Explanation:
The chemical reaction equation will be as follows.
[tex]H_{3}PO_{4} + 3NaOH \rightarrow Na_{3}PO_{4} + 3H_{2}O[/tex]
In this reaction, 1 mole of [tex]H_{3}PO_{4}[/tex] reacts with 3 mole NaOH. So, the number of moles of [tex]H_{3}PO_{4}[/tex] present in 150 ml of 0.1 M solution is calculated as follows.
No. of moles = [tex]\frac{150}{1000 \times 0.1}[/tex]
= 0.015 mol
As it reacts with 3 moles of NaOH. Hence, no.. of moles of NaOH are:
[tex]3 \times 0.015 mol[/tex]
= 0.045 mol
So, moles of NaOH in 400 of 0.2 M NaH is as follows.
No. of moles = [tex]\frac{400}{1000 \times 2}[/tex]
= 0.080 mol
Hence, no. of moles remained after the reaction are as follows.
(0.080 - 0.045) mol
= 0.035 mol NaOH in 550 ml (400 ml + 150 ml)
As molarity is the no. of moles present in liter of solution. Hence, molarity of NaOH is as follows.
Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
= [tex]\frac{0.035}{550}[/tex]
= 0.0636 M
As, [tex][OH^{-}][/tex] = 0.0636 M. Hence, pOH will be 1.20.
As, pH + pOH = 14
pH = 14 - pOH
= 14 - 1.20
= 12.80
Also, [tex][H^{+}] = 10^{-pH}[/tex]
So, [tex][H^{+}] = 10^{-12.80}[/tex]
= [tex]1.58 \times 10^{-13}[/tex] M
Thus, we can conclude that pH of the given solution is 12.80 and its hydrogen ion concentration is [tex]1.58 \times 10^{-13}[/tex] M.
Answer:
pH = 12.80
[H+] = 1.58 * 10^-13 M
Explanation:
Step 1: Data given
Volume of 0.2M NaOH = 400 mL
Volume of 0.1M H3PO4 = 150 mL
Step 2: The balanced equation
H3PO4 + 3NaOH → Na3PO4 + 3H2O
For 1 mol H3PO4 we need 3 mol of NaOH to produce 1 mol Na3PO4 and 3 mol H2O
Step 3: Calculate moles H3PO4
Moles H3PO4 = molarity * volume
Moles H3PO4 = 0.1 M * 0.150 L
Moles H3PO4 = 0.015 moles
Step 4: Calculate moles NaOH
Moles NaOH = 0.2M * 0.400 L
Moles NaOH = 0.08 moles
For 1 mol H3PO4 we need 3 mol of NaOH to produce 1 mol Na3PO4 and 3 mol H2O
0.015 mol H3PO4 will react with 0.045 mol NaOH
Step 5: Calculate moles remaining
H3PO4 will be completely consumed
There will remain 0.08 - 0.045 = 0.035 moles of NaOH
Step 6: Calculate total volume
Total volume = 400 mL + 150 mL = 550 mL = 0.550 L
Step 7: Calculate molarity of the solution
Molarity = moles / volume
Molarity = 0.035 moles / 0.550 L
Molarity = 0.0636 M NaOH
Step 8: Calculate pOH
[OH-] = 0.0636M
pOH = -log [OH-]
pOH = -log(0.0636)
pOH= 1.20
Step 9: Calculate pH
pH = 14.00- pOH
pH = 14.00 - 1.20
pH = 12.80
[H+] = 10^-12.80
[H+] = 1.58 * 10^-13 M
A three level system has energies ε0=0.00J, ε1=5.00x10-21J, and ε2=10.0x10-21J. a. Using the Boltzmann Distribution Law, calcuate the molar internal energy U at T=10.0K. Hint: This means you have to calculate three probabilites at T=10K and use them to calculate the average energy per particle and then calculate the energy of a mole of particles.b. Again using the Bolztmann Distribution Law and the same approach as in part A, calculate the molar internal enerrgy U at T=1000Kc. Using you results from parts A and B, calculate the change in the molar internal energy ΔU when the temperature changes from T=10K to T=1000K.d. Using the Bolzmann Distribution Law probabilities calculated in part A, calculate the molar entropy at T=10K.e. Calculate the change in molar entropy when the temperature of this 3 level system changes from T=10K to T=1000K.
To find the molar internal energy using the Boltzmann Distribution Law, calculate the energy level probabilities at the given temperatures, compute the average energy, then multiply it by Avogadro's number. The change in internal energy and molar entropy is derived from the differences in these quantities between 10K and 1000K.
Explanation:To solve for the molar internal energy U at different temperatures using the Boltzmann Distribution Law, we first need to calculate the probabilities of each energy level at the given temperature T. For a three-level system, these probabilities depend on the energy levels and the temperature through the relation Pi = e(-εi/kBT), where εi is the energy of level i, kB is the Boltzmann constant, and T is the temperature.
For T=10.0K:
Calculate the probability P0, P1, and P2 for each energy level.Find the average energy <E> per particle using the probabilities.Multiply the average energy by Avogadro's number to find the molar internal energy U.Repeat the steps for T=1000K to find U at the higher temperature.
The change in molar internal energy ΔU when the temperature changes from T=10K to T=1000K can be calculated by taking the difference of U values obtained at both temperatures.
To calculate molar entropy S at T=10K, use the probabilities you calculated and the Boltzmann formula S = kB Σ Pi ln(Pi) and sum over all states i.
The change in molar entropy when the temperature changes from T=10K to T=1000K can be calculated by finding the difference in entropy values at both temperatures.
If a snail crawls at a rate of 0.10 cm/s, what is the speed in miles per hour? (1 mi = 5280 ft)
Answer:
0.0022369 mph
Explanation:
a snail crawls at a rate of 0.10 cm/s
1mile = 5280 ft
Scientifically 1 cm/s = 0.022369 mph
therefore 0.10cm/s = 0.10cm/s x 0.022369 mph = 0.0022369 mph
For the decomposition of A to B and C, A(s)⇌B(g)+C(g) how will the reaction respond to each of the following changes at equilibrium?
a. double the concentrations of both products and then double the container volume
b. double the container volume
c. add more A
d. double the concentration of B and halve the concentration of C
e. double the concentrations of both products
f. double the concentrations of both products and then quadruple the container volume
Answer:
a. No change.
b. The equilibrium will shift to the right.
c. No change
d. No change
e. The equilibrium will shift to the left
f. The equilibrium will shift to the right
Explanation:
We are going to solve this question by making use of Le Chatelier´s principle which states that any change in a system at equilibrium will react in such a way as to attain qeuilibrium again by changing the equilibrium concentrations attaining Keq again.
The equilibrium constant for A(s)⇌B(g)+C(g)
Keq = Kp = pB x pC
where K is the equilibrium constant ( Kp in this case ) and pB and pC are the partial pressures of the gases. ( Note A is not in the expression since it is a solid )
We also use Q which has the same form as Kp but denotes the system is not at equilibrium:
Q = p´B x p´C where pB´ and pC´ are the pressures not at equilibrium.
a. double the concentrations of Q which has the same form as Kp but : products and then double the container volume
Effectively we have not change the equilibrium pressures since we know pressure is inversely proportional to volume.
Initially the system will decrease the partial pressures of B and C by a half:
Q = pB´x pC´ ( where pB´and pC´are the changed pressures )
Q = (2 pB ) x (2 pC) = 4 (pB x PC) = 4 Kp ⇒ Kp = Q/4
But then when we double the volume ,the sistem will react to double the pressures of A and B. Therefore there is no change.
b. double the container volume
From part a we know the system will double the pressures of B and C by shifting to the right ( product ) side since the change reduced the pressures by a half :
Q = pB´x pC´ = ( 1/2 pB ) x ( 1/2 pC ) = 1/4 pB x pC = 1/4 Kp
c. add more A
There is no change in the partial pressures of B and C since the solid A does not influence the value of kp
d. doubling the concentration of B and halve the concentration of C
Doubling the concentrantion doubles the pressure which we can deduce from pV = n RT = c RT ( c= n/V ), and likewise halving the concentration halves the pressure. Thus, since we are doubling the concentration of B and halving that of C, there is no net change in the new equilibrium:
Q = pB´x pC´ = ( 2 pB ) x ( 1/2 pC ) = K
e. double the concentrations of both products
We learned that doubling the concentration doubles the pressure so:
Q = pB´x pC´ = ( 2 pB ) x ( 2 pC ) = 4 Kp
Therefore, the system wil reduce by a half the pressures of B and C by producing more solid A to reach equilibrium again shifting it to the left.
f. double the concentrations of both products and then quadruple the container volume
We saw from part e that doubling the concentration doubles the pressures, but here afterward we are going to quadruple the container volume thus reducing the pressure by a fourth:
Q = pB´x pC´ = ( 2 pB/ 4 ) x (2 pC / 4) = 4/16 Kp = 1/4 Kp
So the system will increase the partial pressures of B and C by a factor of four, that is it will double the partial pressures of B and C shifting the equilibrium to the right.
If you do not see it think that double the concentration and then quadrupling the volume is the same net effect as halving the volume.
Final answer:
The decomposition reaction of A into B and C responds to changes at equilibrium by shifting in a direction that opposes the imposed change, as explained by Le Chatelier's principle. The effects of changes in concentration, container volume, and addition of reactants are used to predict shifts in equilibrium.
Explanation:
Response to Changes in Chemical Equilibrium
For the decomposition of a solid A into gases B and C, represented by the reaction A(s) ⇌ B(g) + C(g), here is how the reaction will respond to changes at equilibrium:
a. Doubling the concentrations of both products B and C and then doubling the container volume will initially cause the reaction to shift to the left (toward A), as the increase in volume will decrease the pressure and favor the formation of fewer gas molecules. However, because the concentrations of the gases are also doubled, the net effect may be smaller, and the system will work to re-establish equilibrium according to Le Chatelier's principle.b. Doubling the container volume alone will shift the equilibrium to the left because the system will try to increase the pressure by forming more gas molecules (favoring the formation of B and C).c. Adding more of the solid reactant A will shift the equilibrium to the right, increasing the production of B and C.d. Doubling the concentration of B and halving the concentration of C will lead to a shift in the equilibrium toward the left, as the system tries to counteract the changes and restore equilibrium.e. Doubling the concentrations of both products, B and C, without changing the volume will shift the equilibrium to the left, as the system tries to reduce the concentrations of the products by forming more A.f. Doubling the concentrations of B and C and then quadrupling the container volume will have a more pronounced effect in shifting the equilibrium to the left than simply doubling the volume, as the change in pressure will be more significant.Each of these actions will induce a response from the reaction to maintain the established equilibrium according to Le Chatelier's principle. The reaction will typically shift in the direction that opposes the change imposed on the system.
In the molecular orbital model of benzene, the six p atomic orbitals combine to form how many molecular orbitals?
Answer:
6 molecular orbital.
Explanation:
According to the molecular orbital theory, number of molecular orbital are obtained is equal to the number of atomic orbital combine.
When six p orbitals of benzene combine, the will form six molecular orbital. These six molecular orbital have 3 molecular orbital with lower energy and three molecular orbital with higher energy.
The orbital with lower energies are called bonding molecular orbital and orbital with higher energies are called antibonding molecular orbital.
In the molecular orbital model of benzene, six p atomic orbitals combine to form six molecular orbitals. These result from the overlapping p orbitals of the six carbon atoms in the benzene ring, forming a 'pi electron cloud'.
Explanation:In the molecular orbital model of benzene, the six p atomic orbitals combine to form six molecular orbitals. This happens because benzene has a cyclic structure where the overlapping p orbitals of six carbon atoms form what is known as a 'pi electron cloud'. This pi electron cloud is a result of side-on overlap of p orbitals encompassing all the six carbon atoms. So, essentially, the six overlapping p orbitals produce six molecular orbitals. These six orbitals then distribute themselves into differing energy levels. Three of these orbitals tend to be lower energy (bonding), and the other three are higher energy (anti-bonding).
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A 10.00 mL sample of 8.50 M HNO3 solution is diluted to a new volume of 65.0 mL. What is the concentration of the dilute solution
Answer: 1.31M
Explanation:
V1 = 10mL
C1 = 8.5M
V2 = 65mL
C2 =?
C1V1 = C2V2
10 x 8.5 = C2 x 65
C2 = (10 x 8.5 ) /65
C2 = 1.31M
The concentration of the dilute solution of nitric acid is 1.31 M.
We have 10.00 mL (V₁) of 8.50 M HNO₃ (C₁) and we add water to obtain a dilute solution with a volume (V₂) of 65.0 mL. We can calculate the concentration of the dilute solution (C₂) using the dilution rule.
[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{8.50 M \times 10.00 mL}{65.0 mL} = 1.31 M[/tex]
The concentration of the dilute solution of nitric acid is 1.31 M.
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Three of the following statements are wrong. One is correct. Which of the following statements is correct?
Enzymes require optimum pH for maximal activity
Enzymes are denatured at high temperature but in certain exceptional organisms they are effective even at temperatures 80∘−90∘C
Enzymes are highly specific
Most enzymes are proteins but some are lipids
Answer:
The incorrect statement is "Most enzymes are proteins but some are lipids" the others statements are correct regarding to the Enzymes.
Explanation:
Enzymes are complex macromolecules of globular proteins. However, many enzymes contain certain non-protein substances associated with them for their function, which are known as cofactors and can be organic or inorganic compounds. In addition, the enzymes are thermolabile, presenting their best performance at the ideal temperature. Enzyme activity decreases with increasing and increases with increasing temperature and stops at 0 degrees and above 80 degrees. However, the enzymes of the bacteria that inhabit the hot springs have an ideal temperature of 70 degrees or more. Enzymes also show maximum activity at optimal pH. Varying its activity with increasing or decreasing pH. Enzymes are specific in their action. An enzyme can catalyze only a specific type of reaction or even act on a specific substrate. For example, the enzyme lactase catalyzes the hydrolysis of lactose and no other disaccharides.
What kind of intermolecular forces act between a hydrogen iodide molecule and a dichloroethylene
Final answer:
The intermolecular forces between a hydrogen iodide molecule and a dichloroethylene molecule involve both dispersion forces and dipole-dipole forces.
Explanation:
The intermolecular forces between a hydrogen iodide molecule (HI) and a dichloroethylene molecule (C2H2Cl2) involve both dispersion forces and dipole-dipole forces.
Dispersion Forces: Iodine (I) in HI is a larger atom compared to chlorine (Cl) in dichloroethylene. As a result, HI has stronger dispersion forces due to the larger number of electrons, making the attraction between HI molecules stronger than between dichloroethylene molecules.Dipole-Dipole Forces: Dichloroethylene is a polar molecule with a partial positive charge on the hydrogen atom and partial negative charges on the chlorine atoms. HI is also a polar molecule with a partial positive charge on the hydrogen atom and a partial negative charge on the iodine atom. The positive end of the dipole in HI is attracted to the negative end of the dipole in dichloroethylene, creating dipole-dipole interactions between the two molecules.Overall, the intermolecular forces between a hydrogen iodide molecule and a dichloroethylene molecule include both dispersion forces and dipole-dipole forces.
In a first-order decomposition reaction. 36.8% of a compound decomposes in 7.6 min. How long (in min) does it take for 88.8% of the compound to decompose?
Answer: 36.5 minutes
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant = 100
a - x = amount left after decay process
a) for completion of 36.8 % of reaction
[tex]7.6=\frac{2.303}{k}\log\frac{100}{100-36.8}[/tex]
[tex]k=\frac{2.303}{7.6}\times 0.19[/tex]
[tex]k=0.060min^{-1}[/tex]
b) for completion of 88.8 % of reaction
[tex]t=\frac{2.303}{k}\log\frac{100}{100-88.8}[/tex]
[tex]t=\frac{2.303}{0.060}\log\frac{100}{11.2}[/tex]
[tex]t=\frac{2.303}{0.060}\times 0.95[/tex]
[tex]t=36.5min[/tex]
It will take 36.5 minutes for 88.8% of the compound to decompose.
Final answer:
To accurately determine the time required for 88.8% of a compound to decompose in a first-order reaction, knowing the rate constant is essential. While the question provides a specific data point for decomposition over time, it lacks the rate constant needed for a direct calculation. Therefore, accurately answering this specific question based on the provided context is challenging without additional information.
Explanation:
The question asks for the time required for 88.8% of a compound to decompose in a first-order decomposition reaction given that 36.8% decomposes in 7.6 minutes. In first-order reactions, the time it takes for a certain percentage of the reactant to decompose does not depend on the initial concentration but on the rate constant (k) of the reaction. The integrated rate law for first-order reactions is given by the formula ln([A]0/[A]) = kt, where [A]0 is the initial concentration, [A] is the concentration at time t, and k is the rate constant.
To determine the time required for 88.8% decomposition, we would need to know the rate constant of the reaction. However, with the information provided, we can infer the rate constant by using the time and percent decomposition already given (36.8% in 7.6 minutes), but to find the exact time for 88.8% decomposition without additional specific information (e.g., actual rate constant value) is not directly feasible based on the provided context alone. Knowing the rate constant, we could then apply the formula to find the time for 88.8% of the compound to decompose.
Without the rate constant, an alternative but less precise method involves understanding that the time to reach a certain level of decomposition is related to the half-life of the reaction. Since the time for half of a reactant to decompose (its half-life) in a first-order reaction is constant, we can indirectly estimate times for specific percentages of decomposition if we know the half-life, which again depends on knowing the rate constant.
Draw the structure of the major organic product isolated from the reaction of 1-hexyne with hydrogen chloride (2 mol). Click the "draw structure" button to launch the drawing utility.
Answer:
The structure of the major organic product isolated from the reaction of 1-hexyne with hydrogen chloride (2 mol) is attached below.
Explanation:
Hydracids are added to triple bonds by a mechanism similar to that of the addition to double bonds. The regioselectivity of the addition of H-X to the triple bond follows the rule of Markovnikov, where the Z conformation predominates in the addition of halide to the alkyne, because in the formation of the carbocation it prefers to place the positive charge on the more substituted carbon where the nucleophilic attack of the halide ion will occur.
With respect to the halogenation of the alkene, the same procedure occurs at the time of the formation of the carbocation, joining the nucleophilic ion to the most substituted carbon.
(b) Data has been collected to show that at a given wavelength in a 1 cm pathlength cell, Beer's Law for the absorbance of Co2+ is linear. If a 0.135 M solution of Co2+ has an absorbance of 0.350, what is the concentration of a solution with an absorbance of 0.460?
Answer : The concentration of a solution with an absorbance of 0.460 is, 0.177 M
Explanation :
Using Beer-Lambert's law :
[tex]A=\epsilon \times C\times l[/tex]
where,
A = absorbance of solution
C = concentration of solution
l = path length
[tex]\epsilon[/tex] = molar absorptivity coefficient
From this we conclude that absorbance of solution is directly proportional to the concentration of solution at constant path length.
Thus, the relation between absorbance and concentration of solution will be:
[tex]\frac{A_1}{A_2}=\frac{C_1}{C_2}[/tex]
Given:
[tex]A_1[/tex] = 0.350
[tex]A_2[/tex] = 0.460
[tex]C_1[/tex] = 0.135 M
[tex]C_2[/tex] = ?
Now put all the given values in the above formula, we get:
[tex]\frac{0.350}{0.460}=\frac{0.135}{C_2}[/tex]
[tex]C_1=0.177M[/tex]
Therefore, the concentration of a solution with an absorbance of 0.460 is, 0.177 M
Beer's Law states that there is a linear relationship between the concentration of a substance in a solution and its absorbance. We can use the equation c = A/(εb) to solve for the concentration of a solution with a given absorbance.
Explanation:Beer's Law states that there is a linear relationship between the concentration of a substance in a solution and its absorbance. The equation for Beer's Law is A = εbc, where A is the absorbance, ε is the molar absorptivity, b is the path length of the cell, and c is the concentration.
In this case, we have the absorbance (0.350) and concentration (0.135 M) for a solution of Co2+. We can rearrange the equation to solve for the concentration: c = A/(εb). Plug in the given values to find the molar absorptivity and path length, and then substitute those values into the equation to calculate the concentration of a solution with an absorbance of 0.460.
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Impure crystalline (solid) substances can be purified by recrystallization from a suitable solvent. Arrange the steps of the recrystallization procedure, from start to finish.
Weigh the crystals into a tared flask.
Dissolve the crude substance in a minimum amount of hot solvent.
Filter the hot solution to remove the solid impurities (if present)
Allow the solution to cool slowly so crystals form.
Filter the crystals.
Dry the crystals
Answer:
1) Dissolve the crude substance in a minimum amount of hot solvent.
2) Filter the hot solution to remove the solid impurities (if present)
3) Allow the solution to cool slowly so crystals form.
4) Filter the crystals.
5) Dry the crystals
6) Weigh the crystals into a tared flask.
Explanation:
Recrystallization is a method used in chemistry to obtain crystals having a high degree of purity for the purpose of analytical or synthetic laboratory work.
Draw the structure of the starting material needed to make 2-methylhept-3-yne using sodium amide in liquid ammonia, followed by 1-bromopropane. The starting hydrocarbon must have no more than five carbons. 1) NaNH2, NH3(l) 2) CH3CH2CH2Br + NaBr
The starting material to synthesize 2-methylhept-3-yne under the given conditions is 4-methyl-1-pentyne, which undergoes alkylation with 1-bromopropane to form the desired product.
The question involves a synthetic route to create 2-methylhept-3-yne using specific reagents and limitations on the carbon count of the starting material. The sodium amide in liquid ammonia indicates a need for a strong base, suggesting an elimination reaction. With the addition of 1-bromopropane, we're looking at an alkylation step which introduces the propyl group.
Given the restrictions, the most fitting starting material would be 4-methyl-1-pentyne. This molecule can undergo alkylation at the terminal methyl group with 1-bromopropane to extend the carbon chain by three carbons, leading to the desired product, 2-methylhept-3-yne. The overall strategy leverages a reaction sequence involving an initial alkyne substrate that can be elongated with an alkyl halide using the action of a strong base.
The starting material for the formation of 2-methylhept-3-yne is 3-methylbutyne. The structure of 3-methylbutyne is attached below
The starting material for the formation of 2-methylhept-3-yne is 3-methylbutyne
Stepwise formation of 2-methylhept-3-yne
Step 1:
The first step is the formation of the Acetylide Ion
3-methylbutyne reacts with sodium amide (NaNH₂) in liquid ammonia to form the acetylide ion:
(CH₃)₂CHC≡CH + NaNH₂ → (CH₃)₂CC≡C⁻Na⁺ + NH₃
Step 2:
The second step is the alkylation of the Acetylide Ion
The acetylide ion reacts with 1-bromopropane (CH₃CH₂CH₂Br) through an SN2 mechanism to form 2-methylhept-3-yne:
(CH₃)₂CC≡C⁻Na⁺ + CH₃CH₂CH₂Br → (CH₃)₂CC≡CCH₂CH₂CH₃ + NaBr
Hence, the final product is 2-methylhept-3-yne:
(CH₃)₂CC≡CCH₂CH₂CH₃
Consider a solution that contains 0.274 M potassium chloride and 0.155 M magnesium chloride.
What is the concentration of chloride ions in this solution?
Answer:
Concentration of chloride ions = 0.584M
Explanation:
The step by step calculations is shown as attached below.
In what way are planned economies and command economies similar?
O
A. They both are characteristic of the United States.
O
B. They both rely on private ownership.
C. They both involve strong governmental control.
O
D. They both require open exchange.
SUBMIT
Answer: D
Explanation:
Answer:
I believe the answer is C I took the test a little bit ago
Explanation:
Reaction 1: 2NO(g) + Br2(g) ⇄ 2NOBr(g) ΔH°rxn = −16 kJ Kp = 27 at 25°CReaction 2: CO2(g) ⇄ C(s) + O2(g) ΔH°rxn = +100 kJ Kp = 2.6x10−9 at 25°C8. For which reaction(s) above would an increase in reaction vessel volume at constant temperature result in an increase in product yield? A) Reaction 1 only B) Reaction 2 onlyC) Both reactionsD) Neither reactionFor which reaction(s) above would an increase in temperature at constant volume result in an increase in product yield? A) Reaction 1 only B) Reaction 2 onlyC) Both reactionsD) Neither reaction
Answer:
First question: D)
Second question: B)
Explanation:
A reversible reaction intends to achieve the equilibrium, a state in the velocity of product formation is equal to the velocity of reactants formation. This equilibrium can be disturbed by some alterations in the reaction, and, by Le Chatelier's principle, the reaction will shift in to reestablish the equilibrium.
The equilibrium may be shift for three principal factors: concentration, temperature, and pressure. If a concentration of some of the substance increases, the reaction will shift to consume it; if it decreases, the reaction is shifted to form more this substance.
When the temperature of the system increases, the reaction shifts for the consume of the heat, thus, the endothermic reaction (ΔH°rxn > 0) is favored; if it decreases, the exothermic reaction (ΔH°rxn < 0) is favored. If the direct reaction is endothermic, the inverse is exothermic, and vice versa.
When the pressure of the system increases, the volume decreases, so, the equilibrium shifts for the less gas volume, or the side with fewer moles of gas substance; if the pressure decreases the inverse occurs.
So, if the vessel volume increases, the pressure will decrease, and so, the formation of more gas substances is favored. In reaction 1, in the reactants, there are 3 moles ( 2 of NO + 1 of Br2), and in the products 2 moles, so the reactants are favored, and the product yield decreases. In reaction 2, there is 1 mol on both sides of the reaction, so the pressure doesn't affect the equilibrium.
The increase in temperature favors the endothermic reaction. The reaction 2 has it's direct reaction endothermic, so the product yield increases. In reaction 1, the reactants increases, because the inverse reaction is endothermic.
An increase in temperature at constant volume will result in an increase in product yield in Reaction 1 only.
Explanation:An increase in temperature at constant volume will result in an increase in product yield for Reaction 1 only. This is because the reaction in Reaction 1 is exothermic, meaning it releases heat. When the temperature increases, the reaction will shift in the direction that absorbs heat to counteract the increase in temperature. Since Reaction 1 releases heat, an increase in temperature will favor the formation of products and increase the product yield. For Reaction 2, an increase in temperature will not result in an increase in product yield because the reaction is endothermic, meaning it requires heat as a reactant. Increasing the temperature will shift the reaction in the direction that releases heat, reducing the product yield.
Free radical chlorination of ethane can produce higher halogenation products (dichlorinated, trichlorinated, etc...) in addition to chloroethane. How could the production of higher halogenated products be minimized? Use an excess of chlorine Use an excess of ethane Use equimolar chlorine and ethane It is not possible to minimize the production of higher halogenated products
Answer:Use an excess of ethane
Explanation:
The halogenation of alkanes is a substitution reaction. All the hydrogen atoms in the alkanes could be potentially substituted. How ever the reaction can be controlled by using an excess of either the alkane or the halogen. If the aim (as it is in this question) is to minimize the yield of halogenated alkanes, an excess of the alkane (in this case, ethane) is used.
The work function of palladium (Pd) is 5.22 eV. What is the minimum frequency of light to observe the photoelectric effect from a Pd surface? If light with a 200 nm wavelength is absorbed by the Pd surface, what is the velocity of the emitted electrons?
Answer:
1)4×10^13Hz
2) 9.95×10-9J
Explanation:
From the image attached, it is clear that the work function of the palladium metal must first be obtained in joules. Then, the frequency is obtained from E=hf.
The kinetic energy if the photoelectrons is obtained as the difference between the energy of the photon and the work function of the metal.
Type of bond Average Bond Enthalpy (kJ/mol) O-O 150O=O 500The conversion of ozone to diatomic oxygen is represented by the equation above. Based on the data in the table above, what is the approximate average bond enthalpy for the oxygen-to-oxygen bonds in ozone? a. 0 kJ/mol b. 150 kJ/mol c. 300 kJ/mol d. 500 kJ/mol
Missing information:
The reaction is 2O₃(g) --> 3O₂ (g) ΔH = -300 kJ/mol
Answer:
c
Explanation:
In the molecule of ozone, 3 oxygens are bonded, and, because each one needs to share two pairs of electrons, these 2 bonds are something between a simple and a double bond.
In the reaction of the transformation of the ozone to oxygen gas, these bonds are broken, and a double bond is formed between two oxygen atoms. The sum of the energy of the broken and the formation of the bond is the enthalpy variation of the reaction.
To break a bond, energy must be added to the system, so it's an endothermic reaction and energy is positive, so the formation is exothermic and the energy is negative. Because there're 2 ozone molecules, 4 bonds will be broken, and because there are 3 oxygen molecules, 3 bonds will be formed:
4*E - 3*500 = -300
4E = -300 + 1500
4E = 1200
E = 300 kJ/mol
So, each O3 bond has 300 kJ/mol as an average energy.
Each ozone molecule has 300 kJ/mol as an average bond energy.
The reaction is
[tex]\bold { 2O_3(g) \rightarrow 3O_2 (g)\ \ \ \ \ \ \ \ \ \ \ \Delta H = -300 kJ/mol}[/tex]
In the molecule of ozone, there 2 double bonds are present with continuous variation.
The enthalpy variation of the reaction is the sum of the energy of the broken and the formation of the bond.
Because of 2 ozone molecules, 4 bonds will be broken, and because there are 3 oxygen molecules, 3 bonds will be formed.
[tex]\bold {4\times E - 3 \times 500 = -300}\\\\\bold {4\times E = -300 + 1500}\\\\\bold {E = 300 kJ/mol}[/tex]
Therefore, Each ozone molecule has 300 kJ/mol as an average bond energy.
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In the following reaction, which component acts as an oxidizing agent? 10 I− (aq) + 2 MnO4− (aq) + 16 H+ (aq) → 5 I2 (s) + 2 Mn2+ (aq) + 8 H2O (l)
Answer:
The oxidizing agent is the MnO₄⁻
Explanation:
This is the redox reaction:
10 I⁻ (aq) + 2 MnO₄⁻ (aq) + 16 H⁺ (aq) → 5 I₂ (s) + 2 Mn²⁺ (aq) + 8 H2O (l)
Let's determine the oxidation and the reduction.
I⁻ acts with -1 in oxidation state and changes to 0, at I₂.
All elements in ground state has 0 as oxidation state.
As the oxidation state has increased, this is the oxidation, so the iodide is the reducing agent.
In the permanganate (MnO₄⁻), Mn acts with +7 in oxidation state and decreased to Mn²⁺. As the oxidation state is lower, we talk about the reduction. Therefore, the permanganate is the oxidizing agent because it oxidizes iodide to iodine
g What mass (in g) of a concentrated solution of nitric acid (69.0% HNO3 by mass) is needed to prepare 364.5 g of a 12.2% solution of HNO3 by mass?
Answer:
64.4 g of concentrated nitric acid solution are needed.
Explanation:
First we calculate the mass of HNO₃ contained in 364.5 g of a 12.2% solution:
364.5 gsolution * 12.2 gHNO₃/100gsolution = 44.47 g HNO₃Now we calculate how many grams of the concentrated solution would contain 44.47 grams of HNO₃:
44.47 g HNO₃ * 100 g solution / 69.0 = 64.45 g concentrated solutionSo 64.4 g of concentrated nitric acid solution are needed.
A driver with a nearly empty fuel tank may say she is "running on fumes."
If a 15.0 gallon automobile gas tank had only gasoline vapor remaining in it, what is the farthest the vehicle could travel if it gets 20.0 miles per gallon on liquid gasoline?
Assume the average molar mass of molecules in gasoline is 101 g/mol , the density of liquid gasoline is 0.75 g/mL, the pressure is 747 mmHg , and the temperature is 25 ∘C.
Answer:
The farthest the vehicle could travel (if it gets 20.0 miles per gallon on liquid gasoline) is 1.62 miles.
Explanation:
The automobile gas tank has a volume capacity of 15 gallons which can be converted to liters: 15 × 3.7854 = 56.781 liters
We can find the moles of gasoline by using the ideal gas equation: PV = nRT.
Make n (number of moles) the subject of the formula: n = PV/RT, where:
P = 747 mmHg
V = 56.781 liters
R (universal gas constant) = 0.0821 liter·atm/mol·K
T = 25 ∘C = (273 + 25) K = 298 K
1 atm (in the unit of R) = 760 mmHg
Therefore n = 747 × 56.781/(0.0821 × 760 × 298) = 2.281 mol.
Given that the molar mass of the gasoline = 101 g/mol,
the mass of gasoline = n × molar mass of gasoline = 2.281 mol × 101 g/mol = 230.38 g
the density of the liquid gasoline = 0.75 g/mL
In order to calculate the distance the vehicle can travel, we have to calculate volume of gasoline available = mass of the liquid gasoline ÷ density of liquid gasoline
= 230.38 g ÷ 0.75 g/mL = 307.17 mL = 0.3071 liters = 0.3071 ÷ 3.7854 = 0.0811 gallons
since the vehicle gets 20.0 miles per gallon on liquid gasoline, the distance traveled by the car = gallons available × miles per gallon = 0.0811 × 20 = 1.62 miles.
A substance that is required to run the vehicle and the machine is known as fuel. These substance are as follows:-
PetrolDiesalsKerosineAccording to the data given in the question.
The farthest the vehicle could travel is 1.62 miles.
The volume capacity of 15 gallons which can be converted to liters: [tex]15 * 3.7854 = 56.781 liters[/tex]
The formula used is as follows:-[tex]PV =nRT[/tex]
After putting the value:-
P = 747 mmHg
V = 56.781 liters
R (universal gas constant) = 0.0821 liter·atm/mol·K
[tex]T = 25C = (273 + 25) K = 298 K[/tex]
Therefore, the value of [tex]n = \frac{747 * 56.781}{(0.0821 * 760 * 298} = 2.281 mol.[/tex]
Given that the molar mass of the gasoline = 101 g/mol,
Mass of gasoline = n × molar mass of gasoline
[tex]= 2.281 mol * 101 g/mol = 230.38 g[/tex]
the density of the liquid gasoline = 0.75 g/mL
[tex]= \frac{230.38}{0.75}= 307.17 mL \\\\= 0.3071 liters\\\\= \frac{0.3071}{3.7854} = 0.0811 gallons[/tex]
Since the vehicle gets 20.0 miles per gallon on liquid gasoline, the distance traveled by car = gallons available × miles per gallon =
[tex]0.0811 * 20 = 1.62 miles.[/tex]
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A meter of polyaramide fiber has a diameter of 710.μm , a mass of 0.059g , and takes 0.13s to produce on an industrial spinneret. How would you calculate the mass of fiber that can be spun in 55.min ?Set the math up. But don't do any of it. Just leave your answer as a math expression.Also, be sure your answer includes all the correct unit symbols.
In order to calculate the mass of fiber that can be spun in 55 min we have to determine the rate of production.
Generally speaking, all the different types of rates have in common a variable divided over time. In this case the variable we're interested in is the mass of fiber, so we have to divide the mass of fiber over the time it takes to spin, and then if we want to find the mass of fiber for any given time, we just have to multiply the time and the rate together.
Keep in mind that the time that is given to us to find the rate is in seconds, so we have to convert that to minutes
The math expression would be the following:
[tex]mass\ of \ fiber (g)=time(min)*\frac{0,059 g}{0.13s}*\frac{60 s}{1 min} \\\\mass\ of \ fiber (g)=55 min*\frac{0,059 g}{0.13s}*\frac{60 s}{1 min}[/tex]
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The N-H bond in ammonia is polar because
A: nitrogen occupies more space than hydrogen.
B: hydrogen is much more electronegative than nitrogen.
C: it is a hydrogen bond.
D: it is an ionic bond.
E: nitrogen is much more electronegative than hydrogen.
Answer: option E. nitrogen is much more electronegative than hydrogen.
Explanation:
Answer: E: nitrogen is much more electronegative than hydrogen.
Explanation:
Each N-H bond is polar because N is more electronegative than H. NH3 is overall asymmetrical in its VSEPR shape, so the dipoles don't cancel out and it is therefore polar.
Predict the products for each of the following Diels–Alder reactions: (a) COOH HOOC + (b) HOOC CN + (c) O O + (d) S O O O + (e) CN NC + (f ) O O O O S MeO OMe + 16.43
Answer:
see explanation below
Explanation:
You are not putting all the structures correctly. Luckily I found the question on another place, so the complete structures you can see them in picture 1.
Now, according to the picture 1, we have all the six reactions. The general mechanism is the same for all and then, if you can have the possibility to rearrange the molecule, you can do that too.
Now, the general mechanism as I stated earlier is the same. The double bond from the diene (The one with one double bond at least) attacks the dienophyle (The first double bond), this bond do resonance to the conjugate bond, and the other double bond attacks the diene, and form a new product.
According to this, the product for each reaction, you can see it in picture 2 and 3:
In Diels-Alder reactions, a diene reacts with a dienophile to form a cyclic compound. Only reactions (b) and (e) will result in Diels-Alder adducts.
Explanation:In order to predict the products of the given Diels-Alder reactions, we need to identify the diene and dienophile components. The diene is usually a compound containing two double bonds, while the dienophile is a compound with a double bond. The reaction between the diene and dienophile will form a cyclic compound known as the Diels-Alder adduct. Let's examine each reaction:
(a) COOH + HOOC: Neither COOH nor HOOC contain diene or dienophile functionality, so no Diels-Alder reaction will occur.
(b) HOOC + CN: This reaction involves a diene (HOOC) and a dienophile (CN). The Diels-Alder adduct will be formed.
(c) O + O: Neither O nor O contain diene or dienophile functionality, so no Diels-Alder reaction will occur.
(d) S + OOO: Both S and OOO contain diene functionality, but no dienophile. Therefore, a Diels-Alder reaction will not occur.
(e) CN + NC: This reaction involves a diene (CN) and a dienophile (NC). The Diels-Alder adduct will be formed.
(f) OOO S MeO OMe: Neither OOO nor S MeO OMe contain diene or dienophile functionality, so no Diels-Alder reaction will occur.
The normal boiling point of liquid methyl acetate is 331 K. Assuming that its molar heat of vaporization is constant at 30.6 kJ/mol, the boiling point of CH3COOCH3 when the external pressure is 1.29 atm is
Answer:
T_2=338.9026K
Boiling point of CH3COOCH3 at external pressure is 338.9026K
Explanation:
We are going to use Clausius-Clapeyron Equation:
[tex]ln\frac{P_2}{P_1} =-\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]
Where:
P_2 is the external pressure
P_1 is the atmospheric Pressure=1 atm
ΔH is the heat of vaporization
T_2 boiling point of CH3COOCH3 at external pressure
T_1 normal boiling point of liquid methyl acetate
Now:
[tex]\frac{1}{T_2}=-ln\frac{P_2}{P_1}*\frac{R}{\Delta H}+\frac{1}{T_1} \\\frac{1}{T_2}=-ln\frac{1.29}{1}*\frac{8.314}{30.6*10^3}+\frac{1}{331} \\\frac{1}{T_2}=2.9507*10^-^3\\T_2=\frac{1}{2.9507*10^-^3} \\T_2=338.9026K[/tex]
Boiling point of CH3COOCH3 at external pressure is 338.9026K
To find the new boiling point of methyl acetate, use the Clausius-Clapeyron equation with the provided values for the initial boiling point, molar heat of vaporization, and new external pressure.
Explanation:This question requires us to use the form of the Clausius-Clapeyron equation:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Where T1 and P1 are the initial temperature and pressure (331K and 1 atm respectively, since normal boiling point is defined as the temperature at which the vapor pressure of a liquid equals 1 atm), and P2 is the new pressure (1.29 atm). ΔHvap is the molar heat of vaporization (30.6 kJ/mol, or 30.6 * 10^3 J/mol to convert kilojoules to joules), and R is the gas constant (8.314 J/mol*K). Solving the equation for T2 (the new boiling point temperature), we rearrange to get T2 = 1/(1/T1 + (R/ΔHvap)*ln(P2/P1)). Plugging in the values, you should find the answer.
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Covalent bonds in a molecule absorb radiation in the IR region and vibrate at characteristic frequencies.
A molecular vibration absorbs radiation of wavelength 188 μm. What frequency (in s−1) corresponds to that wavelength? Enter your answer in scientific notation.
Answer:
[tex]Frequency=1.6\times 10^{12}\ Hz[/tex]
Explanation:
The relation between frequency and wavelength is shown below as:
[tex]c=frequency\times Wavelength [/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
Given, Wavelength = 188 μm
Also, 1 μm = [tex]10^{-6}[/tex] nm
So,
Wavelength = [tex]188\times 10^{-6}[/tex] m
Thus, Frequency is:
[tex]Frequency=\frac{c}{Wavelength}[/tex]
[tex]Frequency=\frac{3\times 10^8}{188\times 10^{-6}}\ Hz[/tex]
[tex]Frequency=1.6\times 10^{12}\ Hz[/tex]
Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the percentage of unit cell volume that is occupied in (a) a face- centered cubic lattice, (b) a body-centered cubic lattice, and (c) a diamond lattice.
Answer:
The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05% The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03% The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%Explanation:
The percentage of unit cell volume = Volume of atoms/Volume of unit cell
Volume of sphere = [tex]\frac{4 }{3} \pi r^2[/tex]
a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:
let the side of each cube = a
Volume of unit cell = Volume of cube = a³
Radius of atoms = [tex]\frac{a\sqrt{2} }{4}[/tex]
Volume of each atom = [tex]\frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3[/tex] = [tex]\frac{\pi *a^3\sqrt{2}}{24}[/tex]
Number of atoms/unit cell = 4
Total volume of the atoms = [tex]4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}[/tex]
The percentage of unit cell volume = [tex]\frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6}[/tex] = 0.7405
= 0.7405 X 100% = 74.05%
b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice
Radius of atoms = [tex]\frac{a\sqrt{3} }{4}[/tex]
Volume of each atom =[tex]\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3[/tex] =[tex]\frac{\pi *a^3\sqrt{3}}{16}[/tex]
Number of atoms/unit cell = 2
Total volume of the atoms = [tex]2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}[/tex]
The percentage of unit cell volume = [tex]\frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8}[/tex] = 0.6803
= 0.6803 X 100% = 68.03%
c) Percentage of unit cell volume occupied by atoms in a diamond lattice
Radius of atoms = [tex]\frac{a\sqrt{3} }{8}[/tex]
Volume of each atom = [tex]\frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3[/tex] = [tex]\frac{\pi *a^3\sqrt{3}}{128}[/tex]
Number of atoms/unit cell = 8
Total volume of the atoms = [tex]8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}[/tex]
The percentage of unit cell volume = [tex]\frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16}[/tex] = 0.3401
= 0.3401 X 100% = 34.01%
The percentage of unit cell volume occupied is highest in the face-centered cubic lattice, making it the most efficient packing with the least free space. In contrast, the body-centered cubic and diamond lattices have more free space. Cadmium sulfide crystallizes with cadmium ions occupying half of the tetrahedral holes in a closest-packed sulfide ion array, resulting in a 1:1 ratio and a CdS formula.
Explanation:Calculating Unit Cell Volume Occupancy
To calculate the percentage of the unit cell volume that is occupied in different lattice structures, we consider the arrangement and number of atoms within the unit cell and their geometric relationship. In a face-centered cubic (fcc) lattice, each corner atom is shared by eight unit cells and each face atom is shared by two, resulting in 4 atoms per unit cell. In a body-centered cubic (bcc) lattice, there is one atom entirely within the cell and 8 corner atoms shared by 8 unit cells, totaling 2 atoms per unit cell. The diamond lattice is a more complex structure with 8 atoms per unit cell, each with a fractional part inside the unit cell.
The most efficient packing, meaning the least percentage of free space, is found in the face-centered cubic structure. To calculate this, one needs to consider the volume occupied by atoms—which can be delineated through the atomic radius and the volume of the unit cell based on the edge lengths.
Cadmium sulfide crystallizes in a structure where cadmium ions occupy half of the tetrahedral holes of the sulfide ion closest-packed array. Since each sulfide ion allows for two tetrahedral holes, the ratio of cadmium to sulfide ions is 1:1, so the formula of cadmium sulfide is CdS.
Gaseous compound Q contains only xenon and oxygen. When a 0.100 g sample of Q is placed in a 50.0-mL steel vessel at 0 C, the pressure is 0.230 atm. What is the likely formula of the compound?
The question is incomplete, here is the complete question:
Gaseous compound Q contains only xenon and oxygen. When a 0.100 g sample of Q is placed in a 50.0-mL steel vessel at 0°C, the pressure is 0.230 atm. What is the likely formula of the compound?
A. XeO
B. [tex]XeO_4[/tex]
C. [tex]Xe_2O_2[/tex]
D. [tex]Xe_2O_3[/tex]
E. [tex]Xe_3O_2[/tex]
Answer: The chemical formula of the compound is [tex]XeO_4[/tex]
Explanation:
To calculate the molecular weight of the compound, we use the equation given by ideal gas equation:
PV = nRT
Or,
[tex]PV=\frac{w}{M}RT[/tex]
where,
P = Pressure of the gas = 0.230 atm
V = Volume of the gas = 50.0 mL = 0.050 L (Conversion factor: 1 L = 1000 mL)
w = Weight of the gas = 0.100 g
M = Molar mass of gas = ?
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = Temperature of the gas = [tex]0^oC=273K[/tex]
Putting value in above equation, we get:
[tex]0.230\times 0.050=\frac{0.100}{M}\times 0.0821\times 273\\\\M=\frac{0.100\times 0.0821\times 273}{0.230\times 0.050}=194.9g/mol\approx 195g/mol[/tex]
The compound having mass as calculated is [tex]XeO_4[/tex]
Hence, the chemical formula of the compound is [tex]XeO_4[/tex]
a mammoth skeleton has a carbon-14 content of 12.50 of that found in living organisms
True/False
Correct Question:
A mammoth skeleton has a carbon-14 content of 12.50% of that found in living organisms. When did the mammoth live?
Answer:
17,190 years ago.
Explanation:
The carbon-14 is a carbon isotope that has a half-life of 5730 years. It means that the mass of carbon-14 decreases by half every 5730 years. If we call the initial mass as M and the final mas as m:
m = M/2ⁿ
Where n is the amount of half-live that had past. Thus, in this case, the mass of the skeleton is the final mass, and the mass founded in the living organisms the initial mass, so:
m = 0.1250M, and
0.1250M = M/2ⁿ
2ⁿ = M/0.1250M
2ⁿ = 1/0.1250
2ⁿ = 8
2ⁿ = 2³
n = 3
So it had past 3 half-live or
3*5730 = 17,190 years.
So the mammoth lived 17,190 years ago.
The statement is potentially true, depending on the age of the particular mammoth skeleton. It's expected that a mammoth skeleton, as a remnant of an organism that's been dead for thousands of years, would have a lower carbon-14 content than living organisms. However, the specific amount can vary based on the particular mammoth and its age.
Explanation:The statement is potentially true, but it depends on the specific mammoth and its carbon-14 content. Carbon-14 is a radioactive isotope found in all living organisms. When an organism dies, it stops absorbing carbon-14, and the isotope begins to decay. Over time, the amount of carbon-14 in a dead organism's remains decreases. Thus, it's possible that a mammoth skeleton, which belonged to a species that went extinct thousands of years ago, would have a carbon-14 content lower than that of living organisms today. However, the carbon-14 in a specific mammoth skeleton could also be less than 12.50 percent of that in a present living organism if the skeleton is exceptionally old, or more if the skeleton is relatively recent.
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Predict the major organic product for the reaction below. Draw the product in the planar (overhead) representation, clearly showing stereochemistry by drawing in a wedge or hashed bond per stereocenter.
Answer:
The main organic product for the reaction is shown in the following figure.
Explanation:
This is due to the presence of the substituent -CH3 in the molecule, which makes it impossible to leave the hydrogen to form the double bond in the elimination.
The question involves predicting the major product of a given reaction based on stereochemistry, utilizing concepts from organic chemistry such as skeletal structure and the reactivity of alkenes. Understanding these principles can help predict the product, though without the specified diagram, an exact prediction cannot be made.
Explanation:Your question relates to predicting the major organic product of a given reaction by examining the stereochemistry, which involves the three-dimensional arrangement of atoms in a molecule. Indeed, stereochemistry is a critical aspect of organic chemistry because it influences the properties and reactions of molecules.
Organic chemists often represent large molecules using a skeletal structure or line-angle structure. In this drawing style, carbon atoms are at the ends or bends of lines. Hydrogens attached to carbons are not drawn, and atoms other than carbon and hydrogen are represented by their symbols.
Alkenes are particularly reactive due to the presence of a C=C moiety, a reactive functional group known to undergo addition reactions. For instance, in the halogenation of alkenes, halogens add to the double bond of the alkene instead of replacing hydrogen as they would in an alkane. The stereochemistry of the reaction is significant because the spatial arrangement of atoms can influence the ability of halogens to add across the C=C bond.
Unfortunately, without the specific planar diagram or additional information about the reaction in question, I cannot predict the exact major product; however, this general information about the reactivity and stereochemistry of alkenes might help you predict it.
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