Answer: True
Explanation: the angle of incident Ray equals the angle of the reflected Ray.
From the choices provided below, list the reagent(s) in order that will react with 3-pentanone to form 3-phenyl-2-pentene. (List your answer as a letter (single-step transformation), or series of letters (multi-step transformation) in the order the reagents are used, with no commas separating them. No more than four steps are required for this synthesis.)a. H2 / Pt e. 1. PhMgBr 2. H3O+b. H2SO4, Δ f. H3O+c. OsO4, H2O2 g. RCO3Hd. 1. ch3mgbr2. h30+ h. KOH
Final answer:
To form 3-phenyl-2-pentene from 3-pentanone, phenylmagnesium bromide (PhMgBr) is used to introduce the phenyl group, then acid (H3O+) protonates the alkoxide to form an alcohol, followed by acidic dehydration (H2SO4, Δ) to eliminate water and form the double bond. The answer is e followed by b.
Explanation:
To convert 3-pentanone to 3-phenyl-2-pentene, we'll need to form a new carbon-carbon bond and then eliminate a carbonyl group while introducing a double bond. The first step is the Grignard reaction to introduce the phenyl group. We use phenylmagnesium bromide (PhMgBr) as the Grignard reagent to attack the carbonyl carbon of 3-pentanone and then protonate the resulting alkoxide with acid (H3O+) to give 3-phenyl-3-pentanol. The next step is to eliminate water from the alcohol to form the desired double bond. This is done by using acidic dehydration (H2SO4, Δ) to yield 3-phenyl-2-pentene. Therefore, the correct order of reagents is e followed by b.
The correct answer is f, e, d. The correct sequence of reagents to synthesize 3-phenyl-2-pentene from 3-pentanone is f (PhMgBr), followed by e (H3O+), and then d (H3O+ for hydrolysis and elimination).
To synthesize 3-phenyl-2-pentene from 3-pentanone, we can follow a multi-step transformation process involving the given reagents. Here's the step-by-step
1. Phenylmagnesium bromide (PhMgBr) - Reagent e:
The first step involves the formation of a Grignard reagent by reacting 3-pentanone with phenylmagnesium bromide. This reaction will result in the addition of a phenyl group to the carbonyl carbon of 3-pentanone, forming a tertiary alcohol with a phenyl group attached to the carbon adjacent to the hydroxyl group.
The reaction can be represented as follows:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{COR} + \text{PhMgBr} \rightarrow \text{CH}_3\text{CH}_2\text{CH}(\text{OH})\text{CH}(\text{Ph})\text{R} \][/tex]
Note that R represents the rest of the molecule.
2. Water (H3O+) - Reagent d (Step 2):
The second step is the hydrolysis of the Grignard reaction product. The tertiary alcohol formed in the first step is protonated by water, which is acidic in nature (H3O+), to form a good leaving group (water). This sets up the next step, which is an elimination reaction.
The reaction can be represented as follows:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}(\text{OH})\text{CH}(\text{Ph})\text{R} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{C}=\text{CHCH}(\text{Ph})\text{R} + \text{H}_2\text{O} \][/tex]
This step results in the formation of an alkene, specifically 3-phenyl-2-pentene, through an E1 or E2 elimination mechanism.
The other reagents listed in the question are not suitable for the synthesis of 3-phenyl-2-pentene from 3-pentanone:
- H2 / Pt (Reagent a): This would reduce the ketone to an alcohol, which is not the desired transformation.
- H2SO4, (Reagent b): This would lead to dehydration of the ketone, but not in a controlled manner to give the desired alkene.
- OsO4, H2O2 (Reagent c): This would oxidize the ketone to a diol, which is not on the pathway to the desired product.
- RCO3H (Reagent g): This is a reagent for the oxidation of primary alcohols to carboxylic acids and would not be useful in this synthesis.
- CH3MgBr, H3O+ (Reagent d): Methylmagnesium bromide would add a methyl group to the ketone, which is not what we want for the synthesis of 3-phenyl-2-pentene.
- KOH (Reagent h): While KOH could be used to deprotonate the alpha-hydrogen to form an enolate, which could then react with an electrophile, it is not the reagent of choice for the synthesis of 3-phenyl-2-pentene from 3-pentanone.
Therefore, the correct sequence of reagents to synthesize 3-phenyl-2-pentene from 3-pentanone is f (PhMgBr), followed by e (H3O+), and then d (H3O+ for hydrolysis and elimination).
Rank the following groups in order of decreasing priority:A. bondCH3B. bondCH2CH3C. bondCH2CH2CH3D. bond(CH2)3CH3
Answer:
A.
Explanation:
The priority order from highest to lowest for the given groups according to the Cahn-Ingold-Prelog (CIP) rules is tert-butyl, propyl, ethyl, and methyl, based on the number of carbons and degree of substitution.
Explanation:In the context of organic chemistry, we use the Cahn-Ingold-Prelog (CIP) priority rules to determine the order of substituents for naming compounds using the E/Z system. To rank the given groups by their priority, we consider the atomic number of the atoms directly attached to the double bond or chiral center. The higher the atomic number, the higher the priority.
The given groups are as follows:
-CH3-CH2CH3-CH2CH2CH3-C(CH3)3Based on the CIP rules, the priority order is determined by the number of carbons in the alkyl chain or the degree of substitution. Therefore, we can rank these groups from highest priority to lowest priority as follows:
-C(CH3)3 (tert-butyl)-CH2CH2CH3 (propyl)-CH2CH3 (ethyl)-CH3 (methyl)What is the density of atoms/nm^2 on the (110) plane of a zinc blende lattice with lattice spacing 0.546 nm. Three significant digits, fixed point notation.
Answer:
The density is 10.25 [tex]\frac{atom}{nm^{2} }[/tex]
Explanation:
From the question we are given that the lattice spacing nm
V = [tex](0.546)^{3} }[/tex] × [tex]nm^{3}[/tex]
= 0.1628 [tex]nm^{3}[/tex]
zinc blende lattice has 4 atom per unit cell.
This means that 4 atoms is contained in 0.15056 [tex]nm^{3}[/tex]
Density of the cubic unit in terms of atom is given as =
=26.565 [tex]\frac{atoms}{nm^{3} }[/tex]
For plane (110) the spacing between the cubic unit in the crystal denoted by [tex]d_{hkl}[/tex] is given as = [tex]\frac{a}{\sqrt{1^{2} + 1^{2} +0^{2} } }[/tex] Where a is the lattic spacing = 0.546
= [tex]\frac{0.546}{\sqrt{2} }[/tex] = 0.376 nm
The density of the lattice in (110) plane =26.565 [tex]\frac{atoms}{nm^{3} }[/tex] × 0.386 nm
= 10.25 [tex]nm^{2}[/tex]
se the following key to classify each of the elements below in its elemental form: A. Discrete atoms ... C. Metallic lattice B. Molecules ... D. Extended, three-dimensional network 1. Calcium 2. Helium ... 3. Sulfur 4. Potassium ...
Explanation:
the correct match can as follows:
1. helium ⇒ discrete atoms (helium is an inert gas)
2. oxygen ⇒ molecules (oxygen exists in molecular form as O2)
3 Magnesium ⇒ matalic lattice ( Magnesium is a metal FCC crystal Structure :) )
4 Aluminum⇒ covalent network (since it is situated in the middle of the group and posses amphoteric properties too)
Final answer:
To classify the elements: Calcium and Potassium form metallic lattices (category C), Helium exists as discrete atoms (category A), and Sulfur forms molecules (category B). This is due to how the atoms or molecules arrange themselves in solids, often driven by forces between electrons and nuclei.
Explanation:
To classify each element in its elemental form based on the given key:
Calcium (Ca) would fall into category C, Metallic lattice, as it is a metal and metals typically form extended, repeating three-dimensional patterns.
Helium (He) is correctly classified as A, Discrete atoms, because it exists as individual atoms and does not form bonds easily.
Sulfur (S) in its most common form consists of eight-atom rings, making it B, Molecules.
Potassium (K), like calcium, would also be categorized as C, Metallic lattice, due to its metal characteristics, forming a crystal lattice.
Atoms arrange themselves in these structures based on the net attractive forces between electrons and atomic nuclei, and the structural arrangement significantly affects the properties of the material, such as malleability and ductility for metals. Pure metals are crystalline solids with metal atoms packed in a repeating pattern known as a unit cell.
Construct a simulated proton-decoupled 13C NMR spectrum for tert-butyl alcohol. Drag the resonance signal icon to the appropriate chemical shift positions. Then label the box above each resonance signal with the corresponding number of equivalent carbons. (Not all chemical shift bins will be used.)
Answer:
The ¹³C-NMR Spectrum of tert-butyl alcohol will show only two signals.
(i) Signal at around 31 ppm:
This signal towards upfield is for the carbon atoms which are more shielded and are having rich electron surroundings. The height of peak at y-axis shows the number of carbon atoms as compared to other peaks. In this case it is three times the height of second signal hence, it shows that this peak corresponds to three carbon atoms.
(ii) Signal at around 70 ppm:
This signal towards downfield is for the carbon atom which is more deshielded and is having electron deficient surrounding. As compared to the second signal the height of this peaks corresponds to only one carbon. And the deshielded environment shows that this carbon is directly attached to an electronegative element.
A simulated proton-decoupled 13C NMR spectrum for tert-butyl alcohol would present two distinct peaks reflecting the two types of carbon environments within the molecule. Given that it's a proton-decoupled spectrum, the peaks would exist as singlets, not split into multiplets due to hydrogen atom influence.
Explanation:The student's question concerns the simulated construction of a proton-decoupled 13C NMR spectrum for tert-butyl alcohol. In a substance such as tert-butyl alcohol, the symmetry of the molecule helps in predicting the NMR spectrum. Tert-butyl alcohol (t-BuOH) consists of a central carbon atom attached to three methyl groups (CH3) and one OH group.
When such alcohol is subjected to 13C NMR spectroscopy, the central carbon atom attached to the three methyl groups gives rise to one distinct peak in the spectrum due to the identical chemical environment. The carbon of the OH group will give a separate peak in the spectrum as it's in a different chemical environment.
Owing to the spin-spin coupling effect, in proton-coupled 13C NMR spectra, each peak would be split into a multiplet by any hydrogen atoms bonded to the carbon. But, since we are considering a proton-decoupled spectrum here, such multiplet structure would not be seen, and we would have simple singlets at the relevant shifts due to the two types of carbons present in the molecule.
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Be sure to answer all parts. Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with superheated steam: C(s) + H2O(g) → CO(g) + H2(g) ΔH o rxn = 129.7 kJ (a) Combine the reaction above with the following two to write an overall reaction for the production of methane: CO(g) + H2O(g) → CO2(g) + H2(g) ΔH o rxn = −41 kJ CO(g) + 3H2(g) → CH4(g) + H2O(g) ΔH o rxn = −206 kJ In the overall reaction, include the physical states of each product and reactant. (b) Calculate ΔH o rxn for this overall change. 12.03 kJ (c) Using the value in (b) and calculating ΔH o rxn for the combustion of methane, find the total heat for gasifying 6.28 kg of coal and burning the methane formed. Assume water forms as a gas and the molar mass of coal is 12.00 g/mol.
Final answer:
The overall reaction for converting coal to methane is 2C(s) + 2H2O(g) → CH4(g) + CO2(g). The ΔH°rxn for the overall process is -117.3 kJ. Using the molar mass of coal and the heat of combustion for methane, the total heat for gasifying 6.28 kg of coal and burning the resulting methane is calculated.
Explanation:
To answer the student's question regarding the production of methane from coal via gasification, we combine the given reactions to find the overall reaction:
C(s) + H2O(g) → CO(g) + H2(g) ΔH°rxn = 129.7 kJCO(g) + H2O(g) → CO2(g) + H2(g) ΔH°rxn = −41 kJCO(g) + 3H2(g) → CH4(g) + H2O(g) ΔH°rxn = −206 kJThe H2O(g) and CO(g) appear on both sides of the above equations and can be cancelled out when added together. The overall reaction is:
2C(s) + 2H2O(g) → CH4(g) + CO2(g)
To calculate ΔH°rxn for the overall reaction, we sum the enthalpies of the individual steps:
ΔH°rxn (overall) = 129.7 kJ + (-41 kJ) + (-206 kJ) = -117.3 kJ
Using the heat of combustion for methane, 890.4 kJ/mol, and the assumption that coal is approximated as pure carbon (12.00 g/mol), we can calculate the total heat for gasifying 6.28 kg of coal and then burning the methane produced:
Moles of C in 6.28 kg = (6280 g) / (12.00 g/mol) = 523.33 mol
Total heat from gasification = 523.33 mol × -117.3 kJ/mol = -61411.3 kJ
Total heat from combustion = 523.33 mol × 890.4 kJ/mol = 465774.9 kJ
The total heat for the entire process is the sum of heat from gasification and combustion.
What is the molar mass of a gas if it takes 7.3 min to effuse through a small hole and 6.0 min for the same amount of N2 to effuse through the same hole gas
Answer:
Molar mass of unknown gas = 41.45 g/mol
Explanation:
Time taken for a gas to effuse/diffuse ∝√(Molar Mass of the gas)
Let T₁ be the time taken for the Unknown gas to effuse. T₁ = 7.3 min = 438s
Let T₂ be the time taken for the Nitrogen gas to effuse. T₁ = 6 min = 360s
(T₁/T₂) = √(Molar mass of unknown gas/Molar mass of Nitrogen)
Molar mass of Nitrogen = 2×14 = 28 g/mol
(438/360) = √(Molar Mass of unknown gas/28)
√(Molar mass of unknown gas/28) = 1.21667
molar mass of unknown gas/28 = 1.21667² = 1.4803
Molar mass of unknown gas = 1.4803 × 28 = 41.45 g/mol
Hope this helps!
Which of the following types of molecules always has a dipole moment? Linear molecules with two identical bonds. Trigonal pyramid molecules (three identical bonds). Trigonal planar molecules (three identical bonds equally spaced). Tetrahedral molecules (four identical bonds equally spaced). None has a dipole moment.
Answer:
Trigonal pyramid molecules (three identical bonds)
Explanation:
In trigonal pyramidal molecule like molecule of ammonia , the vector some of intra- molecular dipole moment is not zero because the bonds are not symmetrically oriented . In other molecules , bonds are symmetrically oriented in space so the vector sum of all the internal dipole moment vectors cancel each other to make total dipole moment zero.
The type of molecules that always have a dipole moment is: B. Trigonal pyramid molecules (three identical bonds).
A molecule can be defined as a group of two (2) or more atoms that are chemically bonded together by means of shared electrons.
Also, a molecule form the smallest, fundamental ((basic) unit of a chemical compound or substance and they are capable of taking part in a chemical reaction.
In Chemistry, a dipole moment occurs whenever there is a separation of charge between molecules.
Trigonal pyramidal molecules have three (3) atoms at their trigonal base corners and one (1) atom at the top.
Additionally, a trigonal pyramidal molecule with three (3) identical bonds always have a dipole moment because bond the can't cancel one another.
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In what sense is the wave motion of a guitar string analogous to the motion of an electron in an atom?
Explanation:
A guitar string vibrates when we strikes it. It starts vibrating in several modes simultaneously. It stretches between the saddle and the nut. This distance represents one-half wavelength.
Now if we consider that the string forms a circle, then we have an interpretation of an electron which vibrates in the orbit surrounding the nucleus. We are aware that electrons have wavelength. If the circumference of the orbit happens to be the integer multiple of wavelength , then the orbit is "allowed" since "the electron will retraces its own path."
This explains the line spectrum and not a continuous spectrum.
A line spectrum refers an electron that jumps between the specific energy levels, thus producing only specific colors.
rate = k[AB]2 and k = 0.20 L/mol·s. If the initial concentration of AB is 1.50 M, what is [AB] after 14.0 s?
Answer : The concentration of AB after 14.0 s is, 0.29 M
Explanation :
The expression used for second order kinetics is:
[tex]kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}[/tex]
where,
k = rate constant = [tex]0.20M^{-1}s^{-1}[/tex]
t = time = 14.0 s
[tex][A_t][/tex] = final concentration = ?
[tex][A_o][/tex] = initial concentration = 1.50 M
Now put all the given values in the above expression, we get:
[tex]0.20\times 14.0=\frac{1}{[A_t]}-\frac{1}{1.50}[/tex]
[tex][A_t]=0.29M[/tex]
Therefore, the concentration of AB after 14.0 s is, 0.29 M
One of the hydrates of MnSO4 is manganese(II) sulfate tetrahydrate . A 71.6 gram sample of MnSO4 4 H2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained?
Answer:
48.32 g of anhydrous MnSO4.
Explanation:
Equation of dehydration reaction:
MnSO4 •4H2O --> MnSO4 + 4H2O
Molar mass = 55 + 32 + (4*16) + 4((1*2) + 16)
= 223 g/mol
Mass of MnSO4 • 4H2O = 71.6 g
Number of moles = mass/molar mass
= 71.6/223
= 0.32 mol.
By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4
Number of moles of MnSO4 = 0.32 mol.
Molar mass = 55 + 32 + (4*16)
= 151 g/mol.
Mass = 151 * 0.32
= 48.32 g of anhydrous MnSO4.
A scuba diver that ascends to the surface too quickly can experience decompression sickness, which occurs when nitrogen that dissolves in the blood under high pressure, forms bubbles as the pressure decreases during the ascent. Therefore, an understanding of the gas laws is an important part of a scuba diver's training. In fresh water, the pressure increases by 1 atm 1 atm every 34 ft 34 ft below the water surface a diver descends. If a diver ascends quickly to the surface from a depth of 34 ft 34 ft without exhaling, by what factor will the volume of the diver's lungs change upon arrival at the surface? Assume the atmospheric pressure at the surface of the water is 1 atm 1 atm .
Answer:
2
Explanation:
The pressure at 34ft is the atmospheric pressure (1 atm) plus 1 atm, because at each 34ft the pressure increase by 1 atm, so it will be 2 atm. Let's assume that the lungs will be at a constant temperature, so, by the ideal gas law:
P1*V1 = P2*V2
Where P is the pressure, V is the volume, 1 is the state at 34 ft under the water, and 2 at the water surface, where P2 = 1atm. Thus:
2*V1 = 1*V2
V2 = 2V1
Thus, the volume will change by a factor of 2, so, the volume will double.
A scuba diver's lungs volume will double upon ascending to the surface from a depth of 34 ft without exhaling, due to the decrease in pressure as explained by Boyle's Law.
Explanation:When a scuba diver ascends from a depth of 34 ft in fresh water to the surface, the pressure experienced by the diver changes significantly, which according to Boyle's Law, affects the volume of the diver's lungs. At a depth of 34 ft, the diver experiences a pressure of 2 ata (1 atm from the water plus 1 atm from the atmosphere). Boyle's Law states that at constant temperature, the volume of a gas is inversely proportional to the pressure exerted on it. Therefore, if the diver does not exhale during ascent, the volume of the diver's lungs will double upon arrival at the surface where the pressure is 1 atm. This understanding is critical to avoid potential injury such as lung overexpansion, known as pulmonary barotrauma, and to also mitigate the risks associated with decompression sickness (DCS).
The mole fraction of a non electrolyte (MM 101.1 g/mol) in an aqueous solution is 0.0194. The solution's density is 1.0627 g/mL. Calculate the molarity of the solution.
Answer:
Molarity for the solution is 1.05 mol/L
Explanation:
Mole fraction of solute = 0.0194
Solution's density = 1.0627 g/mL
We must know that sum of mole fraction = 1
Mole fraction of solute + Mole fraction of solvent = 1
0.0194 + Mole fraction of solvent = 1
Mole fraction of solvent = 1 - 0.0194 → 0.9806
Molarity is mol of solute in 1L of solution, so we have to determine solution's volume in L
With molar mass we can determine the mass of solute and solvent and then, the solution's mass
0.0194 mol . 101.1 g/ mol = 1.96 g of non electrolyte solute
0.9806 mol . 18 g/mol = 17.65 g of water
Mass of solution = mass of solute + mass of solvent
1.96 g + 17.65 g = 19.6 g (mass of solution)
Solution's density = Solution's mass / Solution's volume
1.0627 g/mL = 19.6 g / Solution's volume
Solution's volume = 19.6 g / 1.0627 g/mol →18.4 mL
Let's convert the mass from mL to L
18.4mL . 1L / 1000 mL = 0.0184 L
We have the moles of solute, so let's determine molarity
mol/L → 0.0194 mol / 0.0184 L = 1.05 M
Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces. 32P → 32S + e− 15 16 Rate = k [32P], where k = 4.85 ✕ 10−2 day−1 What is the instantaneous rate (in mol/L/d) of production of electrons in a sample with a phosphorus concentration of 0.0031 M?
Answer:
Rate = 1.50x10⁻⁴ mol*day⁻¹*L⁻¹
Explanation:
The equation of the study of biochemical reaction mechanisms is:
³²P → ³²S + e⁻
The rate of that reaction is:
Rate = k [³²P]
where k: is the rate constant and [³²P] is the concentration of ³²P.
Hence, having that: k = 4.85x10⁻² day⁻¹ and, [³²P] = 0.0031 M, the instantaneous rate of production of electrons is:
Rate = k [³²P] = (4.85x10⁻² day⁻¹)(0.0031 mol*L⁻¹) = 1.50x10⁻⁴ mol*day⁻¹*L⁻¹
Therefore, the instantaneous rate of production of electrons is 1.50x10⁻⁴ mol*day⁻¹*L⁻¹.
I hope it helps you!
(6)The first-order rate constant for the decomposition of N2O5, N2O5(g) 2NO2(g) + O2(g)At 70C is 6.810-3s-1. Suppose we start with 0.0250 mol of N2O5(g) in a volume of 1.0 L. (a)How many moles of N2O5 will remain after 2.5 min
Answer:
0.009 moles
Explanation:
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given that:
The rate constant, k = [tex]6.8\times 10^{-3}[/tex] s⁻¹
Initial concentration [tex][A_0][/tex] = 0.0250 mol
Final concentration [tex][A_t][/tex] = ?
Time = 2.5 min = 2.5 x 60 seconds = 150 sec
Applying in the above equation, we get that:-
[tex][A_t]=0.0250e^{-6.8\times 10^{-3}\times 150}\ moles=0.025\times \frac{1}{e^{\frac{51}{50}}}\ moles=\frac{0.025}{2.77319}\ moles=0.009\ moles[/tex]
A metal cylinder holds helium gas at the gift shop. When someone orders a balloon, a small amount of helium is transferred from the tank to the balloon. Use kinetic molecular theory to explain why the helium gas expands to fill the balloon, and why liquid water and paper confetti would not behave in the same way.?
Answer:
Explanation:
Gases are unlike other states of matter in that a gas expands to fill the shape and volume of its container. For a small amount of helium gas transferred from the tank to the balloon, the gas(helium) expands and fills up the entire mass of the balloon.
According to the kinetic theory, gases can also be compressed(that's the reason why the metal cylinder can hold the helium gas) so that a relatively large amount of gas can be forced into a small container.
The kinetic-molecular theory explains why gases are more compressible and expands more than either liquids or solids. Gases can expands to take the shape of the container in which they are in, unlike liquids and solids. This is responsible for in this case, why helium gas introduced to the balloon fills up the balloon.
Write the full ground-state electron configuration for each:
(a) S (b) Kr (c) Cs
Answer:
S: [Ne] 3s² 3p⁴
[Ar] 3d¹⁰ 4s² 4p⁶
[Xe] 6s1
Explanation:
The electronic configuration in ground state
Calculate the energy of attraction between a cation with a valence of 1 and an anion with a valence of 3 the centers of which are separated by a distance of 7.5 nm
The energy of attraction between a cation with a valence of 1 and an anion with a valence of 3 can be calculated using Coulomb's law.
Explanation:The energy of attraction between a cation and an anion can be calculated using Coulomb's law. Coulomb's law states that the energy of attraction is directly proportional to the product of the charges and inversely proportional to the distance between them. In this case, the cation has a valence of +1 and the anion has a valence of -3. The distance between their centers is given as 7.5 nm. The equation to calculate the energy of attraction is:
E = k * (q1 * q2) / r
Where E is the energy of attraction, k is Coulomb's constant (9 x 10^9 Nm^2/C^2), q1 and q2 are the charges of the cation and anion respectively, and r is the distance between their centers.
Substituting the values:
E = (9 x 10^9 Nm^2/C^2) * ((+1) * (-3)) / (7.5 x 10^-9 m)
Simplifying the equation gives:
E = -36 Nm
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An archer pulls her bowstring back 0.370 m by exerting a force that increases uniformly from zero to 264 N. (a) What is the equivalent spring constant of the bow?
Answer:
713.51 N/m
Explanation:
Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.
From hook's law,
F = ke ...........................Equation 1
Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.
Make k the subject of the equation,
k = F/e ............................ Equation 2
Given: F = 264 N, e = 0.37 m.
Substitute into equation 2
k = 264/0.37
k = 713.51 N/m
Hence the spring constant of the bow = 713.51 N/m
The reaction of hydrochloric acid with potassium permanganate is described above. 526.64 g of hydrogen chloride is reacted with 229.19 g of potassium permanganate. Assuming the reaction goes to completion, calculate the mass of each product produced. g of manganese(II) chloride g of water g of chlorine g of potassium chloride
Explanation:
Equation of the reaction:
2KMnO4(aq) + 16HCl(aq) --> 2MnCl2(aq) + 2KCl(aq) + 5Cl2(g) + 8H2O(aq)
To calculate the limiting reagent, we need to calculate the number of moles of the reactants :
KMnO4:
Molar mass = (39 + 55 + (16*4))
= 158 g/mol
Number of moles = mass/molar mass
= 229.19/158
= 1.4506 mol
HCl:
Molar mass = 1 + 35.5
= 36.5 g/mol
Number of moles = 526.64/36.5
= 14.428 mol
By stoichiometry, 2 moles of KMnO4 reacted with 16 moles of HCl
The limiting reagent :
14.428 moles of HCl * 2 moles of KMnO4/16moles of HCl
= 1.8035 moles of KMnO4 is required to react with 14.428 moles of HCl
But there's 1.4506 moles of KMnO4. Therefore, KMnO4 is the limiting reagent.
Mass of the products:
KCl:
2 moles of KMnO4 will produce 2 moles of KCl
Moles of KCl = 1 * 1.4506 mol
= 1.4506 mol
Molar mass = 39 + 35.5 = 74.5 g/mol
Mass of KCl = 74.5 * 1.4506
= 108.07 g
MnCl2:
2 moles of KMnO4 will produce 2 moles of MnCl2
Number of moles of MnCl2 = 1 * 1.4506
= 1.4506 mol
Molar mass = 55 + (35.5*2)
= 126 g/mol
Mass of MnCl2= 1.4506 * 126
= 182.78 g
Cl2:
2 moles of KMnO4 will produce 5 moles of Cl2
Number of moles of Cl2 = 5/2 * 1.4506
= 3.6265 mol
Molar mass of Cl2 = 35.5*2
= 71 g/mol
Mass of Cl2 = 71* 3.6265
= 257.4815 g
H2O:
2 moles of KMnO4 will produce 8 moles of H2O
Number of moles of H2O = 8/2 * 1.4506
= 5.80 mol
Molar mass of H2O =( 1*2) + 16
= 18 g/mol
Mass of H2O = 18*5.80
= 104.44 g
Answer:
(1) mass of KCl =108.025g
(2) mass of MnCl2 =182.7 g
(3) mass of Cl2 =257.37 g
(4) mass of H20(water) =104.4g
Explanation:
we start with determining the limiting factor
molar mass of KMnO4 = 158g/mol
molar mass of HCl = 36.5g/mol
hence,
number of moles of KMnO4= mass /molar mass
number of moles of KMnO4= 229.19/158 = 1.45moles
number of moles of HCl = 526.64/36.5 = 14.43 moles
we chose the lowest number of moles from the reactants as the limiting factor
hence the limiting factor is KMnO4.
calculating the mass of the products:
for KCl
2moles of KMnO4 reacts to give 2moles of KCl
there for KCl contains 1.45moles
(1) mass of KCl = number of moles of KCl x molar mass of KCl = 1.45 x 74.5 = 108.025g
(2) from mole ratio 2 mole of KMnO4 gave 2 moles of MnCl2
therefore 1.45moles KMnO4 will give 1.45 MnCl2
mass of MnCl2 = 1.45moles x molar mass MnCl2 = 1.45 x 126 = 182.7 g
(3) from mole ratio
2moles KMnO4 gave 5moles Cl2
1.45 moles will give (5 x1.45)/2 moles of Cl2
mass of Cl2 = number of moles of Cl2 x molar mass Cl2 = 3.625 x 71 = 257.37 g
(4)from mole ratio
2moles KMnO4 gave 8moles water(H2O)
1.45 moles will give (1.45x 8)/2 moles of H20
mass of H2O produced = number of moles of H20 x the molar mass of H20 = 5.8 x 18 = 104.4g
How fast must a 56.5-g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green light (5400 Å)?
Answer:vv
[tex]v=2.17\times 10^{-26}\ m/s[/tex]
Explanation:
The expression for the deBroglie wavelength is:
[tex]\lambda=\frac {h}{m\times v}[/tex]
Where,
[tex]\lambda[/tex] is the deBroglie wavelength
h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]
m is the mass of = [tex]56.5\ g=0.0565\ kg[/tex]
v is the speed.
Wavelength = 5400 Å = [tex]5400\times 10^{-10}\ m[/tex]
Applying in the equation as:-
[tex]5400\times 10^{-10}=\frac{6.626\times 10^{-34}}{0.0565\times v}[/tex]
[tex]v=\frac{331300000}{10^{34}\times \:1.5255}\ m/s[/tex]
[tex]v=2.17\times 10^{-26}\ m/s[/tex]
Analysis of a sample of a compound, containing only carbon, nitrogen, hydrogen, and oxygen, determined that it contained 20.0% C, 6.7% H, 46.6% N and the balance O. What is the empirical formula of the compound?
Answer:
CH₄N₂O
Explanation:
Let's assume 100 g of the compound, thus, the mass of each substance is it percent multiplied by 100:
C = 20.0 g (20% = 0.20, and 0.20*100 = 20.0)
H = 6.7 g
N = 46.6 g
O = 100 - (20+6.7+46.6) = 26.7 g
The number of moles of each compound is the mass divided by the molar mass. The molar masses are C = 12.011 g/mol, H = 1.00794 g/mol, N = 14.01 g/mol, and O = 15.999 g/mol, so:
nC = 20/12.011 = 1.67 mol
nH = 6.7/1.00794 = 6.65 mol
nN = 46.6/14.01 = 3.33 mol
nO = 26.7/15.999 = 1.67 mol
The empirical formula is the formula with the minimum possible number of the moles of each compound, which must be proportional to the percent of each one. So, we must divide each number of moles for the smallest, 1.67:
C = 1.67/1.67 = 1
H = 6.65/1.67 = 4
N = 3.33/1.67 = 2
O = 1.67/1.67 = 1
So, the empirical formula is CH₄N₂O.
27.4 g of Aluminum nitrite and 169.9 g of ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams of the excess reagent remain after the reaction? Hint: Write a balanced chemical equation for the reaction first. Enter your answer to 1 decimal place.
Answer: The mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For aluminium nitrite:Given mass of aluminium nitrite = 27.4 g
Molar mass of aluminium nitrite = 41 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of aluminium nitrite}=\frac{27.4g}{41g/mol}=0.668mol[/tex]
For ammonium chloride:Given mass of ammonium chloride = 169.9 g
Molar mass of ammonium chloride = 53.5 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of ammonium chloride}=\frac{169.9g}{53.5g/mol}=3.176mol[/tex]
The chemical equation for the reaction of aluminium nitrite and ammonium chloride follows:
[tex]Al(NO_2)_3+3NH_4Cl\rightarrow AlCl_3+3N_2+6H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of aluminium nitrite reacts with 3 moles of ammonium chloride
So, 0.668 moles of aluminium nitrite will react with = [tex]\frac{3}{1}\times 0.668=2.004mol[/tex] of ammonium chloride.
As, given amount of ammonium chloride is more than the required amount. So, it is considered as an excess reagent.
Thus, aluminium nitrite is considered as a limiting reagent because it limits the formation of product.
Excess moles of ammonium chloride = (3.176 - 2.004) mol = 1.172 moles
Calculating the mass of ammonium chloride by using equation 1, we get:
Excess moles of ammonium chloride = 1.172 moles
Molar mass of ammonium chloride = 53.5 g/mol
Putting values in equation 1, we get:
[tex]1.172mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.172mol\times 53.5g/mol)=62.7g[/tex]
Hence, the mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams
Are compounds of these ground-state ions paramagnetic?
(a) Ti²⁺ (b) Zn²⁺ (c) Ca²⁺ (d) Sn²⁺
Answer:
No, (a) Ti²⁺ is only paramagnetic
Explanation:
Paramagnetic are those which has unpaired electrons and diamagnetic are those in which all electrons are paired.
(a) Ti²⁺
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{2}[/tex]
The electrons in 3d orbital = 2 (Unpaired)
Thus, the ion is paramagnetic as the electrons are unpaired.
(b) Zn²⁺
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{10}[/tex]
The electrons in 3d orbital = 10 (paired)
Thus, the ion is diamagnetic as the electrons are paired.
(c) Ca²⁺
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^6[/tex]
The electrons in 3p orbital = 6 (paired)
Thus, the ion is diamagnetic as the electrons are paired.
(d) Sn²⁺
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^2[/tex]
The electrons in 5s orbital = 2 (paired)
Thus, the ion is diamagnetic as the electrons are paired.
The compounds of these ground-state ions (a) Ti²⁺, (b) Zn²⁺, (c) Ca²⁺, and (d) Sn²⁺ are not paramagnetic.
Explanation:The compounds of these ground-state ions are not paramagnetic.
Paramagnetic substances have unpaired electrons in their atoms or ions, which cause them to be attracted to a magnetic field. To determine whether an ion is paramagnetic, we need to consider its electron configuration.
(a) Ti²⁺: It has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d². The 3d² sublevel has two unpaired electrons, so Ti²⁺ is paramagnetic.
(b) Zn²⁺: It has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰. All the sublevels in its electron configuration are filled, so Zn²⁺ is not paramagnetic.
(c) Ca²⁺: It has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s². All the sublevels are filled, so Ca²⁺ is not paramagnetic.
(d) Sn²⁺: It has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰. All the sublevels are filled, so Sn²⁺ is not paramagnetic.
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Write the full electron configuration of the Period 3 element with the following successive IEs (in kJ/mol):
IE₁ = 738
IE₂ = 1450
IE₃ = 7732
IE₄ = 10,539
IE₅ = 13,628
Answer:
Magnesium (Mg)
Electronic Configuration of Magnesium (Mg): [tex]1s^22s^22p^63s^2[/tex]
Explanation:
Ionization Energy:
It is the amount of energy required to remove valance shell electron of an atom in gaseous phase. The more the electron is closer to the nucleus the more energy is required to remove the electron.
Trend:
Increases from left to right in period.
Decreases generally from top to bottom in group.
In our case the data is:
IE₁ = 738
IE₂ = 1450
IE₃ = 7732
IE₄ = 10,539
IE₅ = 13,628
This data matches the Ionization energies of Magnesium (Mg). Magnesium has 12 electrons with 2 electrons in valance shell. It has 12 Ionization energies.
Electronic Configuration of Magnesium (Mg): [tex]1s^22s^22p^63s^2[/tex]
The electron configuration of the Period 3 element with the given successive IEs is;
1s² 2s² 2p⁶ 3s²The ionization energy data given is in synchrony with that of Magnesium, Mg.
In essence, the element in question is Magnesium, Mg.
The electronic configuration of Magnesium is therefore;
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What word best describes the role that the palladium plays in the reaction between propene and hydrogen?
Answer: palladium plays the role of catalyst
Explanation:
Answer: It is a catalyst
Explanation:
In the laboratory you are asked to make a 0.303 m cobalt(II) sulfate solution using 275 grams of water. How many grams of cobalt(II) sulfate should you add?
Answer: 12.92g of CoSO4
Explanation:
Molar Mass of CoSO4 = 59 + 32 + (16x4) = 59 + 32 +64 = 155g/mol
Molarity of CoSO4 = 0.303mol/L
Mass conc. In g/L = Molarity x molar Mass
= 0.303x155 = 46.965g/L
275 grams of water = 0.275L of water
46.965g of CoSO4 dissolves in 1L
Therefore Xg of CoSO4 will dissolve in 0.275L i.e
Xg of CoSO4 = 46.965x0.275 = 12.92g
Therefore 12.92g of CoSO4 is needed
Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abundance), and X-30 (29.974 amu 3.09% abundance). Calculate the atomic weight of X.
a. 29.09 amu
b. 28.09 amu
c. 35.29 amu
d. 86.93 amu
e. 25.80 amu
Answer: The average atomic mass of X is 28.09 amu
Explanation:
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex] .....(1)
For isotope 1:Mass of isotope 1 = 27.979 amu
Percentage abundance of isotope 1 = 92.21 %
Fractional abundance of isotope 1 = 0.9212
For isotope 2:Mass of isotope 2 = 28.976 amu
Percentage abundance of isotope 2 = 4.70 %
Fractional abundance of isotope 2 = 0.0470
For isotope 3:Mass of isotope 3 = 29.974 amu
Percentage abundance of isotope 3 = 3.09 %
Fractional abundance of isotope 3 = 0.0309
Putting values in equation 1, we get:
[tex]\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)][/tex]
[tex]\text{Average atomic mass of X}=28.09amu[/tex]
Hence, the average atomic mass of X is 28.09 amu
The atomic weight of element X is approximately 28.04 amu.
The atomic weight of an element is calculated by taking into account the masses of its isotopes and their abundances. To calculate the atomic weight of element X, we multiply the mass of each isotope by its abundance and sum the results. Using the given data, we can calculate it as follows:
Calculate the weighted masses of each isotope:Therefore, the atomic weight of element X is approximately 28.04 amu.
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The electron geometry and the molecular geometry of ammonia (NH3) are, respectively: The electron geometry and the molecular geometry of ammonia (NH3) are, respectively: tetrahedral, trigonal pyramidal. tetrahedral, tetrahedral. trigonal planar, bent. tetrahedral, bent. none of the above g
Answer: Electronic geometry is tetrahedral and the molecular geometry will be trigonal pyramidal.
Explanation:
According to VSEPR theory:
Number of electron pairs =[tex]\frac{1}{2}[V+N-C+A][/tex]
For [tex]NH_3[/tex]
where,
V = number of valence electrons present in central nitrogen atom = 5
N = number of monovalent atoms = 3
C = charge of cation = 0
A = charge of anion = 0
Number of electron pairs =[tex]\frac{1}{2}[5+3-0+0]=4[/tex]
The number of electron pairs is 4 that means the hybridization will be [tex]sp^3[/tex] but as only 3 atoms are present , one position will be occupied by a lone pair thus electronic geometry is tetrahedral and the molecular geometry will be trigonal pyramidal.
Ammonia (NH3) has tetrahedral electron geometry and a trigonal pyramidal molecular geometry. This is due to the presence of three hydrogen atoms and one lone pair of electrons surrounding the central nitrogen atom. The presence of the lone pair slightly distorts the ideal tetrahedral angle, making the molecule trigonal pyramidal.
Explanation:The electron geometry and the molecular geometry of ammonia (NH3) are respectively tetrahedral and trigonal pyramidal. In ammonia, the electron pair geometry, which includes both bonding electrons and lone pairs, is tetrahedral due to four pairs of electrons around the nitrogen atom (three bonding pairs and one lone pair).
However, the molecular geometry, which describes the spatial arrangement of the atoms, is trigonal pyramidal, as the nitrogen atom is at the apex and the three hydrogen atoms form the base. The ideal bond angles deviate slightly from the idealized angles because the lone pair takes up a larger region of space than the bonding electrons, causing the hydrogen-nitrogen-hydrogen angle (H-N-H) to be slightly smaller than the 109.5° angle in a regular tetrahedral structure.
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Calculate the molality of a 1.06 M sucrose, C12H22O11, with a density of 1.14 g/mL. For sucrose, mol. wt.
Final answer:
To find the molality, multiply the density by 1000 to get the total mass of the solution per liter, subtract the mass of dissolved sucrose, and finally divide the moles of sucrose by the kilograms of water.
Explanation:
To calculate the molality of a 3.1416 M aqueous solution of sucrose with a density of 1.5986 g/mL, we follow these steps:
Firstly, calculate the mass of 1 liter of the solution by multiplying its density by 1000 mL, which gives us 1598.6 grams.Next, since the molarity (M) is given as 3.1416 M, we multiply this by the molar mass of sucrose (342.297 g/mol) to find out how many grams of sucrose are in 1 liter of solution: 3.1416 mol/L * 342.297 g/mol = 1075.4 grams of sucrose.We then subtract the mass of the sucrose from the total mass of the solution to find the mass of the water: 1598.6 g - 1075.4 g = 523.2 g of water, which is equivalent to 0.5232 kg of water.Lastly, we divide the moles of sucrose by the kg of water to get the molality: 3.1416 mol sucrose / 0.5232 kg water = 6.0035 mol/kg, which is the molality of the solution.