Final answer:
Two resonance forms of the triphenylmethyl cation were provided, with the positive charge located at the central carbon atom and one of the ortho positions.
Explanation:
The triphenylmethyl cation is a carbocation formed by removing an electron from the triphenylmethyl radical. It has the formula C19H16+. To write two resonance forms, we need to show the positive charge at different positions in the structure. One possible resonance form is where the positive charge is at the central carbon atom, and the other form is where the positive charge is at one of the ortho positions. The resonance structures can be represented as follows:
C6H5--C+--C6H5
: :
C6H4--C6H5 C6H6--C6H4
Calculate the molality of a solution containing 100.7 g of glycine (NH2CH2COOH) dissolved in 3.466 kg of H2O.
Answer : The molality of a solution is, 0.387 mol/kg
Explanation : Given,
Mass of solute (glycine) = 100.7 g
Mass of solvent (water) = 3.466 kg
Molar mass of glycine = 75.1 g/mole
Formula used :
[tex]Molality=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent}}[/tex]
Now put all the given values in this formula, we get:
[tex]Molality=\frac{100.7g}{75.1g/mol\times 3.466kg}[/tex]
[tex]Molality=0.387mol/kg[/tex]
Thus, the molality of a solution is, 0.387 mol/kg
Answer:
The molality of the solution is 0.387 molal
Explanation:
Step 1: Data given
Mass of glycine = 100.7 grams
Mass of H2O = 3.466 kg
Molar mass of glycine = 75.07 g/mol
Step 2: Calculate moles glycine
Moles glycine = mass glycine / molar mass glycine
Moles glycine = 100.7 grams / 75.07 g/mol
Moles glycine = 1.341 moles
Step 3: Calculate molality of the solution
Molality = moles glycine / mass H2O
Molality = 1.341 moles / 3.466 kg
Molality = 0.387 molal
The molality of the solution is 0.387 molal
How are measurements of paramagnetism used to support electron configurations derived spectroscopically? Use Cu(I) and Cu(II) chlorides as examples.
Answer:
Paramagnetism is dependent on the unpaired electron in the last orbital . In this regard, Cu(I) chloride is paramagnetic whereas Cu(II) chloride is not.
Explanation:
Paramagnetism is the property of materials/components which makes them attracted them weekly to the magnetic field.
It is related to electronic configuration, such that it depends on the unpaired electron in the last orbital possess the property.
On basis of this property, Cu(I) chloride is paramagnetic while Cu(II) chloride is non paramagnetic. This is because Cu(I) chloride contains an unpaired electron in the last orbital whereas Cu(II) chloride does not have any unpaired electron.
Answer:
Explanation:
Paramagnetism is a type of magnetism whereby materials are weakly attracted to an externally applied magnetic field and then form internal, induced magnetic fields in the direction of the applied magnetic field. They are attracted to magnetic fields and have magnetic moment induced by the applied field is linear in the field strength. Paramagnetic materials include elements such as Oxygen,
Aluminium etc. and maybe some compounds like FeO etc.
Paramagnetism occurs due to the presence of unpaired electrons in an atom, so atoms with incompletely filled atomic orbitals are paramagnetic, there are exceptions such as copper exist and this is due to their spin, unpaired electrons have a magnetic dipole moment and act like tiny magnets. They have a magnetic permeability slightly greater than 1. External magnetic field causes the electrons spin to align parallel to the field hence, causing a net attraction. Paramagnetic materials include aluminium, oxygen, titanium, and iron oxide (FeO).
From the example,
Cu(I) and Cu(II)
Electronic configuration
Cu(I) - [Ar] 3d10
Cu(II) - [Ar] 3d9
[Ar] - 1s2 2s2 2p6 3s2 3p6 4s2
Therefore, Cu(I) is Paramagnetic while Cu(II) is not Paramagnetic (diamagnetic).
Suppose a helium-3 nuclide transforms into a helium-4 nuclide by absorbing a proton and emitting a positron. Complete the nuclear chemical equation below so that it describes this nuclear reaction.
Answer: The nuclear equation for the conversion of He-3 nuclide to He-4 nuclide is given above.
Explanation:
Nuclear reaction are defined as the reactions in which nucleus of an atom is involved.
Positron emission is defined as the emission process in which positron particle is emitted. In this process, a proton gets converted to neutron and an electron neutrino particle.
[tex]_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e[/tex]
The chemical equation for the reaction of He-3 with a proton follows:
[tex]_2^3\textrm{He}+_1^1\textrm{H}\rightarrow _2^4\textrm{He}+_{+1}^0e[/tex]
Hence, the nuclear equation for the conversion of He-3 nuclide to He-4 nuclide is given above.
The nuclear equation where a helium-3 nuclide transforms into a helium-4 nuclide by absorbing a proton and emitting a positron is written as ³He + ¹H → ⁴He + e⁺. This ensures that mass and charge are conserved in the reaction.
Explanation:To complete the nuclear chemical equation where a helium-3 nuclide (³He) transforms into a helium-4 nuclide (⁴He) by absorbing a proton (¹H) and emitting a positron (e⁺), we must ensure that both mass and charge are conserved in the reaction. In this case, the equation can be represented as:
³He + ¹H → ⁴He + e⁺
The mass number on the left side of the equation is 3 (from helium-3) plus 1 (from the proton), totaling 4, which matches the mass number of helium-4 on the right side of the equation. The atomic number (number of protons) is also conserved through this reaction: 2 (from helium-3) + 1 (from the proton) equals 2 (from helium-4) + 1 (from the emitted positron), with positrons having a positive charge but no atomic number associated.
If 3.52 L of nitrogen gas and 2.75 L of hydrogen gas were allowed to react, how many litres of ammonia gas could form? Assume all gases are at the same temperature and pressure.
Answer:
V NH3(g) = 1.833 L
Explanation:
balanced reaction:
N2(g) + 3H2(g) → 2NH3(g)assuming STP:
∴ V N2(g) = 3.52 L
∴ V H2(g) = 2.75 L
ideal gas:
PV = RTn∴ moles N2(g) = PV/RT
⇒ mol N2(g) = (1 atm)(3.52 L)/(0.082 atm.L/K.mol)(298 K)
⇒ mol N2(g) = 0.144 mol
∴ moles H2(g) = PV/RT
⇒ mol H2(g) = (1)(2.75)/(0.082)(298) = 0.113 mol (limit reagent)
∴ moles NH3(g) = (0.113 moles H2(g))(2 moles NH3 / 3 mol H2) = 0.075 mol
∴ V NH3(g) = RTn/P
⇒ V NH3(g) = ((0.082 atm.L/K.mol)(298 K)(0.075 mol))/(1 atm)
⇒ V NH3(g) = 1.833 L
Final answer:
From 3.52 L of nitrogen and 2.75 L of hydrogen, 1.83 L of ammonia can be produced, considering hydrogen as the limiting reactant based on the stoichiometry of the balanced chemical equation.
Explanation:
The question involves a stoichiometric calculation based on the reaction between hydrogen and nitrogen gases to form ammonia. Given the balanced chemical equation N2(g) + 3H2(g) → 2NH3(g), we can determine how many litres of ammonia gas could form from 3.52 L of nitrogen gas and 2.75 L of hydrogen gas, assuming all gases are at the same temperature and pressure. Since the reaction consumes nitrogen and hydrogen in a 1:3 ratio to produce ammonia in a 2 moles product per 1 mole of nitrogen ratio, we first identify the limiting reactant. Here, hydrogen gas (H2) is the limiting reactant because we need 3 volumes of hydrogen for every volume of nitrogen, but we have less than that (2.75 L instead of 3.52*3 L). The amount of ammonia produced is therefore determined by the amount of hydrogen available. Since 3 volumes of H2 produce 2 volumes of NH3, 2.75 L of H2 would produce (2.75 L * (2/3)) = 1.83 L of NH3.
Write the full electron configuration of the Period 2 element with the following successive IEs (in kJ/mol):
IE₁ = 801
IE₂ = 2427
IE₃ = 3659
IE₄ = 25,022
IE₅ = 32,822
Answer:
Boron (B) is the element whose IE matches with our data.
Electronic Configuration of boron: [tex]1s^22s^22p^1[/tex]
Explanation:
Ionization Energy (IE):
It is the minimum amount of energy which is required to remove the lose electron. If the electron is closer to the nucleus then greater amount of energy is required to remove the electron.
If we look from left to right in a period, ionization energy increases due stability of valance shell.
From the data given to us:
IE₁ = 801
IE₂ = 2427
IE₃ = 3659
IE₄ = 25,022
IE₅ = 32,822
Boron (B) is the element whose IE matches with our data.
Electronic Configuration of boron: [tex]1s^22s^22p^1[/tex]
Boron has 5 electrons (3 in valance shell) that's why it has 5 Ionization Energies.
Assuming that gasoline is 100% isooctane, that isooctane burns to produce only CO2CO2 and H2OH2O, and that the density of isooctane is 0.792 g/mLg/mL, what mass of CO2CO2 (in kilograms) is produced each year by the annual U. S. gasoline consumption of 4.6×1010L4.6×1010L?
Answer:
1.12×10¹¹ kg of CO₂ are produced with 4.6×10¹⁰ L of isooctane
Explanation:
Let's state the combustion reaction:
C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O
Let's calculate the mass of isooctane that reacts.
Density = Mass / Volume
Density . Volume = Mass
First of all, let's convert the volume in L to mL, so we can use density.
4.6×10¹⁰ L . 1000 mL / 1L = 4.6×10¹³ mL
0.792 g/mL . 4.6×10¹³ mL = 3.64 ×10¹³ g
This mass of isooctane reacts to produce CO₂ and water, so let's determine the moles of reaction
3.64 ×10¹³ g . 1mol / 114 g = 3.19×10¹¹ mol
Ratio is 1:8 so 1 mol of isooctane can produce 8 moles of dioxide
Therefore 3.19×10¹¹ mol would produce (3.19×10¹¹ mol . 8) = 2.55×10¹² moles of CO₂
Now, we can determine the mass of produced CO₂ by multipling:
moles . molar mass
2.55×10¹² mol . 44 g/mol = 1.12×10¹⁴ g of CO₂
If we convert to kg 1.12×10¹⁴ g / 1000 = 1.12×10¹¹ kg
In addition to continuous radiation, fluorescent lamps emit sharp lines in the visible region from a mercury discharge within the tube. Much of this light has a wavelength of 436 nm. What is the energy (in J) of one photon of this light?
Answer:
[tex]4.56 x 10^{-19}J[/tex]
Explanation:
Electromagnetic radiations consist of quanta of energy called photons which have energy, E which is equal to:
E = hν.....................................(1)
where h is the Planck's constant which is [tex]6.626 x10^{-34}Js[/tex] and ν is the frequency of light radiation.
But ν = c/λ ....................................(2)
Putting equation (2) into (1), we have
E = hc/λ..........................................(3)
c is the speed of light (c =[tex]3 x 10^{8}m/s[/tex]) while λ is the wavelength of light.
Wavelength λ = 436nm = [tex]436 x 10^{-9}m[/tex]
Therefore the energy E of one photon of this light, using equation (3) is
[tex]E=\frac{6.626 x 10^{-34} x3 x10^{8} }{436 x 10^{-9} } = 4.56 x 10^{-19} J[/tex]
The energy of one photon of light with a wavelength of 436 nm is 4.546 × 10^-19 J.
Explanation:To calculate the energy of one photon, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 × 10-34 J·s), c is the speed of light (3.00 × 108 m/s), and λ is the wavelength.
Plugging in the given wavelength of 436 nm (which is equal to 4.36 × 10-7 m) into the equation, we have:
E = (6.626 × 10-34 J·s)(3.00 × 108 m/s) / (4.36 × 10-7 m) = 4.546 × 10-19 J
Therefore, the energy of one photon of light with a wavelength of 436 nm is 4.546 × 10-19 J.
Prior to hosting an international soccer match, the local soccer club needs to replace the artifical turf on their field with grass turf. The grass turf will cost $ 9.75 per square meter. If the field is 0.102 km by 0.069 km, how much will it cost the club to add the grass turf to their field?
It will cost the soccer club $68,618.50 to add the grass turf to their field for the international soccer match.
Given: Length = 0.102 km = 0.102 km × 1000 m/km = 102 m
Width = 0.069 km = 0.069 km × 1000 m/km = 69 m
The area of the field: Area = Length × Width
Area = 102 m × 69 m = 7038 sq. meters
The area by the cost of the grass turf per square meter:
Cost = Area × Cost per square meter
Cost = 7038 sq. meters ×$9.75/sq. meter = $68,618.50
Therefore, it will cost the soccer club $68,618.50 to add the grass turf to their field for the international soccer match.
Learn more about Cost, refer to the link:
https://brainly.com/question/14725550
#SPJ12
Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an atomic mass of 122.9042 u . Antimony has an average atomic mass of 121.7601 u . What is the percent natural abundance of each isotope?
Answer:
Percentage abundance of 121 Sb is = 57.2 %
Percentage abundance of 123 Sb is = 42.8 %
Explanation:
The formula for the calculation of the average atomic mass is:
[tex]Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})[/tex]
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope, 121 Sb :
% = x %
Mass = 120.9038 u
For second isotope, 123 Sb:
% = 100 - x
Mass = 122.9042 u
Given, Average Mass = 121.7601 u
Thus,
[tex]121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042[/tex]
[tex]120.9038x+122.9042\left(100-x\right)=12176.01[/tex]
Solving for x, we get that:
x = 57.2 %
Thus, percentage abundance of 121 Sb is = 57.2 %
percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %
Considering the definition of atomic mass, isotopes and atomic mass of an element,
Definition of atomic mass AFirst of all, the atomic mass (A) is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.
Definition of isotopesThe same chemical element can be made up of different atoms, that is, their atomic numbers are the same, but the number of neutrons is different. These atoms are called isotopes of the element.
Atomic mass of an elementOn the other hand, the atomic mass of an element is the weighted average mass of its natural isotopes. In other words, the atomic masses of chemical elements are usually calculated as the weighted average of the masses of the different isotopes of each element, taking into account the relative abundance of each of them.
Percent natural abundance of each isotope of antimonyIn this case, 121 Sb and 123 Sb are the naturally isotopes of antimony. 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an atomic mass of 122.9042 u.
Being the average atomic mass of antomony 121.7601 u, the average mass of antimony can be calculated as:
average mass of antimony= percent natural abundance of 121 Sb× atomic mass of 121 Sb + percent natural abundance of 123 Sb× atomic mass of 123 Sb
Being percent natural abundance of 123 Sb= 1 - percent natural abundance of 121 Sb, and substituting the corresponding values, you get:
121.7601 u= percent natural abundance of 121 Sb× 120.9038 u + (1 - percent natural abundance of 121 Sb)× 122.9042 u
Solving:
121.7601 u= percent natural abundance of 121 Sb× 120.9038 u + 1× 122.9042 u - percent natural abundance of 121 Sb× 122.9042 u
121.7601 u= percent natural abundance of 121 Sb× (-2.0004 u) + 122.9042 u
121.7601 u - 122.9042 u= percent natural abundance of 121 Sb× (-2.0004 u)
-1.1441= percent natural abundance of 121 Sb× (-2.0004 u)
-1.1441÷ (-2.0004 u) = percent natural abundance of 121 Sb
0.5719 ×100 % = percent natural abundance of 121 Sb
57.19% = percent natural abundance of 121 Sb
So:
percent natural abundance of 123 Sb= 1 - percent natural abundance of 121 Sb
percent natural abundance of 123 Sb= 1 - 0.5719
percent natural abundance of 123 Sb= 0.4281
percent natural abundance of 123 Sb= 0.4281× 100
percent natural abundance of 123 Sb= 42.81%
Finally, the percent natural abundance of 121 Sb is 57.19% and the percent natural abundance of 123 Sb is 42.81%
Learn more about isotopes and Atomic mass of an element:
brainly.com/question/14403750?referrer=searchResults
brainly.com/question/10043246?referrer=searchResults
brainly.com/question/15553207?referrer=searchResults
What is the mass of cyclohexane solvent, in kg, if 9.76 mL are used and the cyclohexane has a density of 0.779 g/mL?
Answer:
0.0076kg
Explanation:
To get the mass, we use the relation among density, mass and volume.
Mass = density * volume
Here mass? , density = 0.779g/ml , volume = 9.76ml
Mass = 9.76 * 0.779 = 7.60g
Answer is wanted in kg so we divide by 1000. This is 7.60/1000 = 0.0076kg
The volume of a single tungsten atom is 1.07×10-23 cm3. What is the volume of a tungsten atom in microliters?
Answer: 1.07×10^-20microlitre
Explanation:
1cm3 = 1000microlitres
1.07×10^-23 cm3 of tungsten = 1.07×10^-23 x 1000 = 1.07×10^-20microlitre
The volume of a single tungsten atom in microliters is 1.07x10^-17 µL. This is found by multiplying the given volume in cubic centimeters by the conversion factor of 1,000,000 µL/cm³.
Explanation:The volume of a single tungsten atom is 1.07×10-23 cm3. One cubic centimeter (cm3) is equal to 1,000,000 microliters (µL). To convert the volume from cubic centimeters to microliters, we need to multiply the original value by the conversion factor. Therefore, the volume of a tungsten atom in microliters will be 1.07×10-23 cm3 * 1,000,000 µL/cm3, which equals to 1.07×10-17 µL. Hence, the volume of a tungsten atom in microliters is 1.07×10-17 µL.
Learn more about Conversion of units here:https://brainly.com/question/33699586
#SPJ3
Beta (β) sheets are a type of secondary structure in proteins. A segment of a single chain in an antiparallel β sheet has a length of 80.5 Å . How many residues are in this segment?
Answer:
Explanation:
The structural repeating unit of beta sheet is 7 anstrom/2 aminoacids. So,
[tex]\frac{80.5 Angstrom}{3.5} = 23 aminoacids[/tex]
If a segment of a single chain in an antiparallel beta sheet has a length of 80.5 angstrom, then there will be 23 residues in this segment.
To find the number of residues in an 80.5 Å long segment of an antiparallel beta-pleated sheet, divide the total length by the length of one residue (3.5 Å per residue), which gives approximately 23 residues.
Explanation:The student asked how many residues are in a segment of a single chain in an antiparallel beta-pleated sheet with a length of 80.5 Å. To determine the number of amino acid residues in the segment, we can use the typical amino acid residue length in a ß-pleated sheet, which is approximately 3.5 Å per residue in an extended conformation. This measurement considers the distance hydrogen bonds can span between the carbonyl oxygen and amino hydrogen along the peptide chain in the secondary structure.
By dividing the total length of the chain (80.5 Å) by the length of one residue (3.5 Å), you can calculate the number of residues:
Number of residues = Total length ÷ Length per residue
Number of residues = 80.5 Å ÷ 3.5 Å/residue
Number of residues ≈ 23 residues
This calculation does not account for slight variations that may occur in different proteins or specific contexts, but it provides a general estimate for the number of amino acids in a segment of a beta sheet.
Given that D(H-H) and D(F-F) in H2 and F2 are 436 and 158kJ mol-1, estimate the bond dissociation enthalpy of H-F using a simple additivity rule. Compare the answer with the experimental value of 570kJ mol-1
Explanation:
Equation of the reaction:
H2(g) + F2(g) --> 2HF(aq)
1/2H2(g) + 1/2F2(g) --> HF(aq)
D(H-H) in H2 = 436 kJ/mol
D(F-F) in F2 = 158kJ/mol
ΔH bond breakage (dissociation):
1/2 mol H-H bonds = (1/2 X 436) kJ
= 218 kJ
1/2 mol F-F bonds = (1/2 X 158) kJ = = 80 kJ
Total = 218 + 80 = 298 kJ
ΔH bond formation:
1 mol H-F bonds = - 570 kJ
= DHreactant - DHproduct
ΔH°f = 298 kJ + -570 kJ = -272 kJ
30 mL of 0.25 M acetic acid are titrated with 0.05 M KOH. What is the pH after addition of 75 mL KOH? Group of answer choices
Answer:
The PH of the mixture is 4.74
Explanation:
The number of millimoles of acetic acid is calculated using the formula:
No of millimoles= Molarity * Volume( in ml)
= 0.25M * 30ml = 7.5 moles
Number of millimoles of KOH is calculated using:
Number of millimoles = Molarity * Volume ( in ml)
=0.05M * 75ml
= 3.75 moles
The PH of the solution is derived using:
pH = pKa + log [salt] / acid
= [tex] -log [ 1.8 * 10^5 ] + log [ 3.75 mmoles/ 3.75 mmoles] [/tex]
=4.74
A 50.0 mL sample of waste process water from a food processing plant weighing 50.1 g was placed in a properly prepared crucible that had a tare weight of 17.1234 g. After drying in a 103oC oven it weighed 19.9821 g. After the crucible was placed in a 550oC oven it weighted 18.7777 g. What are the TS and VS (in mg/L)?
Answer:
TS (total solid, mg/L) = 57,174 mg/L
VS (volatile solid, mg/L) = 24,088 mg/L
Explanation:
First step to solve this problem is to know data and questions:
Data:
Sample volume = 50 mL
Sample weight = 50.1 g
P1 (weight empty crucible) = 17.1234 g
P2 (weight after drying a 103º in oven) = 19.9821 g
P3 (weight after incinatrion 550º) = 18.7777 g
Questions:
TS = ?
VS = ?
Formula:
We need to use formulas to calculate total solid (TS) and volatile solid (VS), these are:
[tex]TS =\frac{(P2-P1)*\frac{1,000 mg}{1g} }{Sample volumen (L)}[/tex]
[tex]VS = \frac{(P2-P3)*\frac{1,000 mg}{1g} }{Sample volume (L)}[/tex]
We have to transform mL to L so we will divide mL by 1,000 the sample volume:
[tex]Sample Volumen (L) = 50 mL * \frac{1L}{1,000 mL} = 0.05 L[/tex]
In the formula the value of 1,000 results by the convertion factor to transform grams to miligrams (we have to multiplie by 1,000)
Now we need to replace data on previous formulas and we will get TS and VS expressed in mg/L:
[tex]TS = \frac{(19.9821 g-17.1234g)*\frac{1,000 mg}{1g} }{0.05L} = 57,174mg/L[/tex]
[tex]VS = \frac{(19.9821g-18.7777g)*\frac{1,000mg}{1g} }{0.05L}=24,088 mg/L[/tex]
We divide TS and VS formula by sample volume because the exercise is asking us to express the results in mg/L.
Only certain electron transitions are allowed from one energy level to another. In one-electron species, the change in the quantum number l of an allowed transition must be ±1. For example, a 3p electron can drop directly to a 2s orbital but not to a 2p. Thus, in the UV series, where nfinal = 1, allowed electron transitions can start in a p orbital (l = 1) of n = 2 or higher, not in an s (l = 0) or d (l = 2) orbital of n = 2 or higher. From what orbital do each of the allowed electron transitions start for the first four emission lines in the visible series (nfinal = 2)?
Final answer:
The first four lines of the Balmer series involve electron transitions from 3p to 2s, 4p to 2s, 5p to 2s, and 6p to 2s orbitals.
Explanation:
The Balmer series involves electron transitions from higher energy levels to the second principal energy level (n=2), producing visible spectral lines. For the first four lines of the visible emission spectrum in the Balmer series, the allowed transitions must follow the selection rule Δl = ±1. Therefore, these transitions can only start from orbitals with l=1, which are the p orbitals.
For nfinal = 2, the corresponding ni initial energy levels for the first four visible emission lines are:
3p (n=3, l=1) to 2s (n=2, l=0)4p (n=4, l=1) to 2s (n=2, l=0)5p (n=5, l=1) to 2s (n=2, l=0)6p (n=6, l=1) to 2s (n=2, l=0)Functional groups confer specific chemical properties to the molecules of which they are a part. In this activity, you will identify which compounds exhibit certain chemical properties as well as examples of those six different compounds.
Answer:
This question is incomplete, here is the complete question:
Functional groups confer specific chemical properties to the molecules of which they are a part. In this activity, you will identify which compounds exhibit certain chemical properties as well as examples of those six different compounds.
(a) alcohol: is highly polar and may act as a weak acid (OH)
(b) carboxylic acid: acts as an acid (COOH)
(c) aldehyde: may be a structural isomer of a ketone (C=O)
(d) thiol: forms disulfide bonds (S-H)
(e) amine: acts as a base (NH2)
(f) organic phosphate: contributes negative charge (PO4)
Explanation:
Functional groups are groups of atoms attached to the carbon skeleton of an organic compounds and confers some particular properties. For example:
(a) alcohol: is highly polar and may act as a weak acid (OH)
. In this functional group, oxygen is very electronegative and tend to attract electrons. Water is attracted to this group and because of this, compounds that have OH group will disolve in water. Organic compounds with this functional group are alcohols and carbohydrates for example.
(b) carboxylic acid: acts as an acid (COOH)
. The two electronegative oxygens of this group keep electrons attached to them and do not share them with tha Hydrogen atom. Because of this, the hydrogen atom can dissociates from the molecule, giving an acid.
(c) aldehyde: may be a structural isomer of a ketone (C=O)
. As in the case of -OH group, oxygen of this group is very electronegative. When the group is presents at the end of the carbon skeleton we cal it aldehyde, while when it is in another part of carbon structure we call it ketone.
(d) thiol: forms disulfide bonds (S-H)
. When two thiol groups (S-H) are oxidazed, losting their hydrogen atoms, they form disulfide bonds that gives stability to the molecules that contains them. An example of this is found in proteins when two aminoacids with thiol groups (cysteine, for example) interact and form disulfide bonds that stabilizes the protein.
(e) amine: acts as a base (NH2)
. The nitrogen atom of this groups is slightly electronegative so it tends to pick up hydrogen ions from the surroundings given rise to a base. This group i typicall of aminoacids (glycine, for example)
(f) organic phosphate: contributes negative charge (PO4). In this group, phosphorus easily changes its oxydation state, thus, giving rise to a compound with electronegative oxygens. An important biological molecule that contains this group is ATP
Functional groups are important components of organic molecules, conferring specific chemical properties to the molecules. Common functional groups include hydroxyl, methyl, carbonyl, carboxyl, amino, phosphate, and sulfhydryl. The presence of these groups can greatly affect the chemical properties and function of the molecule.
Explanation:Functional groups are vital components of organic molecules, conferring specific chemical properties upon the molecules in which they occur. These groups are attached to a 'carbon backbone' of chains or rings of carbon atoms, with possible substitutions of elements like nitrogen or oxygen. For example, the C-Cl portion of the chloroethane molecule is a functional group imparting specific chemical reactivity. The nature of the functional groups in an organic molecule are significant determinants of its chemical properties.
Functional groups take part in particular chemical reactions. Some of the important functional groups include hydroxyl, methyl, carbonyl, carboxyl, amino, phosphate, and sulfhydryl. These functional groups may attach to the carbon backbone at various points along the chain or ring structure. The presence of these groups majorly contributes to the varying chemical properties of different organic molecules and their functions in living organisms.
For example, carboxyl groups (-COOH) confer acidic properties to a molecule, while amine groups (NH₂) make a molecule more basic. Molecules with identical molecular formulas may have different structures due to variations in their functional groups, morphing into different types of isomers.
Learn more about Functional Groups here:https://brainly.com/question/36437639
#SPJ12
A solution is made by dissolving 5.61 g of a new polymer in enough water to make 260 mL of solution. At 25.0 oC, the osmotic pressure of the solution is 0.174 atm. What is the molar mass of the polymer in g/mol?
Answer:
3030.2 g/mol is the molar mass for our polymer
Explanation:
Formula for Osmotic pressure → π = M . R .T
where π is pressure (atm)
M is molarity mol/L
R, the Universal Gases Constant (0.082 L.atm/mol.K)
T, Absolute T° (T°C + 273)
25°C + 273 = 298 K
Let's replace the values
0.174 atm = M . 0.082l.atm/mol.K . 298K
0.174 atm / (0.082l.atm/mol.K . 298K) = M
7.12×10⁻³ mol/L
As molarity is mol/L, and we have the volume of solution (in mL we must convert to L) we can find out the moles of our polymer that corresponds to the mass we used.
260mL . 1L/1000mL = 0.260L
7.12×10⁻³ mol/L = mol / 0.260L
7.12×10⁻³ mol . 0.260 = mol → 1.85×10⁻³
These moles refers to te 5.61 g of solute, to if we want to determine the molar mass, we should do:
g/mol → 5.61 g / 1.85×10⁻³ mol = 3030.2 g/mol
Final answer:
The molar mass of the polymer is approximately 28.93 g/mol.
Explanation:
To calculate the molar mass of the polymer, we can use the osmotic pressure formula:
π = MRT
Where π is the osmotic pressure, M is the molar mass, R is the ideal gas constant, and T is the temperature in Kelvin.
Given that the osmotic pressure is 0.174 atm and the temperature is 25.0 °C (which needs to be converted to Kelvin), we can rearrange the formula to solve for M:
M = π / (RT)
Now we can plug in the values:
M = 0.174 atm / (0.0821 L·atm/mol·K * 298 K)
M = 0.00706 mol / 0.024438 L
M = 28.93 g/mol
Therefore, the molar mass of the polymer is approximately 28.93 g/mol.
Today's demand curve for gasoline could shift in response to a change ina.today's price of gasoline. b.the expected future price of gasoline. c.the number of sellers of gasoline. d.All of the above are correct.
Answer:
d.All of the above are correct.
Explanation:
The curve of demand moves left or right continuously. Income, patterns and preferences, related products prices as well as the population size and composition are the key factors causing demand change.
The demand curve for gasoline can shift in response to changes in price, expected future price, and the number of sellers. Option b is the correct option.
Explanation:The correct answer is d. All of the above are correct. The demand curve for gasoline reflects the relationship between the price and quantity demanded, so a change in the price of gasoline can shift the curve. The expected future price of gasoline can also influence current demand, as consumers may adjust their purchasing behavior based on their expectations. Additionally, the number of sellers of gasoline can impact market supply, which in turn affects the equilibrium price and quantity.
Learn more about Demand curve for gasoline here:https://brainly.com/question/33438436
#SPJ11
Complete and balance the molecular equation for the reaction of aqueous sodium carbonate, Na 2 CO 3 Na2CO3 , and aqueous nickel(II) chloride, NiCl 2 NiCl2 . Include physical states. molecular equation:
Na 2 CO 3 ( aq ) + NiCl 2 ( aq ) ⟶ 2 NaCl ( aq ) + NiCO 3 ( s ) Na2CO3(aq)+NiCl2(aq)⟶2NaCl(aq)+NiCO3(s) Enter the balanced net ionic equation for this reaction. Include physical states. net ionic equation:
Answer: The net ionic equation is written below.
Explanation:
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of sodium carbonate and nickel (II) chloride is given as:
[tex]Na_2CO_3(aq.)+NiCl_2(aq.)\rightarrow 2NaCl(aq.)+NiCO_3(s)[/tex]
Ionic form of the above equation follows:
[tex]2Na^{+}(aq.)+CO_3^{2-}(aq.)+Ni^{2+}(aq.)+2Cl^{-}(aq.)\rightarrow NiCO_3(s)+2Na^+(aq.)+2Cl^-(aq.)[/tex]
As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
[tex]Ni^{2+}(aq.)+CO_3^{2-}(aq.)\rightarrow NiCO_3(s)[/tex]
Hence, the net ionic equation is written above.
The balanced molecular equation for the reaction of Na2CO3 and NiCl2 is Na2CO3(aq) + NiCl2(aq) ⟶ 2NaCl(aq) + NiCO3(s). The balanced net ionic equation, excluding the spectator ions, is CO32-(aq) + Ni2+(aq) ⟶ NiCO3(s).
Explanation:The reaction between aqueous sodium carbonate (Na2CO3) and aqueous nickel(II) chloride (NiCl2) results in the formation of sodium chloride (NaCl) and nickel carbonate (NiCO3). Represented as molecular equation, it looks like this: Na2CO3(aq) + NiCl2(aq) ⟶ 2NaCl(aq) + NiCO3(s).
For the net ionic reaction, we exclude the spectator ions, which in this case are Na+ and Cl-. These ions remain unreacted in the solution. Therefore, the net ionic equation will be: CO32-(aq) + Ni2+(aq) ⟶ NiCO3(s).
Learn more about Chemical Reactions here:https://brainly.com/question/34137415
#SPJ3
(a) The first step in ozone formation in the upper atmosphere occurs when oxygen molecules absorb UV radiation of wavelengths ≤ 242 nm. Calculate the frequency and energy of the least energetic of these photons. (b) Ozone absorbs light having wavelengths of 2200 to 2900 Å, thus protecting organisms on Earth’s surface from this high-energy UV radiation. What are the frequency and energy of the most energetic of these photons?
Answer:
a) f = (1.24 × 10^15) Hz and E = (8.214 × 10^-19) J
b) f = (1.36 × 10^15) Hz; E = (9.035 × 10^-19) J
Explanation:
a) The least energetic photons have the highest wavelength. That is, the wavelength of the least energetic photons is equal to the upperlimit of the wavelength inequality given.
λ = 242nm = 2.42 × 10⁻7 m
v = fλ; f = v/λ; v = 3×10^8 m/s
f = (3×10^8)/(2.42×10^-7)
f = (1.24 × 10^15) Hz
E = hf; h = planck's constant = (6.62607004 × 10^-34) Js
E = 6.626 × 10^-34 × 1.24 × 10^15
E = (8.214 × 10^-19) J
b) The photons with the least wavelength in the range provided are the most energetic ones.
λ = (2200 × 10^-10) m = (2.2 × 10^-7) m
v = fλ; f = v/λ; v = 3×10^8 m/s
f = (3×10^8)/(2.2×10^-7)
f = (1.36 × 10^15) Hz
E = hf; h = planck's constant = (6.62607004 × 10^-34) Js
E = 6.626 × 10^-34 × 1.36 × 10^15
E = (9.035 × 10^-19) J
QED!
In ozone formation, least energetic photons having wavelength 242 nm have frequency ≈ 1.24 x 10^15 Hz and energy ≈ 8.2 x 10^-19 J. The most energetic photons that ozone absorbs, with a wavelength of 2200 Å, have a frequency of ≈ 1.36 x 10^15 Hz and energy of ≈ 9.02 x 10^-19 Joules.
Explanation:(a) The frequency (ν) of a photon is given by the formula: ν = c / λ, where c is the speed of light (3.00 x 10^8 m/s) and λ is wavelength. For the least energetic photons (wavelength of 242 nm), we would convert the wavelength to meters (242 nm = 242 x 10^-9 m). Applying the formula: ν = 3.00 x 10^8 m/s / 242 x 10^-9 m, we get ν ≈ 1.24 x 10^15 Hz.
The energy (E) of a photon is given by the formula: E = hν, where h is Planck’s constant (6.63 x 10^-34 Js). So, E = 6.63 x 10^-34 Js x 1.24 x 10^15 Hz, which gives us E ≈ 8.2 x 10^-19 Joules.
(b) For the most energetic of these photons, they have the shortest wavelength (2200 Å = 2200 x 10^-10 m). Using similar calculations as above: ν ≈ 1.36 x 10^15 Hz and E ≈ 9.02 x 10^-19 Joules.
https://brainly.com/question/25626374
#SPJ3
What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3 and the density of alcohol is 790 kg/m3.
Here is the full question
Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 10 cm. The two liquids do not mix.
What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.
Express your answer in two significant figures and include the appropriate units (in cm)
Answer:
ΔH ≅ 3.73 cm
Explanation:
The pressure inside a liquid is known as hydrostatic pressure and which is represent by the formula:
P = ρ × g × h
where;
ρ is the density of the fluid
g is the gravitational constant
h is the height from the surface
From the question above;
For glycerine; we have:
density of glycerine = 1260 kg/m³
gravitational constant = 9.8 m/s²
height = ???
∴
[tex]P_{(g)= 1260kg/m^3}*9.8m/s^2*h_g[/tex] ----- equation (1)
On the other hand for alcohol:
density of alcohol is given as = 790 kg/m³
gravitational constant = 9.8 m/s²
height = 10 cm
∴
[tex]P_{(a)= 790kg/m^3*9.8m/s^2*10[/tex] ----------- equation (2)
if we equate equation 1 and 2 together; we have
[tex]P_{(g)= P_{(a)[/tex]
[tex]1260kg/m^3}*9.8m/s^2*h_g = 790kg/m^3*9.8m/s^2*10cm[/tex]
Making [tex]h_g[/tex] the subject of the formula, we have :
[tex]h_g= \frac{ 790kg/m^3*9.8m/s^2*10cm}{1260kg/m^3*9.8m/s^2}[/tex]
[tex]h_g[/tex] = 6.269 cm
The difference in the height denoted by ΔH can therefore be calculated as:
ΔH [tex]= H_a-H_g[/tex]
ΔH [tex]= 10cm - 6.269cm[/tex]
ΔH = 3.731 cm
ΔH ≅ 3.73 cm (to two significant figures)
A 5.21 mass % aqueous solution of urea (CO(NH2)2) has a density of 1.15 g/mL. Calculate the molarity of the solution. Give your answer to 2 decimal places.
Answer:
Molarity is 0.99 M
Explanation:
5.21% by mass, is a sort of concentration which shows the mass of solute in 100 g of solution.
Molarity is a sort of concentration that indicates the moles of solute in 1 L of solution (mol/L)
Let's find out the volume of solution by density.
Solution density = Solution mass / Solution volume
1.15 g/mL = 100 g / Solution volume
Solution volume = 100 g / 1.15 g/mL → 86.9 mL
We must have the volume of solution in L, so let's convert it.
86.9 mL / 1000 = 0.0869 L
Now, we have to determine the moles of solute (urea)
5.21 g . 1 mol / 60 g = 0.0868 moles
Mol/L = Molarity → 0.0868 moles / 0.0869L = 0.99 M
Answer:
[tex]\large \boxed{\text{1.00 mol/L}}[/tex]
Explanation:
Molar concentration = moles/litres
So, we need both the number of moles and the volume.
1. Volume
Assume a volume of 1 L.
That takes care of that.
2. Moles of urea
(a) Mass of solution
[tex]\text{ Mass of solution} = \text{1000 mL} \times \dfrac{\text{1.15 g solution}}{\text{1 mL}} = \text{1150 g solution}[/tex]
(b) Mass of urea
[tex]\text{Mass of urea} = \text{1150 g solution}\times \dfrac{\text{5.21 g urea}}{\text{100 g solution}} = \text{59.92 g urea}[/tex]
(c) Moles of urea
[tex]\text{Moles of urea} = \text{59.92 g urea} \times \dfrac{\text{1 mol urea}}{\text{60.06 g urea}} = \text{1.00 mol urea}[/tex]
3. Molar concentration
[tex]\text{Molar concentration} = \ \dfrac{\text{1.00 mol}}{\text{1 L}} = \textbf{1.00 mol/L}\\\text{The molar concentration of the urea is $\large \boxed{\textbf{1.00 mol/L}}$}[/tex]
The Tris/Borate/EDTA buffer (TBE) is commonly made as a 5x solution. What volumes of 5x TBE and water are required to make 500 mls of a 0.5x solution which is often used in electrophoresis?
Answer:
50 ml (5x TBE) + 540 ml (water)
Explanation:
To prepare 0.5x TBE solution from 5x TBE solution we need to use the following dilution formula:
C1 x V1 = C2 x V2, where:
- C1, V1 = Concentration/amount (start), and Volume (start)
- C2, V2 = Concentration/amount (final), and Volume (final)
* So when we applied this formula it will be:
5 x V1 = 0.5 x 500
V1= 50ml
- To prepare 0.5x we will take 50ml from 5x and completed with 450ml water and the final volume will going to be 500ml.
How many microliters of original sample are required to produce a final dilution of 10-2 in a total volume of 0.2 mL? 1 microliter is 10-6 L or 10-3 mL.
Answer:
The required volume of the original sample required is 2 micro liter
Explanation:
assuming the original sample concentration is 1N
after final dilution of 10-2 solution concentration becomes 0.01 N
normality of original sample = 1 N
normality of final solution = 0.01 N
volume of original sample= ?
volume of final solution = 0.2 mL
Considering thef formula below :
N1V1 = N2V2
V1 = (N2V2)/N1
= (0.01*0.2)/1
= 0.002 mL
1 milli liter = 1000 micro liter
0.002 mL = 2 micro liter
The original sample required is 2 micro liter
To produce a final dilution of 10^-2 in a total volume of 0.2mL, 2 microliters of original sample are required.
Explanation:To find out how many microliters of original sample are needed to generate a final dilution of 10^-2 in a total volume of 0.2 mL, you can use the formula: V1 = V2 × D2 / D1. Here, V1 is the volume of the original sample needed, V2 is the final volume required (which is 0.2 mL), D2 is the final dilution (which is 10^-2), and D1 is the original dilution (which is 1 for undiluted samples).
So, plugging these values into the formula gives: V1 = 0.2 mL × 10^-2 / 1 = 0.002 mL. Convert this volume from milliliters to microliters (1 mL = 1000 μL), so V1 = 0.002 mL * 1000 = 2 μL.
So, 2 microliters of original sample are required to produce a final dilution of 10^-2 in a total volume of 0.2 mL.
Learn more about Dilution here:https://brainly.com/question/35454012
#SPJ3
Specify the l and ml values for n = 4.
Answer : The specify the l and ml values for n = 4 are:
At l = 0, [tex]m_l=0[/tex]
At l = 1, [tex]m_l=+1,0,-1[/tex]
At l = 2, [tex]m_l=+2,+1,0,-1,-2[/tex]
At l = 3, [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]
Explanation:
There are 4 quantum numbers :
Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as m_l. The value of this quantum number ranges from [tex](-l\text{ to }+l)[/tex]. When l = 2, the value of [tex]m_l[/tex] will be -2, -1, 0, +1, +2.
Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.
As we are given, n = 4 then the value of l and ml are,
l = 0, 1, 2, 3
At l = 0, [tex]m_l=0[/tex]
At l = 1, [tex]m_l=+1,0,-1[/tex]
At l = 2, [tex]m_l=+2,+1,0,-1,-2[/tex]
At l = 3, [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]
When the excited electron in a hydrogen atom falls from to , a photon of blue light is emitted. If an excited electron in falls from , which energy level must it fall to so that a similar blue light (as with hydrogen) is emitted? Prove it.
Answer:
n = 3 for similar blue light
Explanation:
The principle applied here is energy levels and energy changes. There are different energy levels depending on the value of the integer as explained by Max planck - a german physicist in 1900, Max planck claimed that electrons in an atom were presumed to be oscillating with a frequency f, then there enrrgy will be given by the plancks equation ; E =hf, where h is the plancks constant.
In general energy of each level can be written as E =nhf
Draw the partial (valence-level) orbital diagram, and write the symbol, group number, and period number of the element:
(a) [Ar] 4s²3d⁵
(b) [Kr] 5s²4d²
Answer:
a) The element is Manganese (Mn)
b) The element is Zirconium (Zr)
Explanation:
The step by step analysis and explanation is as shown in the attachment
A 3.301 mass % aqueous solution of potassium hydroxide has a density of 1.05 g/mL. Calculate the molality of the solution. Give your answer to 3 decimal places.
Answer: The molality of potassium hydroxide solution is 0.608 m
Explanation:
We are given:
3.301 mass % of potassium hydroxide solution.
This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution
Mass of solvent = Mass of solution - Mass of solute (KOH)
Mass of solvent = (100 - 3.301) g = 96.699 g
To calculate the molality of solution, we use the equation:
[tex]\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams})}[/tex]
Where,
[tex]m_{solute}[/tex] = Given mass of solute (KOH) = 3.301 g
[tex]M_{solute}[/tex] = Molar mass of solute (KOH) = 56.1 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent = 96.699 g
Putting values in above equation, we get:
[tex]\text{Molality of KOH}=\frac{3.301\times 1000}{56.1\times 96.699}\\\\\text{Molality of KOH}=0.608m[/tex]
Hence, the molality of potassium hydroxide solution is 0.608 m
A thermos contains 80.0g of water at 23.4 degrees C. Suppose 0.200 moles of KCl are dissolved in the water. What will be the final temperature of the solution? Assume that there is no energy transfer between the solution and the thermos, and that the specific heat is 4.184J/g*degrees C. Also, the delta H of solvation for KCl at 25 degrees C is 17.1 kJ/mol.
Answer:
32.04°C will be the final temperature of the solution.
Explanation:
Moles of potassium chloride = 0.200 mol
MAs sof KCl= 0.200 mol × 74.5 g/mol= 14.9 g
Enthalpy of solvation of potassium nitrate =
[tex]\Delta H_{solv}=17.1 kJ/mol[/tex]
Energy released when 0.200 moles of KCl is dissolved in water = Q
[tex]Q=17.1kJ/mol\times 0.200 mol=3.42 kJ=3420 J[/tex]
(1 kJ = 1000 J)
Heat released on dissolving 0.200 moles of KCl is equal to heat absorbed by water = Q
Mass of solution , m= 80.0 g +14.9 g = 94.9 g
Specific heat of water = c = 4.184 J/g°C
Initial temperature of the water = [tex]T_1=23.4^oC[/tex]
Final temperature of the water = [tex]T_2=?[/tex]
[tex]Q=m\times c\times (T_2-T_1)[/tex]
[tex]3420 J=94.9g\times 4.184 J/g^oC\times (T_2-23.4^oC)[/tex]
[tex]T_2=32.04^oC[/tex]
32.04°C will be the final temperature of the solution.