Answer:
The reaction requires anti-Markovnikov product without sigmatropic rearrangement.
Explanation:
The reaction is known to begin with the concerted syn addition of B and H across the double bond, with the boron adding to the less substituted carbon atom.
The second step of the reaction involves hydrogen peroxide and a base such as NaOH are added, NaOH deprotonates the hydrogen peroxide.
The resulting NaOOH then attacks the boron and sets up the key migration step, where the carbon-boron bond migrates to the oxygen bound to boron, breaking the weak oxygen-oxygen bond . Then the -OH expelled then returns to form a bond on the boron resulting in a deprotonated alcohol (alkoxide). The alkoxide is then protonated by water or some other comparably acidic species.
Hydroboration is a syn addition that gives an anti-Markovnikov product without sigmatropic rearrangement.
The reasons why hydroboration–oxidation reagents were chosen to effect the transformation will be:
Oxymercuration - demercuration reagents give the Markovnikov product.Hydroboration-oxidation reagents yield the anti-Markovnikov product of addition.The reaction requires anti-Markovnikov product without sigmatropic rearrangement.Addition with acid and water as reagents gives the Markovnikov product.It should be noted that hydroboration–oxidation is vital for the production of alcohol. It's needed for the transformation rather than using other reagents because hydroboration-oxidation reagents yield the anti-Markovnikov product of addition.
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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).
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
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
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]
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
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 4.00 gram sample of a solution of sodium chloride in water was heated until all the water had evaporated. The sodium chloride that remained weighed 1.22 grams. Calculate the percentage of water in the original 4.00 grams of solution.
The weight of the water in the original solution is 2.78 grams. The percentage of water in the original 4.00 grams solution of sodium chloride, once the water was evaporated, is therefore approximately 69.5%.
Explanation:The weight of the water in the original solution can be found by subtracting the weight of the remaining sodium chloride from the total weight of the solution. So, water = total weight - weight of sodium chloride = 4.00 grams - 1.22 grams = 2.78 grams.
The percentage of water in the solution can be found by the following formula:
Percentage = (Weight of water / Total weight) * 100 = (2.78 grams / 4.00 grams) * 100 = 69.5%
So, the percentage of water in the 4.00 gram solution of sodium chloride after the water was evaporated, is approximately 69.5%.
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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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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).
A single penny is 1.52 mm thick. The distance to the next nearest star other than our own (Alpha Centauri) is 4.22 light-years. If it were possible to stack one mole of pennies, how many times would the stack go between the earth and Alpha Centauri? Use the unit factoring method to determine the answer and show your work. You will need to find or look up the appropriate conversion factors to solve the problem. Your answer should be in scientific notation and have the correct number of significant figures in order to get full credit. (Please note that the text editing functions/buttons below for this essay question allows you to show exponents by using the button show as "x2"in the controls. To use it type the number followed by the exponent such as 104, highlight the 4 and hit the x2 button and you will end up with 104 as the result)
Answer:
2.29 × 10⁴ times
Explanation:
A single penny is 1.52 mm thick. The distance covered by 1 mole of pennies (6.02 × 10²³ pennies) is:
6.02 × 10²³ p × (1.52 mm/1 p) = 9.15 × 10²³ mm = 9.15 × 10²³ × 10⁻³ m = 9.15 × 10²⁰ m
The distance to the next nearest star other than our own (Alpha Centauri) is 4.22 light-years. Considering 1 ly = 9.46 × 10¹⁵ m, this distance in meters is:
4.22 ly × (9.46 × 10¹⁵ m/1 ly) = 3.99 × 10¹⁶ m
The times that the stack would go between the earth and Alpha Centauri are:
9.15 × 10²⁰ m / 3.99 × 10¹⁶ m = 2.29 × 10⁴
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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Express your answer as a balanced chemical equation. Identify all of the phases in your answer.
1. Li(s)+N2(g)→Li3N(s)
2. TiCl4(l)+H2O(l)→TiO2(s)+HCl(aq)
3. NH4NO3(s)→N2(g)+O2(g)+H2O(g)
4. Ca3P2(s)+H2O(l)→Ca(OH)2(aq)+PH3(g)
5. Al(OH)3(s)+H2SO4(aq)→Al2(SO4)3(aq)+H2O(l)
6.AgNO3(aq)+Na2SO4(aq)→Ag2SO4(s)+NaNO3(aq)
7. C2H5NH2(g)+O2(g)→CO2(g)+H2O(g)+N2(g)
Explanation:
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
1.[tex]Li(s)+N_2(g)\rightarrow Li_3N(s)[/tex]
The balanced equation is:
[tex]6Li(s)+N_2(g)\rightarrow 2Li_3N(s)[/tex]
2. [tex]TiCl_4(l)+H_2O(l)\rightarrow TiO_2(s)+HCl(aq)[/tex]
The balanced equation is:
[tex]TiCl_4(l)+2H_2O(l)\rightarrow TiO_2(s)+4HCl(aq)[/tex]
3. [tex]NH_4NO_3(s)\rightarrow N_2(g)+O_2(g)+H_2O(g)[/tex]
The balanced equation is:
[tex]2NH_4NO_3(s)\rightarrow 2N_2(g)+O_2(g)+4H_2O(g)[/tex]
4.[tex] Ca3P2(s)+H2O(l)\rightarrow Ca(OH)2(aq)+PH3(g)[/tex]
The balanced equation is:
[tex] Ca_3P_2(s)+6H_2O(l)\rightarrow 3Ca(OH)_2(aq)+2PH_3(g)[/tex]
5. [tex]Al(OH)_3(s)+H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+H_2O(l)[/tex]
The balanced equation is:
[tex]2Al(OH)_3(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+6H_2O(l)[/tex]
6.[tex] AgNO_3(aq)+Na_2SO_4(aq)\rightarrow Ag_2SO_4(s)+NaNO_3(aq)[/tex]
The balanced equation is:
[tex] 2AgNO_3(aq)+Na_2SO_4(aq)\rightarrow Ag_2SO_4(s)+2NaNO_3(aq)[/tex]
7. [tex]C_2H_5NH_2(g)+O_2(g)\rightarrow CO_2(g)+H_2O(g)+N_2(g)[/tex]
The balanced equation is:
[tex]4C_2H_5NH_2(g)+15O_2(g)\rightarrow 8CO_2(g)+14H_2O(g)+2N_2(g)[/tex]
A balanced chemical equation is a representation of a chemical reaction using chemical formulas and symbols, where the number of atoms of each element on the left side (reactants) is equal to the number of atoms of the same element on the right side (products). In other words, it obeys the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction
1. Li(s)+N2(g)→Li3N(s)
The balanced equation is :
[tex](Li(s) + N_2(g) \rightarrow Li_3N(s)\)[/tex]
2. TiCl4(l)+H2O(l)→TiO2(s)+HCl(aq)
The balanced equation is :
[tex]\(TiCl_4(l) + 4H_2O(l) \rightarrow TiO_2(s) + 4HCl(aq)\)[/tex]
3. NH4NO3(s)→N2(g)+O2(g)+H2O(g)
The balanced equation is :
[tex]\(NH_4NO_3(s) \rightarrow N_2(g) + 2O_2(g) + 2H_2O(g)\)[/tex]
4. Ca3P2(s)+H2O(l)→Ca(OH)2(aq)+PH3(g)
The balanced equation is :
[tex]\(Ca_3P_2(s) + 6H_2O(l) \rightarrow 3Ca(OH)_2(aq) + 2PH_3(g)\)[/tex]
5. Al(OH)3(s)+H2SO4(aq)→Al2(SO4)3(aq)+H2O(l)
The balanced equation is :
[tex]\(2Al(OH)_3(s) + 3H_2SO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 6H_2O(l)\)[/tex]
6.AgNO3(aq)+Na2SO4(aq)→Ag2SO4(s)+NaNO3(aq)
The balanced equation is :
[tex]\(2AgNO_3(aq) + Na_2SO_4(aq) \rightarrow Ag_2SO_4(s) + 2NaNO_3(aq)\)[/tex]
7. C2H5NH2(g)+O2(g)→CO2(g)+H2O(g)+N2(g)
The balanced equation is :
[tex]\(2C_2H_5NH_2(g) + 9O_2(g) \rightarrow 4CO_2(g) + 10H_2O(g) + 2N_2(g)\)[/tex]
These equations show the balanced chemical reactions along with the respective phases of the substances involved.
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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.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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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]
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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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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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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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
(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]
g Nitrogen in the atmosphere consists of two nitrogen atoms covalently bonded together (N2). Knowing that nitrogen is atomic number 7, what type of covalent bond holds those two atoms together?
Explanation:
Since, the atomic number of nitrogen is 7 and its electronic distribution is 2, 5. So, in order to attain stability it needs to gain 3 electrons.
Hence, when it chemically combines another nitrogen atom then as both the atoms are non-metals. So, sharing of electrons will take place.
Also, there is no difference in electronegativity of two nitrogen atoms. Hence, compound formed [tex]N_{2}[/tex] is non-polar covalent in nature.
Answer:
non polar bond
Explanation:
What mass of Na2CrO4 is required to precipitate all of the silver ions from 73.6 mL of a 0.150 M solution of AgNO3?
Answer:
We need 0.894 grams of Na2CrO4
Explanation:
Step 1: Data given
Volume of a 0.150 M AgNO3 = 73.6 mL = 0.0736 L
Step 2: Calculate moles of Ag+
Moles Ag+ = moles AgNO3
Moles Ag+ = volume * molarity
moles Ag+ = 0.0736 L x 0.150 M = 0.01104 moles
Step 3: Calculate moles Na2CrO4
2AgNO3 + Na2CrO4 → Ag2CrO4 (s) + 2NaNO3
For 2 moles AgNO3 we need 1 mol Na2CRO4
For 0.01104 moles AgNO3 we need 0.01104/2 = 0.00552 moles Na2CrO4
Step 4: Calculate mass of Na2CrO4
Mass Na2CrO4 = moles * molar mass
Mass Na2CrO4 = 0.00552 moles * 161.97 g/mol
Mass Na2CrO4 = 0.894 grams
We need 0.894 grams of Na2CrO4
To precipitate all the silver ions, 0.893 g of Na2CrO4 is required.
Explanation:To calculate the mass of Na2CrO4 required to precipitate all of the silver ions from the solution, we need to use the stoichiometry of the reaction. The balanced equation for the reaction is:
2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3
From the equation, we can see that 2 moles of AgNO3 react with 1 mole of Na2CrO4 to form 1 mole of Ag2CrO4. The molarity of AgNO3 is given as 0.150 M and the volume is 73.6 mL. First, we need to convert the volume to liters:
73.6 mL × (1 L / 1000 mL) = 0.0736 L
Next, we can use the molarity and volume to calculate the number of moles of AgNO3:
moles of AgNO3 = molarity × volume = 0.150 M × 0.0736 L = 0.01104 moles
Since the reaction ratio is 2 moles of AgNO3 to 1 mole of Na2CrO4, we can calculate the number of moles of Na2CrO4:
moles of Na2CrO4 = (0.01104 moles of AgNO3) / 2 = 0.00552 moles
Finally, we can use the molar mass of Na2CrO4 to calculate the mass:
mass of Na2CrO4 = moles of Na2CrO4 × molar mass of Na2CrO4
From the periodic table, the molar mass of Na2CrO4 is:
molar mass of Na2CrO4 = 22.99 g/mol + (2 × 35.45 g/mol) + 4 × 16.00 g/mol = 161.97 g/mol
Therefore, the mass of Na2CrO4 required to precipitate all of the silver ions is:
mass of Na2CrO4 = 0.00552 moles × 161.97 g/mol = 0.893 g
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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.
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)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.
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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:
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.
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
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.
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