In 1963, the united states passed the clean air act which sets a maximum amount for emissions of pollutants or the presence of pollutants in ambient air.
The Clean Air Act of 1963 was the first piece of federal law to address "regulating" air pollution. This was done by the 1963 legislation, which authorized research into methods for monitoring and regulating air pollution and established a government program inside the U.S. Public Health Service. To reduce air pollution in the United States, significant legislation known as the Clean Air Act was passed. less air pollution overall. The overall amount of pollutants generated may be decreased in a number of methods, including smaller automobiles, the removal of sulfur from coal, and lower combustion temperatures.
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Dark surfaces such as the black paper Response area more light and Response area than the lighter surfaces such as yellow or pink paper. A simple experiment that you could try is to place two bottles out in the sun. Place the black paper under one bottle and white paper under the other. After a few minutes, measure the temperature of the water in each bottle. The glass with the Response area paper on it should be hotter than the other bottle. Lighter surfaces Response area more light, that is why people wear lighter colored clothes in the Response area.
The darker surfaces receive high temperature, and less light, the lighter surface receives low temperature and more light.
What is temperature?
Temperature is a unit of measurement that can be expressed on a variety of scales, including Fahrenheit and Celsius. Temperature shows the direction of the spontaneous movement of heat energy, i.e., from a hotter body (one with a higher temperature) to a colder body (one at a lower temperature).
What is Dark surfaces ?
Black surfaces serve as effective heat radiators and absorbers. Any object's color is determined by the light it reflects. If an object reflects all the light that strikes it, it is said to be white. If an object reflects no or very little light and absorbs practically all of the light, it is said to be black.
Blocking the red, blue, and green lights produces cyan; blocking the green, red, and blue lights produces magenta; and blocking the blue, red, and green lights produces yellow. By switching off several light fixtures, you may create a similar effect.
Therefore, the darker surfaces receive high temperature, and less light, the lighter surface receives low temperature and more light.
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calculate the maximum internal crack length allowable for a ti-6al-4v titanium alloy (table 9.1) component that is loaded to a stress one-half its yield strength. assume that the value of y is 1.50.
The maximum internal crack length allowable for a ti-6al-4v titanium alloy is 4.134 mm.
Calculation:-
Find the allowable internal cracking length as follows:
[tex]2a_{c}\frac{2}{\pi } ( {\frac{k_{ic} }{yo} })[/tex]
= 2/π (55/1.5×455)²
= 0.004134 m
= 4.134 mm
Critical crack length marks the transition from a stable crack growth regime to an unstable crack growth regime leading to catastrophic rupture or failure. The rearrangement gives the critical crack size for a given tensile stress. Ultimate strength is the ability of a material to resist failure and is named.
Critical crack size is the length at which a crack becomes unstable at a given stress. It is very useful in determining the safety of materials. An unstable crack propagates spontaneously without increasing the magnitude of the applied stress once it begins to propagate.
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arrange the following afn species in order of increasing f-a-f bond angles: bf3, bef2, cf4, nf3, of2.
One fluorine atom is added to the phosphorus trifluoride molecule to create this molecule. The molecule has a pyramidal form by nature. In this instance, the bond angle for F-P-F is 109 degrees.
What is nf3's bond angle?
Similar to ammonia, the molecule possesses a trigonal pyramidal structure. The very electronegative fluorines pull the electrons in the N-F bonds towards themselves, lowering interelectronic repulsions, and reducing the bond angle from 107° in ammonia to 101.9° in NF3. As a result, the NF3 "umbrella" closes up.
that has a higher bond angle NF3 or NCl3?
The N-F bond would have a stronger p character than the N-H bond because fluorine is more electronegative than hydrogen. Large bond angles result from more strong character. As a result, NH3 has a larger bond angle than NF3. Now think about NCl3
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You need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a
0.0420 M solution of LICN. What is the final volume (in mL) of this
solution.
The final volume of the solution is 666.7 mL.
What is the final volume?We know that we can be able to obtain the volume of the solution by applying the formula for dilution. When we talk about dilution we are referring to the volume of the water that we have to add to the solution.
Then we have;
Initial volume V1 = 20.0 mL
Final volume V2 = ?
Initial concentration C1 = 1.40 M
Final concentration C2 = 0.0420 M
Then we have
C1V1 = C2V2
V2 = C1V1 /C2
V2 = 20.0 mL * 1.40 M/0.0420 M
V2 = 666.7 mL
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draw the organic product(s) of the following reactions including stereochemistry when it is appropriate. h2 lindlar catalyst
Draw the organic product(s) of the following reactions including stereochemistry when it is appropriate. H₂ lindlar catalyst is given as :
CH₃-C≡C-CH₂-CH₂- CH₃ + H₂ Lindlar catalyst --> CH₃ - C=C - CH₂ -CH₂- CH₃
| |
H H
The lindlar catalyst is used to hydrgenate the bond into the double bonds. H₂ + lindlar catalyst is used to reduce the alkynes which contains the triple bond in to the alkene which contains the double bond. the H₂ + lindlar catalyst doest noty react with the the alkenes. the reaction of H₂ + lindlar catalyst is given as follows :
CH₃-C≡C-CH₂-CH₂- CH₃ + H₂Lindlar catalyst --> CH₃-C=C - CH₂ -CH₂- CH₃
2 - hexyne | |
H H
2 - hexene
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Tungsten atoms produce characteristic x-ray emissions for electron transitions between the K, L and M shells. Given the information below, an electron transitioning from the M to the K shell will produce an x-ray with what characteristic energy?
Shell M: 2 Kev of IE (ionization energy)
shell L: 11 KeV of IE
Shell K: 70 KeV of IE
answer also: which is the closest to the nucleus?
An electron from of the M shell occupying a hole in the K shell at 69 keV expelled as you move from the outside in to the inside will be moving to an energy levels with an ionization energy of 70 keV -2 keV,
the periodic table, what is ionization energy?The minimal energy needed to ionize an atom is sufficient to eject an electron from its ground state. Ionization energy has a periodic trend that rises across the chemical elements form top to bottom as well as from left to right.
Ionization energy: what is the rule?As you move from left to right along a period just on periodic table, the initial ionization energy often rises. This is because when nuclear charge rises, the outermost electron becomes more tightly connected to the nucleus.
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2. A 9.0 g sample of iron (specific heat= 0.44 J/g °C) heated to 60°C eventually cools to 30°C. What is the heat value? Does the sign of "q" indicate heat absorbed or lost?
Answer:
The heat value can be calculated using the equation q = m * c * delta T, where q is the heat value, m is the mass of the iron sample, c is the specific heat of iron, and delta T is the change in temperature. In this case, q = 9.0 g * 0.44 J/g * (60-30) = 804 J.
The sign of q indicates whether heat is being absorbed or lost by the iron sample. In this case, since the temperature of the iron sample decreases from 60°C to 30°C, the sample loses heat and q is negative. Therefore, the heat value of the iron sample in this case is -804 J.
A 9.0 g sample of iron (specific heat= 0.44 J/g °C) heated to 60°C eventually cools to 30°C. 804 J is the heat value. The sign of heat is negative which indicate energy is released.
What is heat?Heat is the energy that moves through one thing to another if the temperatures are different. Heat passes from the hotter to the colder body when two bodies with differing temperatures are brought together. Typically, but not always, this energy transfer results in a rise inside the temperature of a colder body and a fall inside the temperature of a hotter body.
By transitioning through one physical state (and phase) to the other, such as melting from such a solid to a liquid, sublimation from either a solid to a vapour, boiling from such a liquid to a vapour, or shifting through one stable state to another, a substance can absorb heat without increasing in temperature.
q = m × c × ΔT
q = heat
m = mass of the iron sample
c =specific heat of iron
Δ T= change in temperature
q = 9.0 g×0.44 J/g × (60-30) = 804 J
heat = -804 J.
Therefore, a 9.0 g sample of iron (specific heat= 0.44 J/g °C) heated to 60°C eventually cools to 30°C. 804 J is the heat value. The sign of heat is negative which indicate energy is released.
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We can calculate the pH range over which a particular acid and its conjugate base can be used to make an effective buffer. If a buffer solution is created by adding NaOH to a solution of benzoic acid, C6H5COOH. (pka = 4.20). What is effective buffer range for the solution? O 1-7 O 10.80 - 8.80 O 2.20-6.20 O 3.20 - 5.20 O 7-14
If a buffer solution is created by adding NaOH to a solution of benzoic acid, C6H5COOH, the effective buffer range for the solution will be (d) 3.20 - 5.20.
pKa of benzoic acid = 4.20
The effective buffer range of solution = ?
The term "buffer" can also refer to a solution that has reserve acidity or alkalinity and is able to resist a change in pH even after a trace amount of acid or alkali has been added to it.
Any buffer has an effective pH range of one pH unit on either side of the pKa value for the weak acid.
As the pKa is 4.20
then pH range will be
4.20 +- 1
pH range = 3.20 - 5.20
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mannose, one of the eight essential monosaccharides, is biosynthesized as its 6-phosphate derivative from fructose 6-phosphate. no cofactor is required. draw curved arrows to show the movement of electrons in this step of the reaction mechanism.
The 6-phosphate derivative of mannose, one of the eight essential monosaccharides, is biosynthesized from fructose 6-phosphate. The oxygen atom experiences electron migration.
How do monosaccharides work?The most fundamental type of a carbohydrate is a monosaccharide. Larger carbohydrates, such as oligosaccharides or polysaccharides, can be created by joining monosaccharides through glycosidic linkages. A disaccharide is an oligosaccharide that contains just two monosaccharides. An oligosaccharide turns into a polysaccharide when it forms glycosidic connections with more than 20 monosaccharides.
Where can you find mannose?A simple sugar called d-mannose can be found in many fruits. It has to do with glucose. In some human body cells, it also happens naturally.
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What are Quantitative and qualitative properties?
This is a science grade 9 question
Answer:
Explanation:
Quantitative properties are characteristics of a substance that can be measured using a numerical value. Examples of quantitative properties include mass, volume, density, and chemical concentration. These properties can be measured and compared using mathematical techniques and formulas.
Qualitative properties, on the other hand, are characteristics of a substance that cannot be measured using a numerical value. Examples of qualitative properties include color, odor, taste, and texture. These properties are typically described using words or phrases, and cannot be compared using mathematical techniques.
Quantitative and qualitative properties are both important for understanding the characteristics and behavior of a substance. Quantitative properties are useful for making precise measurements and predictions, while qualitative properties provide valuable information about the appearance and sensory properties of a substance. Together, these two types of properties help us to understand the nature of a substance and its behavior in different situations.
Which of the following is a redox reaction?
Answer:
Explanation:
Correct option is ( A ).
in this respective reaction 3MnO2 + 4Al ---- 3Mn + 2Al2O3 ,
MnO2 is acting as an oxidising agent i.e. it oxidises Al to Al2O3.
Hence , it is an Redox reaction.
This data table shows how quickly rope will rot away in freshwater and saltwater. what can you conclude about the rates of the rotting reaction?
A. The reaction occurs more slowly when the concentration of salt is higher.
B. The reaction occurs more quickly when the concentration of salt is higher.
C. The reaction occurs at the same rater, regardless of the concentration salt.
D. There is not enough information to conclude anything about the reaction rates.
One can either conclude that the rope rotted quicker due to the presence of a higher concentration of salt in the ocean or that there is not enough information to make a conclusion. Options B and D.
Rate of reactionsFrom the available data in the table, it took the rope 10 years to rot in a lake water. Whereas, it took the rope 4 years to rot in an ocean water.
Also from the table, the average concentration of salt in the lake water is almost zero, whereas, the average concentration of salt in the ocean water is high.
Two things can be inferred from this reaction:
That the shorter time it took the rope to rot is due to the presence of high salt concentration in ocean water.There is not enough information to make any specific claim.In order to accept the first inference, we will assume that all other factors have been examined and kept constant except the concentration of salt in the two water bodies. This will also include that the same size of rope was used in the two water bodies.
Without the assumptions made in the first inference, there is not enough information to arrive at a specific conclusion.
Thus, both B and D are plausible options.
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what is the best transition metal cation that can react with ethene
Answer:
Explanation:
1+ 2+ 3+
copper(I), Cu+ cadmium, Cd2+ chromium(III), Cr3+
gold(I), Au+ chromium(II), Cr2+ cobalt(III), Co3+
mercury(I), Hg22+ cobalt(II), Co2+ gold(III), Au3+
silver, Ag+ copper(II), Cu2+ iron(III), Fe3+
Which of the following is the conjugate acid of H₂PO4?
hs
OA. HP04-2
OB. PO4³-
OC. H₂PO4
D. H3PO4
The conjugate acid of H₂PO4 is H₃PO₄.
The correct option is D.
What are conjugate acids?Conjugate acids are chemical species that are formed when a base accepts a proton.
A conjugate acid contains one more proton than the base from which it is formed.
Conjugate acids are derived from the Brønsted–Lowry acid–base theory, which states that an acid is a chemical compound that donates a proton to a base.
The conjugate acid of H₂PO4⁻ will contain one more proton and a higher positive charge than H₂PO4⁻. Hence, the conjugate acid of H₂PO4⁻ is H₃PO₄.
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The next 7 questions are related to the titration of 40.0 mL of a 0.0150 M Zn2+ solution with 0.0250 M EDTA in a solution buffered at pH 11. Assume that the temperature is 25 oC and that the formation constant for Zn2+ is 3.13 x 1016 at this temperature.
How many mmols of Zn2+ are present in the solution before the titration begin?
What is the conditional formation constant for Zn at this pH?
What volume of the EDTA solution is needed to reach the equivalence point?Alpha values for EDTA are in table 13-3 on page 299 of the textbook.
What is the pZn of the analyte solution before the titration begins?
What is the pZn of the solution after 15 mL of titrant have been added?
What is the pZn at the equivalence point of the titration?
What is the pZn of the solution after 30 mL of titrant have been added?
Using complex-forming processes, complex metric titrations are primarily used to identify metal ions.
Although there are various complexing agents that can be utilized for this task (cyanide, thiocyanate, fluoride, 1,2-diaminoethane, etc.), in actuality the titrants are usually invariably compounds with the iminodiacetic acid functional groups. The sodium salt, Na₂H₂Y 2H₂O, and ethylenediaminetetraacetic acid are the two that are used the most frequently (better soluble in water).
Most requirements for a good complexometric titrant are mostly met by EDTA. It reacts quickly with most metal ions (except Cr₃⁺), forms sufficiently stable complexes with most metal ions, all of the complexes have perfect 1:1 stoichiometry (independent of the charge of the cation), are water soluble, and are colorless (unless the metal is present).
(unless the metal ion is colored by itself). A typical metal ion's reaction with EDTA (H₄Y) can be expressed as follows:
that is, as a battle between the metal ion and hydrogen ions for interacting with Y₂−. The equilibrium constant for the reaction, KMeY, which measures the stability of binding of Men + with Y₂ is as follows.
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If the nuclide is above the valley of stability, the n/Z ratio is too ____ (high/low) and it will undergo ____ decay to reach the valley of stability.
If the nuclide is below the valley of stability, the n/Z ratio is too ___ (high/low) and it will undergo ___ decay or ___decay to reach the valley of stability.
If the nuclide is above the valley of stability, the n/Z ratio is too high and it will undergo β decay to reach the valley of stability.
If the nuclide is below the valley of stability, the n/Z ratio is too low and it will undergo β+ decay or α decay to reach the valley of stability.
The nuclear reaction is one that changes the structure of the nucleus of an atom. The atomic numbers and mass numbers in a nuclear equation must be balanced. Protons and neutrons are made up of quarks. The two most common modes of natural radioactivity are alpha decay and beta decay. The nuclear disintegration process that emits alpha particles is called alpha decay. All nuclei with 84 or more protons are radioactive, and elements with less than 84 protons have both stable and unstable isotopes. All of these elements can go through nuclear changes and turn into different elements.
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What volume of I2 gas forms when 21 L Cl2 react at STP?
Please help me!
At STP, the volume of one mole of any gas is 22.4 L. As per the given reaction, one mole or 22.4 L of Cl₂ gas gives one mole. Thus, 22.4L of Cl₂ gives 21 L of I₂ gas.
What is molar volume?The molar volume of a substance is the space occupied by one mole of that substance. Each chemical has a molecular weight of 22.4 L when measured under standard temperature and pressure conditions (STP).
At STP, a mole of Cl2 gas holds 22.4 L, while a mole of I2 also contains 22.4 L. According to the reaction, a mole of Cl2 yields a mole of L2. As a result, both are contributing equally to this reaction.
As a result, at STP, 21 L Cl₂ reacts to produce 22.4 L₂.
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FILL IN THE BLANK. The lewis structure of pf3 shows that the central phosphorus atom has ________ nonbonding and ________ bonding electron pair(s)
The lewis structure of [tex]PF_3[/tex] shows that the central phosphorus atom has 1 nonbonding and 3 bonding electron pair(s)
Let us consider the Lewis Structure of [tex]PF_3[/tex]
Here, the central atom is phosphorus P.
We know, The atomic number of Phosphorus is 15.
The so the distribution will be like 2, 8, 5
Here, in the last orbit total of 5 electrons are present, so for completing the octet it needs 3 more electrons.
Hence, The phosphorus atom has 1 non-bonding and 3 bonding electron pairs with fluorine atoms.
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which of the following types of particle is most common in the universe? neutrinos electrons neutrons protons
The universe's most prevalent mass-containing particles are neutrinos. Atomic nuclei emit neutrinos whenever they combine (as in the sun) or disintegrate (as in a nuclear reactor).
A particle is a neutrino. It is one of the so-called fundamental particles, which means that, at least as far as we are aware, it is not composed of any smaller parts. The electron, the most well-known fundamental particle (and the one driving the device you're reading this on right now), belongs to the same family as neutrinos. Neutrinos don't have any charge at all, in contrast to electrons which have a negative charge.
Additionally, neutrinos are exceedingly light and tiny. They have a small amount of mass. Of all the mass-containing subatomic particles, they are the lightest. They are the most abundant massive particle in the universe, and they are also quite common. Neutrinos originate from a variety of sources and are frequently the result of a process known as "decay," in which heavy particles transform into lighter ones.
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5.By what factor must the original volume be increased when the temperature is increased from 25oC to 50oC
Answer:
To determine the factor by which the volume of a substance must be increased when the temperature is increased from 25oC to 50oC, we need to know the substance's coefficient of thermal expansion.
The coefficient of thermal expansion is a measure of how much a substance's volume will increase in response to a given temperature change.
Assuming that we have the coefficient of thermal expansion for the substance in question, we can use the following formula to calculate the factor by which the volume must be increased:
Factor = 1 + coefficient of thermal expansion * (change in temperature)
In this case, the change in temperature is 50oC - 25oC = 25oC. Therefore, the factor by which the volume must be increased is:
Factor = 1 + coefficient of thermal expansion * 25oC
The exact value of the factor will depend on the substance's coefficient of thermal expansion. In general, substances with a high coefficient of thermal expansion will have a larger increase in volume when the temperature is increased.
Example 2:
=
(i) N2(g) + O2(g) →NO [8] ∆H (kJ)
=+90
2[g]
(i) NO [g] +0[g] →NO 2[g] ∆H (kJ)
=-56
(ii) 2NO 2[g] →N₂04[g]
ΔΗ (kJ)=-58
Evaluate AH for the reaction (iv) N2tg +202 [g] → N₂O4[gl
The ΔH of the reaction of the formation of N₂O₄ is equal to - 90 KJ.
What is Hess's law?Hess's law of constant heat summation can be defined as the total enthalpy change while the complete course of a chemical reaction independent of the sequence of steps.
Given, the reactions can be represented as:
N₂ (g) + O₂(g) → NO } ×1 ΔH = + 90 KJ
NO (g) + 1/2 O₂ (g) → NO₂ (g) } × 2 ΔH = -56 KJ
2NO₂ (g) → N₂O₄ (g) } × 1 ΔH = - 58 KJ
The overall reaction of the formation of N₂O₄ (g) can be obtained as:
N₂ (g) +2 O ₂ (g) → N₂O₄ (g)
ΔH = 1 × 90 + 2 ×(-56) + 2 × (-58)
ΔH = = - 90 KJ
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29. Which of the following best describes the properties of protons, neutrons, and electrons?
A. Protons and electrons are inside the nucleus while neutrons are outside the nucleus.
B. Protons, electrons, and neutrons are all inside the nucleus.
C. Protons and neutrons are inside the nucleus while electrons are outside the nucleus.
D. Protons are outside the nucleus while electrons and neutrons are inside the nucleus.
Answer:
C.
Explanation:
a region of vacuum contains both a uniform electric field with magnitude e and a uniform magnetic field with magnitude b.
A region of the vacuum that contains both uniform electric field and a uniform magnetic field is called electromagnetic radiation.
A region of the vacuum that contains uniform electric and magnetic field simultaneously is called an electromagnetic radiation or an Electromagnetic wave.
Wave does not recover any medium to travel because it is made up of electric and magnetic field which also do not require any medium to be in existence.
An Electromagnetic wave travels with the speed of light. The component of the electric field and the magnetic field contributes equally in sense of energy of the wave.
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1. the concentrations of the I-, IO3-, and H3AsO3 using m1v1=m2v2
2. calculate the initial concentration of [H+] in the solition for each experiment
3 . As well as using the [H3AsO3] and measured time to calculate the rate for each experiment
•rate law
•rate constant
the concentrations of the I-, IO3-, and H3AsO3 using m1v1=m2v2 is 2.2867 x 10^6, The initial concentration of [H+] in the solition for each experiment is 0.976 x 10^6, The rate of reaction is 2.7427 x 10^-
substance's concentration is the sum of solute in a provided amount of solution. Molarity is the amount of moles of solvent in one volume of water and is used to express concentrations. A solution's concentration is described as the quantity of solute present in a provided amount of solution. It can be stated as follows: Mass by a solution's mass percentage =mass of solutemass of solution 100 . The rate at which a chemical reaction occurs is defined as commensurate to an increase in the amount of a product per unit time and the reduce in the amount of a reactant per unit time. The rates of reaction could really vary greatly.
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Draw the Lewis structure for the polyatomic nitrate NO−3 anion. Be sure to include all resonance structures that satisfy the octet rule.
It is a monovalent inorganic ion that belongs to the reactive nitrogen species and is called a nitrogen oxoanion.
What is ion ?
An ion with more than one atom is referred to as a polyatomic ion. One such ion is the nitrate ion! It has one nitrogen atom and three oxygen atoms.
As seen in the graphic accompanying this answer, it is an ion in which there is charge separation. The nitrate ion's Lewis structure, which complies with the octet rule, is represented in the diagrams.Nitrates are an ester family of nitric acid (HNO3) and alcohol molecules. By losing a proton from nitric acid, nitrate is a nitrogen oxoanion. main species found at pH 7.3.
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the product of the following reaction has the constitution shown. no stereochemistry is implied. deduce the stereochemistry on the basis of the fact that iodolactonization is normally an anti addition and it was determined experimentally that the ring junction is cis. select the single best answer. 1935 1935c 1935a 1935d 1935b
The stereochemistry on the basis of the fact that iodolactonization is normally an anti addition and it was determined experimentally that the ring junction is cis. is Option A.
Stereochemistry is a branch of chemistry that involves the study of different spatial arrangements of atoms within molecules. Stereochemistry is a systematic presentation of a particular field of science and technology and traditionally requires a brief history course.
Stereochemistry is the visualization of molecules in 3D. Before I learned stereochemistry, I had only studied the two-dimensional structure of molecules so it is very difficult for students. You can explore different ball-and-stick models of molecules for proper visualization in 3D. Stereoisomerism.
Certain substituted derivatives of cycloalkanes exhibit a type of isomerism called stereoisomerism. In this isomerism, two substances have the same molecular formula and composition but differ in the spatial arrangement of their atoms.
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draw the arrow pushing reaction mechanism for the aldol condensation of vanillin and acetone during the first step
Aldol condensation is a reaction that results in the formation of a carbon-carbon bond, an enol or enolate molecule, and an aldehyde or ketone.
Removing the alpha hydrogen, establishing fresh carbon and carbonyl bonds, and changing carbonyl groups into COH.
These reactions include an enol or enolate ion reacting with a carbonyl molecule to produce a -hydroxy aldehyde or -hydroxy ketone, which is then dehydrated to produce a conjugated enone. Let's talk more about cross aldol condensation, the aldol condensation process, and its mechanism.
Thus, the acetaldehyde enolate's nucleophilic addition to benzaldehyde produces the aldol product: We will discuss all the additional potential outcomes and specifics of the crossover aldol reaction.
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calculate the final velocity of each particle when they reach the negative plate, assuming they both started from rest
The final velocity of particle B when it reaches the negative plate is 2.318×10^6 m/s.
According to the work-energy theorem work done by the particle is equal to an increase in kinetic energy
Given
Mass of particle A is m(A)= 5×10^-30 kg
Charge on particle A q(A)= 1.6×10^-19 C
V1=40 volt and V2= -100 volt
Work done by particle A is ∆E=q(A)(V1-V2).
∆E=1.6×10^-19*×(40-(-100))=1.6×10^-19×140 jule
∆E=224×10^-19 joule
Let the velocity of particle A at the negative plate is v(A)
Kinetic energy of particle K.E=1/2×m(A)×v(A)^2
By the work-energy theorem ∆E=K.E
So. 1/2×5×10^-30×v(A)^2=224×10^-19
V(A)^2=8.96×10^12
V(A) = 2.993×10^6 m/s
The final velocity of particle A when it reaches the negative plate is 2.993×10^6 m/s
Now
Mass of partial B is m(B) =2.5×10^-29 kg
Charge of particle B is q(B)= 4.8×10^-19 C
Work done by particle B is ∆E= q(B)×(40-(-100)
∆E=4.8×10^-19×140 =672×10^-19 jule
Let the final velocity of a particle at the negative plate is v(B)
By applying the work energy theorem.
1/2×m(B)×v(B)^2= 672×10^-19
v(B)^2=537.6×10^10
V(B)=2.318×10^6 m/s
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in general, the basic the leaving group, the reactive than carboxylic acid derivative. (select all that apply)
in general, the basic the leaving group, the reactive than carboxylic acid derivative is Acid chloride, Acid anhydride.
Carboxylic acid is a carboxylate gang organic compound. They are abundant and are also synthesised by humans. Carboxylic acids start to denature to form a carboxylate anion with the general formula R-COO-, that can form a variety of useful salts such as shampoos. The IUPAC name of a carboxylic acid is derived from the spelling of the parent alkane's lengthiest carbon chain that includes the carbon - carbon double bond by removing the final -e as well as adding the suffix -oic followed by the term "acid." The sequence is counted starting with the carboxyl group's carbon.
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the complete question is :
in general, the basic the leaving group, the reactive than carboxylic acid derivative. (select all that apply)
Acid chloride
Acid anhydride
Ester
Amide
1. The experiment is set up as shown and left until there is no further change. water purple crystal of potassium manganate(VII) What is observed?
A. a colourless layer below a purple layer
B. a colourless liquid with the purple crystal unchanged
C. a purple layer below a colourless layer
D. a uniformly purple solution
Answer:
C
Explanation:
Based on the information provided, it is likely that the observation will be a purple layer below a colorless layer. Potassium manganate(VII) is a purple crystalline compound that is often used as a laboratory reagent. When it is placed in water, it will dissolve and form a purple solution. However, if the solution is left to stand for a period of time, the purple color will gradually fade and a colorless layer of liquid will form at the bottom of the container, with the purple crystal unchanged at the top. Therefore, the most likely observation in this experiment will be a purple layer below a colorless layer (Option C).