The nucleophilic addition reaction depicted below involves a prochiral ketone carbon atom reacting with a nucleophilic hydride ion source (LiAlH4 or NaBH4) and, subsequently, a proton source (e.g., H2O or dilute aq. HCl). Consequently, the reaction produces a racemic mixture of an alcohol. Finish drawing the structures of the products resulting from nucleophilic attack upon the front and back faces of the carbonyl group, being careful to specify the stereochemistry via wedge-and-dash bonds.
Answers
Answer:
we are given the 3-methyl2 butanone and upon the reduction with LiAIH4 there is formed alcohol, there are two possible side attack,
from the back sidefrom the front side.therefore, whenever the front side attacks then the -CH3 moves back the plain and the H will be above the plane. More so, when attack from the back side the H moves below the plane and -CH3 moves above the plane. -OH is evident in the plane. see the attachment below to view the structure.
Explanation:
Related Questions
For the reaction ? NO + ? O2 → ? NO2 , what is the maximum amount of NO2 which could be formed from 16.42 mol of NO and 14.47 mol of O2? Answer in units of g. 003 1.0 points For the reaction ? C6H6 + ? O2 → ? CO2 + ? H2O 37.3 grams of C6H6 are allowed to react with 126.1 grams of O2. How much CO2 will be produced by this reaction? Answer in units of gram
Answers
Final answer:
The maximum possible mass of NO2 that could be formed from the given amounts of reactants is 755.20 grams, and the amount of CO2 that would be produced from the combustion of 37.3 grams of C6H6 is 126.10 grams.
Explanation:
Calculating the Maximum Amount of NO2 Formed
For the balanced reaction 2 NO + O2 → 2 NO2, the stoichiometry shows a 1:1 ratio between NO and NO2. With 16.42 mol of NO, if O2 is in excess, 16.42 mol of NO2 could theoretically be formed. The molar mass of NO2 is 46.0055 g/mol, so the maximum mass of NO2 would be 16.42 mol × 46.0055 g/mol = 755.20 grams of NO2.
Amount of CO2 Produced from C6H6 Combustion
The balanced reaction for the combustion of benzene (C6H6) is 2 C6H6 + 15 O2 → 12 CO2 + 6 H2O. One mole of C6H6 (78.1134 g/mol) produces 6 moles of CO2. With 37.3 grams of C6H6, there would be 37.3 g / 78.1134 g/mol = 0.4775 mol C6H6 which would yield 0.4775 mol C6H6 × 6 mol CO2/mol C6H6 = 2.865 mol CO2.
The molar mass of CO2 is 44.0095 g/mol, so the mass of CO2 would be 2.865 mol × 44.0095 g/mol = 126.10 grams of CO2.
Final answer:
The maximum amount of NO2 that can be formed from 16.42 mol of NO and 14.47 mol of O2 is 755.1 grams. When reacting 37.3 grams of C6H6 with 126.1 grams of O2, the amount of CO2 produced is 125.96 grams.
Explanation:
Reaction Stoichiometry and Limiting Reactants:
To determine the maximum amount of NO2 that can be formed from 16.42 mol of NO and 14.47 mol of O2, we use the balanced chemical equation 2 NO + O2 → 2 NO2. This equation shows that 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2. We can see that NO will be the limiting reactant because 14.47 mol of O2 would theoretically react with only 28.94 mol of NO, which is more than the 16.42 mol of NO present.
Since NO is the limiting reactant, we can form a maximum of 16.42 mol of NO2. Using the molar mass of NO2 (46.01 g/mol), this is equivalent to 755.1 grams of NO2. For the reaction C6H6 + O2 → CO2 + H2O, first, we would calculate the moles of C6H6 and O2, then use stoichiometry to determine the moles of CO2 produced. The molar mass of C6H6 (78.11 g/mol) means we start with 0.477 moles of C6H6.
From the balanced equation, we can see that for every 2 moles of NO, we need 1 mole of O2 to produce 2 moles of NO2. This gives us the mole ratio of NO to O2 as 2:1.
To find the limiting reagent (the reactant that is completely consumed and determines the maximum amount of product formed), we compare the number of moles of each reactant to the mole ratio.
For NO:
16.42 mol NO * (1 mol O2 / 2 mol NO) = 8.21 mol O2
For O2:
14.47 mol O2 * (2 mol NO / 1 mol O2) = 28.94 mol NO
Since the calculated amount of O2 (28.94 mol) is greater than the actual amount of O2 (14.47 mol), O2 is in excess and NO is the limiting reagent.
Now, we need to calculate the maximum amount of NO2 produced from the limiting reagent, which is NO.
From the balanced equation, we know that 2 moles of NO react to form 2 moles of NO2. Therefore, the mole ratio of NO to NO2 is 2:2.
Using the moles of NO (8.21 mol), we can calculate the moles of NO2:
8.21 mol NO * (2 mol NO2 / 2 mol NO) = 8.21 mol NO2
To convert moles of NO2 to grams, we need to use the molar mass of NO2, which is 46.01 g/mol.
8.21 mol NO2 * 46.01 g/mol = 377.27 g NO2
Therefore, the maximum amount of NO2 that could be formed from 16.42 mol of NO and 14.47 mol of O2 is 377.27 grams of NO2.
For the second question, we have the balanced chemical equation:
C6H6 + 15O2 → 6CO2 + 3H2O
From the balanced equation, we can see that for every mole of C6H6, we need 15 moles of O2 to produce 6 moles of CO2.
To determine the amount of CO2 produced, we first need to find the limiting reagent.
For C6H6:
37.3 g C6H6 * (1 mol C6H6 / 78.11 g C6H6) = 0.477 mol C6H6
For O2:
126.1 g O2 * (1 mol O2 / 32 g O2) = 3.94 mol O2
Since the calculated amount of C6H6 (0.477 mol) is less than the actual amount of O2 (3.94 mol), C6H6 is the limiting reagent.
Using the mole ratio from the balanced equation, we find that 0.477 mol of C6H6 will produce 6 moles of CO2.
To convert moles of CO2 to grams, we need to use the molar mass of CO2, which is 44.01 g/mol.
0.477 mol CO2 * 44.01 g/mol = 21.0 g CO2
Therefore, 37.3 grams of C6H6 reacted with 126.1 grams of O2 will produce 21.0 grams of CO2.
Enter your answer in the provided box. Calculate the pH of 1.00 L of a buffer that is 1.00 M in acetic acid and 1.00 M in sodium acetate after the addition of 0.450 mole of NaOH.
Answers
Answer : The pH of buffer is, 5.17
Explanation : Given,
[tex]pK_a=4.75[/tex]
Concentration of acetic acid = 1.00 M
Concentration of sodium acetate = 1.00 M
Volume of solution = 1.00 L
As, [tex]Moles=Concentration\times Volume[/tex]
So,
Moles of acetic acid = 1.00 mol
Moles of sodium acetate = 1.00 mol
Moles of NaOH added = 0.450 mol
The balanced chemical equilibrium reaction is:
[tex]CH_3COO+NaOH\rightleftharpoons CH_3COONa+H_2O[/tex]
Initial mole 1 0.450 1
At eqm. (1-0.450) 0 (1+0.450)
= 0.55 =1.450
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[CH_3COONa]}{[CH_3COOH]}[/tex]
Now put all the given values in this expression, we get:
[tex]pH=4.75+\log (\frac{1.450}{0.55})[/tex]
[tex]pH=5.17[/tex]
Therefore, the pH of buffer is, 5.17
The pressure of a balloon made of a stretchy material is held constant at
2 atm. If the initial volume is 250 mL at room temperature (25 ˚C), what
would be the final volume at 50 ˚C?
Answers
Answer:
New volume is 271 mL
Explanation:
To determine the volume for a gas, when the pressure remains constant we follow this ratio:
V₁ / T₁ = V₂ / T₂
Remember the Ideal Gases Law → P . V = n . R . T
That's why we propose V / T
We need to determine the Absolute T°
25°C + 273 = 298 K
50°C + 273 = 323 K
We convert the volume from mL to L → 250 mL . 1L / 1000 mL = 0.250L
Now we replace: 0.250L / 298K = V₂ / 323K
V₂ = (0.250L / 298K) . 323K → 0.271 L
In conclussion volume of a gas will be increased, while the temperature is also increased and the pressure remains constant.
Answer:
The final volume is 271 mL
Explanation:
Step 1: Data given
The initial pressure = 2 atm
The pressure will be kept constant
The initial volume = 250 mL = 0.250 L
The initial temperature = 25°C = 298 K
The final temperature = 50 °C = 323 K
Step 2: Calculate the final volume
V1/T1 = V2/T2
⇒with V1 = the initial volume = 0.250 L
⇒with T1 = the initial temperature = 25 °C = 298 K
⇒with V2 = the final volume = TO BE DETERMINED
⇒with T2 = the final temperature = 50 °C = 323 K
0.250 L / 298 K = V2 / 323 K
V2 = (0.250 /298) *323
V2 = 0.271 L = 271 mL
The final volume is 271 mL
What is the partial pressure of carbon dioxide in a container that contains 3.63 mol of oxygen, 1.49 mol of nitrogen, and 4.49 mol of carbon dioxide when the total pressure is 871 mmHg?
Answers
Answer:
Partial pressure of CO₂ is 406.9 mmHg
Explanation:
To solve the question we should apply the concept of the mole fraction.
Mole fraction = Moles of gas / Total moles
We have the total moles of the mixture, if we have the moles for each gas inside. (3.63 moles of O₂, 1.49 moles of N₂ and 4.49 moles of CO₂)
Total moles = 3.63 mol O₂ + 1.49 mol N₂ + 4.49 mol CO₂ = 9.61 moles
To determiine the partial pressure of CO₂ we apply
Mole fraction of CO₂ → mol of CO₂ / Total moles = P. pressure CO₂ / Total P
Partial pressure of CO₂ = (mol of CO₂ / Total moles) . Total pressure
We replace values: (4.49 moles / 9.61 moles) . 871 mmHg = 406.9 mmHg
Answer:
Partial pressure O2 = 329 mmHg
Partial pressure N2 = 135 mmHg
Partial pressure CO2 = 407 mmHg
Explanation:
Step 1: Data given
Number of moles oxygen (O2) = 3.63 moles
Number of moles nitrogen (N2) = 1.49 moles
Number of moles carbon dioxide (CO2) = 4.49 moles
Total pressure = 871 mmHg
Step 2: Calculate total number of moles
Total moles = moles O2 + moles N2 + moles CO2
Total moles = 3.63 + 1.49 + 4.49
Total moles = 9.61 moles
Step 3: Calculate the mol ratio
Mol ratio number of moles compound / total moles
Mol ratio O2 = 3.63 moles / 9.61 moles
Mol ratio O2 = 0.378
Mol ratio N2 = 1.49 moles / 9.61 moles
Mol ratio N2 = 0.155
Mol ratio CO2 = 4.49 moles / 9.61 moles
Mol ratio CO2 = 0.467
Step 4: Calculate partial pressure
Partial pressure = mol ratio * total pressure
Partial pressure O2 = 0.378 * 871 mmHg
Partial pressure O2 = 329 mmHg
Partial pressure N2 = 0.155 * 871mmHg
Partial pressure N2 = 135 mmHg
Partial pressure CO2 = 0.467 * 871 mmHg
Partial pressure CO2 = 407 mmHg
What are two causes of soil loss?
Answers
Answer: The agents of soil erosion are the same as of other types of erosion for example water, ice, wind, and gravity. Soil erosion is more likely where the ground has been disturbed by agriculture, grazing animals, logging, mining, construction, and recreational activities.Basically what I mean is some causes of solid loss is mining, construction
An unknown liquid has a pH lower than 7, conducts electricity poorly, and tastes sour, what kind of solution is the unknown?
Answers
Answer:
Weak Acid solution
Explanation:
Acidic substances usually have a pH of 0.1-6.9 or it is usually less than 7. Acidic substances too have a very unique sour taste.
However it is understood that it conducts electricity poorly which means the acid doesn’t readily dissociate in water. This makes it a weak acid and an example is Acetic acid.
what are the three laws of motion
Answers
Answer:
Newton's three laws of motion may be stated as follows:
Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it.
Force equals mass times acceleration
For every action there is an equal and opposite reaction
Explanation:
pls mark brainliest
Cycloalkanes are (saturated/unsaturated) compounds.
(which one)
Answers
Answer:
Saturated .
Explanation:
The expected first intermediate formed during a halohydrin reaction is:
a. a cyclic oxonium ion.
b. the most stable carbocation with OH on the adjacent carbon.
c. ahalonium ion.
d. the most stable carbanion.
e. the most stable carbocation with X on the adjacent carbon.
Answers
Final answer:
The first intermediate formed during a halohydrin reaction is a halonium ion. This intermediate is key in the process of adding a halogen and an OH group to adjacent carbons in the formation of halohydrins.
Explanation:
The expected first intermediate formed during a halohydrin reaction is c. ahalonium ion. In the presence of a halogen and water, the reaction of an alkene leads to the formation of a halonium ion intermediate. Specifically, the alkene undergoes an addition reaction with a halogen such as bromine or chlorine to form a three-membered ring halonium intermediate. This is followed by the nucleophilic attack by water, which opens the ring and leads to the formation of the halohydrin, with the halogen and an OH group on two adjacent carbons. Halohydrins such as bromohydrins or chlorohydrins are important intermediates in organic synthesis.
2. Incoming wastewater, with BOD5 equal to 200 mg/L, is treated in a well-run secondary treatment plant that removes 90 percent of the BOD. You are to run a five-day BOD test with a standard 300-mL bottle, using a mixture of treated sewage and dilution water (no seed). Assume the initial DO is 9.2 mg/L. a. Roughly what maximum volume of treated wastewater should you put in the bottle if you want to have at least 2.0 mg/L of DO at the end of the test (filling the rest of the bottle with water). (answer in mL)
Answers
Answer:
10.8 ml
Explanation:
The BOD is an empirical test to determine the molecular oxygen used during a specified incubation period (usually five days), for the biochemical degradation of organic matter (carbonaceous demand) and the oxygen used to oxidise inorganic matter.
See attached file
The maximum volume of treated wastewater that will be in the bottle is 10.8 mL.
The given parameters;
wastewater density = 200 mg/Lstandard volume = 300 mLinitial DO = 9.2 mg/LThe dilution factor (P) is calculated as follows;
[tex]200 \ mg/L= \frac{9.2 \ mg/L \ - \ 2\ mg/L}{P} \\\\P = \frac{7.2 \ mg/L}{200 \ mg/L} \\\\P = 0.036[/tex]
The maximum volume of treated wastewater that will be in the bottle to have at least 2.0 mg/L DO;
[tex]0.036 = \frac{V_w}{300 \ mL} \\\\V_w = 0.036 \times 300 \ mL\\\\V_w = 10.8 \ mL[/tex]
Thus, the maximum volume of treated wastewater that will be in the bottle is 10.8 mL.
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A mixture of He
, N2
, and Ar
has a pressure of 13.6
atm at 28.0
°C. If the partial pressure of He
is 1831
torr and that of Ar
is 997
mm Hg, what is the partial pressure of N2
?
Answers
Answer : The partial pressure of nitrogen gas in the mixture is, 9.88 atm
Explanation :
According to the Dalton's Law, the total pressure of the gas is equal to the sum of the partial pressure of individual gases.
Formula used :
[tex]p_T=p_{He}+p_{Ar}+p_{N_2}[/tex]
where,
[tex]p_T[/tex] = total pressure of gas = 13.6 atm
[tex]p_{He}[/tex] = partial pressure of helium gas = 1831 torr = 2.41 atm
[tex]p_{Ar}[/tex] = partial pressure of argon gas = 997 torr = 1.31 atm
Conversion used: (1 atm = 760 torr)
[tex]p_{N_2}[/tex] = partial pressure of nitrogen gas = ?
Now put all the given values in the above formula, we get:
[tex]13.6=2.41+1.31+p_{N_2}[/tex]
[tex]p_{N_2}=9.88atm[/tex]
Thus, the partial pressure of nitrogen gas in the mixture is, 9.88 atm
The Ph scale is logarithmic; how many times stronger is a Ph of 4 versus a Ph of 2?
Answers
Answer:
A pH greater than 7 is basic. The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. For example, pH 4 is ten times more acidic than pH 5 and 100 times (10 times 10) more acidic than pH 6.
write a balanced equation for the reaction between hydrogen peroxide H202 and Fe2+ to produce Fe3+ and H2O in acidic solution
Answers
The balanced chemical equation for the reaction between hydrogen peroxide (H2O2) and Fe2+ in acidic solution to produce Fe3+ and H2O is: 2 H2O2 + 2 Fe2+ + 2 H+ → 2 Fe3+ + 2 H2O
In this reaction, hydrogen peroxide (H2O2) acts as an oxidizing agent, while iron(II) ions (Fe2+) are being oxidized to iron(III) ions (Fe3+). The hydrogen peroxide molecules donate oxygen atoms to the iron(II) ions, causing them to undergo oxidation. The reaction takes place in an acidic solution, which provides the necessary protons (H+) to balance the reaction and ensure the overall charge neutrality.
Balancing the equation is crucial to ensure that the same number of atoms of each element are present on both sides of the reaction arrow, preserving the law of conservation of mass. In this balanced equation, there are two hydrogen (H) atoms, four oxygen (O) atoms, and two iron (Fe) atoms on both sides, demonstrating that mass is conserved in the chemical reaction.
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Final answer:
The balanced chemical equation in an acidic solution for the oxidation of Fe2+ with hydrogen peroxide to form Fe3+ and water is 2 Fe2+ (aq) + H2O2 (aq) + 2 H+ (aq) → 2 Fe3+ (aq) + 2 H2O (l).
Explanation:
The balanced equation for the reaction between hydrogen peroxide (H2O2) and Fe2+ to produce Fe3+ and H2O in an acidic solution is given by:
2 Fe2+ (aq) + H2O2 (aq) + 2 H+ (aq) → 2 Fe3+ (aq) + 2 H2O (l)
In this reaction, Fe2+ is oxidized to Fe3+, and H2O2 (hydrogen peroxide) acts as the oxidizing agent, being reduced to water (H2O). The presence of H+ indicates that the reaction occurs in an acidic solution. To balance the equation, take into account the transfer of electrons, the conservation of mass, and the charge balance on both sides of the equation.
II. Practice An ideal gas occupies 5 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to 3 atm (point B). Then it is allowed to expand isothermally to 1 atm (point C) and at last compressed isobarically to its original state. A. How many moles of gas are being used? B. Find the temperature at point C. C. Find the work done on the gas in each process. D. Find the amount of heat added to/removed from the gas in one cycle.
Answers
Answer:
The process can be represented as shown in the figure below; having got the diagram, we can solve for the questions.
A. the number of moles of gas used
n = PV/ RT = (1.013 *10^5 Pa) * (5.0 *10^-3 m^3) / (8.314 * 300)
n = 5.065 * 10^2 / 2494.2
n = 0.00203 *10^2
n = 0.203 moles
B. Temperature at point C (Tc)
Pa/Ta= Pb/Tb
Tb = Pb *Ta / Pa
Tb = 3 * 300 / 1
Tb = 900 K
Since Tb = Tc = 900 K
C. For process AB,
work done is zero
For process BC,
work done = -nRTbln (Vc/Vb)
W = -(0.203 * 8.314 * 900 ln (3)
W = -(1.518 kJ ln 3
W = -1.67 kJ
For process CA,
W = -P V =-nRT
W = -(0.203 * 8.314 * (-600))
W = 1.01 kJ
Explanation:
A chemist dissolves 751.mg of pure nitric acid in enough water to make up 290.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.
Answers
Answer:
1.4
Explanation:
Mass of pure nitric acid = 751mg
Volume of solution = 290mL
Unknown:
pH of the solution = ?
Solution:
To solve this problem, we need the concentration of the acid in the aqueous form.
This is given by molarity;
Molarity = [tex]\frac{number of moles }{volume}[/tex]
Since the number of moles of nitric acid is unknown, we can easily solve for it.
Number of moles of nitric acid = [tex]\frac{mass}{molar mass}[/tex]
molar mass of HNO₃ = 1 + 14 + 3(16) = 63g/mol
mass of nitric acid = 751mg = 0.751g
Number of moles = [tex]\frac{0.751}{63}[/tex] = 0.012mole
Volume of solution = 290mL = 0.29dm³
Now molarity of the solution = [tex]\frac{0.012}{0.29}[/tex] = 0.041moldm⁻³
Since:
pH = -log [H₃O⁺]
HNO₃ + H₂O → H₃O⁺ + NO₃⁻
1moldm⁻³ 1moldm⁻³ 1moldm⁻³
0.041moldm⁻³ 0.041moldm⁻³ 0.041moldm⁻³
pH = -log[0.041] = 1.4
Write a balanced half-reaction for the oxidation of gaseous arsine AsH3 to aqueous arsenic acid H3AsO4 in acidic aqueous solution. Be sure to add physical state symbols where appropriate.
Answers
Answer:
AsH3(g) + 4H2O(l)------> H3AsO4(aq) +8H^+(aq) + 8e-
Explanation:
Now we need to work through the problem in steps.
Step1: reaction of arsine with water
AsH3(g) + 4H2O(l) -----> H3AsO4(aq) this is the molecular reaction
Step II: oxygen is balanced using hydrogen ions
AsH3(g) + 4H2O(l)------> H3AsO4(aq) +8H^+(aq)
Step III: We specify the number of electrons transferred in the redox reaction
AsH3(g) + 4H2O(l)------> H3AsO4(aq) +8H^+(aq) + 8e-
The balanced half-reaction for the oxidation of gaseous arsine AsH₃ to
aqueous arsenic acid H₃AsO₄ in acidic aqueous solution is
AsH₃(g) + 4H₂O(l)------> H₃AsO₄(aq) +8H+(aq) + 8e-
The first step involves reaction of arsine with waterAsH₃(g) + 4H₂O(l) -----> H₃AsO₄(aq)
The second step involves balancing oxygen using hydrogen ionsAsH₃(g) + 4H₂O(l)------> H₃AsO₄(aq) +8H⁺(aq)
The third and final step involves noting the the number of electrons transferred in the redox reactionAsH₃(g) + 4H₂O(l)------> H₃AsO4(aq) +8H⁺(aq) + 8e⁻
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A sample of an unknown gas effuses in 12.5 min. An equal volume of H2 in the same apparatus under the same conditions effuses in 2.42 min. What is the molar mass of the unknown gas
Answers
Answer:
53.4 gMol-1
Explanation:
Let the mass of the unknown gas be M
Let the molar mass of hydrogen gas be 2×1=2gMol-1
Time for diffusion of unknown gas = 12.5 min
Time for diffusion of hydrogen= 2.42 min
From Graham's law:
t1/t2=√M1/M2
Hence:
2.42/12.5= √2/M
Hence M= 53.4 gMol-1
Answer:
The molar mass of the unknown gas is 40.06 g/mol
Explanation:
Graham's law of effusion states that the rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass.
[tex]\frac{R_{b} }{R_{a} } = \sqrt{\frac{M_{a} }{M_{b} } }[/tex] [tex]= \frac{t_{a} }{t_{b} }[/tex] where R = rate of effusion, M = molar mass and t= time of effusion
⇒ [tex]\sqrt{\frac{2 g/mol}{x g/mol} } = \frac{162secs}{725secs}[/tex]
x g/mol = [tex]\frac{2}{0.223448^{2} }[/tex]
= 40.06 g/mol
Which of the following substances is never a Brønsted-Lowry base in an aqueous solution?Group of answer choicespotassium hydroxide, KOHsodium hydrogen phosphate, Na2HPO4sodium phosphate, Na3PO4ammonium chloride, NH4Clsodium bicarbonate, NaHCO3
Answers
Among the following substances, NH₄Cl is never a Brønsted-Lowry base in an aqueous solution because it is a salt.
What is Brønsted-Lowry base?There are different types of theories of acids and bases like Arrhenius theory, Bronsted Lowry theory and Lewis acids and bases theory. According to Bronsted Lowry theory, acids are the substances which release H+ ions in the solution and bases are the substances which accepts H+ ions in the solution.
This also introduce the concept of conjugate acid base pair. A salt can never be an acid or a base because it is made up of both of them.
Given options are sodium hydrogen phosphate, sodium phosphate, ammonium chloride, sodium bicarbonate and Potassium Hydroxide.
Therefore, Among the following substances, NH₄Cl is never a Brønsted-Lowry base in an aqueous solution because it is a salt.
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The Brønsted-Lowry theory, ammonium chloride (NH₄Cl) is never a base in an aqueous solution. It dissociates into NH₄+ and Cl-, where NH₄+ acts as a Brønsted-Lowry acid. This makes NH₄Cl the correct answer.
The Brønsted-Lowry theory defines a base as a substance that can accept a proton (H+) during a chemical reaction. In an aqueous solution, the behavior of different substances can be predicted based on this definition.
Potassium hydroxide (KOH): This is a strong base as it dissociates completely in water to give OH- ions.Sodium hydrogen phosphate (Na₂HPO₄): This compound can act as a base because the HPO₄²- ion can accept a proton.Sodium phosphate (Na₃PO₄): This compound can also behave as a base, with the PO₄³- ion capable of accepting protons.Ammonium chloride (NH₄Cl): This substance does not act as a Brønsted-Lowry base. In aqueous solution, it dissociates into NH₄+ and Cl-, where NH4+ is an acid (the conjugate acid of the weak base NH₃).Sodium bicarbonate (NaHCO₃): This compound can act as a base because the HCO₃- ion can accept a proton.Based on the Brønsted-Lowry definition, ammonium chloride (NH₄Cl) is never a base in an aqueous solution.
Element X reacts with hydrogen gas at 200°C to form compound Y. When Y is heated to a higher temperature, it decomposes to the element X and hydrogen gas in the ratio of 559 mL of H2 (measured at standard temperature and pressure) for 1.00 g of X reacted. X also combines with chlorine to form a compound Z, which contains 63.89 percent by mass of chlorine. Deduce the identity of X. The symbol of element X is
Answers
Answer:
X is calcium with the symbol Ca
Explanation:
From the first statement:
X reacted with H2 to produce Z i.e
X + H2 —> Y
Y is heated to form X and H2 i.e
Y —> X + H2
Let us obtain moles of H2 produced. This is illustrated below:
Volume of H2 produced = 559 mL = 0.559 L.
I mole of H2 occupy 22.4L at stp
Therefore, b mol of H2 will occupy 0.559 L at stp i.e
b mol of H2 = 0.559/22.4
b mol of H2 = 0.025 mole
Next 0.025 mole of H2 to gram..
Molar Mass of H2 = 2x1 = 2g/mol
Mole of H2 = 0.025 mole
Mass = number of mole x molar Mass
Mass of H2 = 0.025 x 2
Mass of H2 = 0.05g
From the question given, we were told that 1g of X reacted. This means that 1g of X reacted with 0.05g of H2. Now let us determine the mass of X that will react with 1 mol ( i.e 2g) of H2. This is illustrated below:
From the reaction,
It was discovered that 1g of X reacted with 0.05g of H2.
Therefore, P g of X will react with 2g of H2 i.e
P g of X = 2/0.05 = 40g/mol
The molar mass of X is 40g/mol.
Now let us consider the second statement to see if we'll obtained the same result as 40g/mol of X
From the second statement:
X also combines with chlorine to form a compound Z, which contains 63.89 percent by mass of chlorine i.e
X + Cl2 —> Z
Molar Mass of Z (X + 2Cl) = X + (35.5 x 2) = X + 71.
Z contains 63.89% by mass of Cl2.
We can obtain the molar mass of X as follow:
Percentage by mass of Cl2 in the compound Z is given by:
Mass of Cl2/Molar Mass x100
63.89/100 = 71/(X + 71)
Cross multiply to express in linear form
63.89(X + 71) = 100 x 71
Clear the bracket
63.89X + 4536.19 = 7100
Collect like terms
63.89X = 7100 - 4536.19
63.89X = 2563.81
Divide both side by 63.89
X = 2563.81/63.89
X = 40g/mol
Now we can see that in both experiments, the molar mass of X is 40g/mol.
Comparing the value of X i.e 40g/mol with that from the periodic table, X is calcium with the symbol Ca
A student measures the S2- concentration in a saturated aqueous solution of iron(II) sulfide to be 2.29×10-9 M. Based on her data, the solubility product constant for iron(II) sulfide is
Answers
Answer:
Ksp FeS = 5.2441 E-18
Explanation:
FeS ↔ Fe2+ + S2-S S S
∴ Ksp = [Fe2+]*[S2-].....solubility product constant
∴ [S2-] = 2.29 E-9 M = S
⇒ Ksp = (S)(S) = S²
⇒ Ksp = (2.29 E-9)²
⇒ Ksp = 5.2441 E-18
A simple equation relates the standard free‑energy change, ΔG∘′, to the change in reduction potential. ΔE0′. ΔG∘′ = −nFΔE0′ The n represents the number of transferred electrons, and F is the Faraday constant with a value of 96.48 kJ⋅mol^(−1)⋅V^(−1). Use the standard reduction potentials provided to determine the standard free energy released by reducing O2 with FADH2. FADH2 + 1/2O2 → FAD + H2O
given that the standard reduction potential for the reduction of oxygen to water is +0.82 V and for the reduction of FAD to FADH2 is +0.03 V.
Answers
Answer : The value of standard free energy is, -152.4 kJ/mol
Explanation :
The given balanced cell reaction is:
[tex]FADH_2+\frac{1}{2}O_2\rightarrow FAD+H_2O[/tex]
The half reaction will be:
Reaction at anode (oxidation) : [tex]FADH_2\rightarrow FAD+2H^++2e^-[/tex] [tex]E^0_{Anode}=+0.03V[/tex]
Reaction at cathode (reduction) : [tex]\frac{1}{2}O_2+2H^++2e^-\rightarrow H_2O[/tex] [tex]E^0_{Cathode}=+0.82V[/tex]
First we have to calculate the standard electrode potential of the cell.
[tex]E^o=E^o_{cathode}-E^o_{anode}[/tex]
[tex]E^o=(+0.82V)-(+0.03V)=+0.79V[/tex]
Relationship between standard Gibbs free energy and standard electrode potential follows:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
where,
[tex]\Delta G^o[/tex] = standard free energy = ?
n = number of electrons transferred = 2
F = Faraday constant = [tex]96.48kJ.mol^{-1}V^{-1}[/tex]
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 0.79 V
Now put all the given values in the above formula, we get:
[tex]\Delta G^o=-(2)\times (96.48kJ.mol^{-1}V^{-1})\times (0.79V)[/tex]
[tex]\Delta G^o=-152.4kJ/mol[/tex]
Therefore, the value of standard free energy is, -152.4 kJ/mol
To calculate the standard free energy change for the reduction of O2 with FADH2, we first calculate the difference in reduction potentials to find ΔE0′. We then substitute the values into the relation ΔG∘′ = −nFΔE0′, with n=2 and Faraday constant F=96.48 kJ⋅mol‾¹⋅V‾¹. The calculated ΔG∘′ is -152.4384 kJmol‾¹.
Explanation:The standard free energy change, ΔG∘′, is related to the change in reduction potential, ΔE0′, by the equation ΔG∘′ = −nFΔE0′. To find the standard free energy released by the reduction of O2 with FADH2, we first need to find ΔE0′.
ΔE0′ is given by the difference in reduction potentials of the two half reactions involved. In this case, the reduction of O2 to H2O (+0.82 V) and the reduction of FAD to FADH2 (+0.03 V). Therefore, ΔE0′ = E(O2/H2O) - E(FAD/FADH2) = +0.82 V - (+0.03 V) = +0.79 V.
Inserting the values into the equation, and knowing that the number of transferred electrons (n) is 2 and the Faraday constant (F) is 96.48 kJ⋅mol‾¹⋅V‾¹, we get ΔG∘′ = −2 * 96.48 kJ⋅mol‾¹⋅V‾¹ * (+0.79 V) = -152.4384 kJmol‾¹. Thus, the standard free energy released by reducing O2 with FADH2 is -152.4384 kJmol‾¹.
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How many moles of water are present in 15.00 ml of water
that has a density of 0.9956 g/ ml.
MW for Hydrogen 1.00 g/ 1 mole; MW for Oxygen 16.00 g/ 1 mole
Answers
Answer : The moles of water present in solution are, 0.8297 moles.
Explanation :
First we have to calculate the mass of water.
[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}[/tex]
[tex]\text{Mass of water}=0.9956g/mL\times 15.00mL=14.934g[/tex]
Now we have to calculate the moles of water.
[tex]\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}[/tex]
Molar mass of water = (2 × Molecular weight of hydrogen) + Molecular weight of oxygen
Molar mass of water = (2 × 1.00g/mol) + 16.00 g/mol
Molar mass of water = 18.00 g/mol
[tex]\text{Moles of water}=\frac{14.934g}{18.00g/mol}[/tex]
[tex]\text{Moles of water}=0.8297mol[/tex]
Therefore, the moles of water present in solution are, 0.8297 moles.
What is the pH of an aqueous solution at 25.0°C in which [H+] is 0.0025 M?
A) 5.99 B) 2.60 C) -2.60 D) -5.99 E) none of the above
Answers
The pH of an aqueous solution at 25.0°C with [H+] = 0.0025 M is calculated using the pH formula and results in a pH of 2.60, which is answer option B.
Explanation:The pH of an aqueous solution at 25.0°C with a hydrogen ion concentration [H+] of 0.0025 M can be found using the pH formula:
pH = -log [H3O+]
By substituting the given concentration into the formula, the calculation would be:
pH = -log(0.0025) = -log(2.5 x 10-3)
Using a scientific calculator:
pH = 2.60
Therefore, the correct answer is B) 2.60.
Problem PageQuestion The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide () into solid sodium and gaseous dinitrogen. 2. Suppose of dinitrogen gas are produced by this reaction, at a temperature of and pressure of exactly . Calculate the mass of sodium azide that must have reacted. Round your answer to significant digits.
Answers
Answer:
1. 2NaN₃(s) → 2Na(s) + 3N₂(g)
2. 14.5 g NaN₃
Explanation:
The answer is incomplete, as it is missing the required values to solve the problem. An internet search shows me these values for this question. Keep in mind that if your values are different your result will be different as well, but the solving methodology won't change.
" The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen. 2. Suppose 71.0 L of dinitrogen gas are produced by this reaction, at a temperature of 16.0 °C and pressure of exactly 1 atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits. "
1. The reaction that takes place is:
2NaN₃(s) → 2Na(s) + 3N₂(g)2. We use PV=nRT to calculate the moles of N₂ that were produced.
P = 1 atm
V = 71.0 L
n = ?
T = 16.0 °C ⇒ 16.0 + 273.16 = 289.16 K
1 atm * 71.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 289.16 Kn = 0.334 molNow we convert N₂ moles to NaN₃ moles:
0.334 mol N₂ * [tex]\frac{2molNaN_{3}}{3molN_2}[/tex] = 0.223 mol NaN₃Finally we convert NaN₃ moles to grams, using its molar mass:
0.223 mol NaN₃ * 65 g/mol = 14.5 g NaN₃Select all the correct locations on the image. The model shows global atmospheric circulation. Identify the wind directions that are correct.
Answers
Answer:1 2 4
Explanation:
Answer: 1.3.5.6.
Explanation:
At 5 atmospheres of pressure and 70oC, how many moles are present in 1.5 L of O2 gas?
Answers
Answer:
0.27 mol
Explanation:
Given data
Pressure (P): 5 atmTemperature (T): 70°CVolume (V): 1.5 LMoles of O₂ (n): ?Step 1: Convert the temperature to the Kelvin scale
When working with gases, we need to convert all temperatures to the absolute scale, using the following expression.
[tex]K = \° + 273.15\\K = 70 + 273.15 = 343 K[/tex]
Step 2: Calculate the moles of gaseous oxygen
We will apply the ideal gas equation.
[tex]P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{5 atm \times 1.5 L}{\frac{0.0821atm.L}{mol.K} \times 343 K} = 0.27 mol[/tex]
Density is a physical property that relates the mass of a substance to its volume. Calculate the density, in g / mL , of a liquid that has a mass of 0.175 g and a volume of 0.000225 L.
Answers
Answer: Density = 0.77g/ml
Explanation:
From the question,
Mass = 0.175g
Volume = 0.000225L
Convert the volume in Litres to Millilitres by multiplying it by 1000, hence the volume is 0.225ml.
Therefore,
Density = mass / volume
0.175g / 0.225ml = 0.77g/ml
Answer:
Density = Mass/ Volume,
Mass = 0.000175kg
Volume = 0.000225L,
Density = 0.000175/ 0.000225
Density = 0.78kg/L
Explanation:
Density is an important physical property. Density is the mass of a substance per unit volume. Volume is the amount of space an object occupies. Chemical properties- These are properties that can only be observed by changing the identity of the substance.
The presence of intermolecular forces in liquids is observed through several different phenomena. Match each of below statement with its corresponding option.
adhesion, surface tension, capillary action, cohesion, meniscus, viscosity.1. _____ the ability of a liquid to flow up a narrow tube unassisted against gravity.2. _____ the attraction between molecules of the same substance.3. _____ the curvature of the surface of a liquid at the interface with the container.4. _____ the attraction between dissimilar molecules.5. _____ the resistance of a liquid to flow.6. _____ the elasticity of the surface layer of a liquid due to the liquid trying to minimize its surface area.
Options:
a. adhesion,
b. surface tension,
c. capillary action,
d. cohesion,
e. meniscus,
f. viscosity.
Answers
Answer:
1) Capillary action
2) Cohesion
3)Miniscus
4) Adhesion
5) Viscousity
6) Surface tension
Explanation:
Intermolecular forces are forces that holds molecules together in liquid, these is possible by inter-molecular interactions that exist within the liquids.these forces includes forces such as Waals forces and hydrogen bonds.
When there is great inter-molecular forces, there will be high freezing point and boiling point . At a lower inter-molecular forces the boiling point becomes low too,which brings about great fluidity of the liquid. The liquid flow reluctantly where greater force exist in the liquid. Some of those factors used in characterizing which are;
1)adhesion,
2) surface tension,
3)capillary action,
4)cohesion,
5) meniscus,
6)viscosity.
The statement with its corresponding option is as follows:
Capillary action is the ability of a liquid to flow up a narrow tube unassisted against gravity.Capillarity is a phenomenon through which liquids have the ability to rise or fall through a capillary tube.
Cohesion is the attraction between molecules of the same substance.Cohesion is the force of attraction between adjacent particles within the same body.
Meniscus is the curvature of the surface of a liquid at the interface with the container.The meniscus is the up or down curve on the surface of a liquid that occurs in response to the surface of its container.
Adhesion is the attraction between dissimilar molecules.Adhesion is the interaction between the surfaces of different bodies and they are held together by intermolecular forces.
Viscousity is the resistance of a liquid to flow.Viscosity refers to the resistance that some liquids possess during their gradual flow and deformation as a result of shear stresses or tensile stresses.
Surface tension is the elasticity of the surface layer of a liquid due to the liquid trying to minimize its surface area.Surface tension refers to the amount of energy required to increase the surface of a liquid per unit area.
Therefore, we can conclude that fluids have elemental properties that define and differentiate them from other forms of matter, such as viscosity, surface tension, among others.
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Use scientific reasoning to explain what happens if 0.87 mol NaCl is dissolved in pure water creating 3.00 L of solution, what is its molarity of the resulting solution?
Answers
Answer : The molarity of the resulting solution is, 0.29 M
Explanation :
When NaCl dissolved in water then it dissociates to give sodium ions and chloride ions.
Given,
Moles of [tex]NaCl[/tex] = 0.87 mol
Volume of solution = 3.00 L
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
[tex]\text{Molarity}=\frac{\text{Moles of }NaCl}{\text{Volume of solution (in L)}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Molarity}=\frac{0.87mol}{3.00L}=0.29mole/L=0.29M[/tex]
Therefore, the molarity of the resulting solution is, 0.29 M
When toluene is used in free radical bromination, a very small amount of product is formed that contains only carbons and hydrogens and no bromine. Show the structure of that product and the arrow curved mechanism of how it is formed starting from the alkyl radical intermediate of the reaction.(9 pts)
Answers
Answer:
Explanation:
The product formed that contains only carbons and hydrogens after free radical bromination of toluene is 1,2-diphenylethane.....
Please go through the attached file for the diagrams .
Answer:
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Explanation: