Answer:
The right answer to this question is no, all of the liquid will not evaporate, there will be 8.61 ×10⁻⁴ moles or 6.22×10⁻² grams left in the container
At equilibrium the pressure in the container will be the vapor pressure of liquid pentane which is = 100. mm Hg
Explanation:
To solve this we list the known values as follows
vapor pressure of liquid pentane = 100 mmHg = 13.33 KPa
Temperature T = 260 K
Volume of container = 350 mL = 0.00035 m³
The number of moles of liquid pentane = n
The universal gas constant = R = 8.314 J/(mol·K)
Thus From the ideal gas equation PV = nRT →
Thus plugging in the values in the above equateion we have
n = [tex]\frac{PV}{RT} = \frac{(13330)(0.00035)}{(8.314)(260)}[/tex] = 2.16×10⁻³ moles
Hence the number of moles in 0.218 g sample of liquid pentane C₅H₁₂ with molar mass = 72.15 g/mol = 0.218/72.15 = 3.02×10⁻³ moles
Hence the number of moles present in the sample placed in the closed evacuated container = 3.02×10⁻³ moles
However number of moles to completely evaporate at 100 mmHg and 260 K is 2.16×10⁻³ moles hence, 3.02×10⁻³ moles - 2.16×10⁻³ moles, or 8.61 ×10⁻⁴ moles will be left in the container
converting the value in moles to mass we have number of moles, n = mass/(molar mass)
Therefore the mass = number of moles × molar mass = 8.61 ×10⁻⁴ × 72.15 = 6.22 × 10⁻² grams left in the container
The pressure in the container at equilibrium will be vapor pressure of liquid pentane C₅H₁₂, or 100. mm Hg
In the described closed system, all of the liquid pentane will evaporate due to the vapor pressure. The final pressure when equilibrium is reached will be the vapor pressure, which is 100.0 mm Hg.
Explanation:The subject of the question is the vapor pressure of pentane, C5H12, which is a concept from Chemistry.
The given vapor pressure is 100.0 mm Hg at 260 K. In this closed system the liquid and vapor will come to equilibrium at the given temperature, at which point the vapor pressure will be equal to the given 100.0 mm Hg.
The sample mass of 0.218 g doesn't exceed the amount needed for evaporation at the equilibrium pressure. Therefore, all of the liquid pentane will evaporate.
The final pressure in the container when equilibrium is reached will be the vapor pressure, which is 100.0 mm Hg.
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A water bath in a physical chemistry lab is 1.85m long, 0.810m wide and 0.740m deep. If it is filled to within 2.57 in from the top, how many liters of water are in it
Final answer:
The water bath, filled to within 2.57 inches of the top, contains approximately 1014.48 liters of water. This calculation involves converting inches to meters, calculating the volume in cubic meters, and then converting that volume to liters.
Explanation:
To calculate the volume of water in the water bath, we need to take into account the dimensions given and the fact that it is filled to within 2.57 inches from the top. First, we need to convert the depth from which it's filled into meters, as the other dimensions are given in meters.
2.57 inches = 0.06533 meters (since 1 inch = 0.0254 meters)
The actual depth filled with water will be:
0.740 m (total depth) - 0.06533 m = 0.67467 m
Next, we multiply the length, the width, and the newly calculated depth to get the volume in cubic meters:
Volume = 1.85 m × 0.810 m × 0.67467 m = 1.01448 cubic meters
To convert the volume from cubic meters to liters, we use the conversion factor 1 cubic meter = 1000 liters:
Volume = 1.01448 m3 × 1000 L/m3 = 1014.48 liters
Therefore, there are approximately 1014.48 liters of water in the water bath.
Chromium crystallizes with a body-centered cubic unit cell. The radius of a chromium atom is 125 pm . Calculate the density of solid crystalline chromium in grams per cubic centimeter.
Answer:
[tex]\rho=7.15\ g/cm^3[/tex]
Explanation:
The expression for density is:
[tex]\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}[/tex]
[tex]N_a=6.023\times 10^{23}\ {mol}^{-1}[/tex]
M is molar mass of Chromium = 51.9961 g/mol
For body-centered cubic unit cell , Z= 2
[tex]\rho[/tex] is the density
Radius = 125 pm = [tex]1.25\times 10^{-8}\ cm[/tex]
Also, for BCC, [tex]Edge\ length=\frac{4}{\sqrt{3}}\times radius=\frac{4}{\sqrt{3}}\times 1.25\times 10^{-8}\ cm=2.89\times 10^{-8}\ cm[/tex]
Thus,
[tex]\rho=\frac{2\times \:51.9961}{6.023\times \:10^{23}\times \left(2.89\times 10^{-8}\right)^3}\ g/cm^3[/tex]
[tex]\rho=7.15\ g/cm^3[/tex]
The density of solid crystalline chromium will be "7.15 g/cm³".
According to the question,
Radius,
125 pm or [tex]1.25\times 10^{-8} \ cm[/tex]Molar mass of Chromium,
M = 51.9961 g/molFor body-centered,
Z = 2For BCC,
The edge length will be:
= [tex]\frac{4}{\sqrt{3} }\times radius[/tex]
= [tex]\frac{4}{\sqrt{3} }\times 1.25\times 10^{-8}[/tex]
= [tex]2.89\times 10^{-8} \ cm[/tex]
hence
The expression for density will be:
→ [tex]\rho = \frac{Z\times M}{N_a(Edge \ length)^3}[/tex]
By substituting the values, we get
[tex]= \frac{2\times 51.9961}{6.023\times 10^{23}\times (2.89\times 10^{-8})}[/tex]
[tex]= 7.15 \ g/cm^3[/tex]
Thus the above approach is right.
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What is the mass of water necessary to generate 11.2 L of hydrogen gas if calcium metal reacts with water at standard temperature and pressure (STP)?
Answer:
The mass of water is 18 g
Explanation:
The reaction of calcium with water can be represented in the equation below:
Ca + 2H₂O --------->Ca(OH)₂ +H₂
1 Mole of gas at STP = 22.4L
From the displacement reaction above, calculate the mass of water that will produce 22.4L of hydrogen gas at STP.
Mass of 2H₂O = 2(2x1 + 16) = 2X18 = 36 g/mol
Using proportional analysis;
36 g of 2H₂O produced 22.4 L of H₂, then
what mass of 2H₂O will produce 11.2L of H₂ ?
Mathematically,
22.4 L ----------------------------------> 36g
11.2 L -----------------------------------> ?
Cross and multiply, to obtain the expression below
= (11.2 X 36)/22.4
= 18 g
Therefore, the mass of water is 18 g
We create an electron with wavefunction ψ = [ψ1s(r)+3ψ3s(r)]/ √ 10. Find the expected value hEi of the energy in that state. Use the energies of the hydrogen atom to extract your answer in electron volts.
Answer: r=132pm
Explanation:
r = ⟨r⟩=5a0/Z
r=(5(5.29177*〖10〗^(-11) m) x^2)/2
r=132pm
The melting point of sodium metal is 97.8 ∘C and the melting point of sodium chloride is 801 ∘C. What can you infer about the relative strength of metallic and ionic bonding from these melting points?
Answer:
The metallic bond of the sodium metal is weaker than the ionic bond of the sodium chloride
Explanation:
The metallic bond of the sodium metal is weaker than the ionic bond of the sodium chloride. This is seen in the fact that the melting point of the sodium metal which is 97.8°C is lower than the melting point of the sodium chloride which is 801°C. This shows that the inter-atomic bonds of the sodium metal are weaker than that of the inter-ionic bonds of the sodium chloride which will require higher energy to break which is shown by its high melting point of 801°C. The melting point of the sodium metal of 97.8 °C which is lower requires less energy to break its bonds. This is shown by its lower melting point.
Pick the correct statement for the following isotope: a. 42Ca 42 is the mass number and 20 is the atomic number. b. 42 is the number of neutrons and 20 is the number of protons. c. 42 is the number of protons and 20 is the number of electrons. d. 42 is the atomic number and 20 is the number of neutrons.
Answer:
A
Explanation:
To label an element correctly using a combination of the symbol, mass number and atomic number furnishes some important information about the element.
We can obtain these information from the element provided that correct labeling of the element is presented. Firstly, after writing the symbol of the element, the atomic number is placed as a subscript on the left while the mass number of the atomic mass is placed as a superscript on the same left.
Looking at the question asked, we have the element symbol in the correct position as Ca, with 42 also in the correct position which is the mass number. The third number which is 20 is thus the atomic number of the element.
Heavy elements, like uranium, were/are created _______________. a. in supernovae and star collisions b. all of the above c. in the Big Bang. d. through nuclear fusion in mid-sized suns, like ours.
Answer:
a. in supernovae and star collisions
Explanation:
The periodical table contains some heavier elements, which are formed as neutron stars pairs hit eachother and erupt cataclysmically.
The star emitts very large quantities of energy and neutrons during supernova, which allow for the production of heavier elements than iron, such as uranium and gold. All these elements are ejected into space during the supernova explosion.
A 3.78-gram sample of iron metal is reacted with sulfur to produce 5.95 grams of iron sulfide. Determine the empirical formula of this compound.
Answer:FeS
Explanation:
Mass of sulphur=5.95-3.78=2.17g
Fe S
3.78/56. 2.17/32
0.0675/0.0675. 0.0678/0.0675
1:1
Empirical formula=Fe1S1=FeS
An empirical formula is a compound's chemical formula that only specifies the ratios of the elements it contains and not the precise number or arrangement of atoms. The empirical formula is FeS.
The formula of a material expressed with the smallest integer subscript is referred to as an empirical formula for a compound. The empirical formula provides details regarding the ratio of atom counts in the molecule. A compound's empirical formula is directly related to its % content.
Mass of sulphur = 5.95 - 3.78 = 2.17g
Moles of Fe and S are:
Fe = 3.78/56
n = 0.0675
S = 2.17/32
n = 0.0678
Dividing with the smallest number:
0.0675/0.0675 = 1
0.0678/0.0675 = 11
The ratio is:
1:1
Empirical formula = FeS
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Flammable materials, like alcohol, should never be
dispensed or used near
A. an open door.
B. an open flame.
C. another student.
D. a sink
Answer:
b.open flame because it is fundamental end of the alcohol mixes in with the flame then it will become a bigger fire
The safest practice is to never use flammable materials like alcohol near an open flame as it can easily ignite and cause a fire.
Explanation:Flammable materials such as alcohol should never be dispensed or used near an open flame (option B). These substances are highly reactive and can easily ignite, potentially causing a fire. It's crucial for safety purposes to avoid using flammable materials around direct sources of heat or open flames. This rule is applicable not only in chemistry laboratories, but also in any other scenario where flammable substances are present.
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In a neutralization reaction, 24.6 mL of 0.300 M H2SO4(aq) reacts completely with 20.0 mL of NaOH(aq). The products are Na2SO4(aq) and water. What is the concentration of the NaOH solution?
Answer : The concentration of the NaOH solution is, 0.738 M
Explanation :
To calculate the concentration of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=2\\M_1=0.300M\\V_1=24.6mL\\n_2=1\\M_2=?\\V_2=20.0mL[/tex]
Putting values in above equation, we get:
[tex]2\times 0.300M\times 24.6mL=1\times M_2\times 20.0mL[/tex]
[tex]M_2=0.738M[/tex]
Thus, the concentration of the NaOH solution is, 0.738 M
Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25∘C diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Determine △ Hrxn for C(diamond) → C(graphite) with equations from the following list: (1) C(dianond)+O2(g)⟶CO2(g)ΔH=−395.4kJ (2) 2CO2(g)⟶2CO(g)+O2(g)ΔH=566.0kJ (3) C(graphite)+O2(g)→CO2(g)ΔH=−393.5kJ (4) 2CO(g)⟶C(graphite)+CO2(g)ΔH=−172.5kJ
Answer:
-1.9 KJ/mol
Explanation:
In order to solve the problem, we have to rearrange the equations in a way in which all molecules of O₂ and CO₂ are eliminated:
2C(diamond) + 2O₂(g) → 2CO₂(g) ΔH₁= 2 x (-395.4 KJ) ------> we multiply by 2 both reactants and products
2 CO₂(g) → 2CO(g) + O₂(g) ΔH₂= 566.0 KJ
CO₂(g) → C(graphite) + O₂(g) ΔH₃= -1 x (-393.5 KJ) ------> we use reverse rxn
2CO(g) → C(graphite) + CO₂(g) ΔH₄= -172.5 KJ
When we cancel the molecules that appear both in reactants and products, the total reaction is the following:
2C(diamond) → 2C(graphite)
ΔHt= ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ = 2 x (-395.4 KJ) + 566.0 KJ + (-1 x (-393.5 KJ)) - 172.5 KJ
ΔHt= 347.2 KJ
This is for 2 mol of C(diamond) which are converted in 2 mol of C(graphite). To obtain ΔH for the reaction of 1 mol C(diamond) to 1 mol (graphite) we have to divide into 2:
ΔH= -3.8 KJ/2mol= -1.9 KJ/mol
To determine ΔHrxn for the reaction C(diamond) → C(graphite), we can use Hess's law and the given equations. By canceling out common compounds/products and summing the remaining equations, we can calculate ΔHrxn. The enthalpy change can be determined by summing the enthalpy changes of the canceled equations.
Explanation:This type of calculation usually involves the use of Hess's law, which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. For this specific reaction, we can use the given equations to determine the enthalpy change ΔHrxn for C(diamond) → C(graphite).
Step 1: C(diamond) + O2(g) → CO2(g) (ΔH = -395.4 kJ)Step 2: 2CO2(g) → 2CO(g) + O2(g) (ΔH = 566.0 kJ)Step 3: C(graphite) + O2(g) → CO2(g) (ΔH = -393.5 kJ)Step 4: 2CO(g) → C(graphite) + CO2(g) (ΔH = -172.5 kJ)To obtain the ΔHrxn for C(diamond) → C(graphite), we need to cancel out common compounds/products in these equations. From the given equations, we can see that CO2(g) is a common compound in steps 1, 2, and 3, and thus we can cancel it out. Likewise, O2(g) is present in steps 1, 2, and 3, so we can also cancel it out. By summing the canceled equations, we get the final equation:
C(diamond) → C(graphite)
The enthalpy change for this reaction is the sum of the enthalpy changes of the canceled equations:
ΔHrxn = ΔH1 + ΔH2 + ΔH3 + ΔH4 = -395.4 kJ + 566.0 kJ + (-393.5 kJ) + (-172.5 kJ) = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + (-393.5 kJ) + 566.0 kJ + (-172.5 kJ) = -395.4 kJ - 393.5 kJ + 566.0 kJ - 172.5 kJ = -395.4 kJ - 393.5 kJ + 566.0 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ
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Which of the following are true about balanced chemical reactions? (Select all that apply)
a
Atoms of each element are equal on both sides.
b
All chemical formulas have a coefficient greater than one.
c
It is better to change subscripts to balance equations.
d
Coefficients are added to balance the equation.
e
Chemical equations must be balanced to satisfy the Law of Conservation of Mass.
f
Balanced chemical equations help to determine the number of moles needed in the reaction.
Answer: D. There must be as equal number of atoms of each element on both sides of the equation.
Explanation: I just took the test on Plato and this is correct
Arrange the following alkyl bromides in order from most reactive to least reactive in an SN₂ reaction: 1-bromo-2-methylbutane, 1-bromo-3-methylbutane, 2-bromo-2-methylbutane, and 1-bromopentane.
Answer:
1- bromopentane 1- bromo-3-methylbutane 1-bromo2-methylbutane 2-bromo-2-methylbutane
Explanation:
SN square reaction is a concentrated reaction. All the bond making and bond braking occur in one step
The nucleophile attack the back side of the carbon that bears the halide and replace it
At a certain temperature, the Kp for the decomposition of H2S is 0.842.
H2S(g) image from custom entry tool H2(g) + S(g)
Initially, only H2S is present at a pressure of 0.104 atm in a closed container. What is the total pressure in the container at equilibrium?
Answer:
0.1976 atm is the total pressure in the container at equilibrium.
Explanation:
The value of [tex]K_p[/tex] for the given reaction = 0.842
Partial pressure of the hydrogen sulfide = 0.104 atm
Partial pressure of the hydrogen sulfide at equilibrium = (0.104-x) atm
Partial pressure of the hydrogen gas at equilibrium = x
Partial pressure of the sulfur gas at equilibrium = x
[tex]H_2S(g)\rightleftharpoons 2H_2(g)+S(g)[/tex]
Initially
0.104 atm 0 0
At equilibrium
(0.104 -2x) 2x x
The expression of [tex]K_p[/tex] is given by ;
[tex]K_p=\frac{x\times x}{(0.104 -x) }[/tex]
[tex]0.842 =\frac{x^2}{(0.104-x)}[/tex]
Solving for x:
x = 0.0936 atm
Partial pressure of the hydrogen sulfide at equilibrium = (0.104-x) atm
= (0.104-0.0936 ) atm = 0.0104 atm
Partial pressure of the hydrogen gas at equilibrium = x = 0.0936 atm
Partial pressure of the sulfur gas at equilibrium = x = 0.0936 atm
Total pressure in the container will be sum of all the partial pressure of the hydrogen sulfide gas, hydrogen gas and sulfur gas at the equilibrium.
P = 0.0104 atm + 0.0936 atm + 0.0936 atm = 0.1683 atm
0.1976 atm is the total pressure in the container at equilibrium.
The total pressure in the system at equilibrium is calculated using the equilibrium constant Kp and the initial pressure of the system. By setting up a equilibrium constant expression and solving the quadratic equation for 'x', the change in pressure, we find that the total pressure at equilibrium in the system is ~0.101 atm.
Explanation:The decomposition reaction of H2S into H2 and S is a chemical equilibrium problem in which the equilibrium constant is given in terms of pressure, or Kp. We know that at the beginning, only H2S is present at a pressure of 0.104 atm, and at equilibrium, the pressure of H2S decreases by 'x', being 'x' the pressure of H2 and S at equilibrium. According to the expression of the equilibrium constant Kp, Kp = (pH2* pS) / p_H2S = x^2 / (0.104-x).
From here, if we solve for 'x' using the quadratic formula and the given Kp = 0.842, we find two solutions, ~0.0967 atm and ~0.0072 atm. We discard the result 0.0967 because it cannot be higher than the initial pressure of the system, so the correct 'x' is ~0.0072 atm. Therefore, the total pressure will be the sum of the individual pressures at equilibrium. Total Pressure = pH2 + pS + p_H2S = 2x + (0.104-x) = 2*0.0072 + (0.104-0.0072) = ~0.101 atm.
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Which results when an atom has such a strong attraction for electrons that it pulls one or more electrons completely away from another atom?
All people are tax people
-Turbo tax
What potential difference is required to bring the proton to rest?
Answer:
0
Explanation:
Δk = (kf - Ki). Kf is 0 because your stopping it [m(v_i)^2 ]/2q = V
The potential difference required to bring a proton to rest can be calculated by knowing the initial kinetic energy of the proton. This energy, when applied in the opposite direction, would act to decelerate the proton and bring it to rest.
Explanation:The potential difference required to stop a moving proton can be calculated using the equation PEelec = qV, where q is the proton's charge and V is the potential difference (also known as voltage). Given that the proton's charge, q = qe = 1.6×10−¹⁹ Coulombs, we can rearrange the equation to solve for V: V = PEelec/q. If we know the initial kinetic energy of the proton (its energy due to motion, which can be thought of as the electrical energy it gained from accelerating through the potential difference), then the same voltage (potential difference) applied in the opposite direction will bring the proton to rest.
In other words, if the proton gained a certain amount of energy (in electron volts, eV) while accelerating through a potential difference, it will require the same amount of energy in the opposite direction to decelerate and come to rest.
For example, if the proton gained 1 electron Volt (1 eV) of energy from the potential difference, it would require a potential difference of -1 V to bring it to rest. This is because 1 eV = (1 V)(1.6×10−¹⁹ C).
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The average C - H bond energy in CH4 is 415 kJ / mol. Use the following data to calculate the average C - H bond energy in ethane ( C2H6;C - C bond ) , in ethane ( C2H4;C C bond ) , and in ethyne ( C2H2;C C bond ) .
C2H6 ( g ) + H2 ( g ) rightarrow 2CH4 ( g ) ΔHrxn = - 65.07 kj / mol
C2H4 ( g ) + H2 ( g ) rightarrow 2CH4 ( g ) ΔH = - 202.21kJ / mol
C2H2 ( g ) + 3H2 ( g ) rightarrow 2 CH4 ( g ) ΔH = - 376.74kj / mol
Average C - H bond
Average C - H bond energy in ethane = kJ / mol energy in ethane = kj / mol
Average C - H bond energy in ethyne = kJ / mol
Average C-H bond energy: Ethane = 764.93 kJ/mol, Ethene = 627.79 kJ/mol, Ethyne = 528.26 kJ/mol.
To find the average C-H bond energy in ethane (C₂H₆), ethene (C₂H₄), and ethyne (C₂H₂), we can use Hess's law and the given reactions along with the average C-H bond energy in methane (CH₄), which is 415 kJ/mol.
1. **Calculate the average C-H bond energy in ethane (C₂H₆):**
Given reaction:
[tex]\[C2H6 (g) + H2 (g) \rightarrow 2CH4 (g) \quad \Delta H_{\text{rxn}} = -65.07 \, \text{kJ/mol}\][/tex]
This reaction breaks one C-C bond and adds two C-H bonds.
Change in bond energy = Total energy of bonds broken - Total energy of bonds formed
[tex]\[= 1 \times (\text{C-C bond energy}) - 2 \times (\text{C-H bond energy})\][/tex]
From the given reaction, the change in bond energy is -65.07 kJ/mol.
Substituting the known values:
[tex]\[-65.07 \, \text{kJ/mol} = 1 \times (\text{C-C bond energy}) - 2 \times (415 \, \text{kJ/mol})\][/tex]
Solving for the C-C bond energy:
[tex]\[\text{C-C bond energy} = 2 \times 415 - 65.07 \, \text{kJ/mol} = 764.93 \, \text{kJ/mol}\][/tex]
2. **Calculate the average C-H bond energy in ethene (C₂H₄):**
Given reaction:
[tex]\[C2H4 (g) + H2 (g) \rightarrow 2CH4 (g) \quad \Delta H = -202.21 \, \text{kJ/mol}\][/tex]
This reaction breaks one C=C bond and adds two C-H bonds.
Change in bond energy = Total energy of bonds broken - Total energy of bonds formed
[tex]\[= 1 \times (\text{C=C bond energy}) - 2 \times (\text{C-H bond energy})\][/tex]
From the given reaction, the change in bond energy is -202.21 kJ/mol.
Substituting the known values:
[tex]\[-202.21 \, \text{kJ/mol} = 1 \times (\text{C=C bond energy}) - 2 \times (415 \, \text{kJ/mol})\][/tex]
Solving for the C=C bond energy:
[tex]\[\text{C=C bond energy} = 2 \times 415 - 202.21 \, \text{kJ/mol} = 627.79 \, \text{kJ/mol}\][/tex]
3. **Calculate the average C-H bond energy in ethyne (C₂H₂):**
Given reaction:
[tex]\[C2H2 (g) + 3H2 (g) \rightarrow 2CH4 (g) \quad \Delta H = -376.74 \, \text{kJ/mol}\][/tex]
This reaction breaks one C≡C bond and adds four C-H bonds.
Change in bond energy = Total energy of bonds broken - Total energy of bonds formed
[tex]\[= 1 \times (\text{C≡C bond energy}) - 4 \times (\text{C-H bond energy})\][/tex]
From the given reaction, the change in bond energy is -376.74 kJ/mol.
Substituting the known values:
[tex]\[-376.74 \, \text{kJ/mol} = 1 \times (\text{C≡C bond energy}) - 4 \times (415 \, \text{kJ/mol})\][/tex]
Solving for the C≡C bond energy:
[tex]\[\text{C = C bond energy} = 4 \times 415 - 376.74 \, \text{kJ/mol} = 528.26 \, \text{kJ/mol}\][/tex]
So, the average C-H bond energy in ethane [tex](C2H6) is \(764.93 \, \text{kJ/mol}\), in ethene (C2H4) is \(627.79 \, \text{kJ/mol}\), and in ethyne (C2H2) is \(528.26 \, \text{kJ/mol}\).[/tex]
A sample of 42 mL of carbon dioxide gas was placed in a piston in order to maintain a constant 101 kPa of pressure.
If the gas was cooled from 20°C to -60°C, what was its new volume? ( Don't forget to convert Celsius to Kelvin)
The gas laws involve Charles' Law, showing a direct proportionality between volume and temperature for a gas. The new volume of the gas after being cooled from 20°C to -60°C (converted to Kelvin), which is under constant pressure, is calculated to be approximately 30.6 mL.
Explanation:This question involves the concept of the gas laws in Physics, specifically, Charles' Law which states that the volume of a given mass of an ideal gas is directly proportional to its temperature on an absolute scale if pressure and the amount of gas remain constant.
First, let's convert the Celsius temperatures to Kelvin by adding 273.15. This gives us initial temperature T1 = 20°C + 273.15 = 293.15 K and final temperature T2 = -60°C + 273.15 = 213.15 K. The initial volume V1 is 42 mL.
According to Charles' Law, V1/T1 = V2/T2. Substituting the given values, we find the new volume V2 = V1*(T2/T1) = 42 mL * (213.15 K / 293.15 K) = 30.6 mL.
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Answer: that one
Explanation:
Note that 5° of longitude or latitude equals about 563 km. Use this information and a ruler to create a scale to use with this map. Draw and label your scale in the legend of the map.
Answer:
A suitable scale, say 1 cm: 100 km can be used.
Explanation:
Thinking process:
The best way to approach the question will be to consider the requirements. This is a simple case of scaling. In order to achieve the objective, you need to choose a scale that does not consume space and that presents more details at the same time.
For instance, a scale of 1 cm to 100 km will give me lines which are a little more then 5 cm. This can be presented as:
1: 500
This is appreciable for the paper size.
You breathe in 3.0 L of pure oxygen at 298 K and 1,000 kPa to fill your lungs. How many moles of oxygen did you take in? Use the ideal gas law: PV = nRT where R=8.31 L−kPa/mol−K
A. 0.05 MOLE
B. 1.21 MOLES
C. 2.42 MOLES
D. 20.0 MOLES
Answer:
Option B.
Explanation:
Let's replace the data given in the formula of Ideal Gases Law.
P. V = n . R . T
3L . 1000kPa = n . 8.31 L.kPa/mol.K . 298K
(3L . 1000 kPa) / ( 8.31 L.kPa/mol.K . 298K) = n
1.21 moles
The number of moles of oxygen by using ideal gas equation is 1.21 moles. Hence, the correct option is B.
Ideal Gas equation[tex]PV=nRT[/tex]
Here, P is the pressure, V is the volume, n is the number of moles, R is the gas constant, T is the temperature.
We can calculate the number of moles by using the above equation as follows:-
[tex]PV = n R T\\ 1000kPa\times 3L = n \times 8.31 L.kPa/mol.K \times298K\\(3L . 1000 kPa) / ( 8.31 L.kPa/mol.K . 298K) = n\\n=1.21 moles[/tex]
So, the number of moles of oxygen is 1.21 moles.
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A gas mixture of nitrogen and oxygen having a total pressure of 2.50 atm is above 2.0 L of water at 25 °C. The water has 51.3 mg of nitrogen dissolved in it. What is the molar composition of nitrogen and oxygen in the gas mixture? The Henry’s constants for N2 and O2 in water at 25 °C are 6.1×10–4 M/atm and 1.3×10–3 M/atm, respectively.
Answer: The molar composition of nitrogen gas is 0.6 and that of oxygen gas is 0.4
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
Given mass of nitrogen gas = 51.3 mg = 0.0513 g (Conversion factor: 1 g = 1000 mg)
Molar mass of nitrogen gas = 28 g/mol
Volume of solution = 2 L
Putting values in above equation, we get:
[tex]\text{Molarity of nitrogen gas}=\frac{0.0513g}{28g/mol\times 2L}\\\\\text{Molarity of nitrogen gas}=9.16\times 10^{-4}mol/L[/tex]
To calculate the partial pressure, we use the equation given by Henry's law, which is:
[tex]C_{N_2}=K_H\times p_{N_2}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]6.1\times 10^{-4}mol/L.atm[/tex]
[tex]C_{N_2}[/tex] = molar solubility of nitrogen gas = [tex]9.16\times 10^{-4}mol/L[/tex]
Putting values in above equation, we get:
[tex]9.16\times 10^{-4}mol/L=6.1\times 10^{-4}mol/L.atm\times p_{N_2}\\\\p_{N_2}=\frac{9.16\times 10^{-4}mol/L}{6.1\times 10^{-4}mol/L.atm}=1.50atm[/tex]
We are given:
Total pressure of the mixture = 2.50 atm
Partial pressure of oxygen gas = 2.50 - 1.50 = 1.00 atm
To calculate the mole fraction of a substance at 25°C, we use the equation given by Raoult's law, which is:
[tex]p_{A}=p_T\times \chi_{A}[/tex] ......(1)
where,
[tex]p_A[/tex] = partial pressure
[tex]p_T[/tex] = total pressure
[tex]\chi_A[/tex] = mole fraction
For nitrogen gas:We are given:
[tex]p_{N_2}=1.50atm\\p_T=2.50atm[/tex]
Putting values in equation 1, we get:
[tex]1.50atm=2.50\times \chi_{N_2}\\\\\chi_{N_2}=\frac{1.50}{2.50}=0.6[/tex]
For oxygen gas:We are given:
[tex]p_{O_2}=1.00atm\\p_T=2.50atm[/tex]
Putting values in equation 1, we get:
[tex]1.00atm=2.50\times \chi_{O_2}\\\\\chi_{O_2}=\frac{1.00}{2.50}=0.4[/tex]
Hence, the molar composition of nitrogen gas is 0.6 and that of oxygen gas is 0.4
The mole fraction of nitrogen and oxygen is 0.6 and 0.4 respectively.
Data Given;
Total pressure P = 2.50 atmvolume of the water = 2.0Lmass of water = 51.3mg = 0.0513gHenry constant for N2 = 6.1*10^-4 M/atmHenry constant for O2 = 1.3*10^-3 M/atmHenry's LawThe law states that the mass of a dissolved gas in a given volume of solvent at equilibrium will be proportional to the partial pressure of the gas.
Mathematically;
C = KP
c = concentration of the gasK = Henry's constantP = partial pressureThe number of moles Nitrogen dissolved is
[tex]n = mass/ molar mass\\ n = 0.0513/28\\ n = 0.00183127 moles[/tex]
The concentration of Nitrogen in water is
[tex]\frac{0.00183127}{2} *1 = 0.0009156M[/tex]
Applying Henry's law,
[tex]0.0009156 = 6.1*10^-^4 * P\\ P = 1.5atm[/tex]
The partial pressure of nitrogen in the mixture is 1.5atm
The total pressure of the gas is 2.50atm
Partial Pressure of oxygen = total pressure - partial pressure of nitrogen
partial pressure of oxygen = 2.50 - 1.50 = 1
Pressure FractionThe pressure fraction of the gas is the ratio between the partial pressure to the total pressure
pressure fraction of nitrogen = 1.5/2.5 = 0.6
pressure fraction of oxygen = 1/2.5 = 0.4
But partial pressure is equal to molar fraction.
This makes the mole fraction of nitrogen equals 0.6 and mole fraction of oxygen equals 0.4.
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In most chemical reactions the amount of product obtained is
Answer:
called theoretic yield
In most chemical reactions, the amount of product obtained is typically less than the theoretical yield due to various reasons such as incomplete reactions, loss of product during isolation, or the reversible nature of some reactions.
Explanation:In most chemical reactions, the amount of product obtained is usually less than the theoretical yield which is the amount predicted by a stoichiometric calculation based on the number of moles of all reactants present. This is because some reactants may not react to form the product, some products may be lost during the isolation step, or the reaction may not go to completion because it is reversible. For instance, if you mix 10 grams of hydrogen with 10 grams of oxygen in a sealed container, you won't get 20 grams of water, but less, because not all the hydrogen and oxygen will react.
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What is the density of a piece of aluminum that has a mass of 270 grams and a volume of 100.0 cubic centimeters? If this piece was cut in half what would be the density of one of the pieces?
The density of the original piece of aluminum is 2.7 grams per cubic centimeter. If one piece is cut in half, the density of each piece would remain the same.
Explanation:In order to find the density of an object, you divide the mass of the object by its volume. The mass of the aluminum piece is 270 grams and its volume is 100.0 cubic centimeters. So, the density would be 270 grams divided by 100.0 cubic centimeters, which equals 2.7 grams per cubic centimeter.
If the piece of aluminum is cut in half, the mass of each piece would be half of the original mass, so each piece would have a mass of 135 grams. Since the volume of each piece remains the same, 100.0 cubic centimeters, the density of one of the pieces would still be 2.7 grams per cubic centimeter.
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What must be true for diffusion of a solute to occur across a partition that separates two compartments?
Answer:
The following must be true for diffusion of a solute to occur across a partition that separates two compartments
The partition is semi permeable
There is concentration gradient across the partition
The process is known as Osmosis
Explanation:
Diffusion is the migration of particles of a substance, liquid or gas, from a region of high concentration to a region of lower concentration and it only occurs where there is a concentration gradient for example spraying a perfume at a corner of a room, the spray sent spreads across the room as there is a concentration gradient
Osmosis is the movement or diffusion of solvent molecules across a semi permeable membrane from a region of low solute concentration to a region of high solute concentration to create a balanced concentration of solute on both sides of the compartment.
Diffusion requires a concentration gradient and a membrane permeable to the solute; in osmosis, water diffuses to balance solute concentrations.
For the diffusion of a solute to occur across a partition that separates two compartments, there must be a concentration gradient present, where the solute molecules move from an area of higher concentration to an area of lower concentration. Additionally, permeability of the membrane to the solute is crucial, as only materials capable of passing through will diffuse.
In certain situations, such as osmosis, water will move across the membrane to balance the concentration gradient of solutes, when the solutes themselves cannot pass through the membrane. In living systems, osmosis is a dynamic and continuous process that maintains the balance of fluids.
When light is absorbed by chlorophyll that has been extracted into a solvent solution, it _________ releases the energy in the form of ____________
Answer:
Photosystems PSI (P700) and PSII (P680) and photolysis of water releases energy
In the form of movement of electrons which is later extracted as ATP
Explanation:
On absorbing light energy electrons obtained from photolysis of water and those present in and those present in the reaction centres PSI and PSII move to the higher energy level and electron acceptors accept them. They then move down through the electron transport chain to the lower energy level during which ATP is produced which is a chemical form of energy.
When atmospheric carbon dioxide reacts with water in the atmosphere, it forms carbonic acid. This acid can then react with some forms of rock, weakening it and changing its chemical composition. What is this known as?
Answer:
The terms used to describe the given process in Chemical weathering.
Explanation:
Weathering of rock is defined as breaking down of the rock into small pieces.
There are two types pf weathering :
Mechanical weathering : This weathering is due to change in physical parameters : temperature change, pressure change etc.
For example : When water soaked up in cracks or crevices of rocks freezes it expands and physically breaks the rock.
Chemical weathering : This weathering is decomposition of rocks due to action of chemicals.This type of weathering also changes chemical composition of rocks.
For example : when gases like carbon dioxide, sulfur oxide get dissolves in water present in rock to form weak acid which ease up the dissolving of rock in that weak acid.
Based upon the mass of baking soda (NaHCO3) and using an excess of HCl in this experiment, you will 1) determine the mass of CO2 produced (actual yield); 2) calculate a theoretical yield for CO2; and 3) calculate a percent yield for CO2.
Equation of reaction
NaHCO3 + HCl ---------> NaCl + H2O + CO2(g)
1)The mass(actual yield) of CO2 can be gotten by isolating it from other products and getting its mass.
Assume 1.0g of CO2 was gotten as a product of the experiment
2) For the theoretical yield
The mass of NaHCO3 was not stated, but for the purpose of this solution, I would assume 2g of NaHCO3 was used for the experiment. You can substitute any parameter by following the steps I follow
Number of mole of NaHCO3 = Mass of NaHCO3/ Molar Mass of NaHCO3
Number of mole of NaHCO3 = 2/84.01 = 0.0238 moles
1 mole of NaHCO3 yielded 1 mole of CO2
0.0238 moles of NaHCO3 will yield 0.0238 moles of CO2
Mass of CO2 = Number of moles * Molar Mass
Mass of CO2 = 0.0238 * 44 = 1.0472g
Theoretical yield of CO2 = 1.0472 grams
3) Percentage yield of CO2 = Actual yield/Theoretical yield *100%
Percentage yield of CO2 = 1.0/1.0472 *100
Percentage yield of CO2 = 95.49%
Answer:
1) The mass of [tex]CO_{2}[/tex] produced is 1.29g
2) The theoretical yield for [tex]CO_{2}[/tex] is 1.311g
3) The percent yield is 98.4%
Explanation:
Step 1: in an experiment let's allow sodium bicarbonate (baking soda) to react with hydrochloric acid HCl to obtain high yield [tex]CO_{2}[/tex].
[tex]NaHCO_{3}+HCl[/tex] ⇒ [tex]NaCl + H_{2} O + CO_{2}[/tex]
from the balanced equation it can be seen that 1 mole of [tex]NaHCO_{3[/tex] produces 1 mole of [tex]CO_{2}[/tex].
Step 2: Let assume 2.5g of [tex]NaHCO_{3[/tex] were used
mole of [tex]NaHCO_{3[/tex] = 2.5g/molecular mass of [tex]NaHCO_{3[/tex]
molecular mass of[tex]NaHCO_{3[/tex] = 23+1+12+(16×2) = 84g/mol
converting mass of [tex]NaHCO_{3[/tex] into mole we have:
[tex]\frac{2.5g}{84gmol^{-1} }[/tex]= 0.0298 moles
Step 3: since [tex]NaHCO_{3[/tex] : [tex]CO_{2}[/tex] mole is ratio 1:1, then 0.0298 mole of [tex]NaHCO_{3[/tex] will produce 0.0298 mole of [tex]CO_{2}[/tex]
The mass of this amount of [tex]CO_{2}[/tex] = 0.0298 × molecular mass of [tex]CO_{2}[/tex]
= 0.0298 mol × ( 12+(16×2))g/mol = 1.3111g of [tex]CO_{2}[/tex]
therefore; the theoretical yield expected is 1.311g
Step 4: Let assume for the experiment that we obtained 1.29g of [tex]CO_{2}[/tex] as the actual yield
then, percent yield for [tex]CO_{2}[/tex] = [tex]\frac{Actual yield obtained}{theoretical yield that should have been obtained}[/tex] × 100%
= 1.2g/1.311g × 100% =98.4%
What makes a nucleus unstable
Answer:
Unstable Nuclei. ... Too many neutrons or protons upset this balance disrupting the binding energy from the strong nuclear forces making the nucleus unstable. An unstable nucleus tries to achieve a balanced state by given off a neutron or proton and this is done via radioactive decay.
Explanation:
A compound contains carbon, hydrogen, and chlorine. It has a molar mass of 98.95 g/mol. Analysis of a sample shows that it contains 24.27% carbon and 4.07% hydrogen. What is its molecular formula?
Answer:
C2H4Cl2
Explanation:
Firstly, we know that the compound contains only three elements. These are carbon, hydrogen and oxygen. We have the percentage compositions of carbon and hydrogen, thus we need the one for chlorine. To get the one for chlorine, we simply subtract that of carbon and hydrogen from a total of 100%.
Hence percentage composition of chlorine = 100 - 24.27 - 4.07 = 71.66%
Now, we divide the percentage compositions by the atomic masses. The atomic masses of carbon, hydrogen and chlorine are 12, 35.5 and 1 respectively. We go on to the divisions as follows.
C = 24.27/12 = 2.0225
H = 4.07/1 = 4.07
Cl = 71.66/35.5 = 2.02
We then go on to divide each by the smallest which is 2.02
C = 2.0225/2.02 = 1
H = 4.07/2.02 = 2
Cl = 2.02/2.02 = 1
Hence the empirical formula is CH2Cl
Now, since the molecular mass is 98.95, we need to calculate the molecular formula
Hence, [CH2Cl]n = 98.95
[12 + 2(1) + 35.5]n = 98.95
[12 + 2 + 35.5]n = 98.95
49.5n = 98.95
n = 98.95/49.5 = 2
The molecular formula is thus [CH2Cl]2 = C2H4Cl2
One mole of aspartame (C14H18N2O5) reacts with two moles of water to produce one mole of
aspartic acid (C4H7NO4), one mole of methanol (CH3OH) and one mole of phenylalanine.
a. What is the molecular formula of phenylalanine?
b. What mass of phenylalanine is produced from 378 g of aspartame?
Answer:
a. Molecular formula of phenylalanine → C₉H₁₁NO₂
b. 234.2 g of C₉H₁₁NO₂
Explanation:
Let's think the reaction:
C₁₄H₁₈N₂O₅ + 2H₂O → CH₃OH + C₉H₁₁NO₂ + C₄H₇NO₄
In order to work with this reaction, we assume that water is on excess.
Let's determine the moles of aspartame → molar mass = 294 g/mol
Moles = mass / molar mass → 378 g / 294 g/mol = 1.28 moles
As ratio is 1:1, 1 mol of aspartame can produce 1 mol of phenylalanine; therefore 1.28 moles of aspartame must produce 1.28 moles of phenylalanine.
Let's convert the moles to mass → molar mass C₉H₁₁NO₂ = 183 g/mol
Moles . molar mass = Mass → 1.28 mol . 183 g/mol = 234.2 g
A. The molecular formula of phenylalanine is C₉H₁₁NO₂
B. The mass of phenylalanine, C₉H₁₁NO₂ produced from 378 g of aspartame is 212.14 g
A. Determination of the molecular formula of phenylalanine.
To obtain the molecular formula of phenylalanine, we shall write the balanced equation. This is given below
C₁₄H₁₈N₂O₅ + 2H₂O → C₄H₇NO₄ + CH₃OH + C₉H₁₁NO₂
Thus, the molecular formula of phenylalanine is C₉H₁₁NO₂
B. Determination of the mass of phenylalanine, C₉H₁₁NO₂ produced from 378 g of aspartame
C₁₄H₁₈N₂O₅ + 2H₂O → C₄H₇NO₄ + CH₃OH + C₉H₁₁NO₂
Molar mass of C₁₄H₁₈N₂O₅ = (14×12) + (18×1) + (14×2) + (16×5) = 294 g/mol
Mass of C₁₄H₁₈N₂O₅ from the balanced equation = 1 × 294 = 294 g
Molar mass of C₉H₁₁NO₂ = (9×12) + (11×1) + 14 + (16×2) = 165 g/mol
Mass of C₉H₁₁NO₂ from the balanced equation = 1 × 165 = 165 g
From the balanced equation above,
294 g of C₁₄H₁₈N₂O₅ reacted to produce 165 g of C₉H₁₁NO₂
Therefore,
378 g of C₁₄H₁₈N₂O₅ will react to produce = (378 × 165) / 294 = 212.14 g of C₉H₁₁NO₂
Thus, 212.14 g of C₉H₁₁NO₂ were obtained from the reaction.
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