Answer:
Explanation:
Oxidation is defined as the reaction of oxygen and a substrate which could be a metal, non-metal etc. Pure Silicon can be found to be too reactive and hence forms alloys with non-metals.
Therefore, oxidation of silicon will form a layer of silicon dioxide on the surface of the silicon and hence, the crystal Silicon structure is partly lost with the formation of an amorphous SiO2. An example of a feasible oxidation of silicon is thermal oxidation which follows the equation:
Si + 2H2O -> SiO2 + 2H2
Si + O2 -> SiO2
Final answer:
The number of SiO2 molecules per unit volume in nm-3 is calculated to be 22.8 molecules per nm³ based on the density and molar mass of SiO2. This calculation reveals the considerable volume expansion that occurs when the surface of a Si crystal oxidizes to form SiO2, potentially impacting semiconductor properties.
Explanation:
To calculate the number of SiO2 molecules per unit volume in nm-3, we first need to find the molar mass of SiO2. The atomic masses of Si and O are 28.09 and 16, respectively. Thus, the molar mass of SiO2 is 28.09 + 2(16) = 60.09 g/mol.
Given the density of SiO2 is 2.27 g/cm3, we can calculate the number of moles in 1 cm3 as follows:
Number of moles = density / molar mass = 2.27 g/cm3 / 60.09 g/mol = 0.0378 mol/cm3.Since 1 mol contains Avogadro's number of molecules (6.022 x 1023 molecules/mol), the number of molecules in 1 cm3 is 0.0378 mol/cm3 x 6.022 x 1023 = 2.28 x 1022 molecules/cm3.Converting cm3 to nm3, where 1 cm3 equals 1 x 1021 nm3, the number of molecules per nm3 is (2.28 x 1022) / (1 x 1021) = 22.8 molecules/nm3.Regarding the effect of surface oxidation on a Si crystal, the expansion of volume during the transformation from Si to SiO2 implies that the material becomes less densely packed with increased volume. Given that 0.44 Å of Si is used to obtain 1.0 Å of SiO2, this indicates that the oxidation process introduces more space within the structure due to the larger volume of SiO2 compared to Si. This expansion could affect the electrical and mechanical properties of silicon components, particularly in semiconductor applications, where precise control of material properties is essential.
Covalent bonds in a molecule absorb radiation in the IR region and vibrate at characteristic frequencies.
A molecular vibration absorbs radiation of wavelength 188 μm. What frequency (in s−1) corresponds to that wavelength? Enter your answer in scientific notation.
Answer:
[tex]Frequency=1.6\times 10^{12}\ Hz[/tex]
Explanation:
The relation between frequency and wavelength is shown below as:
[tex]c=frequency\times Wavelength [/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
Given, Wavelength = 188 μm
Also, 1 μm = [tex]10^{-6}[/tex] nm
So,
Wavelength = [tex]188\times 10^{-6}[/tex] m
Thus, Frequency is:
[tex]Frequency=\frac{c}{Wavelength}[/tex]
[tex]Frequency=\frac{3\times 10^8}{188\times 10^{-6}}\ Hz[/tex]
[tex]Frequency=1.6\times 10^{12}\ Hz[/tex]
Three of the following statements are wrong. One is correct. Which of the following statements is correct?
Enzymes require optimum pH for maximal activity
Enzymes are denatured at high temperature but in certain exceptional organisms they are effective even at temperatures 80∘−90∘C
Enzymes are highly specific
Most enzymes are proteins but some are lipids
Answer:
The incorrect statement is "Most enzymes are proteins but some are lipids" the others statements are correct regarding to the Enzymes.
Explanation:
Enzymes are complex macromolecules of globular proteins. However, many enzymes contain certain non-protein substances associated with them for their function, which are known as cofactors and can be organic or inorganic compounds. In addition, the enzymes are thermolabile, presenting their best performance at the ideal temperature. Enzyme activity decreases with increasing and increases with increasing temperature and stops at 0 degrees and above 80 degrees. However, the enzymes of the bacteria that inhabit the hot springs have an ideal temperature of 70 degrees or more. Enzymes also show maximum activity at optimal pH. Varying its activity with increasing or decreasing pH. Enzymes are specific in their action. An enzyme can catalyze only a specific type of reaction or even act on a specific substrate. For example, the enzyme lactase catalyzes the hydrolysis of lactose and no other disaccharides.
A driver with a nearly empty fuel tank may say she is "running on fumes."
If a 15.0 gallon automobile gas tank had only gasoline vapor remaining in it, what is the farthest the vehicle could travel if it gets 20.0 miles per gallon on liquid gasoline?
Assume the average molar mass of molecules in gasoline is 101 g/mol , the density of liquid gasoline is 0.75 g/mL, the pressure is 747 mmHg , and the temperature is 25 ∘C.
Answer:
The farthest the vehicle could travel (if it gets 20.0 miles per gallon on liquid gasoline) is 1.62 miles.
Explanation:
The automobile gas tank has a volume capacity of 15 gallons which can be converted to liters: 15 × 3.7854 = 56.781 liters
We can find the moles of gasoline by using the ideal gas equation: PV = nRT.
Make n (number of moles) the subject of the formula: n = PV/RT, where:
P = 747 mmHg
V = 56.781 liters
R (universal gas constant) = 0.0821 liter·atm/mol·K
T = 25 ∘C = (273 + 25) K = 298 K
1 atm (in the unit of R) = 760 mmHg
Therefore n = 747 × 56.781/(0.0821 × 760 × 298) = 2.281 mol.
Given that the molar mass of the gasoline = 101 g/mol,
the mass of gasoline = n × molar mass of gasoline = 2.281 mol × 101 g/mol = 230.38 g
the density of the liquid gasoline = 0.75 g/mL
In order to calculate the distance the vehicle can travel, we have to calculate volume of gasoline available = mass of the liquid gasoline ÷ density of liquid gasoline
= 230.38 g ÷ 0.75 g/mL = 307.17 mL = 0.3071 liters = 0.3071 ÷ 3.7854 = 0.0811 gallons
since the vehicle gets 20.0 miles per gallon on liquid gasoline, the distance traveled by the car = gallons available × miles per gallon = 0.0811 × 20 = 1.62 miles.
A substance that is required to run the vehicle and the machine is known as fuel. These substance are as follows:-
PetrolDiesalsKerosineAccording to the data given in the question.
The farthest the vehicle could travel is 1.62 miles.
The volume capacity of 15 gallons which can be converted to liters: [tex]15 * 3.7854 = 56.781 liters[/tex]
The formula used is as follows:-[tex]PV =nRT[/tex]
After putting the value:-
P = 747 mmHg
V = 56.781 liters
R (universal gas constant) = 0.0821 liter·atm/mol·K
[tex]T = 25C = (273 + 25) K = 298 K[/tex]
Therefore, the value of [tex]n = \frac{747 * 56.781}{(0.0821 * 760 * 298} = 2.281 mol.[/tex]
Given that the molar mass of the gasoline = 101 g/mol,
Mass of gasoline = n × molar mass of gasoline
[tex]= 2.281 mol * 101 g/mol = 230.38 g[/tex]
the density of the liquid gasoline = 0.75 g/mL
[tex]= \frac{230.38}{0.75}= 307.17 mL \\\\= 0.3071 liters\\\\= \frac{0.3071}{3.7854} = 0.0811 gallons[/tex]
Since the vehicle gets 20.0 miles per gallon on liquid gasoline, the distance traveled by car = gallons available × miles per gallon =
[tex]0.0811 * 20 = 1.62 miles.[/tex]
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The work function of palladium (Pd) is 5.22 eV. What is the minimum frequency of light to observe the photoelectric effect from a Pd surface? If light with a 200 nm wavelength is absorbed by the Pd surface, what is the velocity of the emitted electrons?
Answer:
1)4×10^13Hz
2) 9.95×10-9J
Explanation:
From the image attached, it is clear that the work function of the palladium metal must first be obtained in joules. Then, the frequency is obtained from E=hf.
The kinetic energy if the photoelectrons is obtained as the difference between the energy of the photon and the work function of the metal.
(b) Data has been collected to show that at a given wavelength in a 1 cm pathlength cell, Beer's Law for the absorbance of Co2+ is linear. If a 0.135 M solution of Co2+ has an absorbance of 0.350, what is the concentration of a solution with an absorbance of 0.460?
Answer : The concentration of a solution with an absorbance of 0.460 is, 0.177 M
Explanation :
Using Beer-Lambert's law :
[tex]A=\epsilon \times C\times l[/tex]
where,
A = absorbance of solution
C = concentration of solution
l = path length
[tex]\epsilon[/tex] = molar absorptivity coefficient
From this we conclude that absorbance of solution is directly proportional to the concentration of solution at constant path length.
Thus, the relation between absorbance and concentration of solution will be:
[tex]\frac{A_1}{A_2}=\frac{C_1}{C_2}[/tex]
Given:
[tex]A_1[/tex] = 0.350
[tex]A_2[/tex] = 0.460
[tex]C_1[/tex] = 0.135 M
[tex]C_2[/tex] = ?
Now put all the given values in the above formula, we get:
[tex]\frac{0.350}{0.460}=\frac{0.135}{C_2}[/tex]
[tex]C_1=0.177M[/tex]
Therefore, the concentration of a solution with an absorbance of 0.460 is, 0.177 M
Beer's Law states that there is a linear relationship between the concentration of a substance in a solution and its absorbance. We can use the equation c = A/(εb) to solve for the concentration of a solution with a given absorbance.
Explanation:Beer's Law states that there is a linear relationship between the concentration of a substance in a solution and its absorbance. The equation for Beer's Law is A = εbc, where A is the absorbance, ε is the molar absorptivity, b is the path length of the cell, and c is the concentration.
In this case, we have the absorbance (0.350) and concentration (0.135 M) for a solution of Co2+. We can rearrange the equation to solve for the concentration: c = A/(εb). Plug in the given values to find the molar absorptivity and path length, and then substitute those values into the equation to calculate the concentration of a solution with an absorbance of 0.460.
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Consider a solution that contains 0.274 M potassium chloride and 0.155 M magnesium chloride.
What is the concentration of chloride ions in this solution?
Answer:
Concentration of chloride ions = 0.584M
Explanation:
The step by step calculations is shown as attached below.
What kind of intermolecular forces act between a hydrogen iodide molecule and a dichloroethylene
Final answer:
The intermolecular forces between a hydrogen iodide molecule and a dichloroethylene molecule involve both dispersion forces and dipole-dipole forces.
Explanation:
The intermolecular forces between a hydrogen iodide molecule (HI) and a dichloroethylene molecule (C2H2Cl2) involve both dispersion forces and dipole-dipole forces.
Dispersion Forces: Iodine (I) in HI is a larger atom compared to chlorine (Cl) in dichloroethylene. As a result, HI has stronger dispersion forces due to the larger number of electrons, making the attraction between HI molecules stronger than between dichloroethylene molecules.Dipole-Dipole Forces: Dichloroethylene is a polar molecule with a partial positive charge on the hydrogen atom and partial negative charges on the chlorine atoms. HI is also a polar molecule with a partial positive charge on the hydrogen atom and a partial negative charge on the iodine atom. The positive end of the dipole in HI is attracted to the negative end of the dipole in dichloroethylene, creating dipole-dipole interactions between the two molecules.Overall, the intermolecular forces between a hydrogen iodide molecule and a dichloroethylene molecule include both dispersion forces and dipole-dipole forces.
Arrange the following compounds in order of increasing solubility in water: O2, LiCl, Br2, methanol (CH3OH).
The order of increasing solubility in water for O2, Br2, LiCl, and methanol (CH3OH) is: O2 < Br2 < LiCl < Methanol (CH3OH). This is based on whether the molecules are nonpolar or polar and the intermolecular forces present.
Explanation:The solubility of compounds in water ultimate depends on the specific intermolecular interactions. The key principle here is 'like dissolves like', with polar substances dissolving in polar solvents (like water) and nonpolar substances dissolving in nonpolar solvents.
In your list: O2, LiCl, Br2, and methanol (CH3OH), the order of increasing solubility in water would in fact be: O2 < Br2 < LiCl < Methanol (CH3OH)
O2 and Br2 are both nonpolar molecules and hence have a very limited solubility in polar solvents such as water. LiCl is a polar ionic compound, it will easily dissolve in water through ion-dipole interactions. Methanol (CH3OH) is the most soluble because, in addition to being polar, it can form hydrogen bonds with water molecules, which is a stronger intermolecular force. Learn more about Solubility of Compounds here:
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The order of increasing solubility in water is O₂ < Br₂ < LiCl < CH₃OH.
To arrange these compounds in order of increasing solubility in water, we need to consider their polarity and ability to form hydrogen bonds with water. Water is a polar solvent and dissolves polar compounds and ionic compounds well, but non-polar compounds poorly.
O₂ (Oxygen): Non-polar molecule, very low solubility in water.Br₂ (Bromine): Non-polar molecule, low solubility in water, but slightly more soluble than oxygen due to its larger size and polarizability.LiCl (Lithium Chloride): Ionic compound, highly soluble in water due to dissolution into Li+ and Cl- ions.Methanol (CH₃OH): Polar molecule, high solubility in water due to hydrogen bonding with water molecules.Therefore, the order of increasing solubility in water is O₂ < Br₂ < LiCl < CH₃OH.
In a first-order decomposition reaction. 36.8% of a compound decomposes in 7.6 min. How long (in min) does it take for 88.8% of the compound to decompose?
Answer: 36.5 minutes
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant = 100
a - x = amount left after decay process
a) for completion of 36.8 % of reaction
[tex]7.6=\frac{2.303}{k}\log\frac{100}{100-36.8}[/tex]
[tex]k=\frac{2.303}{7.6}\times 0.19[/tex]
[tex]k=0.060min^{-1}[/tex]
b) for completion of 88.8 % of reaction
[tex]t=\frac{2.303}{k}\log\frac{100}{100-88.8}[/tex]
[tex]t=\frac{2.303}{0.060}\log\frac{100}{11.2}[/tex]
[tex]t=\frac{2.303}{0.060}\times 0.95[/tex]
[tex]t=36.5min[/tex]
It will take 36.5 minutes for 88.8% of the compound to decompose.
Final answer:
To accurately determine the time required for 88.8% of a compound to decompose in a first-order reaction, knowing the rate constant is essential. While the question provides a specific data point for decomposition over time, it lacks the rate constant needed for a direct calculation. Therefore, accurately answering this specific question based on the provided context is challenging without additional information.
Explanation:
The question asks for the time required for 88.8% of a compound to decompose in a first-order decomposition reaction given that 36.8% decomposes in 7.6 minutes. In first-order reactions, the time it takes for a certain percentage of the reactant to decompose does not depend on the initial concentration but on the rate constant (k) of the reaction. The integrated rate law for first-order reactions is given by the formula ln([A]0/[A]) = kt, where [A]0 is the initial concentration, [A] is the concentration at time t, and k is the rate constant.
To determine the time required for 88.8% decomposition, we would need to know the rate constant of the reaction. However, with the information provided, we can infer the rate constant by using the time and percent decomposition already given (36.8% in 7.6 minutes), but to find the exact time for 88.8% decomposition without additional specific information (e.g., actual rate constant value) is not directly feasible based on the provided context alone. Knowing the rate constant, we could then apply the formula to find the time for 88.8% of the compound to decompose.
Without the rate constant, an alternative but less precise method involves understanding that the time to reach a certain level of decomposition is related to the half-life of the reaction. Since the time for half of a reactant to decompose (its half-life) in a first-order reaction is constant, we can indirectly estimate times for specific percentages of decomposition if we know the half-life, which again depends on knowing the rate constant.
Predict the major organic product for the reaction below. Draw the product in the planar (overhead) representation, clearly showing stereochemistry by drawing in a wedge or hashed bond per stereocenter.
Answer:
The main organic product for the reaction is shown in the following figure.
Explanation:
This is due to the presence of the substituent -CH3 in the molecule, which makes it impossible to leave the hydrogen to form the double bond in the elimination.
The question involves predicting the major product of a given reaction based on stereochemistry, utilizing concepts from organic chemistry such as skeletal structure and the reactivity of alkenes. Understanding these principles can help predict the product, though without the specified diagram, an exact prediction cannot be made.
Explanation:Your question relates to predicting the major organic product of a given reaction by examining the stereochemistry, which involves the three-dimensional arrangement of atoms in a molecule. Indeed, stereochemistry is a critical aspect of organic chemistry because it influences the properties and reactions of molecules.
Organic chemists often represent large molecules using a skeletal structure or line-angle structure. In this drawing style, carbon atoms are at the ends or bends of lines. Hydrogens attached to carbons are not drawn, and atoms other than carbon and hydrogen are represented by their symbols.
Alkenes are particularly reactive due to the presence of a C=C moiety, a reactive functional group known to undergo addition reactions. For instance, in the halogenation of alkenes, halogens add to the double bond of the alkene instead of replacing hydrogen as they would in an alkane. The stereochemistry of the reaction is significant because the spatial arrangement of atoms can influence the ability of halogens to add across the C=C bond.
Unfortunately, without the specific planar diagram or additional information about the reaction in question, I cannot predict the exact major product; however, this general information about the reactivity and stereochemistry of alkenes might help you predict it.
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In what way are planned economies and command economies similar?
O
A. They both are characteristic of the United States.
O
B. They both rely on private ownership.
C. They both involve strong governmental control.
O
D. They both require open exchange.
SUBMIT
Answer: D
Explanation:
Answer:
I believe the answer is C I took the test a little bit ago
Explanation:
A meter of polyaramide fiber has a diameter of 710.μm , a mass of 0.059g , and takes 0.13s to produce on an industrial spinneret. How would you calculate the mass of fiber that can be spun in 55.min ?Set the math up. But don't do any of it. Just leave your answer as a math expression.Also, be sure your answer includes all the correct unit symbols.
In order to calculate the mass of fiber that can be spun in 55 min we have to determine the rate of production.
Generally speaking, all the different types of rates have in common a variable divided over time. In this case the variable we're interested in is the mass of fiber, so we have to divide the mass of fiber over the time it takes to spin, and then if we want to find the mass of fiber for any given time, we just have to multiply the time and the rate together.
Keep in mind that the time that is given to us to find the rate is in seconds, so we have to convert that to minutes
The math expression would be the following:
[tex]mass\ of \ fiber (g)=time(min)*\frac{0,059 g}{0.13s}*\frac{60 s}{1 min} \\\\mass\ of \ fiber (g)=55 min*\frac{0,059 g}{0.13s}*\frac{60 s}{1 min}[/tex]
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Gaseous compound Q contains only xenon and oxygen. When a 0.100 g sample of Q is placed in a 50.0-mL steel vessel at 0 C, the pressure is 0.230 atm. What is the likely formula of the compound?
The question is incomplete, here is the complete question:
Gaseous compound Q contains only xenon and oxygen. When a 0.100 g sample of Q is placed in a 50.0-mL steel vessel at 0°C, the pressure is 0.230 atm. What is the likely formula of the compound?
A. XeO
B. [tex]XeO_4[/tex]
C. [tex]Xe_2O_2[/tex]
D. [tex]Xe_2O_3[/tex]
E. [tex]Xe_3O_2[/tex]
Answer: The chemical formula of the compound is [tex]XeO_4[/tex]
Explanation:
To calculate the molecular weight of the compound, we use the equation given by ideal gas equation:
PV = nRT
Or,
[tex]PV=\frac{w}{M}RT[/tex]
where,
P = Pressure of the gas = 0.230 atm
V = Volume of the gas = 50.0 mL = 0.050 L (Conversion factor: 1 L = 1000 mL)
w = Weight of the gas = 0.100 g
M = Molar mass of gas = ?
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = Temperature of the gas = [tex]0^oC=273K[/tex]
Putting value in above equation, we get:
[tex]0.230\times 0.050=\frac{0.100}{M}\times 0.0821\times 273\\\\M=\frac{0.100\times 0.0821\times 273}{0.230\times 0.050}=194.9g/mol\approx 195g/mol[/tex]
The compound having mass as calculated is [tex]XeO_4[/tex]
Hence, the chemical formula of the compound is [tex]XeO_4[/tex]
In ionic bonds the atom that contributes an electron and has a positive charge as a result is called the . The atom in ionic bonding that accepts an electron and has a negative charge as a result is called the .
Answer:
a. Cation
b. Anion
Explanation:
Given the reactions below, answer the following questions.
Cl_2(g) + F_2(g) rlhar 2ClF(g) delta G degree_rxn = 115.4 kJ/mol
Cl_2(g) + Br_2(g) rlhar 2ClBr(g) delta G degree_rxn = -2.0 kJ/mol
Calculate the delta G degree_rxn for 2ClF(g) + Br_2(g) rlhar 2ClBr(g) + F_2(g) __________ kJ/mol
Answer:
[tex]\Delta G_{rxn}^{0}=-117.4kJ/mol[/tex]
Explanation:
Gibbs free energy is an additive property.
[tex]2ClF(g)\rightarrow Cl_{2}(g)+F_{2}(g)[/tex] ; [tex]\Delta G_{1}^{0}=-115.4kJ/mol[/tex]
[tex]Cl_{2}(g)+Br_{2}(g)\rightarrow 2ClBr(g)[/tex] ; [tex]\Delta G_{2}^{0}=-2.0kJ/mol[/tex]
----------------------------------------------------------------------------
[tex]2ClF(g)+Br_{2}(g)\rightarrow 2ClBr(g)+F_{2}(g)[/tex] ; [tex]\Delta G_{rxn}^{0}=\Delta G_{1}^{0}+\Delta G_{2}^{0}=(-115.4-2.0)kJ/mol=-117.4kJ/mol[/tex]
So, standard gibbs free enrgy change for the given reaction is -117.4 kJ/mol
The change in free energy is called G (∆G). The [tex]\rm \Delta G^\circ rxn= -117.4\; kj/mol[/tex]
What is Gibbs free energy?Gibb free energy is the measure of the maximum amount of reversible work done in a thermodynamic system.
It means only when the temperature change, but the pressure is constant.
[tex]\rm Cl_2(g) + F_2(g) \longrightarrow 2ClF(g) \;delta\; G\;degree_rxn = 115.4 kJ/mol[/tex]
[tex]\rm Cl_2(g) + Br_2(g) \longrightarrow 2ClBr(g) \;delta\; G \;degree_r_x_n = -2.0 kJ/mol[/tex]
[tex]\rm \Delta G^\circ rxn= \Delta G^\circ1 +G^\circ2[/tex]
[tex]2ClF(g) + Br_2(g) \longrightarrow 2ClBr(g) + F_2(g)[/tex]
[tex]\rm \Delta G^\circ rxn= -115.4\;kj/mol - 2.0 \;kj/mol = -117.4\;kj/mol[/tex]
Thus, the [tex]\rm \Delta G^\circ rxn= -117.4\; kj/mol[/tex].
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The surface tension of water is 7.28 ✕ 10−2 J/m2 at 20°C. Predict whether the surface tension of heptane would be higher or lower than that of water at the same temperature. Explain your answer.
Answer:
Lower
Explanation:
Surface tension occurs because molecules at the surface do not have molecules above them, so they cohere more strongly to their neighbours.
The stronger cohesive forces make it more difficult to move an object through the surface than when it is beneath the surface.
The attractive forces in water are strong because of hydrogen bonding.
A hexane molecule is nonpolar, so the only attractions are the weak London dispersion forces.
The cohesive forces at the surface are much lower than those in water, so the surface tension of hexane is lower than that of water at the sane temperature.
Specify which atoms, if any, bear a formal charge in the Lewis structure given and the net charge for the species. Be sure to answer all parts.
a. Formal charge C:___________.
b. Formal charge N:________.
d. Net charge:________.
This is an incomplete question, here is a complete question.
Specify which atoms, if any, bear a formal charge in the Lewis structure given and the net charge for the species. Be sure to answer all parts.
[tex]:C\equiv N:[/tex]
a. Formal charge C:___________.
b. Formal charge N:________.
d. Net charge:________.
Answer :
a. Formal charge C : (-1) charge
b. Formal charge N : (0) charge
d. Net charge : (-1) charge
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, [tex]CN^-[/tex]
As we know that carbon has '4' valence electrons and nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in [tex]CN^-[/tex] = 4 + 5 + 1 = 10
According to Lewis-dot structure, there are 6 number of bonding electrons and 4 number of non-bonding electrons.
Now we have to determine the formal charge for each atom.
Formula for formal charge :
[tex]\text{Formal charge on C}=4-2-\frac{6}{2}=-1[/tex]
[tex]\text{Formal charge on N}=5-2-\frac{6}{2}=0[/tex]
Thus, the net charge will be = -1 + 0 = -1
The highest concentration of aqueous HCL is ~ 37% by weight, and the density of this solution is 1.18 g/cm3. What is the molarity of HCl of this 37% aqueous HCl solution?
Answer: The molarity of HCl of this 37% aqueous HCl solution is 12.0 M
Explanation:
Given : 37 g of HCl is dissolve in 100 g of solution.
Mass of solute (HCl) = 37 g
Mass of solution = 100 g
Density of solution = 1.18g/ml
Volume of solution =[tex]\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.18g/ml}=84.7ml[/tex]
Molarity is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute
[tex]V_s[/tex] = volume of solution in ml = 1000 ml
[tex]moles of solute =\frac{\text {given mass}}{\text {molar mass}}=\frac{37g}{36.5g/mol}=1.01moles[/tex]
Now put all the given values in the formula of molarity, we get
[tex]Molarity=\frac{1.01moles\times 1000}{84.7ml}=12.0mole/L[/tex]
The molarity of HCl solution is 12.0 M
Reaction 1: 2NO(g) + Br2(g) ⇄ 2NOBr(g) ΔH°rxn = −16 kJ Kp = 27 at 25°CReaction 2: CO2(g) ⇄ C(s) + O2(g) ΔH°rxn = +100 kJ Kp = 2.6x10−9 at 25°C8. For which reaction(s) above would an increase in reaction vessel volume at constant temperature result in an increase in product yield? A) Reaction 1 only B) Reaction 2 onlyC) Both reactionsD) Neither reactionFor which reaction(s) above would an increase in temperature at constant volume result in an increase in product yield? A) Reaction 1 only B) Reaction 2 onlyC) Both reactionsD) Neither reaction
Answer:
First question: D)
Second question: B)
Explanation:
A reversible reaction intends to achieve the equilibrium, a state in the velocity of product formation is equal to the velocity of reactants formation. This equilibrium can be disturbed by some alterations in the reaction, and, by Le Chatelier's principle, the reaction will shift in to reestablish the equilibrium.
The equilibrium may be shift for three principal factors: concentration, temperature, and pressure. If a concentration of some of the substance increases, the reaction will shift to consume it; if it decreases, the reaction is shifted to form more this substance.
When the temperature of the system increases, the reaction shifts for the consume of the heat, thus, the endothermic reaction (ΔH°rxn > 0) is favored; if it decreases, the exothermic reaction (ΔH°rxn < 0) is favored. If the direct reaction is endothermic, the inverse is exothermic, and vice versa.
When the pressure of the system increases, the volume decreases, so, the equilibrium shifts for the less gas volume, or the side with fewer moles of gas substance; if the pressure decreases the inverse occurs.
So, if the vessel volume increases, the pressure will decrease, and so, the formation of more gas substances is favored. In reaction 1, in the reactants, there are 3 moles ( 2 of NO + 1 of Br2), and in the products 2 moles, so the reactants are favored, and the product yield decreases. In reaction 2, there is 1 mol on both sides of the reaction, so the pressure doesn't affect the equilibrium.
The increase in temperature favors the endothermic reaction. The reaction 2 has it's direct reaction endothermic, so the product yield increases. In reaction 1, the reactants increases, because the inverse reaction is endothermic.
An increase in temperature at constant volume will result in an increase in product yield in Reaction 1 only.
Explanation:An increase in temperature at constant volume will result in an increase in product yield for Reaction 1 only. This is because the reaction in Reaction 1 is exothermic, meaning it releases heat. When the temperature increases, the reaction will shift in the direction that absorbs heat to counteract the increase in temperature. Since Reaction 1 releases heat, an increase in temperature will favor the formation of products and increase the product yield. For Reaction 2, an increase in temperature will not result in an increase in product yield because the reaction is endothermic, meaning it requires heat as a reactant. Increasing the temperature will shift the reaction in the direction that releases heat, reducing the product yield.
Draw the structure of the major organic product isolated from the reaction of 1-hexyne with hydrogen chloride (2 mol). Click the "draw structure" button to launch the drawing utility.
Answer:
The structure of the major organic product isolated from the reaction of 1-hexyne with hydrogen chloride (2 mol) is attached below.
Explanation:
Hydracids are added to triple bonds by a mechanism similar to that of the addition to double bonds. The regioselectivity of the addition of H-X to the triple bond follows the rule of Markovnikov, where the Z conformation predominates in the addition of halide to the alkyne, because in the formation of the carbocation it prefers to place the positive charge on the more substituted carbon where the nucleophilic attack of the halide ion will occur.
With respect to the halogenation of the alkene, the same procedure occurs at the time of the formation of the carbocation, joining the nucleophilic ion to the most substituted carbon.
Identify the functional group(s) that appear in betaxolol. This compound is in a class of drugs called beta-blockers, which are used to lower blood pressure, lower heart rate, reduce angina (chest pain), and reduce the risk of recurrent heart attacks Alcohol Ether Arene Carboxylic Acid Aldehyde Ester Amine Alkene KetoneFigure:contains some chemical structures
Answer:
AlcoholEtherAreneAmineExplanation:
In the attached picture you may find the structure of betaxolol.
You can see the alcohol group C-O-H as well as the ether group C-O-C.
The arene -or aromatic ring- can also be seen.
There's also a secondary amine group, C-NH-C.
Classify each of the following particulate level illustrations as a representation of either a pure substance, a homogeneous mixture, or a heterogeneous mixture.Figure:three compounds were drawn
Answer:
The classification and illustrations are attached in the drawing.
Explanation:
It is possible to identify the pure substance observing the figure, since it is the only one that has 2 joined atoms (purple and blue) which forms a single compound.
On the other hand, the homogeneous mixture is identified by noting that its atoms are more united with respect to the heterogeneous mixture, highlighting that in homogenous mixtures the atoms, elements or substances are not visible to the naked eye and are in a single phase, instead in the heterogeneous mixture if they can be differentiated.
Heterogeneous mixture: Visible multiple chemicals with distinct parts.
Homogeneous mixture: Appears as one material, uniform with indistinguishable atoms and elements.
Pure material distinguished by figure with linked blue and purple atoms forming a single compound.
Heterogenous combination of visible compounds. Homogenous mixture comprised of various materials that seem like one substance Each sample portion has the same makeup and properties. The graphic shows that the pure material is the only one with two connected atoms (blue and purple) that form a compound.
The homogeneous mixture is distinguished by its more unified atoms, emphasizing that while in homogeneous mixtures the atoms, elements, and substances are not visible to the eye and are in a single phase, in the heterogeneous mixture they can be distinguished.
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Type of bond Average Bond Enthalpy (kJ/mol) O-O 150O=O 500The conversion of ozone to diatomic oxygen is represented by the equation above. Based on the data in the table above, what is the approximate average bond enthalpy for the oxygen-to-oxygen bonds in ozone? a. 0 kJ/mol b. 150 kJ/mol c. 300 kJ/mol d. 500 kJ/mol
Missing information:
The reaction is 2O₃(g) --> 3O₂ (g) ΔH = -300 kJ/mol
Answer:
c
Explanation:
In the molecule of ozone, 3 oxygens are bonded, and, because each one needs to share two pairs of electrons, these 2 bonds are something between a simple and a double bond.
In the reaction of the transformation of the ozone to oxygen gas, these bonds are broken, and a double bond is formed between two oxygen atoms. The sum of the energy of the broken and the formation of the bond is the enthalpy variation of the reaction.
To break a bond, energy must be added to the system, so it's an endothermic reaction and energy is positive, so the formation is exothermic and the energy is negative. Because there're 2 ozone molecules, 4 bonds will be broken, and because there are 3 oxygen molecules, 3 bonds will be formed:
4*E - 3*500 = -300
4E = -300 + 1500
4E = 1200
E = 300 kJ/mol
So, each O3 bond has 300 kJ/mol as an average energy.
Each ozone molecule has 300 kJ/mol as an average bond energy.
The reaction is
[tex]\bold { 2O_3(g) \rightarrow 3O_2 (g)\ \ \ \ \ \ \ \ \ \ \ \Delta H = -300 kJ/mol}[/tex]
In the molecule of ozone, there 2 double bonds are present with continuous variation.
The enthalpy variation of the reaction is the sum of the energy of the broken and the formation of the bond.
Because of 2 ozone molecules, 4 bonds will be broken, and because there are 3 oxygen molecules, 3 bonds will be formed.
[tex]\bold {4\times E - 3 \times 500 = -300}\\\\\bold {4\times E = -300 + 1500}\\\\\bold {E = 300 kJ/mol}[/tex]
Therefore, Each ozone molecule has 300 kJ/mol as an average bond energy.
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Calculate the solubility of carbon dioxide in water at an atmospheric pressure of 0.400 atm (a typical value at high altitude).
Answer:
1.40*10⁻² M
Explanation:
We have the solubility formula
Solubility,
S = KH*P
where
KH = measure of hardness of water / carbonate hardness = 3.50*10⁻² mol/L.atm
P = atmospheric pressure = 0.400 atm
Hence, we have
S = KH*P
= (3.50*10⁻² mol/L.atm)*(0.400 atm)
= 1.40*10⁻² mol/L
But 1 mol/L = 1 M,
Hence, the answer (1.40*10⁻² mol/L ) is equivalent to
= 1.40*10⁻² M
A 10.00 mL sample of 8.50 M HNO3 solution is diluted to a new volume of 65.0 mL. What is the concentration of the dilute solution
Answer: 1.31M
Explanation:
V1 = 10mL
C1 = 8.5M
V2 = 65mL
C2 =?
C1V1 = C2V2
10 x 8.5 = C2 x 65
C2 = (10 x 8.5 ) /65
C2 = 1.31M
The concentration of the dilute solution of nitric acid is 1.31 M.
We have 10.00 mL (V₁) of 8.50 M HNO₃ (C₁) and we add water to obtain a dilute solution with a volume (V₂) of 65.0 mL. We can calculate the concentration of the dilute solution (C₂) using the dilution rule.
[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{8.50 M \times 10.00 mL}{65.0 mL} = 1.31 M[/tex]
The concentration of the dilute solution of nitric acid is 1.31 M.
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For the decomposition of A to B and C, A(s)⇌B(g)+C(g) how will the reaction respond to each of the following changes at equilibrium?
a. double the concentrations of both products and then double the container volume
b. double the container volume
c. add more A
d. double the concentration of B and halve the concentration of C
e. double the concentrations of both products
f. double the concentrations of both products and then quadruple the container volume
Answer:
a. No change.
b. The equilibrium will shift to the right.
c. No change
d. No change
e. The equilibrium will shift to the left
f. The equilibrium will shift to the right
Explanation:
We are going to solve this question by making use of Le Chatelier´s principle which states that any change in a system at equilibrium will react in such a way as to attain qeuilibrium again by changing the equilibrium concentrations attaining Keq again.
The equilibrium constant for A(s)⇌B(g)+C(g)
Keq = Kp = pB x pC
where K is the equilibrium constant ( Kp in this case ) and pB and pC are the partial pressures of the gases. ( Note A is not in the expression since it is a solid )
We also use Q which has the same form as Kp but denotes the system is not at equilibrium:
Q = p´B x p´C where pB´ and pC´ are the pressures not at equilibrium.
a. double the concentrations of Q which has the same form as Kp but : products and then double the container volume
Effectively we have not change the equilibrium pressures since we know pressure is inversely proportional to volume.
Initially the system will decrease the partial pressures of B and C by a half:
Q = pB´x pC´ ( where pB´and pC´are the changed pressures )
Q = (2 pB ) x (2 pC) = 4 (pB x PC) = 4 Kp ⇒ Kp = Q/4
But then when we double the volume ,the sistem will react to double the pressures of A and B. Therefore there is no change.
b. double the container volume
From part a we know the system will double the pressures of B and C by shifting to the right ( product ) side since the change reduced the pressures by a half :
Q = pB´x pC´ = ( 1/2 pB ) x ( 1/2 pC ) = 1/4 pB x pC = 1/4 Kp
c. add more A
There is no change in the partial pressures of B and C since the solid A does not influence the value of kp
d. doubling the concentration of B and halve the concentration of C
Doubling the concentrantion doubles the pressure which we can deduce from pV = n RT = c RT ( c= n/V ), and likewise halving the concentration halves the pressure. Thus, since we are doubling the concentration of B and halving that of C, there is no net change in the new equilibrium:
Q = pB´x pC´ = ( 2 pB ) x ( 1/2 pC ) = K
e. double the concentrations of both products
We learned that doubling the concentration doubles the pressure so:
Q = pB´x pC´ = ( 2 pB ) x ( 2 pC ) = 4 Kp
Therefore, the system wil reduce by a half the pressures of B and C by producing more solid A to reach equilibrium again shifting it to the left.
f. double the concentrations of both products and then quadruple the container volume
We saw from part e that doubling the concentration doubles the pressures, but here afterward we are going to quadruple the container volume thus reducing the pressure by a fourth:
Q = pB´x pC´ = ( 2 pB/ 4 ) x (2 pC / 4) = 4/16 Kp = 1/4 Kp
So the system will increase the partial pressures of B and C by a factor of four, that is it will double the partial pressures of B and C shifting the equilibrium to the right.
If you do not see it think that double the concentration and then quadrupling the volume is the same net effect as halving the volume.
Final answer:
The decomposition reaction of A into B and C responds to changes at equilibrium by shifting in a direction that opposes the imposed change, as explained by Le Chatelier's principle. The effects of changes in concentration, container volume, and addition of reactants are used to predict shifts in equilibrium.
Explanation:
Response to Changes in Chemical Equilibrium
For the decomposition of a solid A into gases B and C, represented by the reaction A(s) ⇌ B(g) + C(g), here is how the reaction will respond to changes at equilibrium:
a. Doubling the concentrations of both products B and C and then doubling the container volume will initially cause the reaction to shift to the left (toward A), as the increase in volume will decrease the pressure and favor the formation of fewer gas molecules. However, because the concentrations of the gases are also doubled, the net effect may be smaller, and the system will work to re-establish equilibrium according to Le Chatelier's principle.b. Doubling the container volume alone will shift the equilibrium to the left because the system will try to increase the pressure by forming more gas molecules (favoring the formation of B and C).c. Adding more of the solid reactant A will shift the equilibrium to the right, increasing the production of B and C.d. Doubling the concentration of B and halving the concentration of C will lead to a shift in the equilibrium toward the left, as the system tries to counteract the changes and restore equilibrium.e. Doubling the concentrations of both products, B and C, without changing the volume will shift the equilibrium to the left, as the system tries to reduce the concentrations of the products by forming more A.f. Doubling the concentrations of B and C and then quadrupling the container volume will have a more pronounced effect in shifting the equilibrium to the left than simply doubling the volume, as the change in pressure will be more significant.Each of these actions will induce a response from the reaction to maintain the established equilibrium according to Le Chatelier's principle. The reaction will typically shift in the direction that opposes the change imposed on the system.
I 1 mL o a 0.02% w/v isoproterenol hydrochloride solution is diluted to 10 mL with sodium chloride injection be ore intravenous administration, calculate the percent concentration o the diluted solution
Answer:
0.002% w/v
Explanation:
The unit w/v means that mass (in g) per volume (in mL). When the solution is diluted, the mass of the solvent will not change, and the mass can be calculated by the concentration (C) multiplied by the volume (V). So, if 1 is the initial solution, and 2 the diluted solution:
C1*V1 = C2*V2
C1 = 0.02%
V1 = 1 mL
V2 = 10 mL
0.02*1 = C2*10
10C2 = 0.02
C2 = 0.002% w/v
1-propanol (P1° = 20.9 Torr at 25 °C) and 2-propanol (P2° = 45.2 Torr at 25 °C) form ideal solutions in all proportions. Let x1 and x2 represent the mole fractions of 1-propanol and 2-propanol in a liquid mixture, respectively, and y1 and y2 represent the mole fractions of each in the vapor phase. For a solution of these liquids with x1 = 0.450, calculate the composition of the vapor phase at 25 °C.
y1= y2=
To find the vapor phase composition of a mixture of 1-propanol and 2-propanol at 25 °C, we apply Raoult's Law and Dalton's Law using the given vapor pressures and mole fraction. The calculated vapor phase composition is y1 = 0.2744 for 1-propanol and y2 = 0.7256 for 2-propanol.
The question asks for the composition of the vapor phase at 25 °C for a liquid mixture of 1-propanol and 2-propanol with a given mole fraction of 1-propanol (x1 = 0.450). According to Raoult's Law, the partial pressure of each component in an ideal solution is equal to the mole fraction of the component in the liquid phase times the vapor pressure of the pure component. The total vapor pressure (Ptot) of the solution is the sum of the partial pressures. We can then use Dalton's Law to find the mole fractions of the components in the vapor phase.
To calculate the composition of the vapor phase, we will use the following steps:
Calculate the partial pressure of 1-propanol (P1) using the formula P1 = x1 × P1°, where P1° is the vapor pressure of pure 1-propanol.
P1 = x1 × P1° = 0.450 × 20.9 Torr = 9.405 Torr
Calculate the partial pressure of 2-propanol (P2) using the formula P2 = x2 × P2°, where P2° is the vapor pressure of pure 2-propanol and x2 is the mole fraction of 2-propanol in the liquid phase.
Since the mole fractions must sum to 1,
x2 = 1 - x1 = 1 - 0.450 = 0.550.
Now we calculate P2 = x2 × P2° = 0.550 × 45.2 Torr = 24.86 Torr.
Ptot = P1 + P2 = 9.405 Torr + 24.86 Torr = 34.265 Torr
Calculate the mole fraction of 1-propanol in the vapor phase (y1) using y1 = P1 / Ptot.
y1 = P1 / Ptot = 9.405 Torr / 34.265 Torr = 0.2744
Calculate the mole fraction of 2-propanol in the vapor phase (y2) using y2 = P2 / Ptot.
y2 = P2 / Ptot = 24.86 Torr / 34.265 Torr = 0.7256
Thus, the composition of the vapor phase at 25 °C for the solution with x1 = 0.450 is y1 = 0.2744 and y2 = 0.7256.
The normal boiling point of liquid methyl acetate is 331 K. Assuming that its molar heat of vaporization is constant at 30.6 kJ/mol, the boiling point of CH3COOCH3 when the external pressure is 1.29 atm is
Answer:
T_2=338.9026K
Boiling point of CH3COOCH3 at external pressure is 338.9026K
Explanation:
We are going to use Clausius-Clapeyron Equation:
[tex]ln\frac{P_2}{P_1} =-\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]
Where:
P_2 is the external pressure
P_1 is the atmospheric Pressure=1 atm
ΔH is the heat of vaporization
T_2 boiling point of CH3COOCH3 at external pressure
T_1 normal boiling point of liquid methyl acetate
Now:
[tex]\frac{1}{T_2}=-ln\frac{P_2}{P_1}*\frac{R}{\Delta H}+\frac{1}{T_1} \\\frac{1}{T_2}=-ln\frac{1.29}{1}*\frac{8.314}{30.6*10^3}+\frac{1}{331} \\\frac{1}{T_2}=2.9507*10^-^3\\T_2=\frac{1}{2.9507*10^-^3} \\T_2=338.9026K[/tex]
Boiling point of CH3COOCH3 at external pressure is 338.9026K
To find the new boiling point of methyl acetate, use the Clausius-Clapeyron equation with the provided values for the initial boiling point, molar heat of vaporization, and new external pressure.
Explanation:This question requires us to use the form of the Clausius-Clapeyron equation:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Where T1 and P1 are the initial temperature and pressure (331K and 1 atm respectively, since normal boiling point is defined as the temperature at which the vapor pressure of a liquid equals 1 atm), and P2 is the new pressure (1.29 atm). ΔHvap is the molar heat of vaporization (30.6 kJ/mol, or 30.6 * 10^3 J/mol to convert kilojoules to joules), and R is the gas constant (8.314 J/mol*K). Solving the equation for T2 (the new boiling point temperature), we rearrange to get T2 = 1/(1/T1 + (R/ΔHvap)*ln(P2/P1)). Plugging in the values, you should find the answer.
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In the following reaction, which component acts as an oxidizing agent? 10 I− (aq) + 2 MnO4− (aq) + 16 H+ (aq) → 5 I2 (s) + 2 Mn2+ (aq) + 8 H2O (l)
Answer:
The oxidizing agent is the MnO₄⁻
Explanation:
This is the redox reaction:
10 I⁻ (aq) + 2 MnO₄⁻ (aq) + 16 H⁺ (aq) → 5 I₂ (s) + 2 Mn²⁺ (aq) + 8 H2O (l)
Let's determine the oxidation and the reduction.
I⁻ acts with -1 in oxidation state and changes to 0, at I₂.
All elements in ground state has 0 as oxidation state.
As the oxidation state has increased, this is the oxidation, so the iodide is the reducing agent.
In the permanganate (MnO₄⁻), Mn acts with +7 in oxidation state and decreased to Mn²⁺. As the oxidation state is lower, we talk about the reduction. Therefore, the permanganate is the oxidizing agent because it oxidizes iodide to iodine