Determine the ground-state electron configuration and bond order for each of the Period 2 diatomic molecules. Specify which MO is the HOMO and which is the LUMO. List the molecules in order of increasing dissociation and list them in order of bond length. You may need to THINK a bit to get this correct
Answers
Answer:
Explanation:
check the attachment below for correct explanations.
Related Questions
Which statement below correctly describes how amino acids join?
Amino acids join by linking the amino groups of the two compounds together.
Amino acids join by formation of amino bonds between them.
Amino acids join by forming an ester linkage.
Amino acids form bonds via E1 elimination reactions.
Amino acids join by linking the acid group of one amino acid to the amino group of another.
Answers
Answer:
Amino acids join by linking the acid group of one amino acid to the amino group of another.
Explanation:
Amino acids are organic molecules that form the basic molecules for making proteins and there are. An amino acid comprises of an acidic carboxyl (-COOH) functional group and an amino group (-NH2) as well as a side an organic side chain (R group).
In the formation of proteins, several amino acids join together by the formation of peptide bonds between each amino acids to form a long polypeptide. These peptide bods are formed by the linking of the acidic carboxyl group of one amino acid to the amino group of another amino acid, during this process water is removed.
Describe at least two factors that can affect the rate of a chemical reaction.
Answers
Answer: Concentration or pressure of a reactant and temperature
Explanation: The two major factors that can affect the rate of a chemical change are concentration or pressure of a reactant and temperature.
Using the complex based titration system: 50.00 mL 0.00250 M Ca2+ titrated with 0.0050 M EDTA, buffered at pH 11.0 determine (i) first pCa first before initiating the titration process and then (ii) at equivalence when all the Ca2+ is titrated to CaY2-. Please, use your text books and/or lecture notes to find potentially missing information about constants needed to solve the problem.
Answers
Answer:
i) The pCa before initiating the titration is 2.6
ii) The pCa is 6.67
Explanation:
please look at the solution in the attached Word file
How does the molecular motion of a sustance change when it goes from liquid to solid?
A) The molecules lose energy, and their relative motion decreases
B) The molecules gain energy,and their relative motion decreases
C)The molecules gain energy, and their relative motion increases
D) The molecules lose energy, and their relative motion increases
Answers
Answer:
B) The molecules gain energy, and their relative motion decreases
The change in the molecular motion should be option B.
Change in molecular motion:The molecular motion refers to the movement of constituent particles or molecules in a specific direction. It should be impacted by heat and temperature. When there is the transformation from liquid to solid so the change in the molecular motion of the substance should be that the molecular gained the energy and there should be a decrease in the relative motion.
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17-61 Nucleophilic aromatic substitution provides one of the common methods for making phenols. (Another method is discussed in Section19-17 .) Show how you would synthesize the following phenols, using benzene or toluene as your aromatic starting material, and explain why mixtures of products would be obtained in some cases. p-nitrophenol 2,4,6-tribromophenol
Answers
Answer:
Explanation:
find the solution below
Calcium hydride reacts with water to form hydrogen gas according to the unbalanced equation below: CaH2(s) + H2O(l) --> Ca(OH)2(aq) + H2(g) This reaction is sometimes used to inflate life rafts, weather balloons, and the like, where a simple, compact means of generating H2 is desired. How many grams of calcium hydride are needed to generate 15.0 L of hydrogen gas at 25 degrees C and 825 torr of pressure?
Answers
Answer:
28 grams CaH₂(s) is required for production of 15L H₂(g) at 25°C and 825Torr.
Explanation:
CaH₂(s) + H₂O(l) => Ca(OH)₂(s) + H₂(g)
Using ideal gas law, PV = nRT
=> moles H₂(g) = PV/RT = [(825/760)Atm](15L)/(0.08206L·Atm·mol⁻¹·K⁻¹)(298K) = 0.6659 mol H₂(g)
From stoichiometry of given equation,
=> 0.6659 mol H₂(g) requires 0.6659 mole CaH₂(s)
Converting moles to grams, multiply by formula weight,
=> 0.6659 mole CaH₂(s) = 0.6659 mole CaH₂(s) x 42g/mole = 27.966 grams CaH₂(s) ≅ 28 grams CaH₂(s) (2 sig. figs.)
Which of the following acid/base pairs would be best for making a buffer with a pH of 8.00?
1.Acid Formula Ka1 Ka2 Ka3 Acetic acid CH3COOH 1.8×10-5
2.Ascorbic acid H2C6H6O6 7.9×10-5 1.6×10-12
3.Carbonic acid H2CO3 4.2×10-7 4.8×10-11
4.Hydrocyanic acid HCN 4.0×10-10
5.Hydrofluoric acid HF 7.2×10-4
6.Hypochlorous acid HClO 3.5×10-8
7.Phosphoric acid H3PO4 7.5×10-3 6.2×10-8 4.8×10-13
8.Sulfurous acid H2SO3 1.7×10-2 6.4×10-8
Answers
Answer:
7. Phosphoric acid H₃PO₄ 7.5×10⁻³ 6.2×10⁻⁸ 4.8×10⁻¹³
3. Carbonic acid H₂CO₃ 4.2×10⁻⁷ 4.8×10⁻¹¹
Explanation:
Their blend will result to the closest pH of 8.00
Draw the mechanism of the slow step that occurs in both first-order substitution and first-order elimination reactions for (R)-3-bromo-2,3-dimethylpentane in methanol with heat applied. Provide curved arrows in Box 1 to depict the flow of electrons and draw the intermediate in Box 2.
Answers
Answer:
see explaination
Explanation:
We are given the (R)-3-bromo-2,3-dimethylpentane and asking to draw the curved arrow which is the showing the mechanism for first-order substitution and first-order elimination reactions. We know the formation of carbocation is the rate determining step in the first-order substitution and first-order elimination reactions.
So in the (R)-3-bromo-2,3-dimethylpentane there is –Br gets removed and formed the tertiary carbocation which is more stable, so the curved arrows in Box 1 to depict the flow of electrons and intermediate in Box 2.
Check attachment
The absorbance features observed in the visible spectrum for curcumin is a result of an allowed optical excitation of an electron from the π-HOMO to the π*-LUMO. What is the energy change for this electronic excitation based on the spectrum? (Hint: Energy and wavelength are related by the equation E = hc/λ.) h = 4.136 × 10-15 eV⋅ s c = 2.998 × 108 m/s
Answers
Answer:
2.3 ev or 3.68 ×10^-19J
Explanation:
The spectrum is shown in the image attached
h= 4.136 × 10-15 eV⋅ s
c = 2.998 × 108 m/s
λmax= 550×10^-9 (from the spectrum attached)
E=hc/λmax
E= 4.136 × 10^-15 × 2.998 × 10^8/550×10^-9
E= 2.3 ev or 3.68 ×10^-19J
The energy change for the electronic excitation is : 3.68 * 10⁻¹⁹J
Given data :
h = 4.136 * 10⁻¹⁵ eV⋅ s
c = 2.998 * 108 m/s
λmax = 550 * 10⁻⁹ ( Obtained from image attached below )
Applying the energy and wavelength relationship equation
E = hc / λmax
= ( 4.136 * 10⁻¹⁵ * 2.998 * 108 ) / 550 * 10⁻⁹
= 2.3 ev ≈ 3.68 * 10⁻¹⁹J.
Hence we can conclude that the energy change for the electronic excitation is 2.3 ev ≈ 3.68 * 10⁻¹⁹J.
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Missing data related to your question is attached below
: If a 250. mL sample of the above buffer solution initially has 0.0800 mol H2C6H5O7 - and 0.0600 mol HC6H5O7 2- , what would be the new concentration of HC6H5O7 2- after 25.0 mL of 0.125 M NaOH is added to the buffer?
Answers
Answer: New concentration of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] is 0.23 M.
Explanation:
The given data is as follows.
Moles of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] = 0.06 mol
Moles of [tex]H_{2}C_{6}H_{5}O_{7}[/tex] = 0.08 mol
Therefore, moles of [tex]OH^{-}[/tex] added are as follows.
Moles of [tex]OH^{-}[/tex] = [tex]0.125 \times \frac{25}{1000}[/tex]
= 0.003125 mol
Now, new moles of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] = 0.06 + 0.003125
= 0.063125
Therefore, new concentration of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] will be calculated as follows.
Concentration = [tex]\frac{0.063125}{0.275}[/tex]
= 0.23 M
Thus, we can conclude that new concentration of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] is 0.23 M.
Consider the perbromate anion. What is the central atom? Enter its chemical symbol. How many lone pairs are around the central atom? What is the ideal angle between the bromine-oxygen bonds? Compared to the ideal angle, you would expect the actual angle between the bromine-oxygen bonds to be ...
Answers
Answer:
See explanation
Explanation:
The central atom in the perbromate ion is bromine. The chemical symbol of bromine is Br. There are no lone pairs around the central bromine atom. The ion is tetrahedral in shape hence we expect a bond angle of 109°. 27 which is the ideal tetrahedral bond angle. The actual bond angle of the prebromate ion is 109.5°. The perbromate ion is BrO4^-
The observed bond angle is very close to the ideal value because of the absence of lone pairs of electrons from the central atom in the ion.
The perbromate anion, BrO4-, has Bromine (Br) as its central atom and two lone pairs of electrons. This configuration results in a square planar molecular structure, presenting ideal Bromine-Oxygen bond angles of 90° and 180°. These angles are expected to be virtually accurate due to minimization of lone pair-bonding pair repulsions.
Explanation:The perbromate anion, represented by the chemical formula BrO4-, has bromine (Br) as its central atom. Based on the octet rule, the central bromine atom is surrounded by four oxygen atoms and has two lone pairs of electrons. Given this arrangement, the perbromate anion exhibits an octahedral electron-pair geometry, but due to the presence of the two lone pairs, its molecular structure is square planar. The ideal angle between the Bromine-Oxygen bonds in a square planar structure is 90° or 180°. Since the lone pairs occupy the positions minimizing their interactions with the bonded oxygen atoms, the actual angle in the perbromate anion is expected to closely match this ideal angle.
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A papermaking factory releases waste into a local lake. Soon, many fish in the lake begin to die. How might this hurt the economy of the area?
a Local fishermen will lose their jobs.
B. The papermaking factory will make less paper.
C. This won’t harm the economy in any way.
Answers
Answer:
Local fishermen will lose their jobs
The decarboxylation of lysine catalyzed by lysine decarboxylase has a kcat value of 500 s-1 at 298K, and loss of CO2 is the rate-determining step. What is the free energy of activation for the CO2 loss step? The half-life for the uncatalyzed reaction under the same conditions is 4 billion years (1017 seconds). How much does the enzyme lower the free energy of activation for this reaction? Show your work.
Answers
Answer:
The decrease in free energy is 113.299kJ
Explanation:
K for enzyme catalyzed reaction = 500s^-1
Temperature (T) =298k
ΔG =?
ΔG = - 2.303 RT log k
ΔG = (-2.303)(8.314)(298) log 500
ΔG = - 15399.9 J
ΔG catalyzed = - 15. 399kJ
The first order reaction is given as:
t1/2= 0.693/k
or k= 0.693/t1/2
0.693/10^17
Therefore,
K= 0.693 × 10^-17
Now,
K= 0.693 × 10^-17
T= 298k
ΔG uncatalyzed =?
ΔG uncatalyzed = - 2.303 RT log k
ΔG uncatalyzed = (-2.303)(8.314)(298) log0.693 × 10^-17
= 97908.1J
ΔG uncatalyzed = 97.9081kJ
Therefore,
The decrease in free energy is:
ΔG uncatalyzed - ΔG catalyzed
97.908 - (-15.399)
= 113.299KJ
The decrease in free energy is 113.299kJ
Final answer:
The free energy of activation for the CO2 loss step can be calculated using the Arrhenius equation. The enzyme lowers the free energy of activation by comparing the activation energies of the catalyzed and uncatalyzed reactions.
Explanation:
The free energy of activation for the CO2 loss step can be calculated using the Arrhenius equation:
k = Ae^(-Ea/RT)
Where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant (8.314 J/mol·K), and T is the temperature in Kelvin.
Since the rate-determining step is the loss of CO2, we can use the kcat value (500 s-1) as the rate constant for this step. To find the activation energy, we need to rearrange the Arrhenius equation:
Ea = -RT ln(k/A)
Now we can substitute the given values into the equation:
Ea = -(8.314 J/mol·K)(298 K) ln(500 s-1/A)
To calculate the value of A, we can use the half-life for the uncatalyzed reaction:
t1/2 = ln(2)/(kuncat)
Replacing kuncat with the appropriate value, we can solve for A:
A = e^(ln(2)/(kuncat) - ln(2)/(kcat))/t1/2 = e^(ln(kcat/kuncat))/t1/2
Finally, we can substitute the values of kcat, kuncat, and t1/2 into the equation to find A.
To calculate how much the enzyme lowers the free energy of activation, we can compare the activation energies of the uncatalyzed and catalyzed reactions:
∆∆G (ΔEa) = ∆Ga - ∆Ga,uncat
Where ∆Ga is the activation energy of the catalyzed reaction and ∆Ga,uncat is the activation energy of the uncatalyzed reaction.
Consider the reaction: NO2(g) + CO(g) ⇌ NO(g) + CO2(g) Kc = 0.30 at some temperature. If the initial mixture has the concentrations below, the system is_______.
Answers
This is an incomplete question, here is a complete question.
Consider the reaction: [tex]NO_2(g)+CO(g)\rightleftharpoons NO(g)+CO_2(g)[/tex]
Kc = 0.30 at some temperature.
If the initial mixture has the concentrations below, the system is_______.
Chemicals Concentration (mol/L)
- NO₂ 0.024
- CO 0.360
- NO 0.180
- CO₂ 0.120
Possible answers:
1) not at equilibrium and will remain in an unequilibrated state.
2) not at equilibrium and will shift to the left to achieve an equilibrium state.
3) not at equilibrium and will shift to the right to achieve an equilibrium state.
4) at equilibrium
Answer : The correct option is, (2) not at equilibrium and will shift to the left to achieve an equilibrium state.
Explanation:
Reaction quotient (Qc) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.
First we have to determine the value of reaction quotient (Qc).
The given balanced chemical reaction is,
[tex]NO_2(g)+CO(g)\rightleftharpoons NO(g)+CO_2(g)[/tex]
The expression for reaction quotient will be :
[tex]Q_c=\frac{[NO][CO_2]}{[NO_2][CO]}[/tex]
In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.
Now put all the given values in this expression, we get
[tex]Q_c=\frac{(0.180)\times (0.120)}{(0.024)\times (0.360)}=2.5[/tex]
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
There are 3 conditions:
When [tex]Q>K[/tex] that means product > reactant. So, the reaction is reactant favored.
When [tex]Q<K[/tex] that means reactant > product. So, the reaction is product favored.
When [tex]Q=K[/tex] that means product = reactant. So, the reaction is in equilibrium.
The given equilibrium constant value is, [tex]K_c=0.30[/tex]
From the above we conclude that, the [tex]Q>K[/tex] that means reactant < product. So, the reaction is reactant favored that means reaction must shift to the reactant or left to be in equilibrium.
Hence, the correct option is, (2) not at equilibrium and will shift to the left to achieve an equilibrium state.
You mix sodium metal with nitric acid. What salt is produced?
Answers
Answer:
By mixing 2 Moles of Sodium Metal with 2 Moles of nitric acid, Sodium Nitrate is formed.
Explanation:
2Na+ + HNO3 produces 2NaNO3 + H2
You have a friend who wants to move to Hawaii because, "Hawaii has better weather". Based on your knowledge from the text, is this an accurate statement? Why or why not?
Answers
Answer:
"Hawaii has better weather" is an opinion.
Therefore, you would need to support it with facts in order to deem it accurate or not.
If the text suggests that Hawaii has nice weather, then the statement would be accurate.
If the text hints that Hawaii does not have ideal weather, the statement would be inaccurate.
Final answer:
The statement that 'Hawaii has better weather' is subjective; Hawaii has a tropical type A climate with wet and dry areas due to the rain shadow effect. While it's warm, Kauai receives over 460 inches of rain annually, and snow can occur on mountain peaks in winter.
Explanation:
Whether or not Hawaii has 'better weather' is subjective and depends on personal preferences. However, based on the text provided, the statement that Hawaii has better weather because it has a tropical type A climate may not be entirely accurate for everyone. While Hawaii does have a warm and tropical climate, there are variations across the islands. For example, the island of Kauai is one of the wettest places on Earth, receiving over 460 inches of rain per year. Moreover, the rain shadow effect caused by Mount Wai'ale'ale leads to heavy rainfall on the windward side and dry conditions on the leeward side, creating a semi-desert environment. Additionally, it's noteworthy that snow can be found on the tops of Hawaii's highest mountains in the winter.
The diverse climate conditions in Hawaii mean that the weather can vary significantly from one part of the island to another, which might be pleasant for some but not for others. Therefore, whether Hawaii has 'better weather' is based on an individual's weather preferences and what they consider to be better.
research to obtain more information about where we get our fuel supply.Evaluate and communicate how technology affects the supplies of nonrenewable resource.
Answers
Answer:
hola como estas
Explanation:
The pressure in a bicycle tire is __55.0__psi at __30.0__˚C in Phoenix. You take the bicycle up to Flagstaff, where the temperature is _5.0___˚C. What is the pressure, in kPa, in the tire?(Volume and Amount of moles is held constant)
Answers
Answer:
[tex]p_2=347.9kPa[/tex]
Explanation:
Hello,
In this case, we use the Gay-Lussac's law which allows us to understand a gas' pressure-temperature behavior as a directly proportional relationship:
[tex]\frac{p_1}{T_1}= \frac{p_2}{T_2}[/tex]
Whereas it is convenient to use the pressure in kPa and the temperature in kelvins in order to compute the required resulting pressure, therefore:
[tex]p_1=55.0psi*\frac{6.89476kPa}{1psi} =379.2kPa\\T_1=30.0+273.15=303.15K\\T_2=5.0+273.15=278.15K[/tex]
Thus, we obtain:
[tex]p_2= \frac{p_1T_2}{T_1}=\frac{379.2kPa*278.15K}{303.15K}\\ \\p_2=347.9kPa[/tex]
Best regards.
Answer:
The pressure in the tire at 5.0 °C is 347.91 kPa
Explanation:
Step 1: Data given
The pressure in a bicycle tire is 55.0 psi
Temperature = 30.0 °C = 303 K
Temperature decreases to 5.0 °C = 278 K
Volume and Amount of moles are held constant
Step 2: Calculate the pressure at the new temperature
P1/T1 = P2 / T2
⇒with P1 = the initial pressure of the bicycle tire is 55.0 psi
⇒with T1 = the initial temeprature = 303 K
⇒with P2 = the pressure at the new temperature
⇒with T2 = the decreased temperature = 278 K
55.0 psi / 303 K = P2 / 278 K
P2 = (55.0 psi / 303 K) * 278 K
P2 = 50.46 psi
Step 3: Convert pressure from psi to kPa
50.46 psi = 50.46 * 6.895 = 347.91 kPa
The pressure in the tire at 5.0 °C is 347.91 kPa
Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that is a weak acid. mol of is added to of a solution.
Answers
Answer: seen below
Explanation:
HCH3CO2 + NaOH --------------> CH3CO2- + H2O
Acid specie- HCH3CO2
base- NaOH
Neutral- Na+
CH3COOH + KOH ----------> CH3COOK + H20
Acid- CH3COOH
Base- KOH
Neutral- K+
If the vapor pressure of an aqueous solution containing 6.00 moles of a nonvolatile solute has a vapor pressure of 19.8 torr, and given that the vapor pressure of water at room temperature is 23.7 torr, how many total moles are present in solution? Your answer should have three significant figures.
Answers
Answer:
36.5 mol
Explanation:
The vapor pressure of a solution of a non volatile solute in water is given by Raoult´s law:
P H₂O = χ H₂O x P⁰ H₂O
where χ H₂O is the mole fraction of water in the solution and P⁰ H₂O
In the turn the mole fraction is given by
χ H₂O = mol H₂O / total # moles = mol H₂O /ntot
Thus
P H₂O = mole H₂O / n tot x P⁰ H₂0
now the mol of H₂O is equal n tot - 6 mol solute
Plugging the values given in the question and solving the resultant equation
19.8 torr = ( ntot - 6 ) x 23.7 torr / n tot
19.8 ntot = 23.7 ntot - 142.2
ntot = 36.5 ( rounded to 3 significant figures )
A concentrated binary solution containing mostly species 2 (but x2 ≠ 1) is in equilib- rium with a vapor phase containing both species 1 and 2. The pressure of this two- phase system is 1 bar; the temperature is 25°C. At this temperature, 1 = 200 bar and P2sat = 0.10 bar. Determine good estimates of x1 and y1. State and justify all assumptions.
Answers
Answer:
x1= 4.5 × 10^-3, y1= 0.9
Explanation:
A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2
Take the basis as the pressure of the 2 phase system is 1 bar. The assumption are as follows:
1. The vapor phase is ideal at pressure of 1 bar
2. Henry's law apply to dilute solution only.
3. Raoult's law apply to concentrated solution only.
Where,
Henry's constant for species 1 H= 200bar
Saturation vapor pressure of species 2, P2sat= 0.10bar
Temperature = 25°C= 298.15k
Apply Henry's law for species 1
y1P= H1x1...... equation 1
y1= mole fraction of species 1 in vapor phase.
P= Total pressure of the system
x1= mole fraction of species 1 in liquid phase.
Apply Raoult's law for species 2
y2P= P2satx2...... equation 2
From the 2 equations above
P=H1x1 + P2satx2
200bar= H1
0.10= P2sat
1 bar= P
Hence,
P=H1x1 + (1 - x1) P2sat
1bar= 200bar × x1 + (1 - x1) 0.10bar
x1= 4.5 × 10^-3
The mole fraction of species 1 in liquid phase is 4.5 × 10^-3
To get y, substitute x1=4.5 × 10^-3 in equation 1
y × 1 bar = 200bar × 4.5 × 10^-3
y1= 0.9
The mole fraction of species 1 in vapor phase is 0.9
What is Binary solution?
A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2
If we consider the pressure of the 2 phase system is 1 bar.
The assumption are as follows:
The vapor phase is ideal at pressure of 1 bar. Henry's law apply to dilute solution only. Raoult's law apply to concentrated solution only.Where, these values are given:
Henry's constant for species 1 H= 200bar
Temperature = 25°C= 298.15K
P₂sat= 0.10 bar
Apply Henry's law for species 1
y₁P= H₁x₁.......... (i)
where y₁= mole fraction of species 1 in vapor phase, P= Total pressure of the system ,x₁= mole fraction of species 1 in liquid phase.
Apply Raoult's law for species 2
y₂P= P₂sat. x₂...........(ii)
From (i) and (ii)
P=H₁x₁ + P₂sat. x₂
200bar= H₁
0.10= P₂sat
1 bar= P
Hence,
P=H₁x₁ + (1 - x₁) P₂sat
1bar= 200bar × x₁ + (1 - x₁) 0.10bar
x₁= [tex]4.5 * 10^{-3}[/tex]
The mole fraction of species 1 in liquid phase is [tex]4.5 * 10^{-3}[/tex]
To get y, substitute x₁=[tex]4.5 * 10^{-3}[/tex] in (i)
y × 1 bar = 200bar × [tex]4.5 * 10^{-3}[/tex]
y₁= 0.9
The mole fraction of species 1 in vapor phase is 0.9.
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It is difficult to prepare an amide from a carboxylic acid and an amine directly, since an acid-base reaction occurs which renders the amine nitrogen non-nucleophilic. Typically, in such an amide synthesis, the carboxylic acid OH group is first transformed into a better, nonacidic leaving group. In practice, amides are often prepared by treating the carboxylic acid with dicyclohexylcarbodiimide (DCC). The amine is then added and nucleophilic acyl substitution occurs easily because dicyclohexylurea is a good leaving group. This method of amide bond formation is a key step in the laboratory synthesis of peptide bonds (amide bonds) between protected amino acids. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answers
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution.
To a flask, 15.0 mL of 1.25 M hydrofluoric acid is added. Then, 3.05 M KOH is used to titrate the acid sample. Write the balanced net ionic equation for the acid-base reaction.
Answers
Answer:
HF + OH- = F- + H2O
Explanation:
Since hydrofluoric acid does not ionize in aqueous solution, the fluoride ion is still present as part of the product
Answer:
H+ (aq) + OH-(aq) → H2O(l)
Explanation:
Step 1: Data given
Volume of hydrofluoric acid = 15.0 mL = 0.015 L
Molarity = 1.25 M
Molarity of KOH = 3.05 M
Step 2: The unbalanced equation
HF(aq) + KOH(aq) → KF(aq) + H2O(l)
This equation is already balanced
Step 3: The net ionic equation
The net ionic equation shows only those elements, compounds, and ions that are directly involved in the chemical reaction.
The elements, compounds, and ions that do not take part in the chemical reaction are called spector ions.
H+ (aq) + F-(aq) + K+(aq) + OH-(aq)→ K+(aq) +F-(aq) + H2O(l)
We'll remove all the spector ions.
H+ (aq) + OH-(aq) → H2O(l)
A metal salt with the formula MSO4 crystallizes from water to form a solid with the composition MSO4⋅3H2O. At 298 K, the equilibrium vapor pressure of water above this solid is 14.7 Torr. MSO4⋅3H2O(s)↽−−⇀MSO4(s)+3H2O(g) What is the value of ΔG for the reaction when the vapor pressure of water is 14.7 Torr?
Answers
To find the value of ΔG for the reaction MSO4⋅3H2O(s)↽−−⇀MSO4(s)+3H2O(g) at the given vapor pressure of water, use the equation ΔG = -RTln(K), where ΔG is the change in Gibbs free energy, R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant.
Explanation:The reaction can be represented as: MSO4·3H2O(s) <--> MSO4(s) + 3H2O(g)
The equilibrium vapor pressure of water above the solid is 14.7 Torr.
Since we are given the equilibrium condition, we can use the equation ΔG = -RTln(K), where ΔG is the change in Gibbs free energy, R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant.
By plugging in the given values, we can calculate the value of ΔG at the equilibrium condition.
The value of [tex]\Delta G[/tex] for the reaction when the vapor pressure of water is 14.7 Torr is approximately 2.005 kJ/mol.
The value of [tex]\Delta G[/tex] for the reaction at 298 K when the vapor pressure of water is 14.7 Torr is given by the equation:
[tex]\[ \Delta G = -RT \ln \left( \frac{P_{H_2O}^3}{P_{H_2O}^{eq}} \right) \][/tex]
where:
- [tex]\( \Delta G \)[/tex] is the change in Gibbs free energy,
- [tex]R[/tex] is the universal gas constant (8.314 J/(mol·K)),
- [tex]T[/tex] is the temperature in Kelvin (298 K in this case),
- \[tex]\( P_{H_2O}^3 \)[/tex] is the partial pressure of water raised to the power of the moles of water in the reaction (which is 3),
- [tex]\( P_{H_2O}^{eq} \)[/tex] is the equilibrium vapor pressure of water (14.7 Torr).
First, we need to convert the equilibrium vapor pressure of water from Torr to atmospheres to match the units of the gas constant [tex]R[/tex]. The conversion factor is 1 atm = 760 Torr.
[tex]\[ P_{H_2O}^{eq} = \frac{14.7 \text{ Torr}}{760 \text{ Torr/atm}} = 0.01934 \text{ atm} \][/tex]
Now we can plug in the values into the equation:
[tex]\[ \Delta G = -(8.314 \text{ J/(mol·K)}) \times (298 \text{ K}) \times \ln \left( \frac{(0.01934 \text{ atm})^3}{(0.01934 \text{ atm})} \right) \][/tex]
[tex]\[ \Delta G = -8.314 \times 298 \times \ln \left( (0.01934)^2 \right) \][/tex]
[tex]\[ \Delta G = -8.314 \times 298 \times \ln \left( 0.000373 \right) \][/tex]
[tex]\[ \Delta G = -8.314 \times 298 \times (-7.936) \][/tex]
[tex]\[ \Delta G = 8.314 \times 298 \times 7.936 \][/tex]
[tex]\[ \Delta G = 2004.8 \text{ J/mol} \][/tex]
[tex]\Delta G = 2.005 kJ/mol[/tex]
The dehydrogenation of benzyl alcohol to make the flavoring agent benzaldehyde is an equilibrium process described by the equation: C6H5CH2OH(g) ⇆ C6H5CHO(g) + H2(g) At 523 K, the value of its equilibrium constant is K = 0.558. (a) Suppose that 1.20 g of benzyl alcohol is placed into a 2.00 L vessel and heated to 523 K. What is the partial pressure of benzaldehyde when equilibrium is attained? (b) What fraction of benzyl alcohol is dissociated into products at equilibrium?
Answers
Answer:
pC6H5CHO = 0.180 atm
Fraction dissociated = 0.756
Explanation:
Step 1: Data given
Temperature = 523 K
the value of its equilibrium constant is K = 0.558
Mass of benzyl alcohol = 1.20 grams
Molar mass of benzyl alcohol = 108.14 g/mol
Volume = 2.00 L
heated to 523 K
Step 2: The balanced equation
C6H5CH2OH(g) ⇆ C6H5CHO(g) + H2(g)
Step 3: Calculate moles benzyl alcohol
Moles benzyl alcohol = Mass / molar mass
Moles benzyl alcohol = 1.20 grams / 108.14 g/mol
Moles benzyl alcC6H5CH2OHohol = 0.0111 moles
Step 4: Initial moles
Moles C6H5CH2OH = 0.0111 moles
Moles C6H5CHO = 0 moles
Moles H2O = 0 moles
Step 5: moles at the equilibrium
Moles C6H5CH2OH = 0.0111 - X moles
Moles C6H5CHO = X moles
Moles H2O = X moles
Step 6: Calculate the total number of moles at equilibrium
Total number of moles = (0.0111 - X moles) + X moles + X moles
Total number of moles = 0.0111 + X moles
Step 7: Calculate the total pressure at the equilibrium
p*V = n*R*T
p = (n*R*T) / V
⇒with p = the total pressure at the equilibrium = TO BE DETERMINED
⇒with n = the total number of moles = 0.0111 + X moles
⇒with R = the gas constant = 0.08206 L*atm / mol * K
⇒with T = the temperature = 523 K
⇒with V = the volume of the vessel = 2.00 L
p = (0.0111 - X moles ) * 0.08206*523 / 2.00
p = 21.46(0.0111 - X moles)
Step 8: Define the equilibrium constant K
K = 0.558 = (pC6H5CHO)*(pH2) / (pC6H5CH2OH)
0.558 = (X / (0.0111 + X)*P)² / ((0.0111-X)/(0.0111+X)*P)
0.558 = (X²(21.46 * (0.0111+X))) / ((0.0111 + X) (0.0111-X))
X = 0.00839
Step 9: Calculate the equilibrium partial pressure
pC6H5CHO = X / (0.0111 + X) * (21.46 * (0.0111 +X))
pC6H5CHO = 0.180 atm
Step 10: What fraction of benzyl alcohol is dissociated into products at equilibrium?
Fraction dissociated = Δn / n°
Fraction dissociated = X / 0.0111
Fraction dissociated = 0.00839 / 0.0111
Fraction dissociated = 0.756
What is the correct formula for the compound formed by CA2+ and NO2-
Answers
Answer:
Ca(NO2)2
Explanation:
Analyze and solve this partially completed galvanic cell puzzle. There are 4 electrodes each identified by a letter of the alphabet, A through D. The values in the partially completed grid are measured cell potentials for a cell consisting of electrode #1 and electrode #2. You may assume that each galvanic cell was properly constructed with the appropriate metals and solutions and that all the measured values in the grid are accurate.
electrode #1 ?
C
B
D
A
electrode #2?
Ecell(volts)
Ecell(volts)
Ecell(volts)
Ecell(volts)
C
0
0.91
0.62
0.26
B
0.91
0
1.53
D
0.62
1.53
0
0.36
0 volts
0.10 volts
0.26 volts
0.36 volts
0.55 volts
0.62 volts
0.65 volts
0.88 volts
0.98 volts
1.17 volts
1.27 volts
1.79 volts
1.89 volts
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is [tex]E_{cell}__{AC}} = 0.94[/tex]
Explanation:
From the question we are told that
the cell voltage for AD is [tex]E_{cell}__{AD}} = 1.56V[/tex]
From the data give we can see that
[tex]E_{cell}__{AD}} - E_{cell}__{BD}} = E_{cell}__{AB}}[/tex]
i.e [tex]1.56 - 1.53 = 0.03[/tex]
In the same way we can say that
[tex]E_{cell}__{AD}}-E_{cell}__{CD}} = E_{cell}__{AC}}[/tex]
=> [tex]E_{cell}__{AC}}=1.56- 0.62[/tex]
[tex]E_{cell}__{AC}} = 0.94[/tex]
Galvanic cells are the voltaic cells that generate an electric current from redox reactions. The cell potential of cells A and C will be 0.94 volts.
What are the cell potentials?The cell potential is the estimation of the gained or lost electrons by the species on the electrode of the electrochemical cell.
Given, the cell voltage for cells A and D is 1.56 V.
Cell potential between A and B cells is calculated as:
[tex]\begin{aligned} \rm E_{cell}_{(AB)} &= \rm E_{cell}_{(AD)} - \rm E_{cell}_{(BD)}\\\\&= 1.56 - 1.53\\\\&= 0.03 \end{aligned}[/tex]
Similarly, cell potential between A and C cells is calculated as:
[tex]\begin{aligned} \rm E_{cell}_{(AC)} &= \rm E_{cell}_{(AD)} - \rm E_{cell}_{(CD)}\\\\&= 1.56 - 0.62\\\\&= 0.94 \end{aligned}[/tex]
Therefore, the cell potential between A and C cell is 0.94 volts.
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An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin. ΔS = nothing J/K Request Answer Part E An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin. ΔS = nothing J/K Request Answer Part F An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin. ΔS = nothing J/K Request Answer Part G Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.
Answers
Question:
Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.
Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.
Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.
Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.
Answer:
D) 85 J/K
E) - 50 J/K
F) 62.5 J/K
G) 12.5 J/K
Explanation:
Let's make use of the entropy equation: ΔS = [tex] \frac{Q}{T} [/tex]
Part D)
Given:
T = 20°C = 20 +273 = 293K
Q = 25.0 kJ
Entropy change will be:
ΔS = [tex] \frac{25*1000}{293} [/tex]
= 85 J/K
Part E)
Given:
T = 500K
Q = -25.0 kJ
Entropy change will be:
ΔS = [tex] \frac{-25*1000}{500} [/tex]
= - 50 J/K
Part F)
Given:
T = 400K
Q = 25.0 kJ
Entropy change will be:
ΔS = [tex] \frac{25*1000}{400} [/tex]
= 62.5 J/K
Part G:
Given:
T1 = 400K
T2 = 500K
Q = 25.0 kJ
The net entropy change will be:
ΔS = [tex] (\frac{25*1000}{400}) + (\frac{-25*1000}{500}[/tex]
= 12.5 J/K
Answer:
A) The change in entropy [tex]\delta S = 0.085J/K[/tex]
B) The change in entropy [tex]\delta S = -50J/K[/tex]
C) The change in entropy [tex]\delta S = 62.5J/K[/tex]
D) The net change in entropy [tex]\delta S = 12.5J/K[/tex]
Explanation:
A)
[tex]T = 20^oC + 273k\\\\T = 293k[/tex]
expression for change in entropy,
[tex]\delta S = \frac{\delta Q}{T}\\\\\delta S = \frac{25}{293}\\\\\delta S = 0.085J/K[/tex]
B) [tex]\delta S = \frac{\delta Q}{T}\\\\\delta S = \frac{-25*10^3}{500}\\\\\delta S = -50J/K[/tex]
The negative sign indicates the heat lost into the surrounding
C)
[tex]\delta S = \frac{\delta Q}{T}\\\\\delta S = \frac{25*10^3}{400}\\\\\delta S = 62.5J/K[/tex]
The entropy remains constant in the adiabatic process because no heat is given to the system in this process.
D)
[tex]\delta S_1 = \frac{\delta Q_1}{T_1}\\\\\delta S_1 = \frac{25*10^3}{400}\\\\\delta S_1 = 62.5J/K[/tex]
similarly,
[tex]\delta S_2 = \frac{\delta Q_2}{T_2}\\\\\delta S_2 = \frac{25*10^3}{500}\\\\\delta S_2 = 50J/K[/tex]
Therefore the net change in entropy is,
[tex]\delta S = \delta S_1 - \delta S_2\\\\\delta S = 62.5 - 50\\\\\delta S = 12.5J/K[/tex]
Entropy is not a conserved quantity because it can be created but cannot be destroyed.
The net change in entropy is calculated by difference of the change in entropy at two different temperatures.
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If 1 mol of gas is placed into a balloon under standard temperature and pressure (273 K and 1 atm), what volume would the balloon be?
Answers
Answer:
[tex]V=22.4L[/tex]
Explanation:
Hello,
In this case, considering the ideal gas equation:
[tex]PV=nRT[/tex]
It is possible to compute the volume the gas would have for the given STP conditions as:
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{1mol*0.082\frac{atm*L}{mol*K}*273K}{1atm}\\\\V=22.4L[/tex]
Which correspond to the standard volume as well.
Best regards.
Answer:
The volume of the balloon would be 22.386 L
Explanation:
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P * V = n * R * T
In this case:
P= 1 atmV= ?n= 1 molR= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]T= 273 KReplacing:
1 atm* V= 1 mol* 0.082 [tex]\frac{atm*L}{mol*K}[/tex]*273 K
Solving:
[tex]V=\frac{1 mol*0.082\frac{atm*L}{mol*K} *273 K}{1 atm}[/tex]
V=22.386 L
The volume of the balloon would be 22.386 L
Consider the dissociation of strong versus weak acids: HCl(aq)+H2O(l)→H3O+(aq)+Cl−(aq)HF(aq)+H2O(l)⇌H3O+(aq)+F−(aq) The first reaction is not reversible, but the second one is. So, only the conjugate of the weak acid, F−, can react with H3O+. Now consider the solubility of insoluble salts: AgCl(s)⇌Ag+(aq)+Cl−(aq)AgF(s)⇌Ag+(aq)+F−(aq) The addition of acid has no effect on silver chloride. But for the second reaction, H3O+ will react with F−, decreasing its concentration and driving the equilibrium to the right. Thus, salts that contain the conjugate of a weak acid become more soluble as the acidity of the solution increases.
Answers
Answer:
The solubility of those salts increases which contains conjugate of weak acid. Conjugate of weak acid refers to strong base such as sodium hydroxide and potassium hydroxide etc.
Explanation:
The solubility of salts in strong acidic solution increases due to the presence of conjugate of weak acid which is actually a strong base. So if the salts contain strong base, it readily react with strong acid that is present in the solution.