Answer:
The temperature at which the vapor pressure would be 0.350 atm is 201.37°C
Explanation:
The relationship between variables in equilibrium between phases of one component system e.g liquid and vapor, solid and vapor , solid and liquid can be obtained from a thermodynamic relationship called Clapeyron equation.
Clausius- Clapeyron Equation can be put in a more convenient form applicable to vaporization and sublimation equilibria in which one of the two phases is gaseous.
The equation for Clausius- Clapeyron Equation can be expressed as:
[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]
where ;
[tex]P_1[/tex] is the vapor pressure at temperature 1
[tex]P_ 2[/tex] is the vapor pressure at temperature 2
∆Hvap = enthalpy of vaporization
R = universal gas constant
Given that:
[tex]P_1[/tex] = 1 atm
[tex]P_ 2[/tex] = 0.350 atm
∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol
[tex]T_1[/tex] = 282 °C = (282 + 273) K = 555 K
R = 8.314 J/mol/k
Substituting the above values into the Clausius - Clapeyron equation, we have:
[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]
[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{T_2 - 555}{555T_2} \end {pmatrix} }[/tex]
[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]
[tex]- 1.0498= 3427.953 \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]
[tex]\dfrac{- 1.0498}{3427.953}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]
[tex]- 3.06246906 \times 10^{-4}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]
[tex]\dfrac{1}{T_2} = \begin {pmatrix} \dfrac{1}{555}+ (3.06246906 \times 10^{-4} ) \end {pmatrix} }[/tex]
[tex]\dfrac{1}{T_2} = 0.002108048708[/tex]
[tex]T_2 = \dfrac{1}{0.002108048708}[/tex]
[tex]\mathbf{T_2 }[/tex] = 474.37 K
To °C ; we have [tex]\mathbf{T_2 }[/tex] = (474.37 - 273)°C
[tex]\mathbf{T_2 }[/tex] = 201.37 °C
Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C
The temperature of the liquid at the given vapor pressure is 201.5 ⁰C.
The given parameters;
boiling point temperature, = 282 ⁰Cvapor pressure, P₂ = 0.35 atmenthalpy of vaporization, ∆Hvap = 28.5 kJ/molThe temperature of the liquid will be determined by applying Clausius- Clapeyron Equation;
[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )[/tex]
where;
R is ideal gas constant = 8.314 J/mol.kT₁ is the initial temperature in Kelvin = 282 + 273 = 555 KP₁ is the initial pressure = 1 atm[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )\\\\ln(\frac{0.35}{1} ) = \frac{28.5 \times 10^3}{8.314} (\frac{T_2 - 555}{555T_2} )\\\\-1.049 = 6.176- \frac{3427.95}{T_2} \\\\\frac{3427.95}{T_2} = 6.176 + 1.049\\\\\frac{3427.95}{T_2} = 7.225\\\\T_2 = \frac{3427.95}{7.225} \\\\T_2 = 474.5 \ K\\\\T_2 = 474.5 - 273 = 201.5 \ ^0C[/tex]
Thus, the temperature of the liquid at the given vapor pressure is 201.5 ⁰C.
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When NaOH is added to water, the (OH) = 0.04 M. What is the [H30*]?
What is the PH of the solution?
Answer:
[H₃O⁺] = 2.5 × 10⁻¹³ M
pH = 12.6
Explanation:
Step 1: Given data
Concentration of OH⁻: 0.04 M
Step 2: Calculate the concentration of H₃O⁺
Let's consider the self-ionization of water reaction.
2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)
The ionic product of water is:
Kw = [OH⁻] × [H₃O⁺] = 10⁻¹⁴
[H₃O⁺] = 10⁻¹⁴ / [OH⁻]
[H₃O⁺] = 10⁻¹⁴ / 0.04
[H₃O⁺] = 2.5 × 10⁻¹³ M
Step 3: Calculate the pH
The pH is:
pH = -log [H₃O⁺] = -log 2.5 × 10⁻¹³ = 12.6
What is a chemical formula, and what does it tell you?
Explanation:
chemical formula tells you the specific elements included in the compound and the number of atoms of each. The letters in a chemical formula are the symbols for the specific elements. So for example, H means hydrogen.
Explanation:
A chemical formula is a way of presenting information about the chemical properties of atoms that contribute a particular chemical compound or molecule, using chemical elements.
how do elasticity ang flexibility differ?
Answer:
the object will regain its original form as soon as the deforming force is removed. ... Flexibility means the object is just easily deformed, and will stay that way.
Explanation:
A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 198 picometers and its molar mass is 195.08 g/mol. Calculate the density of the metal in g/cm3.
Answer:
7.38 g/cm³ is the density of the metal
Explanation:
In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).
To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.
Volume of the unit cellVolume = a³
a = √8×r
(r = 198x10⁻¹²m)
a = 5.6x10⁻¹⁰ m
Volume = 1.756x10⁻²⁸ m³
1m = 100cm → 1m³ = (100cm)³:
1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =
1.756x10⁻²² cm³ → Volume of the unit cell in cm³Mass of the unit cell:There are 4 atoms of gold:
4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold
As 1 mole weighs 195.08g:
6.64x10⁻²⁴ moles of gold × (195.08g / mol) =
1.296x10⁻²¹g is the mass of the unit cellDensity of the metal:1.296x10⁻²¹g / 1.756x10⁻²² cm³ =
7.38 g/cm³ is the density of the metalThe density of the metal is 7.40 g/cm³
In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.
It is expressed by using the formula:
[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]
where;
[tex]\rho[/tex] = density of the metalZ = atoms coordination no = 4 (for FCC)Molar mass (M) = 195.8 g/molAvogadro's constant (NA) = 6.022 × 10²³ /mola = edge lengthFor face-centered cubic FFC;
The edge length [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]
[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]
[tex]\mathbf{a =560.0285 \ pm }[/tex]
a = 5.60 × 10⁻⁸ cm
Replacing it into the previous equation, we have:
[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]
[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]
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1. Why is it not possible to resolve the compound CH3-NH-CH2-CH3 into a pair of enantiomers?
2. Which one of the following is not affected (or is least affected) by the lone pair of electrons on an amine's nitrogen?
a. solubility in alcohols and in water.
b. hydrogen-bond formation.
c. melting point.
d. dipole moment.
e. basicity.
3. Which of the following compounds is most basic?
a. cyclohexyl amine.
b. p-nitroaniline.
c. 2,6-dimethylaniline.
d. p-methoxyaniline.
d. aniline.
Answer:
1. In the compound, H3C-NH-CH2-CH3, there are no chiral centers present, chiral centers refer to the configuration in which carbon is attached with four different groups. In the molecules, as there are no chiral centers, therefore the molecule is optically inactive, that is, it will not demonstrate pair of an enantiomer is one of the essential characteristics of optically active compounds is the possession of enantiomeric pairs.
2. On the nitrogen of aniline, the lone pair of electrons can produce hydrogen bonds, play an essential function in basicity, play an essential role in dipole moment or polarity, and wit the increase in solubility there is an increase in the formation of the hydrogen bond, eventually increasing to boiling point. However, the melting point is not affected. As the melting point is the characteristic of the packing efficacy of a molecule and does not rely upon the anilinic nitrogen's lone pairs.
3. With the increase in the tendency to donate an electron, basicity increases. However, if the electron is taking part in resonance, the donation will not take place easily, and the compound will be the least basic. Apart from cyclohexyl amine, in all the other given compounds, the lone pair of nitrogen takes part in the process of delocalization or conjugation. Thus, cyclohexyl amine will be most basic as the lone pairs are easily available for donation.
Read the article. Use your understanding to answer the questions that follow. What type of source is this article? primary or secondary and how do you know
Answer: C
Explanation:
The article was sourced from the Oak National Laboratory
Which reasons did you include in your response? Check all of the boxes that apply.
1. The article does not present original research.
and
3. The article has references to primary sources.
Answer:
C
Explanation:
Which reasons did you include in your response? Check all of the boxes that apply.
The article does not present original research.
The article summarizes other research.
The article has references to primary sources.
What subatomic particles surround the nucleus? Question 1 options: protons neutrons atoms electrons
Answer:
Electrons "surround"
Explanation:
Protons and neutrons "make up" the nucleus so they are contained "within" the nucleus meaning that electrons would "surround" the nucleus as they orbit around the nucleus
Answer:
Electrons
Explanation:
Protons and nuetrons are present inside the nucleus of an atom while the electron revolve around the nucleus in different energy levels.
The half-life of radium-226 is 1620 years. What percentage of a given amount of the radium will remain after 900 years
Answer:
68%
Explanation:
Since we need a percentage we can use any number we want for our initial value.
5(1/2)^900/1620 = 3.40
(3.40 / 5)*100 = 68%
To make sure lets use a different initial amount
1(1/2)^900/1620 = 0.68
(0.68/1) * 100 = 68%
To solve this question, we'll assume the initial amount of radium-226 to be 1.
Now, we shall proceed to obtaining the percentage of radium-226 that will after 900 years. This can be obtained as illustrated below:
Step 1Determination of the number of half-lives that has elapsed.
Half-life (t½) = 1620 years
Time (t) = 900 years
Number of half-lives (n) =?[tex]n = \frac{t}{t_{1/2}}\\\\n = \frac{900}{1620}\\\\n = \frac{5}{9}[/tex]
Step 2:Determination of the amount remaining
Initial amount (N₀) = 1
Number of half-lives (n) = 5/9
Amount remaining (N) =?[tex]N = \frac{N_{0} }{2^{n}}\\\\N = \frac{1}{2^{5/9}}[/tex]
N = 0.68Step 3Determination of the percentage remaining.
Initial amount (N₀) = 1
Amount remaining (N) = 0.68
Percentage remaining =?Percentage remaining = N/N₀ × 100
Percentage remaining = 0.68/1 × 100
Percentage remaining = 68%Therefore, the percentage amount of radium-226 that remains after 900 years is 68%
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The element europium exists in nature as two isotopes: has a mass of u and has a mass of u. The average atomic mass of europium is u. Calculate the relative abundance of the two europium isotopes.
Answer:
Problem Details
The element europium exists in nature as two isotopes: 151Eu has a mass of 150.9196 amu, and 153Eu has a mass of 152.9209 amu. The average atomic mass of europium is 151.96 amu. Calculate the relative abundance of the two europium isotopes.
answer:
151Eu = 48%, 153Eu = 52%
Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation you submit is balanced.
_______ → Ba(ClO)2 + H2O(l)
Answer:
2HClO(aq) + Ba(OH)₂(aq) → Ba(ClO)₂(aq) + 2H₂O(l)
Explanation:
The reaction corresponds to a neutralization reaction between an acid and a base, as follows:
2HClO(aq) + Ba(OH)₂(aq) → Ba(ClO)₂(aq) + 2H₂O(l)
From the equation above we have that the acid HClO reacts with the base Ba(OH)₂ to obtain a salt Ba(ClO)₂ and water.
In the balanced reaction, we have that 2 moles of HClO react with 1 mol of Ba(OH)₂ to produce 1 mol of Ba(ClO)₂ and 2 moles of water.
I hope it helps you!
At a constant temperature, a sample of a gas in a balloon that originally had a volume of 5.00 L and pressure of 626 torr has its volume changed to 6.72 L. Calculate the new pressure in torr.
Answer:
466 torr
Explanation:
Step 1: Given data
Initial pressure (P₁): 626 torrInitial volume (V₁): 5.00 LFinal pressure (P₂): ?Final volume (V₂): 6.72 LConstant temperatureStep 2: Calculate the final pressure
Since we have a gas changing at a constant temperature, we can calculate the final pressure using Boyle's law.
P₁ × V₁ = P₂ × V₂
P₂ = P₁ × V₁ / V₂
P₂ = 626 torr × 5.00 L / 6.72 L
P₂ = 466 torr
at 89 ∘C∘C , where [Fe2+]=[Fe2+]= 3.60 MM and [Mg2+]=[Mg2+]= 0.310 MM . Part A What is the value for the reaction quotient, QQQ, for the cell?
Answer:
8.6×10^-2
Explanation:
The reaction is;
Mg(s) + Fe^2+(aq) -----> Mg^2+(aq) + Fe(s)
This implies that;
Q = [Mg^2+]/[Fe^2+]
But;
[Fe2+]= 3.60 M
[Mg2+]= 0.310 M
Q= [0.310 M]/[3.60 M]
Q= 0.086
Q= 8.6×10^-2
Using these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)/Cu(s) -0.40 V -0.76 V ‑0.25 V +0.34 V Calculate the standard cell potential for the cell whose reaction is Ni2+(aq) + Zn(s) →Zn2+(aq)+ Ni(s)
Answer: The standard cell potential for the cell is +0.51 V
Explanation:
Given : [tex]E^0_{Ni^{2+}/Ni}=-0.25V[/tex]
[tex]E^0_{Zn^{2+}/Zn}=-0.76V[/tex]
The given reaction is:
[tex]Ni^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Ni(s)[/tex]
As nickel is undergoing reduction, it acts as cathode and Zinc is undergoing oxidation, so it acts as anode.
[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]
where both [tex]E^0[/tex] are standard reduction potentials.
Thus putting the values we get:
[tex]E^0_{cell}=-0.25-(-0.76)[/tex]
[tex]E^0_{cell}=0.51V[/tex]
Thus the standard cell potential for the cell is +0.51 V
An ice cube at 0.00C with a mass of 8.32g is placed Into 55g of water, initially at 25C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (answer must be in 3 sig figs)
Answer:
The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0. It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it
Thats all i know
Which of the following are characteristics of ionic
compounds?
Answer:
Ionic compounds have high melting points.
Ionic compounds are hard and brittle.
Ionic compounds dissociate into ions when dissolved in water.
Solutions of ionic compounds and melted ionic compounds conduct electricity, but solid materials do not.
Explanation:
You are a paleontology professor working at a dig site looking for fossils. You come across a deposit that is emitting radiation. Upon further testing you find that the sample is changing from carbon (atomic number 6) into nitrogen (atomic number 7) as radiation is emitted. What type of radiation is it?
Answer:
β particles
Explanation:
The most common radioactive isotope of carbon is C-13.
The unbalanced nuclear equation is
[tex]\rm _{6}^{13}C \longrightarrow \, ? + \, _{7}^{13N}[/tex]
Let's write the question mark as a nuclear symbol.
[tex]\rm _{6}^{13}C} \longrightarrow \, _{Z}^{A}X+ \, _{7}^{13}N[/tex]
The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.
Then
13 = A + 13, so A = 13 - 13 = 0, and
6 = Z + 7, so Z = 6 - 7 = -1
Then, your nuclear equation becomes
[tex]\rm _{6}^{13}C \longrightarrow \, _{-1}^{0}M + \, _{7}^{13}N[/tex]
The particle with "zero" mass and a charge of -1 is an electron, so the balanced nuclear equation is
[tex]\rm _{6}^{13}C \longrightarrow \, _{-1}^{0}e + \, _{7}^{13}N[/tex]
The radiation consists of β particles (electrons)
Answer:
I think think that the one above me is beta radiation
Explanation:
Read the following statement:
Energy cannot be created or destroyed.
Does the statement describe a scientific law? (3 points)
a
No, because it universally applies to all objects
b
No, because it is not true in all circumstances
c
Yes, because it universally applies to all objects
d
Yes, because it is not true in all circumstances
Answer:
C. yes, because it is universally applies to all objects
Does a reaction occur when aqueous solutions of potassium hydroxide and chromium(III) bromide are combined
Explanation:
Potassium hydroxide = KOH
Chromium(iii)bromide = CrBr3
Yes! A reaction occurs. This is given by the balanced equation;
3 KOH + CrBr3 → 3 KBr + Cr(OH)3
How has the work of chemists affected the environment over the years?
Answer:
Chemistry is one of the causes for global warming, and in some cases it can even cause certain illnesses.
Answer:
Chemists have both hurt the environment and helped the environment by their actions.
Explanation:
<3
A student puts a glass of water in the freezer. Later, he notices ice forming on the surface of the water. Which property of water best explains why ice forms on its surface? A. It is made of polar molecules. B. It has low surface tension. C. It has weak adhesion. D. It is densest as a solid.
ℯ ℴ ℴ ℴℯℯ
it has a weak adhesion
whats the ph for a solution poh4 9.78 concentration of solution
Answer:
4.22
Explanation:
According to the question, the pOH of the solution is 9.78. You may recall that pOH is the hydroxide concentration of a solution.
Also pOH = -log[OH^-]. Hence the pOH is obtained from the hydroxide ion concentration.
Finally, pH + pOH =14
Hence;
pH = 14-pOH
pH= 14-9.78 = 4.22
pH= 4.22
The intermolecular forces present in CH 3NH 2 include which of the following? I. dipole-dipole II. ion-dipole III. dispersion IV. hydrogen bonding
Answer:
I. dipole-dipole
III. dispersion
IV. hydrogen bonding
Explanation:
Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.
London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.
Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.
Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.
Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.
Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.
Answer:
The intermolecular forces present in CH_3NH_2 includes
II. (ion-dipole) and IV. (hydrogen bonding)Explanation:
The intermolecular forces present in CH_3NH_2 includes II. (ion-dipole) and IV. (hydrogen bonding)
It is a polar molecule due to NH polar bond and it can form Hydrogen bond also due to NH bond.
Interaction will be dipole- dipole and Hydrogen dispersion forces can always be taken into account.
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Why must beta particles be used to detect leaks in a pipe?
Why must beta particles be used to detect leaks in a pipe?
A. Beta particles will not be absorbed by the soil like gamma radiation, but won't pass through the pipe before reaching the leak like alpha radiation would.
B. Beta particles will not cause an electrical discharge like alpha particles when they interact with the metal pipe, or contaminate the water like gamma radiation.
C. Beta particles are not used, only alpha particles are used because they are not harmful to humans.
D. Beta particles will not be absorbed by the soil like alpha particles, but won't pass through the pipe before reaching the leak like gamma radiation would.
Should be D, because alpha particles are absorbed by soil and gamma isn't
Beta particles will not be absorbed by the soil like alpha particles, but won't pass through the pipe before reaching the leak.
What is Beta particle?This type of particle is a high-speed electron and is derived from the process of beta decay.
It is used to detect leaks in pipe because it will not be absorbed by the soil like alpha particles, but won't pass through the pipe before reaching the leak like gamma radiation would.
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Consider the bond dissociation energies listed below in kcal/mol. CH3-Br 70 CH3CH2-Br 68 (CH3)2CH-Br 68 (CH3)3C-Br 65 These data show that the carbon-bromine bond is weakest when bromine is bound to a __________.
Answer:
The answer is "Tertiary carbon".
Explanation:
Accent to the results, the carbon-bromine bond is weak, whenever, the bromine is connected to tertiary carbon so, bonding energy is separation for methyl-carbon, which is connected to the bromine = 70 kcal/mol and for the primary energy to the secondary energy is= 68 kcal/mol, and for tertiary CO2 = 65 kcal/mol.
The stronger the energy dissociating connection and the weaker, its power dissociation connection and its weaker bond becomes connecting with a tertiary carbon, that's why "Tertiary carbon" is the correct answer.
Briefly describe how you can use spectra as evidence of nuclear fusion in stars.
Spectra can be used as an evidence that nuclear fusion occurs in the star through the ability of the elements generated to absorb light at specific wavelength.
The internal part of the star is made up of hydrogen gas, with a little helium.
The helium atom is formed through the nuclear fusion reaction that occurs in the core of the star.
Nuclear fusion reaction involves the combination of two or more atoms which leads to the formation of a new atom with the emission of great energy. This reaction leads to formation of helium elements in the core of the stars.
To detect if nuclear fusion reaction is occurring in the stars, a spectrum is used.
Because each element emits or absorbs light only at specific wavelengths, the chemical composition of stars can be determined using a spectrum.
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In a fixed cylinder are 3moles of oxygen gas at 300Kelvin and 1.25atm. What is the volume of the container?
Answer:
The volume of the container is 59.112 L
Explanation:
Given that,
Number of moles of Oxygen, n = 3
Temperature of the gas, T = 300 K
Pressure of the gas, P = 1.25 atm
We need to find the volume of the container. For a gas, we know that,
PV = nRT
V is volume
R is gas constant, R = 0.0821 atm-L/mol-K
So,
[tex]V=\dfrac{nRT}{P}\\\\V=\dfrac{3\ mol\times 0.0821\ L-atm/mol-K \times 300\ K}{1.25\ atm}\\\\V=59.112\ L[/tex]
So, the volume of the container is 59.112 L
You wish to construct a galvanic cell with the anode consisting of a Ni electrode in a 1.0 M Ni(NO3)2 solution. What would be the highest standard cell potential if used as the cathode in this galvanic cell?
Answer:
Au^3+(aq) +3e ------> Au(s). 1.50 V
Explanation:
When we construct the galvanic cell, our intention is to produce energy by spontaneous electrochemical reactions. In order to have a spontaneous electrochemical reaction, E°cell must be positive. The more positive the value of E°cell, the more spontaneous the reaction is.
E°cell= E°cathode - E°anode
If E°cathode= 1.50 V
E°anode= -0.25 V
E°cell= 1.50 -(-0.25)
E°cell= 1.75 V
Hence the process; Au^3+(aq) +3e ------> Au(s) yields the highest standard cell potential
Predict the most likely bond type for the following.
a. Cu (Copper)
b. KCl (Potassium Chloride)
c. Si (Silicon)
d. CdTe (Cadmium Telluride)
e. ZnTe (Zinc Telluride)
Answer:
The three major types of bond are ionic, polar covalent, and covalent bonds. Ionic occurs majorly between metals and non-metals, which allows sharing of electrons to form an ionic compound. Whereas covalent bonding calls for complete transfer of electrons between atoms. Polar covalent bonds have unequaly shared electron-pair between two atoms.
Explanation:
a. Cu (Copper)- ionic bonding
b. KCl (Potassium Chloride) - ionic bonding
c. Si (Silicon) - covalent bonding
d. CdTe (Cadmium Telluride) - polar covalent bonding
e. ZnTe (Zinc Telluride)- polar covalent bonding
Calculate the pH of a solution that is 0.210 M in nitrous acid (HNO2) and 0.290 M in potassium nitrite (KNO2). The acid dissociation constant of nitrous acid is 4.50 × 10-4.
Answer:
pH = 3.49
Explanation:
We have a buffer system formed by a weak acid (HNO₂) and its conjugate base (NO₂⁻ coming from KNO₂). We can calculate the pH of a buffer ssytem using the Henderson-Hasselbach equation.
pH = pKa + log [base] / [acid]
pH = -log Ka + log [NO₂⁻] / [HNO₂]
pH = -log 4.50 × 10⁻⁴ + log 0.290 M / 0.210 M
pH = 3.49
The pH of the solution containing 0.210 M nitrous acid (HNO₂) and 0.290 M potassium nitrite (KNO₂) is 3.49
We'll begin by calculating the the pKa of acid. This can be obtained as follow:
Acid dissociation constant (Ka) = 4.50×10¯⁴
pKa =?pKa = –Log Ka
pKa = –Log 4.50×10¯⁴
pKa = 3.35Finally, we shall determine the pH of the solution.pKa = 3.35
Concentration of HNO₂, [HNO₂] = 0.210 M
Concentration of KNO₂, [KNO₂] = 0.290 M
pH =?The pH of the solution can obtain by using the Henderson-Hasselbach equation as illustrated below:
pH = pKa + log [base] / [acid]pH = pKa+ log [NO₂⁻] / [HNO₂]
pH = 3.35 + log (0.290 / 0.210)
pH = 3.49Thus, the pH of the solution is 3.49
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Compound A is an alkene that was treated with ozone to yield only (CH3CH2CH2)2C=O. Draw the major product that is expected when compound A is treated with a peroxy acid (RCO3H) followed by aqueous acid (H3O+).
Answer:
2,2,3,3-tetrapropyloxirane
Explanation:
In this case, we have to know first the alkene that will react with the peroxyacid. So:
What do we know about the unknown alkene?
We know the product of the ozonolysis reaction (see figure 1). This reaction is an oxidative rupture reaction. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If [tex](CH_3CH_2CH_2)_2C=O[/tex] is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.
What is the product with the peroxyacid?
This compound in the presence of alkenes will produce peroxides. Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product 2,2,3,3-tetrapropyloxirane. (see figure 2)