temperature (37.0°C), you use the ideal gas law (PV = nRT) along with the conversion of the temperature to Kelvin. The calculation shows that there are approximately 0.0785 moles of air in this volume at the given temperature.
Explanation:
To calculate the number of moles of air in a 2.00-L volume at 37.0°C (body temperature), we can use the ideal gas law, which is PV = nRT.
In this formula, P represents pressure, V represents volume, n represents the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 37.0 + 273.15 = 310.15 K
We assume that the pressure (P) is 1.00 atm, as this is a common assumption for problems at body temperature. The volume (V) is given as 2.00 L. Substituting the values into the ideal gas law equation, we get:
n = PV / RT
n = (1.00 atm) × (2.00 L) / (0.0821 L·atm/mol·K × 310.15 K)
Now, we can find the number of moles:
n = 2.00 / (0.0821 × 310.15)
n ≈ 0.0785 moles
Therefore, there are approximately 0.0785 moles of air in a 2.00-L volume at body temperature.
KNO3(s) --->ž K+(aq) + NO3-(aq)This reaction was carried out in a Styrofoam insulated calorimeter and the following data were recorded:Mass of solid KNO3 dissolved10.1 gMass of aqueous solution (c = 4.18 J/gºC)100. gT initial30.0ºCT final21.6ºCMolar mass of KNO3101 g/molIf the mass of KNO3 solid dissolved were doubled while all other experimental conditions were kept the same, what change would occur in delta T, J per reaction, J/g of KNO3, and kJ/mol KNO3?Larger delta T, larger J/reaction, larger J/g, no change kJ/molLarger delta T, no change J/reaction, no change J/g, no change kJ/molLarger delta T, larger J/reaction, no change J/g, no change kJ/molLarger delta T, larger J/reaction, larger J/g, larger kJ/mol
Answer:
Larger ΔT, larger J, no change J/g, no change kJ/mol is the correct answer.
Explanation:
When one sees data in this question one tends to think that to be solved we need to perform calculations, and that is not the case here.
What we need is to remembember what are properties which depend on quantities (extensive properties) and the concepts and formulas for heat.
Δ Hrxn = - Q cal
Q cal = m x c x ΔT where
m = mass of water in the calorimeter
c= specific heat of water, and
ΔT = change in temperature
ΔT is directly proportional to Q
ΔH rxn is an extensive quantity dependent on the amount of KNO₃, as is J/ reaction.
J/g , kJ/mol are intensive properties the moment they are defined as the heat per gram, and heat per mol released or absorbed.
mass of KNO₃ ⇒ doubles heat of reaction , doubles ΔT
Therefore,
c) Larger ΔT, larger J, no change J/g, no change kJ/mol is the correct answer.
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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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
SiH4 Draw the molecule by placing atoms on the grid and connecting them with bonds. Do not identify the charge on each of these species. Include all lone pairs of electrons. To change the symbol of an atom, double-click on the atom and enter the letter of the new atom. +- CHONSPFBrClIXMore
Answer:
Attached image of the Lewis structure.
Explanation:
To draw the Lewis structure of SiH₄, we need to consider the octet rule: atoms gain, lose or share electrons to have 8 electrons in their valence shell. H is an exception to this rule because it is completed with 2 electrons (duet).
Si is a semimetal and H a nonmetal, and they form covalent bonds, that is, they share pairs of electrons to be complete.
Si has 4 valence electrons, so it forms 4 covalent bonds to reach the octet.
Each H has 1 valence electron, so each H forms 1 covalent bond to reach the duet.
The resulting structure can be seen in the attached picture.
To draw the Lewis structure of SiH4, the total number of valence electrons in the molecule is determined. The central atom is Silicon (Si) and each Hydrogen atom is bonded to Silicon with a single bond. The Lewis structure is represented by connecting the atoms with single bonds.
Explanation:To draw the Lewis structure of SiH4, we need to determine the total number of valence electrons in the molecule. Silicon (Si) is in group 14 of the periodic table and has 4 valence electrons. Hydrogen (H) is in group 1 and has 1 valence electron. Since there are 4 hydrogen atoms, we have a total of 4 valence electrons. Therefore, the total number of valence electrons in SiH4 is 4 + 4 = 8.
In the Lewis structure, the central atom is silicon. Each hydrogen atom will be bonded to silicon with a single bond. Since each hydrogen atom needs 2 electrons to complete its outer shell, the silicon atom will share its 4 valence electrons with the 4 hydrogen atoms, resulting in 4 single bonds.
The Lewis structure of SiH4 can be represented as follows:
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Draw a highly magnified view of a sealed, rigid container filled with a gas. Then draw what it would look like if you cooled the gas significantly but kept the temperature above the boiling point of the substance in the container. Also draw what it would look like if you heated the gas significantly. Finally, draw what each situation would look like if you evacuated enough of the gas to decrease the pressure by a factor of 2.
Answer:
In order to fully understand the issue of the ideal gas or perfect gas, we must pay attention to the following, an ideal or perfect gas does not really exist, it is a hypothetical gas whose sharing of the variables of pressure, volume and temperature can be fully described by the ideal gas equation.
The molecules that make up an ideal gas do not usually attract or repel each other, and their volume is negligible compared to the volume of the container that contains it. Although in our nature the case of an ideal gas does not exist, the differences between the behavior of a real gas in temperature and pressure margins do not substantially alter the calculations, so we can make use of the equation with all the security, to solve various gas exercises.
Explanation:
The collisions that occur between the molecules and with the molecules and with the walls is elastic because the moment is preserved, in addition to the kinetic energy.
It can be synthesized that a gas is ideal when all collisions that occur between atoms or molecules are completely elastic and there are no attractive forces that are intermolecular.
In ideal gases the kinetic energy is proportional to its temperature. The gases approach an ideal gas if they are mono atomic gases, if it is under pressure and also at room temperature.
The amount of gas in a body is measured in moles. One mole of any type of gas reaches 22.4 liters, in normal condition, 0 ° Celsius and 1 of the pressure atmosphere. That volume is called normal molar volume.
Ideal gases have an equation called the Ideal Gas Equation and is based on three main laws that are Boyle's law, Gay-Lussac's law, Charles's law and also Avogadro's law.
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
Which of the following gases would have the greatest kinetic energy at 300 K?
A. N2
B. NH3
C. Ar
D. All of them would have the same kinetic energy
Answer:
D. All of them would have the same kinetic energy
Explanation:
The expression for the kinetic energy of the gas is:-
[tex]K.E.=\frac{3}{2}\times K\times T[/tex]
k is Boltzmann's constant = [tex]1.38\times 10^{-23}\ J/K[/tex]
T is the temperature
Since, kinetic energy depends only on the temperature. Thus, at same temperature, at 300 K, all the gases which are [tex]N_2,\ NH_3\ and\ Ar[/tex] will posses same value of kinetic energy.
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.
Identify the correct acid and conjugate base pair in this equation: HNO2 + HS - --> NO2- + H2S
Answer:
Acid: HNO₂
Conjugate base: NO₂⁻
Base: HS⁻
Conjugate acid: H₂S
Explanation:
HNO₂ + HS⁻ → NO₂⁻ + H₂S
This is the reaction of nitrous acid with hydrosulfide, to generate nitrite and sulfide acid.
HNO₂ is a weak acid that release the proton to the HS⁻, so the HS⁻ is the base.
Proton that is caught by the HS⁻, produce the H₂S so, if the HS⁻ is the base (accepts a proton from other compound), the H₂S acts like the conjugate acid.
In nitrous acid, which is behaving as an acid (a weak one), the nitrite becomes in the conjugate strong base.
In the equation, HNO2 serves as the Brønsted-Lowry acid, giving up its proton to become its conjugate base, NO2-. HS- is the Brønsted-Lowry base and after accepting a proton, it becomes H2S, its conjugate acid.
Explanation:In the equation HNO2 + HS- --> NO2- + H2S, the acid and its conjugate base can be identified using the concept of the Brønsted-Lowry acid and base. The Brønsted-Lowry acid is a substance that donates a proton (H+) and the Brønsted-Lowry base is a substance that accepts a proton (H+).
Therefore, in this equation, HNO2 is acting as the Brønsted-Lowry acid because it donates a proton to become its conjugate base NO2-. Similarly, HS- is the Brønsted-Lowry base that accepts a proton to become H2S, the conjugate acid. So the acid-conjugate base pair here is HNO2 and NO2-.
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A certain metal alloy is composed of 10% tin, 16% antimony, and 74% lead. If you were to have 500 g of the alloy, how many grams of antimony would be found in this sample?
a. 370g
b.50g
c.80g
d. 16g
Answer:
c. 80 g
Explanation:
metal alloy:
wt% = (mass compound/mass alloy)×100for antimony:
⇒ 16% = (g antimony / g alloy)×100
⇒ 0.16 = g antimony / g alloy
∴ g alloy = 500 g
⇒ (0.16)×(500 g) = g antimony
⇒ g antimony = 80 g
Draw the α-keto acid product of the transamination of an α-keto acid with an amino acid that has the side chain X (double click an existing atom and type X).
Answer:
On the attached picture.
Explanation:
Hello,
In this case, one must remember that transamination is a biochemical reaction that transfers an amino group to the carbonyl group of a ketoacid to form a new amino acid and a new ketoacid.
For this example, on the attached picture you will see the corresponding chemical reaction in which to the initial ketoacid, both the amino and an additional hydrogen are transferred from the amino acid of side chain X to form the requested α-ketoacid of side chain X.
Best regards.
A student placed 11.0 g of glucose (C6H12O6 ) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached.
The flask was then shaken until the solution was uniform. A 55.0 mL sample of this glucose solution was diluted to 0.500 L .
How many grams of glucose are in 100. mL of the final solution?
To calculate the mass of glucose in 100 mL of the final solution, we can use the equation C1V1 = C2V2.
Explanation:To determine the number of grams of glucose in 100 mL of the final solution, we need to use the concept of dilution. The initial solution was prepared by dissolving 11.0 g of glucose in enough water to reach the 100 mL mark. Then, a 55.0 mL sample of this solution was diluted to 0.500 L. We can use the equation:
C1V1 = C2V2
Where C1 and V1 represent the concentration and volume of the initial solution, and C2 and V2 represent the concentration and volume of the final solution.
From this, we can calculate the concentration of the initial solution, and then use it to find the mass of glucose in 100 mL of the final solution.
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The gas OF2 can be produced from the electrolysis of an aqueous solution of KF, as shown in the equation below.
OF2(g) + 2 H+(aq) + 4 e- → H2O(l) + 2 F-(aq) E° = +2.15 V
Using the given standard reduction potential, calculate the amount of OF2 that is produced, and the electrode at which the OF2 is produced, upon the passage of 0. 480 faradays through an aqueous KF solution.
A) 6.48 g of OF2 at the anode.
B) 26.0 g of OF2 at the anode.
C) 6.48 g of OF2 at the cathode.
D) 26.0 g of OF2 at the cathode
Answer:
A) 6.48 g of OF₂ at the anode.
Explanation:
The gas OF₂ can be obtained through the oxidation of F⁻ (inverse reaction of the reduction presented). The standard potential of the oxidation is the opposite of the standard potential of the reduction.
H₂O(l) + 2 F⁻(aq) → OF₂(g) + 2 H⁺(aq) + 4 e⁻ E° = -2.15 V
Oxidation takes place in the anode.
We can establish the following relations:
1 Faraday is the charge corresponding to 1 mole of e⁻.1 mole of OF₂ is produced when 4 moles of e⁻ circulate.The molar mass of OF₂ is 54.0 g/mol.The mass of OF₂ produced when 0.480 F pass through an aqueous KF solution is:
[tex]0.480F.\frac{1mole^{-} }{1F} .\frac{1molOF_{2}}{4mole^{-} } .\frac{54.0gOF_{2}}{1molOF_{2}} =6.48gOF_{2}[/tex]
Using the data provided here, the mass of the compound produced is 6.48 g of OF2
What is electrolysis?Electrolysis refers to the breaking u p of a molecule by the passage of direct current through it. The equation of the reaction is; H₂O(l) + 2 F⁻(aq) → OF₂(g) + 2 H⁺(aq) + 4 e⁻ E° = -2.15 V.
Now;
1 mole of OF2 is realeased by the passage of 4 F of electricity
x moles of OF2 is produced by the passage of 0.480F
x = 0.12 moles
Mass of OF2 = 0.12 moles * 54 g/mol = 6.48 g of OF2
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Red #40 has an acute oral LD50 of roughly 5000 mg dye/1 kg body weight. This means if you had a mass of 1 kg, ingesting 5000 mg of Red #40 will have a 50% of killing you. Determine the amount of dye a 70 kg person would need to eat to reach his or her LD50. How many moles of dye would that be? The molar mass of Red #40 is 496.42 g/mol. Using the calculated concentration of Red #40 dye in the sports drink, how many of liters of sports drink would it take to reach this LD50?
Answer:
350 g dye
0.705 mol
2.9 × 10⁴ L
Explanation:
The lethal dose 50 (LD50) for the dye is 5000 mg dye/ 1 kg body weight. The amount of dye that would be needed to reach the LD50 of a 70 kg person is:
70 kg body weight × (5000 mg dye/ 1 kg body weight) = 3.5 × 10⁵ mg dye = 350 g dye
The molar mass of the dye is 496.42 g/mol. The moles represented by 350 g are:
350 g × (1 mol / 496.42 g) = 0.705 mol
The concentration of Red #40 dye in a sports drink is around 12 mg/L. The volume of drink required to achieve this mass of the dye is:
3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L
A 70 kg person would need to eat 350,000 mg of Red #40 dye to reach the LD50. This is equivalent to 704.01 mol of Red #40 dye. The number of liters of sports drink needed depends on the concentration of Red #40 dye in the drink.
Explanation:To determine the amount of dye a 70 kg person would need to eat to reach the LD50, we can use the information provided. The LD50 for Red #40 is 5000 mg dye/1 kg body weight. So for a 70 kg person, the LD50 would be:
(5000 mg dye/1 kg body weight) * 70 kg = 350,000 mg of Red #40 dye.
To calculate the number of moles of dye, we divide the mass of dye by its molar mass. The molar mass of Red #40 is 496.42 g/mol, so:
(350,000 mg) / (496.42 g/mol) = 704.01 mol of Red #40 dye.
Finally, to determine the number of liters of sports drink needed to reach this LD50, we need to know the concentration of Red #40 dye in the sports drink.
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A survey of online students found that 52.6 percent believed they benefitted most from:_______a. Interacting with other students on the discussion boards.b. Disagree with other students on the discussion boards.c. None of the above
Answer:
The answer is a): Interacting with other students on the discussion boards.
Explanation:
In a survey of online students who identified some tips that made them successful in online courses, and virtual university programs which targeted adults, 52.6 % of them believed that the greatest benefit they derived was due to their interaction with other students (especially classmates) on online discussion boards.
The question is related to the benefits of online learning in Education, specifically for college students. Without additional options, we draw on general benefits reported by students, including discussion board interactions, diverse content delivery methods, and online community support as valuable aspects of online education.
Explanation:The subject of this question is Education, specifically focusing on the benefits of online learning environments for college students. To answer the student's question about what 52.6 percent of online students believed they benefitted most from, we need additional context or options other than those provided in the question. However, past surveys and research can offer insight into common benefits reported by students in online learning settings, such as:
Interacting with other students on the discussion boards.Having access to multiple modes of content delivery, such as screencasts and online resources, which can enhance understanding.Engaging in community-building activities and receiving support through online forums.These activities not only help in understanding the subject matter but also in building a sense of community and improving technical skills. The prevalence of technology in education has made these aspects more accessible than ever before, changing the landscape of how students engage with peers and course materials.
A hot air balloon is filled to a volume of 44.5 L at 758 torr.
What will be the volume of the balloon if the pressure decreases to 748 torr under constant temperature?
A. 45.1 L
B. 43.9 L
C. 44.5 L
D. 49.0 L
Answer:
The volume of the balloon, if the pressure decreases to 748 torr, will be 45.1 L
Explanation:
Step 1: Data given
Volume of the balloon = 44.5 L
Pressure in the balloon = 758 torr
Final pressure = 748 torr
Step 2: Calculate final volume
P1*V1 = P2*V2
⇒ with P1 = the initial pressure = 758 torr
⇒ with V1 = the initial volume = 44.5 L
⇒ with P2 = the final pressure = 748 torr
⇒ with V2 = the final volume = TO BE DETERMINED
V2 = (P1*V1)/P2
V2 = (758 * 44.5) /748
V2 = 45.1 L
The volume of the balloon, if the pressure decreases to 748 torr, will be 45.1 L
Using Boyle's Law, we can determine the new volume of the balloon when the pressure decreases from 758 torr to 748 torr while keeping the temperature constant. After performing the calculation (P1 * V1) / P2, we find that the new volume is approximately 45.1 L. So the correcct option is A.
Explanation:To solve this problem, we can use Boyle's Law, which states that the pressure (P) of a gas is inversely proportional to its volume (V) at constant temperature and number of moles. Mathematically, it can be expressed as P1 * V1 = P2 * V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume, respectively.
The initial conditions are a pressure (P1) of 758 torr and a volume (V1) of 44.5 L. We are looking for the final volume (V2) when the pressure (P2) decreases to 748 torr. By rearranging Boyle's Law, we can solve for V2: V2 = (P1 * V1) / P2.
Plugging in the known values, we have V2 = (758 torr * 44.5 L) / 748 torr. This calculation gives us V2 ≈ 45.1 L, which corresponds to option A).
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Determine the value of Kc for the following reaction if the equilibrium concentrations are as follows:
[N2]eq = 1.5 M,[H2]eq = 1.1 M,[NH3]eq = 0.47 M.
N2(g) + 3 H2(g) ⇌ 2 NH3(g)
Answer:
0.11
Explanation:
The given equilibrium reaction and the equilibrium concentrations are shown below as:-
[tex]\begin{matrix}&N_2&+&3H_2&\rightleftharpoons &2NH_3\\Concentration\ at\ equilibrium:-&1.5&&1.1&&0.47\end{matrix}[/tex]
The Kc of an equilibrium reaction measures relative amounts of the products and the reactants present during the equilibrium.
It is the ratio of the concentration of the products and the reactants each raised to their stoichiometric coefficients. The concentration of the liquid and the gaseous species does not change and thus is not written in the expression.
The equation is as follows:-
[tex]N_2_{(g)} +3H_2_{(g)}\rightarrow 2NH_3_{(g)}[/tex]
The expression for the Kc is:-
[tex]K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]
Thus, applying the values as:-
[tex]K_c=\frac{0.47^2}{1.5\times 1.1^3}=0.11[/tex]
The value of the equilibrium constant (Kc) for the reaction N2(g) + 3 H2(g) ⇌ 2 NH3(g) with the given equilibrium concentrations is approximately 0.11.
To determine the value of the equilibrium constant (Kc) for the reaction N2(g) + 3 H2(g) ⇌ 2 NH3(g), where the equilibrium concentrations are [N2]=1.5 M, [H2]=1.1 M, and [NH3]=0.47 M, we would use the equilibrium expression for Kc:
Kc = [NH3]2 / ([N2] × [H2]3)
Inserting the given equilibrium concentrations, we get:
Kc = (0.47 M)2 / (1.5 M × (1.1 M)3)
Kc = (0.2209 M2) / (1.5 M × 1.331 M3)
Kc = 0.2209 M2 / 1.9965 M4
Kc ≈ 0.11
The value of the equilibrium constant for this reaction is therefore approximately 0.11.
Endings (suffixes) in the naming process can tell us a lot about a chemical. For example, a compound named "Ethyl propanoate" would be categorized as a(n):
Answer:
Oxyanion
Explanation:
Nomenclature is a set of rules designed to name chemical compounds, in this rules the use of suffix is common to determine the functional group of a molecule, in the case of "Ethyl propanoate" the -ate suffix is used to design an oxyanion like sulfate SO42- or nitrate NO3-.
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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]