Explanation:
Let us assume that the concentration of [[tex]OH^{-}[/tex] and [tex]H^{+}[/tex] is equal to x. Then expression for [tex]K_{w}[/tex] for the given reaction is as follows.
[tex]K_{w} = [OH^{-}][H^{+}][/tex]
[tex]K_{w} = x^{2}[/tex]
[tex]6.8 \times 10^{-15} = x^{2}[/tex]
Now, we will take square root on both the sides as follows.
[tex]\sqrt{6.8 \times 10^{-15}} = \sqrt{x^{2}}[/tex]
[tex][H^{+}] = 8.2 \times 10^{-8}[/tex] M
Thus, we can conclude that the [tex]H_{3}O^{+}[/tex] concentration in neutral water at this temperature is [tex]8.2 \times 10^{-8}[/tex] M.
Answer: The concentration of [tex]H_3O^+[/tex] in neutral water is [tex]8.2\times 10^{-8}M[/tex]
Explanation:
The chemical equation for the ionization of water follows:
[tex]2H_2O\rightleftharpoons H_3O^++OH^-[/tex]
The expression of [tex]K_w[/tex] for above equation, we get:
[tex]K_w=[H_3O^+]\times [OH^-][/tex]
We are given:
[tex]K_w=6.8\times 10^{-15}[/tex]
[tex][H^+]=[OH^-]=x[/tex]
Putting values in above equation, we get:
[tex]6.8\times 10^{-15}=x\times x\\\\x=8.2\times 10^{-8}M[/tex]
Hence, the concentration of [tex]H_3O^+[/tex] in neutral water is [tex]8.2\times 10^{-8}M[/tex]
Consider the Diels–Alder reaction of cyclopentadiene with p‑benzoquinone. Diels-Alder reaction between cyclopentadiene and p-benzoquinone Select the correct statement about the stoichiometry of this reaction.
A. Multiple molar equivalents of cyclopentadiene and p‑benzoquinone react to form a long polymer chain.
B. One molar equivalent of p‑benzoquinone can react with only one molar equivalent of cyclopentadiene.
C. Two molar equivalents of p‑benzoquinone can react with one molar equivalent of cyclopentadiene.
D. Two molar equivalents of cyclopentadiene can react with one molar equivalent of p‑benzoquinone.
Answer:
The answer is (D) Two molar equivalents of cyclopentadiene can react with one molar equivalent of p‑benzoquinone.
Explanation:
The reaction between p-benzoquinone and cyclopentadiene is a 1,4-cyclo-addition reaction, will follow the general 4n+2 π electrons rule of reaction between a dienophile and a diene.
Hence, two moles of cyclopentadiene added to p-benzoquinone will result in two possible isomers of cis endo and exo adducts (open and closed rings).
Only reaction conditions can be varied to effect yield increase.
Final answer:
The correct statement regarding the stoichiometry of the Diels-Alder reaction of cyclopentadiene with p-benzoquinone is that one molar equivalent of p-benzoquinone reacts with one molar equivalent of cyclopentadiene, highlighting its [4+2] cycloaddition character and specificity.
Explanation:
Considering the Diels–Alder reaction of cyclopentadiene with p-benzoquinone, the correct statement about the stoichiometry of this reaction is: B. One molar equivalent of p-benzoquinone can react with only one molar equivalent of cyclopentadiene. The Diels-Alder reaction is characterized as a [4+2] cycloaddition, where a conjugated diene (cyclopentadiene) interacts with a dienophile (p-benzoquinone) to form a six-membered cyclic compound. This reaction is highly selective and stereospecific, generally taking place between one mole of diene and one mole of dienophile to produce a distinct cyclohexene derivative without creating long polymer chains or requiring multiple moles of either reactant.
The reaction's selectivity and efficiency contribute to its widespread use in synthetic organic chemistry, particularly in the synthesis of complex molecules and natural products. The process does not favor the formation of polymers or the interaction of uneven molar equivalents of reactants, making option B the correct choice for its stoichiometry.
Consider the reaction below in a closed flask. At 200 o C, the equilibrium constant (Kp) is 2.40 × 103 . 2 NO (g) N2 (g) + O2 (g) If 36.1 atm of NO (g) is added to the closed flask at 200 o C, what is the approximate partial pressure of O2 at equilibrium?
Explanation:
Since, the given reaction is as follows.
[tex]2NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g)[/tex]
Initial: 36.1 atm 0 0
Change: 2x x x
Equilibrium: (36.1 - 2x) x x
Now, expression for [tex]K_{p}[/tex] of this reaction is as follows.
[tex]K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}[/tex]
As the initial pressure of NO is 36.1 atm. Hence, partial pressure of [tex]O_{2}[/tex] at equilibrium will be calculated as follows.
[tex]K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}[/tex]
[tex]2.40 \times 10^{3} = \frac{x \times x}{(36.1 - 2x)^{2}}[/tex]
x = 18.1 atm
Thus, we can conclude that partial pressure of [tex]O_{2}[/tex] at equilibrium is 18.1 atm.
Answer:
partial pressure O2 = 17.867 atm
Explanation:
Step 1: Data given
Temperature = 200 °C
Kp = 2.40 *10^3
Pressure NO = 36.1 atm
Step 2: The balanced equation
2 NO ⇔ N2 + O2
Step 3: The initial pressure
pNO = 36.1 atm
pN2 = 0 atm
pO2 = 0 atm
Step 4: the pressure at the equilibrium
For 2 moles NO we'll have 1 mol N2 and 1 mol O2
pNO = 36.1 - 2X atm
pN2 = X atm
pO2 = X atm
Step 5: Calculate partial pressures
Kp = pN2 * pO2 / (pNO)²
2.40*10³ = x²/(36.1 - 2x)²
48.99 = x/(36.1-2x)
x = 1768.5 -97.98x
x = 17.867
x = partial pressure O2 = 17.867 atm
Predict the sign of ΔS° for 2NO2(g) LaTeX: \longrightarrow⟶ N2O4(g) CaCO3(s) + 2HCl(aq) LaTeX: \longrightarrow⟶ CaCl2(aq) + H2O(l) +CO2(g) Ag+(aq) + Cl-(aq) LaTeX: \longrightarrow⟶ AgCl(s)
Answer: a. [tex]2NO_2(g)\rightarrow N_2O_4(g)[/tex]: [tex]\Delta S[/tex] is negative
b. [tex]CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)[/tex] : [tex]\Delta S[/tex] is negative
c. [tex]Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)[/tex]: [tex]\Delta S[/tex] is negative
Explanation:
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa
a) [tex]2NO_2(g)\rightarrow N_2O_4(g)[/tex]
In this reaction 2 moles of gaseous reactants are converting to 1 mole of gaseous products. The randomness will decrease and hence entropy will also decrease. Thus [tex]\Delta S[/tex] is negative.
b) [tex]CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)[/tex]
In this reaction solid reactants are converting to aqueous products. The randomness will increase and hence entropy will also increase. Thus [tex]\Delta S[/tex] is positive.
c) [tex]Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)[/tex]
In this reaction aqueous reactants are converting to solid products. The randomness will decrease and hence entropy will also decrease. Thus [tex]\Delta S[/tex] is negative
The sign of ΔS° for the given reactions can be determined based on the change in the number of moles of gas. In general, an increase in the number of moles of gas results in a positive ΔS°, while a decrease in the number of moles of gas results in a negative ΔS°.
Explanation:ΔS° represents the change in entropy. To predict the sign of ΔS° for a reaction, we can consider the number of moles of gas formed or consumed. In general, an increase in the number of moles of gas will result in a positive ΔS°, indicating an increase in entropy. On the other hand, a decrease in the number of moles of gas will result in a negative ΔS°, indicating a decrease in entropy.
In the reaction 2NO2(g) → N2O4(g), the number of moles of gas decreases from 2 to 1. Therefore, ΔS° for this reaction is expected to be negative.
In the reaction CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g), the number of moles of gas increases from 0 to 1. Therefore, ΔS° for this reaction is expected to be positive.
In the reaction Ag+(aq) + Cl-(aq) → AgCl(s), the number of moles of gas remains the same. Therefore, ΔS° for this reaction is expected to be close to zero.
When you decrease the diameter of the stationary phase particles and make them more uniform, in HPLC, which term or terms of the van Deemter Equation is or are minimized?
Answer:
C-term
Explanation:
A correlation between plate height and mobile phase velocity is known as the van Deemter equation.
[tex]H = A + \frac{B}{u}+ C*u[/tex]
Where;
H is the plate height
A is the eddy diffusion term
B is the longitudinal diffusion term
C is the resistance to mass transfer coefficient
u is the linear velocity
Here C-term describes the mass transfer of sample components between the stationary phase and the mobile phase during separation. One of the major factor affecting C-term is the particle size, changing the diameter equally changes the particle size.
The answer is C-term
E. Both A and Cux are terms of the van Deemter Equation.
The van Deemter equation describes the relationship between the height equivalent to a theoretical plate (HETP) and the linear velocity (u) of the mobile phase in chromatography. The equation is:
[tex]\[H = A + \frac{B}{u} + C \cdot u\][/tex]
When the diameter of the stationary phase particles is decreased and made more uniform in High-Performance Liquid Chromatography (HPLC), the effects on the terms of the van Deemter equation are as follows:
1. Eddy Diffusion (A term):
- Eddy diffusion occurs due to the multiple pathways available for the analyte molecules through the packed column. Smaller and more uniform particles reduce the multiple pathways and hence minimize the A term.
2. Longitudinal Diffusion (B/u term):
- Longitudinal diffusion is the spreading of the analyte band due to diffusion along the length of the column. It is generally more significant at low flow rates and is not directly affected by the particle size.
3. Mass Transfer (C * u term):
- The mass transfer term relates to the time it takes for the analyte molecules to equilibrate between the stationary and mobile phases. Smaller particles decrease the distance the analytes must travel in and out of the stationary phase, thereby reducing the C.u term.
Therefore, decreasing the diameter of the stationary phase particles and making them more uniform primarily minimizes the A and [tex]\(C \cdot u\)[/tex] terms.
The correct answer is: E. Both A and Cux
Complete Question:
When you decrease the diameter of the stationary phase particles and make them more uniform, in HPLC, which term or terms of the van Deemter Equation is or are minimized?
A. A
B. B/ux
C. Cux
D. Both A and B/ux
E. Both A and Cux
F. Both B/ux and Cux
G. None of the terms
H. All of the terms
A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. After 25.0 mL of base is added, the pH of the solution is 3.42.Estimate the pKa of the weak acid.
Final answer:
The estimated pKa of the weak monoprotic acid is 3.42, as this value is equal to the pH at the halfway point to the equivalence point during a titration with a strong base where the amounts of the weak acid and its conjugate base are equal.
Explanation:
The question addresses the titration of a weak monoprotic acid with a strong base and involves finding the pKa of the acid using pH measurements.
To estimate the pKa of the weak acid, we use the information that at the halfway point of the titration (when half the equivalent amount of base has been added), the pH of the solution equals the pKa of the acid. This is because, at the halfway point, the concentrations of the acid (HA) and its conjugate base (A-) are equal.
Given that 25.0 mL of 0.100 M NaOH is the halfway point since it takes 50.0 mL of NaOH to reach the equivalence point, and the pH at this stage is 3.42, we can directly say that the pKa of the weak acid is approximately 3.42.
This is because at the halfway point of the titration, the amount of acid that has been neutralized by the base is equal to the amount of acid that remains un-neutralized. In such a scenario, according to the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA]) and since [A-] = [HA] at the halfway point, we get:
Thus, the estimated pKa value of the weak acid is 3.42.
Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution, You can assume the volume of the solution doesn't change shen th s sove m Be sure your answer has the correct number of significant digits
Calculating the final molarity of iodide anion in the solution, convert the grams of iron (II) iodide to moles, convert the volume of the silver nitrate solution to liters, use the stoichiometry of the reaction, calculate the moles of iodide anion, and divide the moles by the final volume of the solution.
Explanation:To calculate the final molarity of iodide anion in the solution, we need to use the stoichiometry of the reaction between iron (II) iodide and silver nitrate. Given that there is 0.981 g of iron (II) iodide and 150 mL of a 35.0 mM aqueous solution of silver nitrate, we can determine the moles of iron (II) iodide and the moles of iodide anion produced. Finally, we can calculate the final molarity of iodide anion by dividing the moles by the final volume of the solution.
Convert the grams of iron (II) iodide to moles by dividing by its molar mass: 0.981 g / (55.85 g/mol + 2 (126.9 g/mol)) = 0.007 mol. Convert the volume of silver nitrate solution to liters: 150 mL = 0.150 L. Use the stoichiometry of the reaction 2 AgNO3 + FeI2 → 2 AgI + Fe(NO3)2. Since the volume of the solution does not change, the moles of iodide anion produced will be the same as the moles of iron (II) iodide used. Calculate the moles of iodide anion: 0.007 mol I-. Calculate the final molarity of iodide anion by dividing the moles by the final volume: 0.007 mol / 0.150 L = 0.0467 M (rounded to the correct number of significant digits, which is 3). Learn more about Calculating the final molarity of iodide anion in a solution here:
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When making a solution of sodium hydroxide and water, a student weighed out an certain amount of sodium hydroxide pellets and dissolved them in an certain amount of water. However, the sodium hydroxide concentration of the resulting solution was lower than the concentration that the student thought they made. What was the problem?
Hygroscopic nature of NaOH is the main reason for the lower concentration of the sodium hydroxide in the solution.
Explanation:
It is a well known fact that Sodium hydroxide pellets are hygroscopic in nature, which clearly means that the sodium hydroxide pellets absorbs moisture from the air, so that it becomes deliquescent. When the NaOH crystals are weighed, the crystals absorb moisture from the surroundings, and so the weight of the crystals may change, so the concentration of the solution was lower than the required one.So it is not possible to prepare NaOH solution under normal room temperature. So, while preparing the solution of NaOH we have to be more careful.The main reason for the lower concentration of the sodium hydroxide in the solution is Hygroscopic nature of NaOH.
Nature of Sodium hydroxide:It's obviously true that Sodium hydroxide pellets are hygroscopic in nature, which plainly implies that the sodium hydroxide pellets retains dampness from the air, so it becomes deliquescent. Whenever the NaOH precious stones are gauged, the gems assimilate dampness from the environmental factors, thus the heaviness of the gems might change, so the centralization of the arrangement was lower than the expected one. So it is absurd to expect to get ready NaOH arrangement under ordinary room temperature. In this way, while setting up the arrangement of NaOH we must be more cautious.
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Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO 3 are produced by the reaction of 6.0 grams of O 2 with 7.0 grams of S. What is the % yield of SO 3 in this experiment
Answer: 95%
Explanation:
We first write down the balanced reaction equation as shown in the image attached. Next we identify the limiting reactant from the given data. Oxygen is the limiting reactant in this case and it is the mass of oxygen reacted that is used in the estimation of the theoretical yield as shown in the image attached. The percentage yield is then calculated according to the formula shown in the image attachment.
Consider the reaction C2H4(g) + H2O(g)CH3CH2OH(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.90 moles of C2H4(g) react at standard conditions. S°surroundings = J/K
Answer: The value of [tex]\Delta S^o[/tex] for the surrounding when given amount of ethene gas is reacted is 238.80 J/K
Explanation:
Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate entropy change is of a reaction is:
[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}][/tex]
For the given chemical reaction:
[tex]C_2H_4(g)+H_2O(g)\rightarrow CH_3CH_2OH(g)[/tex]
The equation for the entropy change of the above reaction is:
[tex]\Delta S^o_{rxn}=[(1\times \Delta S^o_{(CH_3CH_2OH(g))})]-[(1\times \Delta S^o_{(C_2H_4(g))})+(1\times \Delta S^o_{(H_2O(g))})][/tex]
We are given:
[tex]\Delta S^o_{(CH_3CH_2OH(g))}=282.7J/K.mol\\\Delta S^o_{(C_2H_4(g))}=219.56J/K.mol\\\Delta S^o_{(H_2O(g))}=188.82J/K.mol[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(1\times (282.7))]-[(1\times (219.56))+(1\times (188.82))]\\\\\Delta S^o_{rxn}=-125.68J/K[/tex]
Entropy change of the surrounding = - (Entropy change of the system) = -(-125.68) J/K = 125.68 J/K
We are given:
Moles of ethene gas reacted = 1.90 moles
By Stoichiometry of the reaction:
When 1 mole of ethene gas is reacted, the entropy change of the surrounding will be 125.68 J/K
So, when 1.90 moles of ethene gas is reacted, the entropy change of the surrounding will be = [tex]\frac{125.68}{1}\times 1.90=238.80J/K[/tex]
Hence, the value of [tex]\Delta S^o[/tex] for the surrounding when given amount of ethene gas is reacted is 238.80 J/K
Phenolphthalein is an indicator that turns from colorless (acidic form) to magenta (basic form) and has a pKa of 9.40. What is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration ([magenta phenolphthalein]/[colorless phenolphthalein]) at a pH of 11?
Answer:
40:1 is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration.
Explanation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[magenta(Php)]}{[Php]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of phenolphthalein = 9.40
[tex][magenta(Php)][/tex] = concentration of magenta phenolphthalein
[tex][Php][/tex] = concentration of colorless phenolphthalein
pH = 11
Putting values in above equation, we get:
[tex]11=9.40+\log(\frac{[magenta(Php)]}{[Php]})[/tex]
[tex]\log(\frac{[magenta(Php)]}{[Php]})=11-9.40=1.6[/tex]
[tex]\frac{[magenta(Php)]}{[Php]}=10^{1.6}=39.81 :1 \approx 40:1[/tex]
40:1 is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration.
Consider the reaction: A <=> B. Under standard conditions at equiliubrium, the concentrations of the compounds are [A] = 1.5 M, and [B] = 0.5 M. Keq' for this reaction is ____ and ∆G°' is _____. (You should not need a calculator for this.)
Answer:
See explanation below
Explanation:
In this case, let's write the equation again:
A <------> B Keq = ?
As we are using standard conditions, we can assume we have a temperature of 0 °C (273 K) and 1 atm.
To get the equilibrium constant we only do the following:
Keq = [B] / [A]
However, the problem is asking the reverse equilibrium constant (because of the ' in Keq'), so, we have to do the reverse division:
Keq' = [A]/[B]
Replacing the given values of A and B:
Keq' = 1.5/0.5 = 3
We have the equilibrium constant, we can calculate now the gibbs free energy with the following expression:
ΔG°' = -RTlnKeq'
As Keq' is > 1, the negative logaritm will result into a negative result or a number < 0, so, calculating this we have:
ΔG°' = -8.31 * 273 ln3
ΔG°' = -2.492.34 J
Give the characteristics of a strong acid:
A. ionizes completely in aqueous solution.
B. has equilibrium far to the right.
C. has a polar bond.
D. has a waker bond to hydrogen.
E. all of the above
Answer:
Not D
Explanation:
Consider the mechanism. Step 1: A + B − ⇀ ↽ − C equilibrium Step 2: C + A ⟶ D slow Overall: 2 A + B ⟶ D Determine the rate law for the overall reaction, where the overall rate constant is represented as
The rate law for the overall reaction derived from a two-step reaction mechanism is determined by the slowest (rate-determining) step. In this case, the rate law would be rate = k * K_eq * [A]^2[B], where k is the rate constant for the slow step, K_eq is the equilibrium constant for the fast step, and [A] and [B] represent the concentrations of A and B, respectively.
Explanation:In the context of chemistry, one step in a multistep reaction mechanism is often significantly slower than the others. This slow step is known as the rate-determining step or the rate-limiting step. The reaction cannot proceed faster than this slowest step.
In your specific example, the given mechanism consists of two steps. The rate law for each step is generally expressed in terms of the concentration of the reactants involved in that step.
For Step 1 (A + B ⇄ C), assuming that the reaction reaches an equilibrium, the concentrations of A, B, and C would remain constant over time, and won't affect the overall rate. Therefore, we would ignore this step when deriving the rate law for the overall reaction.
On the other hand, Step 2 (C + A ⟶ D) is the slow step, and thus determines the rate of the overall reaction. The rate law for this step would be rate = k * [C][A]. But since C is also a product of Step 1, we need to express C in terms of A and B. From equilibrium of Step 1, we know [C] = K_eq*[A][B] (where K_eq is equilibrium constant).
Substituting this in rate law of step 2, we get rate = k * K_eq * [A]^2[B] for the overall reaction.
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The rate law for the overall reaction 2A + B ⟶ D, considering Step 2 is the rate-determining step, is calculated by using the equilibrium from Step 1 to express the concentration of intermediate C in terms of A and B. After substitution and simplification, the rate law for the overall reaction is rate = k [A]²[B], indicating second-order dependence on A and first-order dependence on B.
Explanation:To determine the rate law for the overall reaction, we must look at the mechanism provided. Since Step 2, which involves the conversion of C and A to D, is the rate-determining step, the rate law for the overall reaction will reflect this slowest step. However, since C is an intermediate that we cannot measure directly, we must use the equilibrium established in Step 1 to express the concentration of C in terms of the concentrations of A and B.
Assuming Step 1 is at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, and thus:
Rate of forward reaction (k₁ [A][B]) = Rate of reverse reaction (k₁₁ [C])
We can rearrange this to solve for [C]:
[C] = k₁/k₁₁ [A][B]
Now, since the rate-determining step is Step 2, we write the rate law based on this step:
rate = k₂ [C][A]
Substituting in the expression for [C] gives us:
rate = k₂ (k₁/k₁₁ [A][B])[A]
rate = (k₂ * k₁/k₁₁) [A]²[B]
Thus, after simplifying and combining the rate constants into a single overall rate constant (k), we have:
rate = k [A]²[B]
This shows that the reaction is second order with respect to A and first order with respect to B.
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Consider the reaction 2 SO2(g) + O2(g) <=> 2 SO3(g), which is exothermic as written. What would be the effect on the equilibrium position of removing O2(g)?
Answer:
Removing O₂, means removing one of the reactants and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.
Explanation:
The principle that explains how changes in temperature, Concentration and Pressure of reactants or products of a reaction at equilibrium affect the equilibrium position of the reaction is the Le Chatelier's principle.
The Principle explains that a system/process if a system/process which is at equilibrium is disturbed/perturbed/constrained by one or more changes (in concentration, pressure or temperature), the system would shift the equilibrium position to counteract the effects of this change.
Removing O₂, means removing one of the reactants (changing its concentration) and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.
Final answer:
Removing O₂(g) from the equilibrium of the exothermic reaction 2 SO₂(g) + O₂(g) <=> 2 SO₃(g) shifts the equilibrium towards the reactants' side, increasing the concentration of SO₂(g) and decreasing the concentration of SO₃(g).
Explanation:
The question revolves around the effect of removing O₂(g) from the equilibrium system of the reaction 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), which is exothermic. According to Le Châtelier's Principle, when a change is made to a system in equilibrium, the system adjusts to counteract that change. In this case, removing O₂(g) (oxygen) would decrease its concentration, prompting the equilibrium to shift towards the reactants' side to increase the concentration of O₂(g) and restore equilibrium. This means that the concentration of SO₂(g) will increase, and the concentration of SO₃(g) will decrease as the reaction shifts left to form more reactants.
PLEASE ANSWER ASAP:
The kinetic theory of heat states that heat is the result of the movement of particles in a system. Is this statement True or False
Answer:
TrueExplanation:
Heat is a kind of energy.
The kinetic theory relates the heat with the movement of the particles: the more the particles move, the larger the kinetic energy of the system. The kinetic theory states that heat is the kinetic energy of the particles, atoms or molecules, in a substance, that is transferred from a substance at higher temperature to other substance at lower temperature.
Based on that principle, the kinetic theory explains the changes of phases of the substances in terms of the motion of the particles: the hotter an object the faster the particles move, the more energetic the particles are, and they occupy more space. Thus, when a solid is heated, the particles move faster and it can pass to liquid or gaseous state.
Given the following proposed mechanism, predict the rate law for the overall reaction. A2 + 2 B → 2 AB (overall reaction) Mechanism A2 ⇌ 2 A fast A + B → AB slow
Answer:
The rate of the over all reaction is ;
[tex]R=K[A_2]^{1/2}[B][/tex]
Explanation:
Step 1 : [tex]A_2\rightleftharpoons 2 A[/tex] fast
Step 2 : [tex]A + B\rightarrow AB[/tex] slow
Equilibrium constant of the reaction in step 1:
[tex]K_1=\frac{[A]^2}{[A_2]}[/tex]....[1]
Overall reaction:
[tex]A_2 + 2 B \rightarrow 2 AB[/tex]
When there is a chemical reaction which taking place in more than 1 step than the rate of the over all reaction is determined by the slowest step occurring during that process;
Here step 2 is slow step, so the rate of the reaction will be;
[tex]R=k[A][B][/tex]..[2]
Putting value of [A] from [1] in [2]:
[tex]R=k\times \sqrt{K_1\times [A_2]}\times [B][/tex]
[tex]K=k\times (K_1)^{1/2}[/tex]
K = rate constant of the reaction
The rate of the over all reaction is ;
[tex]R=K[A_2]^{1/2}[B][/tex]
The overall rate law for the given reaction mechanism is first order in A* (generated from A2) and first order in B, leading to an overall rate law of rate = k[A*][B], with k being the rate constant.
Explanation:The rate law for a chemical reaction can be determined based on the mechanism of the reaction and the rate-determining step. Considering the given mechanism, where A2 dissociates into 2 A rapidly, and then A reacts with B to form AB slowly, the slow step is the rate-determining step. Therefore, the reaction is first order with respect to A* (generated from A2) and first order with respect to B. This is because the slow step involves one molecule of A* and one molecule of B. The overall reaction rate would be expressed as rate = k[A*][B], where k is the rate constant. If we assume the steady state approximation applies, A2 rapidly reaches a steady concentration of A*, and thus its concentration does not directly figure into the rate law for the slow step.
An element, X, can form a chloride (XCl3) and an iodide (XI3). The chloride can be converted quantitatively into the iodide when heated and exposed to excess iodine. If 0.760 grams of XCl3 are heated with iodine, 1.610 g of XI3 are produced. What is the chemical symbol for this element
Answer:
La is the chemical symbol for this element.
Explanation:
[tex]2XCl_3+3I_2\rightarrow 2XI_3+3Cl_2[/tex]
Let the molar mass of [tex]XCl_3[/tex] be M.
Let the molar mass of [tex]XI_3[/tex] be M'.
Moles of [tex]XCl_3=\frac{0.760 g}{M}=n[/tex]
Moles of [tex]XI_3=\frac{1.610g}{M'}=n'[/tex]
According to reaction , 2 moles of [tex]XCl_3[/tex] gives 2 moles of [tex]XI_3[/tex], then n moles [tex]XCl_3[/tex] will give:
[tex]\frac{1}{1}\times n=n[/tex] moles of [tex]XI_3[/tex]
[tex]n=n'[/tex]
[tex]\frac{0.760 g}{M}=\frac{1.610g}{M'}[/tex]
Atomic mass of iodine = 127 g/mol
Atomic mass of chlorine = 35.5 g/mol
Atomic mass of X = x
[tex]\frac{0.760 g}{x+3\times 35.5 g/mol}=\frac{1.610g}{x+3\times 127 g/mol}[/tex]
Solving for x:
x = 138.9 g/mol
The value of atomic mass of X corresponds to compound named lanthanum.
La is the chemical symbol for this element.
Answer:
The chemical symbol for this element is La
Explanation:
Step 1: Data given
Mass of XCl3 = 0.760 grams
Mass of XI3 = 1.610 grams
Molar mass Cl = 35.45 g/mol
Step 2: The balanced equation
2XCl3 + 3I2 → 2XI3 + 3Cl2
The mol ratio for XCl3: XI3 = 2:2 or 1:1
Step 3: Calculate moles
Moles = mass / molar mass
Moles XCl3 = mass XCl3 / molar mass XCl3
Moles XCl3 = 0.760 grams / (X + 3*35.45 g/mol)
Moles XI3 = 1.610 grams / (X + 3*126.9 g/mol)
0.760 grams / (X + 3*35.45 g/mol) = 1.610 grams / (X + 3*126.9 g/mol)
0.760 / (X + 106.35) = 1.610 / (X + 380.7)
0.760 (X +380.7) = 1.610 ( X + 106.35)
0.760X + 289.3 = 1.610X + 171.2
118.1 = 0.85X
X = 138.9 g/mol
If we look for an element with atomic mass of 138.9 g/mol we find Lanthanum (La)
2LaCl3 + 3I2 → 2LaI3 + 3Cl2
Consider the following metals: Ag, Au, Mg, Ni, and Zn.
Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank?
Steel is mostly iron, so use −0.447 V as the standard reduction potential for steel.
Answer:
Mg and Zn
Explanation:
In cathodic protection, the sacrificial anode corrodes instead of the cathode which it protects. The anode is usually higher than the cathode in the electrochemical series. This also means that the reduction potential of the sacrificial anode is more negative than that of the cathode. Consider the reduction potentials of the metals listed in the question:
Mg=-1.185V
Zn= -0.7618V
Ag= +0.7996V
Au= +1.629V
Ni= -0.251V
The reduction potential of the cathode stated in the question is -0.447V hence only magnesium and zinc can function as sacrificial anode.
The metals that can be used as sacrificial anodes in the cathodic protection of an underground steel storage tank are silver (Ag), magnesium (Mg), nickel (Ni), and zinc (Zn).
Explanation:The metals that can be used as sacrificial anodes in the cathodic protection of an underground steel storage tank are the ones with a lower standard reduction potential (E°) than steel. In this case, we need to compare the reduction potentials of steel to the standard reduction potentials of the given metals.
Silver (Ag) has a standard reduction potential of -0.799 V, which is lower than the E° for steel (-0.447 V), so it can be used as a sacrificial anode.Gold (Au) has a standard reduction potential of +1.40 V, which is higher than the E° for steel, so it cannot be used as a sacrificial anode.Magnesium (Mg) has a standard reduction potential of -2.37 V, which is significantly lower than the E° for steel, so it can be used as a sacrificial anode.Nickel (Ni) has a standard reduction potential of -0.257 V, which is lower than the E° for steel, so it can be used as a sacrificial anode.Zinc (Zn) has a standard reduction potential of -0.763 V, which is lower than the E° for steel, so it can be used as a sacrificial anode.Therefore, the metals that could be used as sacrificial anodes in the cathodic protection of an underground steel storage tank are silver (Ag), magnesium (Mg), nickel (Ni), and zinc (Zn).
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Your standard iron solution is 0.1511 M Fe(II), your dichromate solution is 0.0181 M dichromate, and it took 26.84 mL of your standard iron solution to titrate the excess dichromate in your unknown. What is the chemical oxygen demand of the sample, in units of mg O2/L? (Report your answer with 4 sig figs)
Answer:
[tex]COD=2030\frac{mg}{L}[/tex]
Explanation:
Hello,
In this case, considering the given information the redox reaction is:
[tex]Fe^{+2}+Cr_2O_7^{-2}\rightarrow Fe^{+3}+Cr^{+3}[/tex]
Which properly balanced turns out:
[tex]Fe^{+2}\rightarrow Fe^{+3}+1e^-\\Cr^{+6}_2O_7^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \downarrow\\6Fe^{+2}+Cr_2O_7^{-2}+14H^+\rightarrow 6Fe^{+3}+2Cr^{+3}+7H_2O[/tex]
In such a way, one sees a 6 to 1 molar relationship between the standard iron (II) solution and the dichromate, therefore, by using the following equation it is possible to determine the oxygen as shown below:
[tex]n_{Fe^{+2}}=n_{Cr_2O_7^{-2}}[/tex]
Thus, the moles of iron (II) solution are:
[tex]n_{Fe^{+2}}=0.1511\frac{molFe^{+2}}{L_{sln}}*0.02684L_{sln}=0.00406molFe^{+2}[/tex]
Moreover, the moles of dichormate result:
[tex]n_{Cr_2O_7^{-2}}=0.00406molFe^{+2}*\frac{1molCr_2O_7^{-2}}{6molFe^{+2}}=6.76x10^{-4}}molCr_2O_7^{-2}[/tex]
Thereby, the volume of the sample is:
[tex]V_{sample}=\frac{n_{Cr_2O_7^{-2}}}{M_{Cr_2O_7^{-2}}}=\frac{6.76x10^{-4}molCr_2O_7^{-2}}{0.0181\frac{molCr_2O_7^{-2}}{L_{sln}} } =0.0373L_{sample}[/tex]
Finally, the chemical oxygen demand result:
[tex]COD=6.76x10^{-4}molCr_2O_7^{-2}*\frac{7molO}{1molCr_2O_7^{-2}}*\frac{16gO}{1molO}*\frac{1000mgO}{1gO}*\frac{1}{0.0373L_{sln}} \\COD=2030\frac{mg}{L}[/tex]
Best regards.
Calculate the pH for each case in the titration of 50.0 mL of 0.220 M HClO ( aq ) with 0.220 M KOH ( aq ) . Use the ionization constant for HClO . What is the pH before addition of any KOH ?
Answer:
Before adding any KOH, the pH is 4.03
Explanation:
Step 1: Data given
Volume of a 0.220 M HClO = 50.0 mL = 0.050 L
Molarity of KOH = 0.220 M
The ionization constant for HClO is 4.0*10^–8
Step 2: The balanced equation
HClO + KOH → KClO + H2O
Step 3: pH before any addition of KOH
When no KOH is added, we only have HClO, a weak acid.
To calculate the pH of a weak acid, we need the Ka
Ka = [H+] / [acid]
4.0*10^-8 = [H+]² / 0.220
[H+]² = (4.0*10^-8 ) * 0.220
[H+]² = 8.8*10^-9
[H+] = √( 8.8*10^-9)
[H+] = 9.38*10^-5 M
pH = -log [H+]
pH = -log(9.38*10^-5)
pH = 4.03
Before adding any KOH, the pH is 4.03
Final answer:
To calculate the pH before the addition of KOH in the titration of HClO with KOH, we can use the ionization constant for HClO (Ka). The initial concentration of HClO can be used to calculate the concentration of H+, which in turn can be used to calculate the pH using the pH formula.
Explanation:
The pH before the addition of any KOH can be calculated using the ionization constant for HClO. HClO is a weak acid, so we can use the expression for the acid dissociation constant, Ka, to calculate the pH. The expression for Ka for HClO is:
Ka = [H+][ClO-] / [HClO]
Since we know the initial concentration of HClO, we can assume that the concentration of H+ is equal to the initial concentration of HClO. Therefore, we can rewrite the expression for Ka as:
Ka = [H+]² / [HClO]
Now we can calculate the concentration of H+ using the given initial concentration of HClO:
[H+] = sqrt(Ka * [HClO])
Finally, we can use the concentration of H+ to calculate the pH using the pH formula:
pH = -log[H+]
A reaction was performed in which 0.55 g of 2-naphthol was reacted with a slight excess of 1-bromobutane to make 0.32 g of 2-butoxynaphthalene. Calculate the theoretical yield and percent yield for this reaction.
Answer:
Theoretical yield=0.764 g.
Percent yield 42%.
Explanation:
Hello,
In this case the undergoing chemical reaction is shown on the attached picture. In such a way, starting by the given mass of 2-naphtol, one computes the theoretical yielded grams of 2-butoxynaphthalene as shown below, considering their 1 to 1 molar relationship:
[tex]m_{2-butoxynaphthalene}=0.55g2-naphtol*\frac{1mol2-naphtol}{144.17g2-naphtol}*\frac{1mol2-butoxynaphthalene}{1mol2-naphtol}*\frac{200.28g2-butoxynaphthalene}{1mol2-butoxynaphthalene}=0.764g2-butoxynaphthalene[/tex]
Moreover, the percent yield turns out:
[tex]Y=\frac{0.32g}{0.764g}*100\% =42\%[/tex]
Best regards.
Without the balanced chemical equation and molar masses, the theoretical yield of 2-butoxynaphthalene cannot be calculated. Normally, it involves stoichiometric calculations from the starting material. Percent yield is obtained by dividing actual yield by theoretical yield and multiplying by 100%.
Explanation:To calculate the theoretical yield and percent yield for the given reaction, we need to perform a few stoichiometric calculations. However, without the balanced chemical equation and the molar masses, we cannot calculate the theoretical yield for the specific reaction of 2-naphthol and 1-bromobutane to form 2-butoxynaphthalene. If we had the balanced equation, we would calculate the moles of 2-naphthol used, and using stoichiometry, find the moles of 2-butoxynaphthalene that could be formed. Multiplying by the molar mass would give us the theoretical yield in grams. To find percent yield, we would then divide the actual yield (0.32 g) by the theoretical yield and multiply by 100.
Generally speaking, percent yield is the ratio of the actual yield to the theoretical yield multiplied by 100%. It indicates the efficiency of a chemical reaction and can be decreased by incomplete reactions, side reactions, or product recovery losses.
Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia: N 2 (g) 3H 2 (g) 2NH3 (g) If the reaction yield is 87.5%, how many moles of N 2 are needed to produce 3.00 mol of NH 3
Answer:
We need 1.714 moles N2
Explanation:
Step 1: Data given
The reaction yield = 87.5 %
Number of moles NH3 produced = 3.00 moles
Step 2: The balanced equation
N2(g)+ 3H2(g) →2NH3(g)
Step 3: Calculate moles N2
For 2 moles NH3 produced we need 1 mol N2 and 3 moles H2
This means, if the yield was 100%, for 3.00 moles NH3 produced , we need 1.5 moles N2
For a 87.5 % yield:
we need more N2, increased by a ratio of 100/87.5.
100/87.5 * 1.5 = 1.714 moles N2
Compositional analysis of a certain lipid shows that it has exactly one mole of fatty acid per mole of inorganic phosphate. Which of the lipids listed below is the lipid with this composition?
a. galactolipid
b. glycerophospholipid
c. sphingolipid
d. triacylglycerol
The sphingolipid is the lipid with this composition.
Option: C
Explanation:
A biomolecule which is soluble in non-polar solvents like hydrocarbons but remain insoluble in water is understood as a lipid. Its function is: energy storage, signaling and working as structural components of cell membranes. A class of lipids, which is structured from sphingoid base backbone, a group of aliphatic amino alcohols including sphingosine, is known as Sphingolipids. It was discovered in brain extracts during 1870s and named after the mythological sphinx for their mysterious existence.Complicated sphingolipids in animal cells' plasma membrane, particularly nerve cells, have a structural role and are thought to shield the cell surface from unfavorable external factors.The complex lipid with one mole of fatty acid per mole of inorganic phosphate has been Sphingolipid. Thus, option C is correct.
The lipids are biomolecules with the ability to bind with the nonpolar surfaces. The lipids have been bound with other biomolecules to form complex lipids.
The galactolipid has been the complex of fatty acid with galactose.Glycerophospholipid has been the composed of glycerophosphate along with the long chain constituted fatty acid and small alcohols.Sphingolipid has been the sphingosine backbone lipid with fatty acid. The sphingosine has been constituted of long amino acid chain. It has been present in cell membrane with one mole of fatty acid per mole phosphate.Triacylglycerol has been acidified fatty acid with three fatty acid to one glycerol.Thus, the complex lipid with one mole of fatty acid per mole of inorganic phosphate has been Sphingolipid. Thus, option C is correct.
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Some COCl2 is placed in a sealed flask and heated to 756 K. When equilibrium is reached, the flask is found to contain COCl2 (7.40×10-4 M), CO (3.76×10-2 M), and Cl2 (1.78×10-2 M). What is the value of the equilibrium constant for this reaction at 756 K?
Answer:
[tex]9.044\times 10^{-3}[/tex] is the value of the equilibrium constant for this reaction at 756 K.
Explanation:
[tex]COCl_2\rightleftharpoons CO+Cl_2[/tex]
Equilibrium concentration of [tex]COCl_2[/tex]
[tex][COCl_2]=7.40\times 10^{-4} M[/tex]
Equilibrium concentration of [tex]CO[/tex]
[tex][CO]=3.76\times 10^{-2} M[/tex]
Equilibrium concentration of [tex]Cl_2[/tex]
[tex][Cl_2]=1.78\times 10^{-4} M[/tex]
The expression of an equilibrium constant can be written as;
[tex]K_c=\frac{[CO][Cl_2]}{[COCl_2]}[/tex]
[tex]=\frac{3.76\times 10^{-2}\times 1.78\times 10^{-4}}{7.40\times 10^{-4}}[/tex]
[tex]K_c=9.044\times 10^{-3}[/tex]
[tex]9.044\times 10^{-3}[/tex] is the value of the equilibrium constant for this reaction at 756 K.
You replicate the CEC analysis of a secondary alcohol you performed in the lab using a reverse phase TLC plate producing the TLC plate below. Based on this TLC plate and the mnemonic in your notes, what can you conclude about the stereochemistry of the alcohol? (1 pts)
Answer:
Answer for the question:
You replicate the CEC analysis of a secondary alcohol you performed in the lab using a reverse phase TLC plate producing the TLC plate below. Based on this TLC plate and the mnemonic in your notes, what can you conclude about the stereochemistry of the alcohol? (1 pts)
is given below which explains the best option for the answer.
Explanation:
The enantiomer of the alcohol cannot be determined.
Which compartment has the higher osmotic pressure? Which compartment has the higher osmotic pressure? 1%% (m/vm/v) starch solution 10%% (m/vm/v) starch solution
Answer : The compartment that has the higher osmotic pressure is, 10 % (m/v) starch solution.
Explanation :
Formula used for osmotic pressure :
[tex]\pi=\frac{nRT}{V}\\\\\pi=\frac{wRT}{MV}[/tex]
where,
= osmotic pressure
V = volume of solution
R = solution constant = 0.0821 L.atm/mole.K
T= temperature of solution = [tex]25^oC=273+25=298K [/tex]
M = molar mass of solute
w = mass of solute
Now we have to determine the osmotic pressure for the following solution.
For 1 % (m/v) starch solution :
1 % (m/v) starch solution means that 1 grams of starch present in 100 mL or 0.1 L of solution.
Molar mass of starch = 692.7 g/mol
[tex]\pi=\frac{(1g)\times (0.0821Latm/moleK)\times (298K)}{(692.7g/mol)\times (0.1L)}[/tex]
[tex]\pi=0.353atm[/tex]
For 10 % (m/v) starch solution :
10 % (m/v) starch solution means that 10 grams of starch present in 100 mL or 0.1 L of solution.
Molar mass of starch = 692.7 g/mol
[tex]\pi=\frac{(10g)\times (0.0821Latm/moleK)\times (298K)}{(692.7g/mol)\times (0.1L)}[/tex]
[tex]\pi=3.53atm[/tex]
From this we conclude that, 10 % (m/v) starch solution has the higher osmotic pressure as compared to 1 % (m/v) starch solution.
Hence, the compartment that has the higher osmotic pressure is, 10 % (m/v) starch solution.
The 10% (m/v) starch solution will have a higher osmotic pressure compared to the 1% (m/v) starch solution because osmotic pressure increases with solute concentration.
Explanation:The osmotic pressure is a property that depends on the solute concentration in a solution. In comparing a 1% (m/v) starch solution to a 10% (m/v) starch solution, the solution with the higher solute concentration is the one with higher osmotic pressure. Therefore, the 10% (m/v) starch solution will have a higher osmotic pressure because it has a greater concentration of starch molecules.
The osmotic pressure of a solution can be calculated using the formula Π = MRT, where Π is the osmotic pressure, M is the molarity of the solution, R is the gas constant, and T is the temperature in Kelvin. As solute concentration increases, so does the osmotic pressure, provided that temperature and the gas constant remain the same. Consequently, hypertonic solutions will have higher osmotic pressure than hypotonic solutions.
A chemist determined by measurements that 0.050 moles of aluminum participated in a chemical reaction. Calculate the mass of aluminum that participated in the chemical reaction. Be sure your answer has the correct number of significant digits.
Answer:
13.5 g
Explanation:
This question is solved easily if we remember that the number of moles is obtained by dividing the mass into the atomic weight or molar mass depending if we are referring to elements or molecules.
Therefore, the mass of aluminum in the reaction will the 0.050 mol Al times the atomic weight of aluminum.
number of moles = n = mass of Al / Atomic Weight Al
⇒ mass Al = n x Atomic Weight Al = 0.050 mol x 27 g mol⁻¹
= 13.5 g
We have three significant figures in 0.050 and therefore we should have three significant figures in our answer.
Mass of aluminum that participated in the chemical reaction is 1.35 grams.
Number of Moles:
Number of moles is defined as the ratio of given mass to the molar mass.
Given:
Moles of Aluminum = 0.050 moles
To find:
Mass of Aluminum=?
As we know, Molar mass of Aluminum = 27g/mol
On substituting the values:
[tex]\text{Number of Moles}=\frac{\text{Given mass}}{\text{Molar mass}} \\\\ \text{Given mass}= \text{Number of Moles}*\text{Molar mass}\\\\ \text{Given mass}= 0.050*27\\\\ \text{Given mass}=1.35\text{ grams}[/tex]
Thus, the mass of aluminum that participated in the chemical reaction is 1.35 grams.
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Spilling room-temperature water over your skin on a hot day will cool you down. Spilling vegetable oil (of the same temperature as the water) over your skin on a hot day will not. Explain the difference.
Explanation:
Spilling room temperature water over your skin on a hot day will cook the body down. This is because water has a low heat of vapourization( water can evaporate into gas molecules easily). This evaporation causes a cooling effect on the skin surface and it requires heat energy as liquid is converted to gas. This is a typical example of an endothermic process. BUT spilling vegetable oil of the same temperature as the water over the skin will not cool the body down because oil has a high heat of vapourization and doesn't evaporate because it is a viscous liquid and the molecules are tightly bound to each other. I hope this helps.
This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH_3CO_2H An analytical chemist has determined by measurements that there are 0.054 moles of oxygen In a sample of acetic acid. How many moles of hydrogen are in the sample? Be sure your answer has the correct number of significant digits.
Answer:
0.108mol of Hydrogen
Explanation:
The formula for the compound is: CH3COOH
From the formula of the compound,
There are 2moles of oxygen and 4moles of Hydrogen.
If for every 2moles of oxygen, 4moles of Hydrogen is present.
Then, for 0.054 moles of oxygen = (0.054 x 4)/2 = 0.108mol of Hydrogen is present
Answer:
For 0.027 moles CH3COOH we have 0.108 moles H ≈ 1.1 *10^-1 moles H
Explanation:
Step 1: Data given
Acetic acid = CH3COOH
Number of moles oxygen in the sample = 0.054 moles
Step 2: calculate moles CH3COOH
In 1 mol CH3COOH we have 2 moles O
For 0.054 moles Oxygen we have 0.054/2 = 0.027 moles CH3COOH
Step 3: Calculate moles H
In 1 mol CH3COOH we have 4 moles H
For 0.027 moles CH3COOH we have 4*0.027 = 0.108 moles H ≈ 1.1 *10^-1 moles
A generic Lewis structure is given where Y represents the central atom and the outer atoms are represented by X. The outer atoms are not necessarily all the same element. The central Y atom has one lone pair of electrons. The Y atom is double bonded to an X atom with two lone pairs of electrons. The Y atom is also single bonded to an X atom with three lone pairs of electrons. Which compound is represented by the generic Lewis structure?
The Lewis structure in the question represents the molecule Oxygen difluoride (OF2), where 'Y' is Oxygen with one lone pair of electrons and 'X' is Fluorine, one of which is double bonded to Oxygen while other is single bonded, with two and three lone pairs of electrons respectively.
Explanation:The given generic Lewis structure shows the molecule Oxygen difluoride (OF2), which fulfills the conditions specified. Here, 'Y' can represent an Oxygen atom (O), which has one lone pair of electrons. 'X' can represent Fluorine atoms (F), one of which is double bonded to Oxygen (giving it two lone pairs of electrons), and the other is single bonded to Oxygen (giving it three lone pairs of electrons).
In the Lewis structure for Oxygen difluoride, the bonding and lone pairs of electrons in the molecule are represented through lines and dots respectively. It looks like this:
O = F
|
F
Each Fluorine atom interacts with eight valence electrons: the six in the lone pairs and the two in the single or the double bond. The Oxygen atom also interacts with eight valence electrons according to the Octet Rule, with two from its lone pair, four from its double bond with one Fluorine atom, and two from its single bond with the other Fluorine atom.
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The compound represented by the generic Lewis structure is [tex]\(\text{SO}_2\).[/tex]
To determine the compound, we need to analyze the information given about the Lewis structure:
1. The central atom Y has one lone pair of electrons.
2. The Y atom is double bonded to an X atom, which has two lone pairs of electrons.
3. The Y atom is also single bonded to an X atom, which has three lone pairs of electrons.
Let's start by identifying the possible elements for X and Y based on the valence electrons and the types of bonds and lone pairs described:
- The Y atom has one lone pair and forms a double bond and a single bond with X atoms. This means Y must have at least 4 valence electrons (2 in the lone pair + 1 in each of the two bonds). Y also forms a total of 3 bonds (1 double bond and 1 single bond), which means it can have a maximum of 6 valence electrons (3 bonds x 2 electrons per bond). Therefore, Y must be an element from Group 16 (chalcogens), which typically have 6 valence electrons. The only chalcogen that can form a double bond with an X atom and still have a lone pair is sulfur (S), as oxygen (O) would not have a lone pair if it formed two double bonds, and the heavier chalcogens (Se, Te) are less likely to form double bonds.
- The X atom that is double bonded to Y has two lone pairs, which means it has at least 4 valence electrons (2 in the double bond + 2 in the lone pairs). This X atom cannot have more than 6 valence electrons because it is not forming more than 2 bonds. Therefore, X must be an element from Group 16 (chalcogens) as well. Given that Y is sulfur, the only chalcogen that can form a double bond with sulfur and have two lone pairs is oxygen (O).
The X atom that is single bonded to Y has three lone pairs, which means it has at least 5 valence electrons (1 in the single bond + 2 in the lone pairs). This X atom cannot have more than 7 valence electrons because it is not forming more than 1 bond. Therefore, X must be an element from Group 17 (halogens), which typically have 7 valence electrons. The most common element from this group that forms a single bond with sulfur and has three lone pairs is chlorine (Cl).
However, since we are looking for a compound where both X atoms are the same element, and considering the generic Lewis structure, we can conclude that both X atoms must be oxygen (O), as oxygen can form both single and double bonds with sulfur.
Putting it all together, the central Y atom is sulfur (S), and the outer X atoms are both oxygen (O), leading to the compound[tex]\(\text{SO}_2\),[/tex]which is sulfur dioxide. The Lewis structure of [tex]\(\text{SO}_2\)[/tex] shows sulfur double bonded to one oxygen atom and single bonded to another oxygen atom, with the appropriate number of lone pairs on each atom as described in the question.