provide the structure of the major organic product of the reaction below. if there is more than one product, be sure to indicate stereochemistry if necessary by using the wedge and dash tool appropriately. there is a scheme of a reaction, where a six-carbon ring, with a ch2 group double-bonded to the first carbon, reacts with hbr in the presence of roor, when heated. draw the molecule on the canvas by choosing buttons from the tools (for bonds), atoms, and advanced template toolbars. the single bond is active by default.
Answers
The major organic product of the reaction between the six-carbon ring with a CH2 group double-bonded to the first carbon and HBr in the presence of ROOR when heated is 2-bromo-6-carbon ring with a CH2Br group attached to the first carbon.This type of reaction is known as an electrophilic addition. The stereochemistry is not relevant in this reaction.
The reaction you are describing is the addition of hydrogen bromide (HBr) to a six-carbon ring with a CH2 group double-bonded to the first carbon. This type of reaction is known as an electrophilic addition.
When HBr is added to the double bond of the CH2 group, the ring becomes protonated, forming a carbocation intermediate. The bromide ion then attacks the carbocation, forming a new bond to the first carbon and resulting in the formation of a 7-membered ring with a bromine atom attached to the first carbon.
The major organic product of this reaction is 2-bromocycloheptane. The structure of this compound consists of a seven-membered ring with a bromine atom attached to the first carbon and a methyl group attached to the fourth carbon.
The bromine atom has a wedge and dash notation to indicate its stereochemistry, with the wedge indicating that it is coming out of the plane of the page and the dash indicating that it is going into the plane of the page.
It is important to note that the reaction conditions described in the question, heating in the presence of a radical initiator (ROOR), suggest that this reaction is a free-radical addition rather than an electrophilic addition.
The mechanism of the reaction is still similar, but instead of a carbocation intermediate, a radical intermediate is formed. Nonetheless, the major organic product of the reaction would still be 2-bromocycloheptane.
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Related Questions
why do atoms exchange or share electrons during bonding?(1 point)
Answers
Atoms exchange or share electrons during bonding to obtain stability by completing their valence shells and also to achieve a lower energy state.
Atoms exchange or share electrons during bonding because of the force of attraction between two oppositely charged ions or particles called the electrostatic force. This force of attraction results in the formation of a bond, holding two atoms together within a compound.
The electrons are either shared or exchanged because they determine the chemical reactivity of an atom and are responsible for forming bonds between atoms. Atoms bond with each other to complete their outer shells and obtain stability, which is usually achieved by acquiring eight electrons in their valence shells. This is known as the octet rule.
The main types of chemical bonds that atoms form include covalent bonds and ionic bonds. Covalent bonds involve the sharing of electrons between atoms, while ionic bonds involve the transfer of electrons from one atom to another. Ionic bonding occurs between atoms that have a large difference in electronegativity, whereas covalent bonding occurs between atoms with a small difference in electronegativity.
In conclusion, atoms exchange or share electrons during bonding to obtain stability by completing their valence shells and also to achieve a lower energy state.
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The use of a pattern or shapes repeated in a recurring and regular arrangement is called
Answers
Please i need help with this question.
Please show workings where necessary and proof/ explain your point
Answers
Answer:
C
Explanation:
Please see the attached picture for the explanation.
Le Châtelier's principle:
When a stress is applied to a system at equilibrium, the system will respond by shifting in the direction that opposes/ minimizes the effect of the stress.
• This stress could be the addition or removal of a product or reactant, changes in the concentration of the products or reactants, or a change in temperature.
☆ Why can heat be written on the same side of the products when the reaction is exothermic?
• Exothermic means that heat is released, thus heat can be treated as a product.
• I like to remember it as Exothermic → Exit
Estimate the value of K sp
for silver iodide using the following standard reduction potentials as needed. AgI(s)+e −
→Ag(s)+Γ −(aq)
;E ∘
=−0.1522 V
Ag ∘
(aq)+e −
→Ag(s);E ∘
=0.7996 V
1 2
(a)+2e −
→21 −
(aq);E ∘
=0.5355 V
Answers
The estimated value of the solubility product constant (Ksp) for silver iodide (AgI) is approximately 3.55 x 10^39.
How to estimate the value of the solubility product constant (Ksp) for silver iodide (AgI)?
To estimate the value of the solubility product constant (Ksp) for silver iodide (AgI), we can use the Nernst equation and the given standard reduction potentials. The overall reaction for the dissolution of AgI can be written as follows:
AgI(s) ⇌ Ag+(aq) + I-(aq)
The reduction half-reaction for the formation of Ag(s) from Ag+(aq) is:
Ag+(aq) + e- → Ag(s) (Reduction half-reaction)
The oxidation half-reaction for the formation of I-(aq) from I2(aq) is:
1/2 I2(aq) + e- → I-(aq) (Oxidation half-reaction)
By combining these two half-reactions, we can construct the overall reaction and determine the value of Ksp for AgI.
AgI(s) ⇌ Ag+(aq) + I-(aq)
To find the value of Ksp, we need to calculate the equilibrium constant (K) using the Nernst equation:
K = [Ag+(aq)]/[I-(aq)]
Using the standard reduction potentials given, we can calculate the overall standard cell potential (E°cell) for the reaction:
E°cell = E°(Ag+(aq)/Ag(s)) + E°(I2(aq)/I-(aq))
E°cell = (0.7996 V) + (0.5355 V)
E°cell = 1.3351 V
Next, we can use the relationship between the standard cell potential and the equilibrium constant:
E°cell = (0.0592 V/n) * log(K)
Where n is the number of electrons involved in the overall reaction. In this case, n = 2 since two electrons are involved in the overall reaction.
Substituting the values:
1.3351 V = (0.0592 V/2) * log(K)
Simplifying:
2.6702 = 0.0296 * log(K)
Taking the antilogarithm:
K = antilog(2.6702/0.0296)
K = antilog(90.203)
K ≈ 3.55 x 10^39
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Q23) quickly please
23) The cooling a food with the formation of the first ice crystals called a. Super cooling b. Initial freezing point c. Critical point d. Eutectic temperature
Answers
The process of cooling a food until the first ice crystals form is referred to as supercooling. Supercooling occurs when a substance is cooled below its freezing point without undergoing solidification.
This phenomenon is commonly observed in liquids that are pure and free from impurities. Supercooling allows the formation of ice crystals to be delayed until a nucleation site is present or an external disturbance occurs.
Supercooling refers to the state where a substance remains in a liquid phase below its freezing point. When a food is rapidly cooled, it can reach a temperature below the freezing point without solidifying. This occurs because the process of nucleation, which initiates the formation of ice crystals, is hindered. In the absence of nucleation sites or external disturbances, the food remains in a supercooled state.
However, as soon as a nucleation site is introduced or an external disturbance occurs, the supercooled food rapidly crystallizes and forms ice. This process is commonly observed when touching an already supercooled liquid, causing it to freeze instantly. Supercooling can be utilized in various applications, such as in the production of supercooled drinks or in cryopreservation techniques.
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Sugar could be classified as which of the following?Question 5 options:compoundsolution (homogenous)elementmechanical mixture (heterogenous)
Answers
Compound, according to our last session.
Explain the term precipitation and provide three examples
Answers
Answer:
Precipitation is any product of the condensation of atmospheric water vapor that falls under gravitational pull from cloudsExplanation:
Examples of Precipitation:
Rain
Drizzle
Ice Pellets
Hail
Snow
Ice Crystal
in the area below, write a series of chemical reactions that leads to the formation of tropospheric ozone in photochemical smog
Answers
The formation of tropospheric ozone in photochemical smog involves a series of chemical reactions primarily driven by sunlight.
Here is a simplified representation of the reactions:
Nitrogen Dioxide (NO2) reacts with sunlight (UV radiation) to produce Nitric Oxide (NO) and an Oxygen atom (O).
NO2 + UV radiation → NO + O
Nitric Oxide (NO) reacts with Oxygen molecules (O2) to form Nitrogen Dioxide (NO2) again.
NO + O2 → NO2
The Oxygen atom (O) produced in reaction 1 reacts with an Oxygen molecule (O2) to form Ozone (O3)
.O + O2 → O3Ozone (O3) can react with Nitric Oxide (NO) to form Nitrogen Dioxide (NO2) and Oxygen (O2).
O3 + NO → NO2 + O2
These reactions are part of a complex set of photochemical reactions involving various pollutants such as nitrogen oxides (NOx), volatile organic compounds (VOCs), and sunlight. The primary precursors for tropospheric ozone formation in photochemical smog are Nitrogen Dioxide (NO2) and volatile organic compounds (VOCs). The reactions outlined above represent a simplified version of the process and do not capture the full complexity of atmospheric chemistry involved in the formation of tropospheric ozone.
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CH4 (g) + O2 (g) → CO2 (g) + H2O (g) to. If 28.6 g of CH4 is reacted with 57.6 g of O2, calculate the number of grams of CO2 produced. b. If you actually get 32.1 g of CO2, calculate the percent yield. c. Calculate the number of moles of excess reagent remaining at the end of the reaction.
Answers
Answer:
See explanation
Explanation:
First we must obtain the limiting reactant. The equation of the reaction is;
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
Number of moles of CH4= 28.6 g/16g/mol = 1.8 moles
Since the reaction is 1:1, 1.8 moles of CO2 was produced
Number of moles of O2 = 57.6 g/32 g/mol = 1.8 moles
2 moles of O2 produced 1 mole of CO2
1.8 moles of O2 produced 1.8 × 1/2 = 0.9 moles of CO2
Hence O2 is the limiting reactant
Mass of CO2 produced = 0.9 moles × 44 g/mol = 39.6 g
%yield = 32.1g/39.6 g × 100
%yield = 81.1%
According to the reaction equation;
2moles of O2 reacts with 1 mole of CH4
1.8 moles of O2 reacts with 1.8 × 1/2 =0.9 moles of CH4
Number of moles of CH4 left = 1.8 moles - 0.9 moles
Number of moles of CH4 left = 0.9 moles
6) Using the rules for significant figures, the sum of 0.027 gram
and 0.0023
gram should be expressed as
A) 0.030 gram B) 0.0293 gram C) 0.029 gram D) 0.03 gram
I need help
Answers
Answer:
it would be C
Explanation:
Answer: C
Explanation:
both numbers have 2 S.F. so the answer should have 2 too.
14 Hydrogen and iodine can react reversibly to produce hydrogen iodide. The equation is shown.
H₂(g) + 12(g) 2HI(g)
4.00 mol of hydrogen gas and Xmol of iodine vapour are mixed in a sealed container of volume
1.00 dm³ at a temperature of 460 K. The system is allowed to reach equilibrium.
The equilibrium mixture contains 2.00 mol of hydrogen iodide. The equilibrium constant, Kc, for
the reaction at 460 K is 4.0.
What is the value of X?
A 0.50 mol
B 1.17 mol
C 1.33 mol
D 2.50 mol
Answers
The concentration of the iodine at equilibrium from the calculation is 5.33 M
What is the equilibrium constant?
The equilibrium constant allows for the prediction of the direction in which a reaction will proceed to establish equilibrium when concentrations or pressures of reactants and products change.
We know that;
H₂(g) + \(I_{2}\)(g) ⇄2HI(g)
I 4 m 0
C -x -x +2x
E 4 - x m - x 2
It the follows that;
2x = 2
x = 1
Then equilibrium concentration of hydrogen = 3 M
Thus we have that;
4 = 3 * [ \(I_{2}\)]/\(2^2\)
16 = 3 * [ \(I_{2}\)]
[ \(I_{2}\)] = 5.33 M
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An eraser is an example of —
Group of answer choices
an insulator
a closed circuit
a conductor
an open circuit
Answers
Answer:
Insulator
Explanation:
An eraser can't conduct electricity, which makes it an insulator.
The article suggests a lot of components to getting a gas station in space. Which part do you think would be hardest to implement and why?
Answers
The development and deployment of a reliable and cost-effective fuel supply chain would likely be the most challenging component of establishing a gas station in space.
As outlined in the article, creating a gas station in space would require a number of technical and logistical advancements, including the development of reusable spacecraft, efficient refueling methods, and a safe and reliable fuel supply chain. However, of all the components, the fuel supply chain may pose the greatest challenge.
This would involve not only the development of a suitable fuel for space travel, but also the creation of efficient and safe methods for storing and transporting that fuel. Additionally, the cost of developing and deploying such a complex supply chain would likely be significant. Overall, while there are many challenges to creating a gas station in space, the fuel supply chain is likely to be the most difficult to implement.
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describe the temperature, moisture and air pressure associated with a continental polar air mass.
Answers
The variance in the US continental region is brought on by the shift in daytime and nighttime weather patterns.
Continental polar air mass -Cold, dry, and stable air masses are found in the continental polar (cP) or continental arctic (cA) regions. Radiative cooling causes these air masses to form over northern Canada and Alaska. They travel south, then east via the Plains and the Rockies.
During the winter, a continental polar air mass can develop over the land. It comes from northern Canada or Alaska in the Northern Hemisphere. It transports dry weather to the United States as it goes south. Low humidity and temperature are both present.
These factors contributed to the polar air mass:
Breezeextreme humiditythe evening's low temperatureDuring the colder months of the year, continental polar air typically forms over vast land masses.
A cool breeze blows across the upper section of the area, while a warm breeze blows through the lower part.
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For I2, the harmonic oscillator approximation (force constant k=170 N/m) assumes that the vibrational energy levels are equally spaced. At 300 K, calculate the ratio of molecules in the first excited state compared to the ground state. Also, calculate the proportion of molecules in the second excited state relative to the ground state. The mass of iodine (I) is 126.9 amu. b) At what temperature the population in the first excited state will be half of that in the ground state?
Answers
For I₂ under the harmonic oscillator approximation at 300 K, the ratio of molecules in the first excited state (n=1) to the ground state (n=0) can be calculated using the Boltzmann distribution equation. The vibrational energy levels are assumed to be equally spaced, and the proportion of molecules in the first excited state compared to the ground state can be determined.
Additionally, the temperature at which the population in the first excited state is half of that in the ground state can be found by setting up an equation based on the Boltzmann distribution.
The ratio of molecules in the first excited state to the ground state for I₂ at 300 K using the harmonic oscillator approximation, we can use the Boltzmann distribution equation. The energy levels of a harmonic oscillator are given by E = (n + 1/2)hν, where n is the vibrational quantum number, h is Planck's constant, and ν is the vibrational frequency.
a) Ratio of molecules in the first excited state to the ground state:
Since the energy levels are equally spaced, the energy difference between adjacent levels is ΔE = hν. Using the formula ΔE = √(k/μ), where k is the force constant and μ is the reduced mass, we can calculate the vibrational frequency (ν) for I₂.
μ = m1 * m2 / (m1 + m2) (reduced mass equation)
Plugging in the mass of iodine (I) as 126.9 amu and applying the reduced mass equation, we can find μ.
Once we have ν, we can calculate the energy difference (ΔE) and determine the ratio of molecules in the first excited state (n=1) to the ground state (n=0) using the Boltzmann distribution equation.
b) Temperature at which the population in the first excited state is half of that in the ground state:
To find this temperature, we can set up an equation where the ratio of molecules in the first excited state (n=1) to the ground state (n=0) is 1:2. We can solve for the temperature (T) using the Boltzmann distribution equation.
With the given force constant (k=170 N/m) and the calculated vibrational frequency (ν), we can determine the temperature at which the population in the first excited state will be half of that in the ground state.
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For zero order reaction, the unit of rate constant is equal to the unit of the rate of the reaction. It is sM or L⋅smol. For first order reaction, the unit of rate constant is equal to s−1 or min−1.
Answers
A zero order reaction's rate constant is measured in molL1s1. The unit of the rate constant is the same as the unit of reaction rate for zero order reactions.
Does a linear reaction have the same unit of velocity and rate constant?As a result, in the no reaction, the velocity is dependent of the reactant concentration and the units proportional rate and rate constant, which are mol/L/time, are equivalent.
What is the zero order response rate?A zero order process is a chemical process where the reaction rate is unaffected by the the amount of the reactants; that is, the rate is unaffected whether the reactant concentration rises or falls.
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Question 5 (5 points)
True or False. Open clusters are groups of up to a few thousand stars that are loosely held together by gravity and
tend to appear blue.
Answers
Answer:
have a p000000000000000000
Helium gas takes 10 seconds to effuse from a porous container of 1500ml under specific conditions of temperature and pressure. How long will SO2 take to effuse from a container of 0. 3dm under similar conditions?
Answers
The 40 seconds long will SO₂ take to effuse from a container of 0. 3dm under similar conditions.
What is graham's Law?
According to Graham's law, the square root of a gas's molecular weight has an inverse relationship to the rate of effusion or diffusion of that gas.
r₁ / r₂ = √(M₂ / M₁)
Where,
r₁ = rate of effusion for gas 1
r₂ = rate of effusion for gas 2
M₁ = molar mass of gas 1
M₂ = molar mass of gas 2
As given,
X ml of Helium gas takes 10 seconds to effuse.
Apply Law,
r₁ / r₂ = √ (M₂ / M₁)
[Note: 1 superscript is use for He gas, and 2 superscript is use for SO₂.]
(X/10) / (t/X) = √ (64/ 4)
t/10 = √16
t/10 = 4
t = 40 sec
Hence, the 40 seconds long will SO2 take to effuse from a container of 0. 3dm under similar conditions.
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Ammonia is produced by the Haber process in which nitrogen and hydrogen combine according to the reaction equation below. Yet, the reaction only results in a 15.0 % yield. How many grams of nitrogen are needed to produce 27.5 grams of ammonia, given an excess of ammonia gas
Answers
To produce 27.5 grams of ammonia with a 15.0% yield, approximately 91.71 grams of nitrogen would be required.
To determine the number of grams of nitrogen needed to produce 27.5 grams of ammonia, we first need to calculate the molar mass of ammonia. The molar mass of ammonia (NH3) is calculated as follows:
Molar mass of NH3 = (1 * atomic mass of N) + (3 * atomic mass of H)
= (1 * 14.01 g/mol) + (3 * 1.01 g/mol)
= 14.01 g/mol + 3.03 g/mol
= 17.04 g/mol
Since the reaction yield is given as 15.0%, we can calculate the theoretical yield of ammonia by dividing the given mass by the percent yield:
Theoretical yield of ammonia = (27.5 g) / (0.15)
= 183.33 g
According to the balanced equation of the Haber process, the stoichiometric ratio between nitrogen (N2) and ammonia (NH3) is 1:2. This means that for every 2 moles of ammonia produced, we need 1 mole of nitrogen. Therefore, we can calculate the moles of nitrogen needed as follows:
Moles of nitrogen = (183.33 g NH3) * (1 mol N2 / 2 mol NH3) * (1 mol / 28.01 g)
≈ 3.273 mol
Finally, we can convert the moles of nitrogen to grams:
Grams of nitrogen = (3.273 mol) * (28.01 g / mol)
≈ 91.71 g
Therefore, approximately 91.71 grams of nitrogen are needed to produce 27.5 grams of ammonia with a 15.0% yield
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-Guys I need help asap pls someone
How can acetaminophen cause liver damage
Answers
Dr. Ke-Qin Hu, a well-known expert in liver disease and director of UCI Health Liver and Pancreas Services, warns that taking or more four grams of acetaminophen in a 24-hour period could result in severe harm.
How do I find out if my liver is healthy?Proteins, liver enzymes, & bilirubin are just a few of the things that a liver blood test analyzes in your blood. This can aid in examining the condition of your liver and looking for indications of inflammation of damage. Infections with the liver, such as hepatitis B and C, can have an impact on your liver.
What foods promote the restoration of the liver?There are numerous nutrients that include certain elements or antioxidants that have been demonstrated to support liver function. The following are a few examples: grapefruit, berries, cranberries, fatty salmon, olive oil, even cruciferous vegetables as broccoli or Brussels sprouts.
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Pleaseee help me on this
Answers
hc and co are high and co2 and o2 are low. this could be caused by a
Answers
HC and CO are high and CO₂ and O₂ are low. This could be caused by a rich mixture.
A) rich mixture
If HC (hydrocarbons) and CO (carbon monoxide) levels are high, while CO₂ (carbon dioxide) and O₂ (oxygen) levels are low, it suggests a condition known as a "rich mixture" in the combustion process. A rich mixture refers to an air-fuel mixture in which there is an excess of fuel compared to the amount of air required for complete combustion.
When the fuel-air mixture is rich, it means that there is more fuel available relative to the available oxygen for combustion. This imbalance can occur due to several reasons, such as:
1. Incorrect fuel-to-air ratio: The air-fuel mixture may be adjusted incorrectly, with too much fuel being supplied relative to the amount of air. This can occur due to a malfunctioning fuel injection system.
2. Malfunctioning sensors: The sensors responsible for measuring the oxygen and fuel levels in the exhaust gases, such as the oxygen sensor or air-fuel ratio sensor, may be faulty or contaminated. This can result in inaccurate readings and improper adjustment of the fuel mixture.
3. Clogged air intake or fuel injectors: If the air intake or fuel injectors are clogged, it can disrupt the proper mixing of fuel and air, leading to a rich mixture.
The consequences of a rich mixture include:
High HC levels: A rich mixture results in incomplete combustion, leading to unburned hydrocarbon molecules being released into the exhaust gases. This increases the HC levels.
High CO levels: In a rich mixture, there is an excess of fuel. As a result, some of the fuel does not undergo complete combustion and is converted into carbon monoxide (CO). This leads to elevated CO levels.
Low CO₂ levels: Since there is incomplete combustion in a rich mixture, the amount of carbon dioxide (CO₂) produced is reduced.
Low O₂ levels: A rich mixture consumes most of the available oxygen for combustion, resulting in lower levels of oxygen (O₂) in the exhaust gases.
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The complete question is:
HC and CO are high and CO₂ and O₂ are low. This could be caused by a ____?
A) rich mixture
B) lean mixture
C) defective ignition component
D) clogged EGR passage
find the atomic numbers of the as yet undiscovered next two members of the series.
Answers
pls help will give brainiest
Answers
Answer: A
Explanation:
Which nonmetal would an ion of Ca +2 most likely ionically bond within a 1:1 ratio?
A: Nitrogen
B: Fluorine
C: Sulfur
D: Carbon
Please explain.
Answers
Answer:C: sulfur
Explanation:
Use the periodic table to answer the question.
Which comparison is correct?
A. PH3 is a stronger acid than NH3.
B. HF is a stronger acid than HCl.
C. H2S is a stronger acid than HCl.
D. NH3 is a stronger acid than HF.
Answers
Based on the periodic table, the correct comparison is: (B) HF is a stronger acid than HCl.
To compare the acid strength of different compounds, we can consider their ability to donate protons (H⁺ ions) in solution. Looking at the given options:
A. PH₃ is a stronger acid than NH₃.
This comparison is incorrect. NH₃ (ammonia) is a weak base, not an acid. PH₃ (phosphine) is also a weak base, but it can act as a weak acid in certain reactions. However, PH₃ is not stronger than NH₃ as an acid.
B. HF is a stronger acid than HCl.
This comparison is correct. HF (hydrofluoric acid) is a stronger acid than HCl (hydrochloric acid) because fluorine (F) is more electronegative than chlorine (Cl). The stronger the electronegativity of the element bonded to hydrogen, the stronger the acid.
C. H₂S is a stronger acid than HCl.
This comparison is incorrect. HCl (hydrochloric acid) is a stronger acid than H₂S (hydrogen sulfide). Sulfur (S) is less electronegative than chlorine (Cl), so HCl is a stronger acid.
D. NH₃ is a stronger acid than HF.
This comparison is incorrect. NH₃ (ammonia) is a weaker acid than HF (hydrofluoric acid). Ammonia is a weak base, while HF is a weak acid. HF is stronger because fluorine (F) is more electronegative than nitrogen (N).
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ACELLUS LAB-ACID BASE TITRATION
a student overshot the equivalence point when he was using NaOh to determine the concentration of an unknown acid, HA. What ions are present in the purple solution in the flask?
Answers
Answer: A-, NA+ And OH-
Explanation: Thank me later
please helpppppppp
helpppppppppppppppppppppppppppp
Answers
Select the correct answer from each drop-down menu. At chemical equilibrium, the amount of because .
Answers
Answer:
The answer that completes the question are in BOLD:
At chemical equilibrium, the amount of PRODUCT AND REACTANT REMAIN CONSTANT because the RATES OF THE FORWARD AND REVERSE REACTIONS ARE EQUAL.
Explanation:
In a reversible chemical reaction, an equilibrium is said to be achieved when the rates of the forward reaction is equal to that of the reverse reaction. A reversible reaction is one in which products are formed from reactants simultaneously with the formation of reactants from products.
The combination of two or more substances called REACTANTS gives rise to another substance called PRODUCT, which can in turn give rise to Reactants again. With time, the rate at which the reactants give rise the products, which is called the FORWARD REACTION will be equal to the rate at which the products give rise to the reactants, which is called REVERSE REACTION. At this point, the chemical reaction is said to be in a STATE OF EQUILIBRIUM.
When the rate at which both reaction occurs becomes equal i.e. at an equilibrium state, the concentration of both the reactants and the products becomes constant i.e. no longer changes. Hence, the amount of the reactants forming the products is the same as the amount of products forming the reactants.
N.B: At chemical equilibrium, the amount of the reactants and products does not necessarily equals zero (0). It simply means that there is no net change in the concentration/amount of both reactants and products.
which one of the following is considered a strong acid?
Acetic acid · Tartaric acid · Lactic acid · Sulphuric acid
Answers
Sulphuric acid is considered a strong acid. An acid that ionizes completely or almost completely in water is known as a strong acid. Sulfuric acid is classified as a strong acid since it has a dissociation constant of greater than 1. Option c. is correct.
The dissociation of H2SO4 in water can be expressed in the following equation:H2SO4 + H2O ⇌ HSO4– + H3O+
Sulfuric acid is a strong acid because it ionizes completely when it is dissolved in water to produce H+ ions. As a result, it is an excellent proton donor, and its strength as an acid is determined by its ability to donate protons in a given medium. Sulfuric acid is a clear, colorless, and odorless liquid that is extremely corrosive and can cause burns. Option c. is correct.
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