Answer:Temperature increases
Explanation: As the gas in the container is an ideal gas so it should follow the ideal gas equation, the equation of state.
We know ideal gas equation to be PV=nRT where
P=pressure
V=Volume
T=Temperature
R=Real gas constant
n=Number of moles
since the gas is insulated such that no heat goes into or out of the system .
When we compress the ideal gas using a piston, Thermodynamically it means that work is done on the system by the surroundings.
Now as the ideal gas is been compressed so the volume of the gas would decrease and slowly a time will reach when no more gas can be compressed that is there cannot be any further decrease in volume of the gas.
From the equation PV=nRT
Once there is no further compression is possible hence volume becomes constant so pressure of the ideal gas becomes directly proportional to the temperature as n and R are constants. Also as the pressure and volume are inversely related so an decrease in volume would lead to an increase in pressure.
As the ideal gas is compressed so the pressure of the gas would increase since the gas molecules have smaller volume available after compression hence the gas molecules would quite frequently have collisions with other gas molecules or piston and this collision would lead to increase in speed of the gas molecules and so the pressure would increase .
The increase in pressure would lead to an increase in temperature as show by the above ideal gas equation because the pressure and temperature are directly related.
So here we can say that work done on the system by surroundings leads to increase in temperature of the system.
Final answer:
When an ideal gas in a piston is adiabatically compressed, its temperature increases due to the increased internal energy resulting from work done on the gas.
Explanation:
When an ideal gas contained in a piston is compressed and the process is adiabatic, meaning no heat flows into or out of the system, the temperature of the gas increases. This is because the work done on the gas during the compression decreases its volume but increases its internal energy.
As the particles of the gas are forced closer together, they collide more frequently and with greater energy, leading to a rise in temperature. The change in temperature during an adiabatic process can be described by the adiabatic condition, which in the case of an ideal gas relates pressure, volume, and temperature in a specific way.
Monosodium glutamate (MSG), a flavor enhancer, has been blamed for "Chinese restaurant syndrome," the symptoms of which are headaches and chest pains. MSG has the following composition by mass: 35.51 percent C, 4.77 percent H, 37.85 percent O, 8.29 percent N, and 13.60 percent Na. What is its molecular formula if its molar mass is about 169 g?
Answer:
The molecular formula of mono sodium glutamate is [tex]C_5H_8O_4N_1Na_1[/tex]
Explanation:
Molar mass of sodium glutamate,M = 169 g/mol
let the molecular formula be [tex]C_aH_bO_cN_dNa_e[/tex]
Percentage of carbon in the M.S.G. =35.52 %
[tex]35.51\%=\frac{a\times 12 g/mol}{169 g/mol}[/tex]
a = 5
Percentage of Hydrogen in the M.S.G. = 4.77 %
[tex]4.77\%=\frac{b\times 1 g/mol}{169 g/mol}[/tex]
b = 8
Percentage of oxygen in the M.S.G. =37.85 %
[tex]8.29\%=\frac{c\times 16 g/mol}{169 g/mol}[/tex]
c = 3.99 ≈ 4
Percentage of nitrogen in the M.S.G. = 8.29 %
[tex]4.77\%=\frac{d\times 14 g/mol}{169 g/mol}[/tex]
d = 1
Percentage of sodium in the M.S.G. =13.60 %
[tex]13.60\%=\frac{c\times 23g/mol}{169 g/mol}[/tex]
e = 0.99 ≈ 1
The molecular formula be :[tex]C_aH_bO_cN_dNa_e=C_5H_8O_4N_1Na_1[/tex]
The molecular formula of Monosodium glutamate is C2H4O4N2Na2
To determine the molecular formula of monosodium glutamate based on its elemental composition, we'll first find the empirical formula, and then calculate the molecular formula.
Find the moles of each element:
Carbon (C): 35.51%
Hydrogen (H): 4.77%
Oxygen (O): 37.85%
Nitrogen (N): 8.29%
Sodium (Na): 13.60%
Calculate the moles of each element using their molar masses:
Moles of C = (35.51/100) * 169 g / (12.01 g/mol) ≈ 5.97 moles
Moles of H = (4.77/100) * 169 g / (1.01 g/mol) ≈ 7.90 moles
Moles of O = (37.85/100) * 169 g / (16.00 g/mol) ≈ 8.37 moles
Moles of N = (8.29/100) * 169 g / (14.01 g/mol) ≈ 9.99 moles
Moles of Na = (13.60/100) * 169 g / (22.99 g/mol) ≈ 9.43 moles
Find the smallest whole number ratio of moles.
Divide all moles by the smallest number of moles (approximately 5.97).
Empirical Formula:
C1H1.32O1.4N1.68Na1.58
Round the subscripts to whole numbers (since you can't have fractions of atoms):
CH2O2N2Na2
The empirical formula of Monosodium glutamate is CH2O2N2Na2.
To find the molecular formula, you need to determine the molar mass of the empirical formula and compare it to the given molar mass (169 g/mol).
The empirical formula mass is 85 g/mol (approximately), which is half of the molar mass.
Therefore, the molecular formula is twice the empirical formula:
Molecular Formula: C2H4O4N2Na2
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f 23.2 grams of an aqueous solution of nickel(II) iodide, NiI2, contains 5.47 grams of nickel(II) iodide, what is the percentage by mass of nickel(II) iodide in the solution?
Answer:
the percentage by mass of Nickel(II) iodide = 23.58%
Explanation:
% by mass of solute = (mass of solute/mass of solution) x 100%
% by mass of NiI2 = (mass of NiI2/mass of solution) x 100%
% by mass of NiI2 = (5.47 grams/23.2 grams) x 100% = 23.58% m/m
10. In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of 27 °C. The mass of the gas was 0.472 g. What was the molar mass of the gas?
To find the molar mass of a gas collected over water, apply the ideal gas law to determine the moles of gas. Then, divide the mass of the gas by the number of moles.
Explanation:Calculating the Molar Mass of a GasTo determine the molar mass of a gas from an experiment in a general chemistry laboratory, you first need to use the ideal gas law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature in Kelvin.
The student collected a gas with a volume of 265 mL, which is equivalent to 0.265 L, at a pressure of 753 torr. Since 1 atm is equal to 760 torr, the pressure in atmospheres is 753 torr / 760 torr/atm = 0.9911 atm. The temperature must be converted from Celsius to Kelvin; thus, 27 °C is equal to 300.15 K (27 + 273.15 = 300.15 K).
To solve for n (the number of moles), you rearrange the ideal gas law to n = PV / RT. With the previously mentioned values and the gas constant R as 0.0821 L·atm/K·mol, n = (0.9911 atm × 0.265 L) / (0.0821 L·atm/K·mol × 300.15 K). After calculation, the number of moles of gas n is found.
Once n is calculated, the molar mass (M) can be found using the formula M = mass of gas (g) / number of moles (mol). Therefore, with the mass of the gas being 0.472 g, we calculate M = 0.472 g / n moles. By plugging in the value of n from our previous calculation, we can determine the molar mass.
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Final answer:
To calculate the molar mass of a gas, the ideal gas law is used with the given pressure, volume, and temperature. The molar mass is found to be approximately 44.22 g/mol after doing the conversions and calculations.
Explanation:
The student's question asks about determining the molar mass of a collected gas sample. To find out the molar mass of the gas, we use the ideal gas law formula (PV = nRT), where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature. Here's the step-by-step solution:
Convert the given pressure from torr to atm (753 torr / 760 torr/atm).
Convert the volume from mL to L (265 mL / 1000 mL/L).
Convert the temperature from Celsius to Kelvin (27 °C + 273.15).
Solve the ideal gas law equation for the number of moles (n).
Use the mass of the gas and the number of moles calculated to find the molar mass (molar mass = mass/n).
The detailed calculations show:
Solve for n: PV = nRT → n = PV / RT = (0.991 atm × 0.265 L) / (0.0821 L·atm/(mol·K) × 300.15 K) ≈ 0.01067 mol.
Finally, determine the molar mass: molar mass = mass/n = 0.472 g / 0.01067 mol ≈ 44.22 g/mol.
Hence, the molar mass of the gas is approximately 44.22 g/mol.
In which of the following bonds, does the Cl atom have the highest electron density? LaTeX: O-Cl O − C l LaTeX: N-Cl N − C l LaTeX: Br-Cl B r − C l LaTeX: H-Cl H − C l LaTeX: S-Cl
hey there!:
Electron density on Cl atom depends on electronegativity difference between Cl and other bonded atom. If the electrnegativity difference is more then Cl has greater electron density, that menas if the bonded atom has less electronegativity then bonded electrons are more attracted by Cl and it has greater electron density.
Among the five atoms which are bonded to Cl atom H has low electronegativity. So in H-Cl two bonded electrons are closer to Cl atom as it has greater electronegativity than H. This results more electron density on Cl atom.
Hence in H-Cl bond Cl atom have the highest electron density
Hope this helps!
The bond in which the chlorine atom has the highest electron density is H − C l.
The polarity of a bond depends on the magnitude of electronegativity difference between the atoms in the bond. The greater the electronegativity difference between the atoms in a bond the more the polarity of the bond.
The magnitude of electron density on each atom in a bond depends on its electronegativity. The more electronegative an atom is, the more it is able to accommodate larger electron density. Looking at the options, hydrogen is far less electronegative than chlorine so a large magnitude of electron density resides on the chlorine atom in HCl.
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A heliox deep-sea diving mixture delivers an oxygen partial pressure of 0.10 atm when the total pressure is 12.5 atm . What is the partial pressure of helium in this mixture?
Answer : The partial pressure of helium in this mixture is, 12.4 atm.
Explanation : Given,
Partial pressure of oxygen = 0.10 atm
Total partial pressure = 12.5 atm
Now we have to calculate the partial pressure of helium.
According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.
[tex]P_T=p_{He}+p_{O_2}[/tex]
where,
[tex]P_T[/tex] = total partial pressure = 12.5 atm
[tex]P_{O_2}[/tex] = partial pressure of oxygen = 0.10 atm
[tex]P_{He}[/tex] = partial pressure of hydrogen = ?
Now put all the given values is expression, we get the partial pressure of the helium gas.
[tex]12.5atm=p_{He}+0.10atm[/tex]
[tex]p_{He}=12.4atm[/tex]
Therefore, the partial pressure of helium in this mixture is, 12.4 atm.
The partial pressure of helium in a deep-sea diving mixture of heliox, given a total pressure of 12.5 atm and an oxygen partial pressure of 0.10 atm, is 12.4 atm.
Explanation:The subject of this question pertains to the principles of gas laws and partial pressures in a gas mixture, specifically applicable to scuba diving environments where deep-sea divers use a unique mixture of gases like heliox (mixture of helium and oxygen).
When the total pressure is 12.5 atm and the oxygen has a partial pressure of 0.10 atm, the partial pressure of helium in the mixture can be found via subtraction. Hence, the partial pressure of helium in this mixture can be found by subtracting the partial pressure of oxygen from the total pressure. That is, 12.5 atm - 0.10 atm = 12.4 atm. In other words, helium, as part of this heliox mixture, offers a partial pressure of 12.4 atm.
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(T) (F) The Great Decoupling means that well-being and wealth can increase while the use of resources and production of pollution can decrease.
In this type of inhibition, the inhibitor can bind to the enzyme regardless of whether the substrate has already bound to the enzyme or not. b) noncompetitive c) uncompetitive d) All of the above a) competitive e) None of the above
Answer:
The correct option is : b) noncompetitive
Explanation:
There are three main types of inhibition:
1. Competitive: In this inhibition, the inhibitor molecule competes with the substrate to bind on the active site of the enzyme.
2. Uncompetitive: In this inhibition, the inhibitor molecule binds to the enzyme- substrate activated complex and thus, does not compete with the substrate to bind on the active site of the enzyme.
3. Non-competitive: In this inhibition, the inhibitor molecule can bind to both the enzyme molecule or to the enzyme-substrate activated complex.
Therefore, In non- competitive inhibition, the inhibitor molecule binds to the enzyme regardless of whether the substrate molecule is bound to the enzyme active site or not.
The pH of saliva is normally in the range of 6.4 to 7.0. However, when a person is ill, the person's saliva becomes more acidic. (a) When Marco is sick, he tests the pH of his saliva and finds that it is 5.3. What is the hydrogen ion concentration of his saliva
Answer:
Concentration of hydrogen ion, [tex][H^+]=5.0118*10^{-6} M[/tex]
Explanation:
pH is defined as the negative logarithm of hydrogen ion's concentration.
The lower the value of pH, the higher the acidic the solution is.
The formula for pH can be written as:
[tex]pH=-log[H^+][/tex]
Given,
pH of the saliva of Marco = 5.3
To calculate: Hydrogen ion concentration in the saliva
Thus, applying in the formula as:
[tex]pH=-log[H^+][/tex]
[tex]5.3=-log[H^+][/tex]
So,
[tex]log[H^+]=-5.3[/tex]
[tex][H^+]=10^{(-5.3)}[/tex]
[tex][H^+]=5.0118*10^{-6} M[/tex]
The hydrogen ion concentration of Marco's saliva is approximately 5.01 x 10^-6 M.
Explanation:The pH scale ranges from 0 to 14, where anything below 7 is acidic and above 7 is alkaline. The hydrogen ion (H+) concentration of a solution can be determined using the pH value. A change of one unit on the pH scale represents a ten-fold change in the concentration of H+ ions. Given that Marco's saliva has a pH of 5.3, we can calculate the hydrogen ion concentration as follows:
Convert the pH value to a H+ concentration by taking the inverse logarithm of the pH value. In this case, H+ concentration = 10⁻⁵(-pH).Substitute the pH value (5.3) into the equation to find the H+ concentration: H+ concentration = 10∧5.3⇒10∧6≈10⁻⁶Calculate the value using a calculator or scientific notation. The H+ concentration is approximately 5.01 x 10⁻⁶M.Therefore, the hydrogen ion concentration of Marco's saliva is approximately 5.01 x 10⁻⁶M.
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The reaction of hydrogen peroxide with iodine, H2O2(aq)+I2(aq)⇌OH−(aq)+HIO(aq) is first order in H2O2 and first order in I2. If the concentration of H2O2 was increased by half and the concentration of I2 was quadrupled, by what factor would the reaction rate increase?
Answer:So we can say that the rate of reaction increases by factor of 6.
Explanation:
The rate law for any given reaction
A+B⇄C+D
Rate law for the above reaction is:
R=K[A]ᵃ[B]ᵇ
a and b are the order of reaction and it is an experimentally determined quantity.
K is the rate constant and it is constant for a given reaction
[A] and [B] are the concentrations of the reactants.
R is the rate of reaction
For the given reaction :
H₂O₂(aq.)+I₂(aq.)⇆OH⁻(aq.)+HIO(aq.)
Also it is given for the reaction that order with respect to H₂O₂ is 1 and order with respect to I₂ is also 1
The rate law can be written as :
R=K[H₂O₂]¹[I₂]¹
k=rate constant
When we increase the concentration of H₂O₂ by half which meansthat new concentration of H₂O₂ will be= 3/2[H₂O₂].
When we increase the concentration of I₂ by 4 which means that new concentration of I₂ will be= 4[I₂]
Putting the values of our new concentration in the rate law:
R=K3/2[H₂O₂]¹4[I₂]¹
R=6K[H₂O₂]¹[I₂]¹=New rate
So as we can see that the new rate of the reaction using the new concentration is 6 times the older rate of reaction.
So we can say that the rate of reaction increases by factor of 6.
A student makes a mistake while preparing a vitamin C sample for titration and adds the potassium iodide solution twice. How will the larger quantity of KI influence the amount of NBS needed to titrate the sample? a. Increase it. b. Decrease it. c. Will have no effect on it.
The added KI does not have any impact
The reaction invovles Titration of vitaminc ( Ascorbic acid)
ascorbic acid + I₂ → 2 I⁻ + dehydroascorbic acid
the excess iodine is free reacts with the starch indicator, forming the blue-black starch-iodine complex.
This is the endpoint of the titration. since alreay excess KI is added ( the source of Iodine), it does not have an influence.
Answer B
Hope this helps!
Final answer:
Adding excess potassium iodide (KI) to a vitamin C sample for titration with N-bromosuccinimide (NBS) results in increased production of iodine, which requires more sodium thiosulfate (Na₂S₂O₃) for back titration. Therefore, the amount of NBS needed to titrate the sample will increase.
Explanation:
When a student mistakenly adds a potassium iodide (KI) solution twice while preparing a vitamin C sample for titration with N-bromosuccinimide (NBS), this will lead to the addition of excess KI in the solution. The presence of additional KI will reduce the titrand more, producing a stoichiometric amount of iodine (I₂). This increased amount of I₂ will then be determined by back titration using sodium thiosulfate (Na₂S₂O₃) as a reducing titrant. Adding excess KI does not change the amount of vitamin C in the sample, but it does result in an increased production of I₂ which needs to be titrated with sodium thiosulfate. Therefore, the larger quantity of KI will increase the amount of NBS needed to titrate the sample since there is more I₂ produced that needs to be accounted for during the titration process.
1. Wine is approximately 12% ethanol C2H5OH by volume . Ethanol has a molar mass of 46.06g/mol and a density 0.789g/mL. How many moles of ethanol are present in a 750mL bottle of wine ?
Hey there!:
Total volume of wine = 750ml
volume ℅ of ethanol = 12 %
volume of ethanol = (12ml/100ml)*750ml = 90ml
Density of Ethanol = 0.789 g/ml
Mass of Ethanol = 0.789 g/ml × 90ml = 71.01 g
Molar mass of ethanol = 46 g/mol Nº of mole of ethanol = Mass/molar mass
=> 71.01 g /46(g/mol)= 1.5437 moles
Hope this helps!
1.541 moles of ethanol are present in a 750 mL bottle of wine.
How to find the number of moles ?Number of moles = [tex]\frac{\text{Mass}}{\text{Molar mass}}[/tex]
What is Density ?The substance per unit volume is called Density. SI unit of density is kg/m.
It is expressed as:
Density = [tex]\frac{\text{Mass}}{\text{Volume}}[/tex]
Volume of ethanol = 12%
= [tex]\frac{12}{100}[/tex]
= 0.12
Volume of ethanol = 0.12 × 750
= 90
Density of ethanol = [tex]\frac{\text{Mass of ethanol}}{\text{Volume of ethanol}}[/tex]
0.789 g/mL = [tex]\frac{\text{Mass of ethanol}}{90}[/tex]
Mass of ethanol = 0.789 × 90
= 71.01 g
Now put the value in above formula we get
Number of moles = [tex]\frac{\text{Mass}}{\text{Molar mass}}[/tex]
= [tex]\frac{71.01\ g}{46.06\ \text{g/mol}}[/tex]
= 1.541 mol
Thus from the above conclusion we can say that 1.541 moles of ethanol are present in a 750 mL bottle of wine.
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What is the composition of Gilsonite?
Answer:
Roughly C100 H140 N3 O
Explanation:
Gilsonite is a bituminous product that resembles shiny black obsidian.
It contains more than 100 elements.
Its mass composition varies but is approximately 84 % C, 10 % H, 3 % N, and 1 % O.
Its empirical formula is roughly C100 H140 N3 O.
At the beginning of an experiment, the concentration of nitrogen dioxide in a reaction vessel was 0.1103 mol/L. After 65.0 s, the concentration decreased to 0.1076 mol/L. What is the average rate of decomposition of NO2 during this time interval, in mol/(L ∙ s)?
Answer:
-4.15*[tex]10^{-5}mol/(L.s)[/tex]
Explanation:
Average rate = [tex]\frac{final concentration -initial concentration}{change in time} = \frac{0.1076mol/L-0.1103mol/L}{65s}=-4.15.10^{-5}mol/(L.s)[/tex]
The rate is negative because it is a decomposition and our focus is the reactant which is depleting.
We give 70 J as heat to a diatomic gas, which then expands at constant pressure. The gas molecules rotate but do not oscillate. By how much does the internal energy of the gas increase?
Answer:
50 Joule
Explanation:
Diatomic gas
Q = Heat given = 70 J
n = number of moles
Cp = specific heat at constant pressure
ΔT = Change in temperature
R = Gas constant
Change in energy
ΔE = Q-w
⇒ΔE = n(Cp)ΔT-nRΔT
As it is a diatomic gas Cp = (7/2)R
Putting the value of Cp in the above equation we get
Q = (7/2)RΔT
ΔE = (5/2)RΔT
Dividing the equations we get
ΔE/Q = 5/7
⇒ΔE = (5/7)Q
⇒ΔE = (5/7)×70
⇒ΔE = 50 J
∴ The internal energy change is 50 Joule
Noble gas compounds like KrF, XeCl, and XeBr are used in excimer lasers. Draw an approximate molecular orbital diagram appropriate for these molecules (they will be quite similar). Give the ground-state electron configuration of KrF and predict whether the cationic analog (KrF) is likely to have a stronger bond.
Answer:
Here's what I get.
Explanation:
The MO diagrams of KrBr, XeCl, and XeBr are shown below.
They are similar, except for the numbering of the valence shell orbitals.
Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.
However, the MO diagrams are approximately correct.
The ground state electron configuration of KrF is
[tex](1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \, (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}[/tex]
KrF⁺ will have one less electron than KrF.
You remove the antibonding electron from the highest energy orbital, so the bond order increases.
The KrF bond will be stronger.
Final answer:
The molecular orbital diagram for noble gas compounds like KrF can be explained by the ground-state electron configuration and the expected bond strength of its cationic form.
Explanation:
Noble Gas Compounds Molecular Orbital Diagram:
The ground-state electron configuration of KrF is [Kr] 5s² 5p⁶. The cationic analog of KrF is likely to have a stronger bond due to the removal of an electron, leading to a decrease in repulsion between the nuclei and shared electrons.
Comparing Bond Strength in KrF vs. KrF⁺:
In KrF, the valence electrons fill the σ bonding and σ non-bonding orbitals, and the π bonding orbitals are empty.
When KrF loses an electron to become KrF⁺, it removes an electron from the σ non-bonding orbital (lone pair).
Losing this electron strengthens the bond because it removes electron density that opposes the bonding interaction in the σ bonding orbital. This is similar to losing a lone pair in other molecules.
Therefore, KrF⁺ is likely to have a stronger bond than KrF.
Copper crystallizes with a face-centered cubic lattice and has a density of 8.93 g/cm3.
a.) Calculate the mass of one unit cell of copper (in grams) b.) Calculate the volume of the copper unit cell (in cm3). c.) Calculate the edge length of the unit cell (in cm). d.) Calculate the radius of a copper atom (in pm).
Answer:
For a: The mass of one unit cell of copper is [tex]1.0553\times 10^{-22}g[/tex]
For b: The volume of copper unit cell is [tex]4.726\times 10^{-23}cm^3[/tex]
For c: The edge length of the unit cell is [tex]3.615\times 10^{-8}cm[/tex]
For d: The radius of a copper atom 127.82 pm.
Explanation:
For a:We know that:
Mass of copper atom = 63.55 g/mol
According to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of atoms.
If, [tex]6.022\times 10^{23}[/tex] number of atoms occupies 63.55 grams.
So, 1 atom will occupy = [tex]\frac{63.55g}{6.022\times 10^{23}atom}\times 1 atom=1.0553\times 10^{-22}g[/tex]
Hence, the mass of one unit cell of copper is [tex]1.0553\times 10^{-22}g[/tex]
For b:Copper crystallizes with a face-centered cubic lattice. This means that 4 number of copper atoms are present in 1 units cell.
Mass of 4 atoms of copper atom = [tex]1.0553\times 10^{-22}g/atom \times 4atoms=4.2212\times 10^{-22}g[/tex]
We are given:
Density of copper = [tex]8.93g/cm^3[/tex]
To find the volume of copper, we use the equation:
[tex]\text{Density of copper}=\frac{\text{Mass of copper}}{\text{Volume of copper}}[/tex]
Putting values in above equation, we get:
[tex]8.93g/cm^3=\frac{4.2212\times 10^{-22}}{\text{Volume of copper}}\\\\\text{Volume of copper}=4.726\times 10^{-23}cm^3[/tex]
Hence, the volume of copper unit cell is [tex]4.726\times 10^{-23}cm^3[/tex]
For c:Edge length of the unit cell is taken as 'a'
Volume of cube = [tex]a^3[/tex]
Putting the value of volume of unit in above equation, we get:
[tex]\sqrt[3]{4.726\times 10^{-23}}cm^3=3.615\times 10^{-8}cm[/tex]
Hence, the edge length of the unit cell is [tex]3.615\times 10^{-8}cm[/tex]
For d:The relation of radius and edge length for a face-centered lattice follows:
[tex]a=r\sqrt{8}[/tex]
Putting values in above equation, we get:
[tex]3.615\times 10^{-8}=r\sqrt{8}\\\\r=1.2782\times 10^{-8}cm[/tex]
Converting cm to pm, we get:
[tex]1cm=10^{10}pm[/tex]
So, [tex]1.2782\times 10^{-8}cm=127.82pm[/tex]
Hence, the radius of a copper atom 127.82 pm.
The wavelength of the red-pink line emitted by a laboratory sample of excited hydrogen is 656 nm. Taking a spectrum of a glowing nebula, you find that the same red-pink line of hydrogen appears at 662 nm. You conclude that the nebula
A. Is 1% hotter than hydrogen in the laboratory sample.
B. Is moving towards us at about 1% the speed of light.
C. Is 1% cooler than hydrogen in the laboratory sample.
D. Is moving away from us at about 1% the speed of light
Answer:
Choice D) The nebula is moving away from the observer.
Explanation:
Is the emission here a result of electron transition or thermal radiation?
The energy difference between two atomic energy levels is discrete. As a result, emissions due to electron transition exist as discrete lines.On the contrary, the thermal radiation of objects above 0 degree Kelvins exists as a continuous frequency spectrum.The red-pink emission here is as a line rather than a continuous spectrum. In other words, the red-pink line observed is a result of electron transition. The energy difference will be constant. That should be the same case on the earth as it is in space at the nebula.
Also, this energy difference does not depend on the temperature of the hydrogen. Only that at higher temperature, low-energy radiations will be less prominent. The wavelength will still be 656 nm when the light was emitted from the nebula.
The wavelength observed on the earth is longer than the wavelength emitted. The Doppler's effect is likely to be responsible. As the star moves away from the earth, the distance that light from the star needs to travel keep increasing. Consider two consecutive peaks from the star. When compared with the first peak, the second peak will need to travel a few more kilometers and will need a few more fractions of a second to get to the earth. It would appear to an observer on the earth that the frequency of the light is lower than it actually is. Accordingly, the wavelength will appear to be longer than it was when emitted from the star.
Conversely, the wavelength will appear shorter if the source is moving toward to observer. For this star, the wavelength appears to be longer than it really is. In other words, the star is moving away from the earth.
The ratio between the speed at which the star moves away from the earth and the speed of the light can be found using the equation: (Source: AstronomyOnline)
[tex]\displaystyle \frac{v}{c} = \frac{\Delta \lambda}{\lambda_0} \approx 0.009[/tex].
The gas cloud is moving away from Earth at about 1% the speed of light due to the Doppler effect.So,option D is correct.
The gas cloud is moving away from Earth at about 1% the speed of light. This can be inferred from the observed shift in the wavelength of the hydrogen emission line from 656.3 nm to 656.6 nm.
The shift in wavelength is due to the Doppler effect, indicating the motion of the source relative to the observer.
The reaction of ethyl acetate with sodium hydroxide, CH3COOC2H5(aq)+NaOH(aq)⇌CH3COONa(aq)+C2H5OH(aq) is first order in CH3COOC2H5 and first order in NaOH. If the concentration of CH3COOC2H5 was increased by half and the concentration of NaOH was quadrupled, by what factor would the reaction rate increase?
Answer:
so the reaction rate increases by a factor 6.
Explanation:
For the given equation the reaction is first order with respect to both ester and sodium hydroxide
So we can say that the rate law is
[tex]Rate(initial)=K[NaOH][CH_{3}COOC_{2}H_{5}][/tex]
now as per given conditions the concentration of ester is increased by half it means that the new concentration is 1.5 times of old concentration
The concentration of NaOH is quadrupled means the new concentration is 4 times of old concentration.
The new rate law is
[tex]Rate(final)=K[1.5XNaOH][4XCH_{3}COOC_{2}H_{5}][/tex]
the final rate = 6 X initial rate
so the reaction rate increases by a factor 6.
The reaction rate of ethyl acetate with sodium hydroxide would increase by a factor of 6 if the concentration of ethyl acetate is increased by half and the concentration of NaOH is quadrupled, as it is first order in both reactants.
The reaction of ethyl acetate with sodium hydroxide is:
CH₃COOC₂H₅(aq) + NaOH(aq) ⇄ CH₃COONa(aq) + C₂H₅OH(aq)This reaction is first order concerning both CH₃COOC₂H₅ and NaOH. The rate law for this reaction can be written as:
Rate = k[CH₃COOC₂H₅][NaOH]If the concentration of CH₃COOC₂H₅ is increased by half, its new concentration becomes 1.5 times its initial concentration. If the concentration of NaOH is quadrupled, its new concentration becomes 4 times its initial concentration. Therefore, the rate of the reaction increases by a factor of:
1.5 (from CH₃COOC₂H₅) x 4 (from NaOH) = 6So, the reaction rate would increase by a factor of 6.
Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides. (a) H2SO4 (b) Ca(OH)2 (c) BrOH (d) ClNO2 (e) TiCl4 (f) NaH'
Answer :
Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.
Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.
Rules for Oxidation Numbers are :
The oxidation number of a free element is always zero.The oxidation number of a monatomic ion equals the charge of the ion.The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.The oxidation number of oxygen (O) in compounds is usually -2.The oxidation number of a Group 17 element in a binary compound is -1.The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.Now we have to determine the oxidation state of the elements in the compound.
(a) [tex]H_2SO_4[/tex]
Let the oxidation state of 'S' be, 'x'
[tex]2(+1)+x+4(-2)=0\\\\x=+6[/tex]
Hence, the oxidation state of 'S' is, (+6)
(b) [tex]Ca(OH)_2[/tex]
Let the oxidation state of 'Ca' be, 'x'
[tex]x+2(-2+1)=0\\\\x=+2[/tex]
Hence, the oxidation state of 'Ca' is, (+2)
(c) [tex]BrOH[/tex]
Let the oxidation state of 'Br' be, 'x'
[tex]x+(-2)+1=0\\\\x=+1[/tex]
Hence, the oxidation state of 'Br' is, (+1)
(d) [tex]ClNO_2[/tex]
Let the oxidation state of 'N' be, 'x'
[tex]-1+x+2(-2)=0\\\\x=+5[/tex]
Hence, the oxidation state of 'N' is, (+5)
(e) [tex]TiCl_4[/tex]
Let the oxidation state of 'Ti' be, 'x'
[tex]x+4(-1)=0\\\\x=+4[/tex]
Hence, the oxidation state of 'Ti' is, (+4)
(f) [tex]NaH[/tex]
Let the oxidation state of 'Na' be, 'x'
[tex]x+(-1)=0\\\\x=+1[/tex]
Hence, the oxidation state of 'Na' is, (+1)
Consider these two cases. Case 1: An electron jumps from energy level 4 to energy level 2 in an atom. Case 2: An electron jumps from energy level 4 to energy level 1 in an atom. For case 1, what happens when an electron jumps from energy level 4 to energy level 2 in an atom?
When an electron jumps from higher energy level to a lower energy level it radiates or gives out energy in the form of radiation.
What is an energy level?
Electrons present in an atom revolve in different orbits which are stationary states and are also called as energy levels. The energy levels are numbered as integers which are also called as principal quantum numbers.
Energy of the stationary state is given as E= -R[tex]_h[/tex] 1/n² where R[tex]_h[/tex] is the Rydberg's constant. When an electron is excited, and it moves from lower to higher energy levels there is absorption of energy, while when it moves from higher energy level to lower energy level it radiates or gives out energy in the form of radiation.
They can also be defined as the distances between electron and nucleus of an atom . Electrons present in K energy level have least energy .Energy level diagrams are studied to understand nature of bonding , placement of electrons in orbits and and elemental behavior under certain conditions.
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Consider the reaction: 3 A + 5 B --> 4 C. The rate of the reaction is found to be .7511 M/s. What is the rate of change of B? Be sure to consider whether the substance is disappearing or appearing in your answer.
Answer:
-3.7555 M/s is the rate of change of B.
Explanation:
3A + 5B → 4C
Given that rate of the reaction ,R= 0.7511 M/s
Rate of the reaction is defined as change in concentration of any one of the reactant or product with respect to time.
[tex]R=\frac{-1}{3}\frac{dA}{dt}=\frac{-1}{5}\frac{dB}{dt}=\frac{1}{4}\frac{dC}{dt}[/tex]
[tex]R=0.7511 M/s=\frac{-1}{5}\frac{dB}{dt}[/tex]
[tex]\frac{dB}{dt}=5\times 0.7511 M/s=-3.7555 M/s[/tex]
The negative sign indicates the concentration of reactant B is decreasing with progress in time. This mean reactant B is disappearing.
-3.7555 M/s is the rate of change of B.
Which isomer would you expect to undergo E2 elimination faster, trans-1-bromo-4-tert-butylcyclohexane or cis-1-bromo-4-tert-butylcyclohexane? Draw each molecule in its more stable chair conformation, and explain your answer.
Answer: Cis-1-bromo-4-tert-butylcyclohexane would undergo faster elimination reaction.
Explanation:
The two primary requirements for an E-2 elimination reaction are:
1.There must be availability of β-hydrogens that is presence of hydrogen on the carbon next to the leaving group.
2.The hydrogen and leaving group must have a anti-periplanar position .
Any substrate which would follow the above two requirements can give elimination reactions.
For the structure of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane to be stable it must have the tert-butyl group in the equatorial position as it is a bulky group and at equatorial position it would not repel other groups. If it is kept on the axial position it would undergo 1,3-diaxial interaction and would destabilize the system and that structure would be unstable.
Kindly find the structures of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane in attachment.
The cis- 1-bromo-4-tert-butylcyclohexane has the leaving group and β hydrogens in anti-periplanar position so they can give the E2 elimination reactions easily.
The trans-1-bromo-4-tert-butylcyclohexane does not have the leaving group and βhydrogen in anti periplanar position so they would not give elimination reaction easily.
so only the cis-1-bromo-4-tert butyl cyclohexane would give elimination reaction.
Trans-1-bromo-4-tert-butylcyclohexane is expected to undergo E2 elimination faster than cis-1-bromo-4-tert-butylcyclohexane due to less steric hindrance.
Explanation:In determining the rate of E2 elimination, the trans-1-bromo-4-tert-butylcyclohexane would undergo E2 elimination faster than the cis-1-bromo-4-tert-butylcyclohexane. This is due to the larger degree of steric hindrance in the case of the cis isomer.
In trans-1-bromo-4-tert-butylcyclohexane, the bromine is at the equatorial position while the tert-butyl group is axial. It forms a structure that allows the compound to experience less steric hindrance with bromine in a more favorable position for leaving.
In comparison, cis-1-bromo-4-tert-butylcyclohexane has a bromine and tert-butyl group both at equatorial positions. This causes steric hindrance, and in turn, slows down the E2 elimination rate. Despite the more stable conformation, the bromine is not well-oriented for a leaving group in E2 elimination.
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The standard reduction potentials of lithium metal and chlorine gas are as follows: Reaction Reduction potential (V) Li+(aq)+e−→Li(s) −3.04 Cl2(g)+2e−→2Cl−(aq) +1.36 In a galvanic cell, the two half-reactions combine to 2Li(s)+Cl2(g)→2Li+(aq)+2Cl−(aq) Calculate the cell potential of this reaction under standard reaction conditions.
Under standard conditions :
E(cell) = E(cathode) - E(anode)
Note : cathode has the larger numeric value and anode has the smaller. Therefore
E(cell) = +1.36V - ( -3.04V)
= 1.36 + 3.04
= +4.40V
Glucose prefers an open-chain conformation in aqueous solution(T/F)
Answer:
False
Explanation:
Glucose is a monosachharide carbohydrate, with the molecular formula C₆H₁₂O₆.
Glucose molecule can exist in two forms-
1. Open chain form
2. cyclic form
The open chain form of the glucose is an unbranched 6 carbon atom chain. The carbon 1 of the molecule is an aldehyde group and the rest of the five carbon atoms have one hydroxyl group each.
The cyclic form of the glucose can be-
a. Pyranose: The pyranose form is a 6-membered cyclic ring, which consists of 5 carbon atoms and 1 oxygen atom in the ring.
b. Furanose: The furanose form is a 5- membered cyclic ring, which consists of 4 carbon atoms and 1 oxygen atom in the ring.
In an aqueous solution, 99% glucose molecule exists in the cyclic pyranose form as it is energetically more stable.
Therefore, in aqueous solution, the glucose molecule does not prefer the open-chain conformation.
Therefore, the statement is false.
___________ minerals are dark-colored because they contain _____________. a. Mafic; iron and/or magnesium b. Felsic; iron and/or magnesium c. Mafic; calcium, sodium, or potassium d. Felsic; calcium, sodium, or potassium
Answer:
A. Mafic; iron and/or magnesium
Explanation:
Let's find the answer by naming some minerals and their chemistry.
Mafic minerals are dark-colored whereas felsic minerals are light-colored, thats way mafic rocks are dark-colored because they are mainly composed by mafic minerals and the other way around for felsic rocks.
But remember that mafic minerals as amphiboles, pyroxenes or biotites, involve in their chemical structure iron and/or magnesium. Although calcium and sodium can be incorporated in amphiboles and clinopyroxenes, they are not involved in orthopyroxenes and biotites. On the other hand, although potassium is involved in biotite and in some extent in amphiboles, this element is not involved in pyroxenes.
So in conclusion, mafic minerals are usually dark-colored because they involve iron and/or magnesium in their chemical structures.
Consider the following equilibrium system involving SO2, Cl2, and SO2Cl2 (sulfuryl dichloride): SO2(g) + Cl2(g) ⇌ SO2Cl2(g) Predict how the equilibrium position would change if the temperature remains constant: a. Cl2 gas were added to the system. b. SO2Cl2 were removed from the system. c. SO2 were removed from the system.
Answer:
(a) forward direction
(b) forward direction
(c) backward direction.
Explanation:
Given , the chemical reaction in equilibrium is,
SO₂(g) + Cl₂(g) ⇄ SO₂Cl₂ (g)
The direction of the reaction by changing the concentration can be determined by Le Chatelier's principle,
It states that ,
When a reaction is at equlibrium , Changing the concentration , pressure, temperature disturbs the equilibrium , and the reaction again tries to attain equilibrium by counteracting the changes.
(a)
For the reaction , Cl₂ is added to the system , i.e. , increasing the concentration of Cl₂ ,Now, according to Le Chatelier , The reaction will move in forward direction , to reduce the increased amount of Cl₂.
Hence, reaction will go in forward direction.
(b)
Removing SO₂Cl₂ from the system ,i.e. , decreasing the concentration of SO₂Cl₂ , according to Le Chatelier , the reaction will move in forward direction , to increase the amount of reduced SO₂Cl₂.
Hence, reaction will go in forward direction.
(c)
Removing SO₂ from the system , i.e. decreasing the concentration of SO₂ , according to Le Chatelier , the reaction will move in backward direction , to increase the amount of reduced SO₂.
Hence, reaction will go in backward direction.
If 25 kg of ice at 0C is combined with 4 kg of steam at 100C, what will be the final equilibrium temperature (in C) of the system? Latent heat of fusion of ice = 3.34 x 105 J.kg-1 Latent heat of steam = 2.23 x 106 J.kg-1 Specific heat of water = 4180 J.kg-1.K-1
Try the suggested option; answer is marked with red colour (18.4953 °C).
All the details are in the attached picture.
The equilibrium temperature of the system is required.
The equilibrium temperature of the mixture is [tex]18.48^{\circ}\text{C}[/tex].
[tex]m_i[/tex] = Mass of ice = 25 kg
[tex]m_s[/tex] = Mass of steam = 4 kg
[tex]T_i[/tex] = Temperature of ice = [tex]0^{\circ}\text{C}[/tex]
[tex]T_s[/tex] = Temperature of steam = [tex]100^{\circ}\text{C}[/tex]
[tex]L_f[/tex] = Latent heat of fusion = [tex]3.34\times 10^5\ \text{J/kg}[/tex]
[tex]L_v[/tex] = Latent heat of vaporization = [tex]2.23\times 10^6\ \text{J/kg}[/tex]
[tex]c_w[/tex] = Specific heat of water = [tex]4180\ \text{J/kg}^{\circ}\text{C}[/tex]
The heat balance of the system will be
[tex]m_iL_f+m_ic_w(T-T_i)=m_sL_v+m_sc_w(T_s-T)\\\Rightarrow m_iL_f+m_ic_wT-m_ic_wT_i=m_sL_v+m_sc_wT_s-m_sc_wT\\\Rightarrow T=\dfrac{m_ic_wT_i+m_sL_v+m_sc_wT_s-m_iL_f}{m_ic_w+m_sc_w}\\\Rightarrow T=\dfrac{25\times 4180\times 0+4\times 2.23\times 10^6+4\times 4180\times 100-25\times 3.34\times 10^5}{25\times 4180+4\times 4180}\\\Rightarrow T=18.49^{\circ}\text{C}[/tex]
The equilibrium temperature of the mixture is [tex]18.49^{\circ}\text{C}[/tex].
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Be sure to answer all parts and enter your answers in scientific notation. Write the 1 g equivalents using the following prefixes: (a) mega−, (b) kilo−, (c) deci−, (d) centi−, (e) milli−, (f) micro−, (g) nano−, (h) pico−. (a) 1 g = × 10 megagrams (b) 1 g = × 10 kilograms (c) 1 g = × 10 decigrams (d) 1 g = × 10 centigrams (e) 1 g = × 10 milligrams (f) 1 g = × 10 micrograms (g) 1 g = × 10 nanograms (h) 1 g = × 10 picograms
Explanation:
Gram is unit which is used to expressed mass of a substance. Many other units can be used to express the mass of a substance. All the units are inter changeable.
Conversion factors used to convert the units are:
(a) Gram to mega gram
[tex]1g=1\times 10^{-6}Mg[/tex]
(b) Gram to kilo gram
[tex]1g=1\times 10^{-3}kg[/tex]
(c) Gram to deci gram
[tex]1g=1\times 10^{1}1dg[/tex]
(d) Gram to centi gram
[tex]1g=1\times 10^{2}cg[/tex]
(e) Gram to mili gram
[tex]1g=1\times 10^{3}mg[/tex]
(f) Gram to micro gram
[tex]1g=1\times 10^{6}\mu g[/tex]
(g) Gram to nano gram
[tex]1g=1\times 10^{9}ng[/tex]
(h) Gram to pico gram
[tex]1g=1\times 10^{12}pg[/tex]
Hence, the conversion factors for gram equivalents are given above.
To convert 1 gram to different metric prefixes in scientific notation, we use the powers of ten specific to each prefix, ensuring the numeric value is between 1 and 1000. The result is megagrams (1e-6), kilograms (1e-3), decigrams (1e1), centigrams (1e2), milligrams (1e3), micrograms (1e6), nanograms (1e9), and picograms (1e12).
Explanation:The task involves converting the mass of 1 gram into various metric prefixes and expressing them in scientific notation, where the numeric value is greater than one but less than 1000. The prefixes include mega, kilo, deci, centi, milli, micro, nano, and pico. Below are the conversions using the appropriate powers of ten to express 1 gram in the specified units:
(a) 1 g = 1 × 10-6 megagrams (Mg)(b) 1 g = 1 × 10-3 kilograms (kg)(c) 1 g = 1 × 101 decigrams (dg)(d) 1 g = 1 × 102 centigrams (cg)(e) 1 g = 1 × 103 milligrams (mg)(f) 1 g = 1 × 106 micrograms (μg)(g) 1 g = 1 × 109 nanograms (ng)(h) 1 g = 1 × 1012 picograms (pg)Consider the following reaction in chemical equilibrium: 2BrNO(g)⇌2NO(g)+Br2(g) What is the effect of adding additional Br2(g) to the reaction mixture? What is the effect of adding additional BrNO? Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.
Answer:The reaction would shift to the left on adding Br2(g)
The reaction would shift to right on adding BrNO(g)
Explanation:
The concept can be explained on the basis of LeChateliers principle.
LeChateliers principle explain the effect on equilibrium when a reaction system at equilibrium is subjected to any disturbance or change in conditions then the reaction shifts in such a way so as to reduce the effect of that change or disturbance and again establish the equilibrium.
For example if we add more reactants in a given reaction at equilibrium, the reaction would shift in such a way so that it can reduce the effect of increasing the concentration of reactants and hence the reaction would favor that direction in which it can reduce the concentration of reactants.So when we increase the concentration of reactants the reaction would move towards the formation of more products and so the concentration of reactants would be less in the reaction.
Likewise if we increase the concentration of products so the reaction would shift in such a way so that it can oppose the increased concentration of products so the reaction moves towards more formation of reactants that is the products decompose to form reactants and reaction moves backwards.
In this reaction:
2BrNO(g)⇄2NO(g)+Br₂(g)
When we add more amount of Br₂(g) the reaction would proceed in such a ways so that oppose the increased concentration of Br₂(g).Hence the reaction would move towards left that is backwards.
When we add more amount of BrNO(g)the reaction would proceed in such a ways so that oppose the increased concentration of BrNO(g).Hence the reaction would move towards right that is forward.
Diethyl ether is produced from ethanol according to the following equation: 2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l) Calculate the percent yield if 65.2 g of ethanol reacts to produce 17.2 g of ether.
Answer : The percent yield is, 32.79 %
Explanation :
First we have to calculate the moles of [tex]CH_3CH_2OH[/tex].
[tex]\text{Moles of }CH_3CH_2OH=\frac{\text{Mass of }CH_3CH_2OH}{\text{Molar mass of }CH_3CH_2OH}=\frac{65.2g}{46.07g/mole}=1.415mole[/tex]
Now we have to calculate the moles of [tex]CH_3CH_2OCH_2CH_3[/tex]
The balanced chemical reaction will be,
[tex]2CH_3CH_2OH(l)\rightarrow CH_3CH_2OCH_2CH_3(l)+H_2O(l)[/tex]
From the balanced reaction, we conclude that
As, 2 moles of [tex]CH_3CH_2OH[/tex] react to give 1 mole of [tex]CH_3CH_2OCH_2CH_3[/tex]
So, 1.415 moles of [tex]CH_3CH_2OH[/tex] react to give [tex]\frac{1.415}{2}=0.7075[/tex] mole of [tex]CH_3CH_2OCH_2CH_3[/tex]
Now we have to calculate the mass of [tex]CH_3CH_2OCH_2CH_3[/tex]
[tex]\text{Mass of ether}=\text{Moles of ether}\times \text{Molar mass of ether}[/tex]
[tex]\text{Mass of }ether=(0.7075mole)\times (74.12g/mole)=52.44g[/tex]
The theoretical yield of ether, [tex]CH_3CH_2OCH_2CH_3[/tex] = 52.44 g
Now we have to calculate the percent yield of [tex]CH_3CH_2OCH_2CH_3[/tex]
[tex]\%\text{ yield of ether}=\frac{\text{Actual yield of ether}}{\text{Theoretical yield of ether}}\times 100=\frac{17.2g}{52.44g}\times 100=32.79\%[/tex]
Therefore, the percent yield is, 32.79 %