provide the structure of the major fragment that results when the molecular ion of ch3ch2ch2ch2oh undergoes fragmentation via alpha cleavage. i ii iii iv v

The strength of a Lewis base is determined by its electron-donating ability, which is influenced by the electronegativity of the atoms it contains.

The electronegativity of an atom determines how strongly it attracts electrons towards itself. Phosphorus has a lower electronegativity than nitrogen, and fluorine has a higher electronegativity than hydrogen.

As a result, the electrons in a P-F bond are more strongly held by the fluorine atoms, making phosphorus trifluoride (PF₃) a weaker Lewis base than phosphine (PH₃).

Similarly, the electrons in an N-F bond are more strongly held by the fluorine atoms, making nitrogen trifluoride (NF₃) a weaker Lewis base than ammonia (NH₃). In short, the higher the electronegativity of the atoms in a molecule, the stronger its Lewis acidic character, and the weaker its Lewis basic character.

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If an atom contained only four energy levels (such as in the figure below), the possible different emission lines could it emit is 6.

Emission line is the spectrum that is produced when low pressure gas is ignited only at certain colors or wavelengths. Each different element emits a different set of lines.

Emissions are defined as the result of burning fossil fuels such as coal, natural gas and oil which are dispersed into the air, depending on the composition of the fuel and the type and size of the boiler.

Emissions are one of the contributors to air pollution which can have an impact on human health and the surrounding environment.

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The freezing point of the aqueous solution formed with 0.8100 m FeCl₃ is -5.67624 °C.

To find the freezing point of an aqueous solution, it is required to use the equation:

ΔT = [tex]\rm K_f[/tex] × m × i

In which:

ΔT = freezing point depression,

[tex]\rm K_f[/tex] = the freezing point depression constant for water

m = molality of the solute

i =  van't Hoff factor.

The van't Hoff factor (i) for FeCl₃ in this situation is 3.400, and you're asked to determine the freezing point depression (T) of an aqueous₃ solution with a molality (m) of 0.8100 m.

First, find the freezing point depression:

ΔT = [tex]\rm K_f[/tex] × m × i

ΔT = 1.860 °C/m × 0.8100 m × 3.400

Now, place the values and find:

ΔT = 1.860 °C/m × 0.8100 m × 3.400

ΔT= 5.67624 °C

The freezing point depression is 5.67624 °C.

To calculate the freezing point of the solution, it is required to minus the freezing point depression from the freezing point of pure water (0 °C):

Freezing point = 0 °C - ΔT

= 0 °C - 5.67624 °C

= -5.67624 °C

Thus, the freezing point is -5.67624 °C.

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If 1801,6 grams of sugar was used in the alcohol fermentation reaction, The grams of ethanol is produced is the correct option is  D. 460.78 g.

The balanced reaction of the sugar fermentation is :

C₆H₁₂O₆  --->  2C₂H₅OH  +  2CO₂

The mass of the sugar = 1801.6 g

The molar mass of the sugar 180.16 g /mol

The moles of sugar = mass /molar mass

The moles of the sugar = 1801.6 / 180.15

The moles of the sugar = 10 mol

1 mole of sugar produces 2 moles of ethanol

The moles of ethanol = 2 × 10

                                    = 20 mol

The mass of ethanol produces = moles × molar mass

                                                    = 20 × 46.07

                                                    = 460.78

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The percent increase in density would be 10%.

For an ideal gas, the relationship between pressure and volume is described by the equation of state: PV = nRT, where

When a gas is compressed isothermally, its temperature remains constant. Therefore, the equation of state can be rearranged as:

V/n = RT/P

From this equation, we can see that the ratio of volume to number of moles is proportional to the inverse of pressure. Therefore, if the pressure of the gas is increased by a certain percentage, the volume per mole of gas will decrease by the same percentage, and the density (mass per unit volume) will increase by the same percentage.

So, the percent increase in density would be 10%.

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If we  have the  sample of 6.45 × 10⁻³ moles of the H₂O, the number of hydrogen atoms the sample contain is 7.7 × 10²¹   atom of H.

1 mole of substance = 6.022 × 10²³ atoms

The moles of water = 6.45 × 10⁻³ moles of the water

18 g of water = 1 mole =  6.45 × 10⁻³ moles of water

1 g of hydrogen  = 1 mole = 6.45 × 10⁻³ atoms of hydrogen

The moles of  H atom = (2 ×  6.45 × 10⁻³ )  6.022 × 10²³ atom of H

The moles of the H atom = 7.7 × 10²¹   atom of H

Thus, the number of the oxygen atom is 7.7 × 10²¹   atom of H.

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The minutes will be required to the electroplate 26.3 g of the chromium by passing the constant current of 4.80  through  solution containing CrCl₃ is 1,675.33 minutes .

The mass of the chromium = 26.3 g

The moles of the chromium  = mass / molar mass

The moles of the chromium =  26.3 / 51.996

                                              = 0.5034 moles.

The current = 4.80 amp

The number of Coulombs required = moles ×faraday constant

                                                            = 0.5034 × 96500

                                                             = 484,986 C.

The time required is as follows :

t = Q / I

t = 484,986 / 4.80

t = 100520 seconds

t =  1,675.33 minutes

The time required is 1,675.33 minutes .

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Answer:

Explanation:

"No", Limiting how much companies can charge will distort markets, they argue, causing shortages and exacerbating supply chain problems while only temporarily reducing inflation.

Answer: True

When ATP is broken down, energy is released.

Inertia of an object is its tendency to  stay in its current state. Hence it is the resistance to change its motion. Thus, option c is correct.

According to Newtons; first law of motion every object tend to continue in the state of rest or motion until an external force act on it. This tendency is called inertia.

The impacts of inertia can be well explained based on on real life examples. If we might had the experience of moving to front in a vehicle which suddenly breakes. It is the tendency of the vehicle to stay in motion.

Similarly the car when started from rest initially tends to stay at rest and that's why we feel it moves back ward. Similarly the car will move forward when it is suddenly stops. Hence, option c is correct.

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The molar enthalpy of the reaction for the combustion of C6H6 is 3146.87 kJ/mol.

To calculate the molar enthalpy of the reaction, we need to find the amount of heat produced per mole of C6H6.

First, find the moles of C6H6: n = m/M where,

n = 300g / 78.11g/mol

= 3.84 moles

Next, divide the heat produced (12050 kJ) by the number of moles:

ΔH = q/n where,

ΔH = 12050 kJ / 3.84 moles = 3146.87 kJ/mol

So, The molar enthalpy of the reaction for the combustion of C6H6 is 3146.87 kJ/mol.

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Answer:  The correct answer is (A) AB will appear between Oil and Water in the graduated cylinder

Explanation:

Density is a measure of how much mass is contained in a given volume. Substances with a higher density will have more mass in a given volume, and will therefore be heavier.

When we fill the graduated cylinder with the above-mentioned substances, we will notice that Water, Oil and AB will be on top of each other. Since the density of AB is 0.88 g/mL and the density of Oil is 0.80 g/mL, this means that AB will have more mass in the same volume than oil and thus AB will be heavier than oil, so AB will appear between oil and water in the graduated cylinder.

Mercury is much denser than water, oil and AB, so it will be at the bottom of the graduated cylinder.

The concentration of the following in milligrams per liter will be: 0.01000 M Ca₂⁺ = 400.8 mg/ L, 0.02000 M H₂SO₄ = 1966.56 mg/L, 1.000 M HCO₃ = 61010 mg/L, 0.02000 M SO₄² = 1920.12 mg/L.

To convert the concentration of a solution from moles per liter (M) to milligrams per liter (mg/L), we need to multiply the molar concentration by the molar mass of the solute in grams.

0.01000 M Ca²⁺: The molar mass of Ca₂⁺ is 40.08 g/mol, so the concentration in mg/L is 0.01000 M × 40.08 g/mol × 1000 mg/g = 400.8 mg/L

0.02000 M H₂SO₄: The molar mass of H₂SO₄ is 98.08 g/mol, so the concentration in mg/L is 0.02000 M × 98.08 g/mol × 1000 mg/g = 1966.56 mg/L

1.000 M HCO₃-: The molar mass of HCO₃⁻ is 61.01 g/mol, so the concentration in mg/L is 1.000 M × 61.01 g/mol × 1000 mg/g = 61010 mg/L

0.02000 M SO₄²⁻: The molar mass of SO₄²⁻ is 96.06 g/mol, so the concentration in mg/L is 0.02000 M × 96.06 g/mol × 1000 mg/g = 1920.12 mg/L

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The overall reaction is thermodynamically favorable as the products have higher stability than the reactants.

The three steps of an electrophilic aromatic substitution reaction are: (a) formation of an electrophile, (b) reaction with an aromatic ring to form an intermediate, and (c) loss of a leaving group to reform the aromatic system. The first step involves attacking an electron-rich aromatic ring with an electrophile, which is an electron-deficient species. This attack forms a cationic intermediate, which is a species with a positive charge. The second step involves the reaction of the intermediate with a nucleophile, resulting in the formation of a new carbon-carbon bond and the release of a leaving group. The final step involves the loss of the leaving group, resulting in the formation of a new aromatic system. The overall reaction can be represented as follows: Ar-H + E+ --> Ar-E+ --> Ar-X + E- --> Ar-Y, where Ar-H is the starting aromatic compound, E+ is the electrophile, Ar-E+ is the intermediate, Ar-X is the product with the leaving group, and Ar-Y is the final product after the leaving group is lost. The net reaction can be written as Ar-H + E+ --> Ar-Y. The overall reaction is thermodynamically favorable as the products have higher stability than the reactants.

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This sample have 3 set of directions of cleavage and not at 90°.

Cleavage is defined as the tendency of minerals to split along crystallographic planes as a result of structural locations of atoms and ions in the crystal, creating planes of relative weakness.

Cleavage is termed as an excellent diagnostic property. Cleavage, distinguishes some amphiboles from other similar minerals. The term cleavage basically refers to the way a mineral cleaves, or breaks, in preferred directions. Cleavage directions are used to represent planes of weak bonding in the mineral's atomic structure.

Calcite is an example which have 3 directions of cleavage but not at 90° and galena also have 3 set of cleavage but at 90°.

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D20 exchange is a method used to determine the presence of certain protons in a molecule.

It involves exchanging a deuterium (D) nucleus for a hydrogen (H) nucleus in the molecule. By measuring the amount of exchange that occurs, it is possible to determine which protons are present and how strongly they are bound.

The formula for D20 exchange is:

[D]/[H] = k1/k2

Where [D] is the concentration of deuterium in the molecule, [H] is the concentration of hydrogen in the molecule, and k1 and k2 are the rate constants of the exchange reaction.

In this case, the presence of OH and NH2 protons is confirmed by D20 exchange, since both of these protons are known to exchange deuterium nuclei. CH protons, on the other hand, do not exchange deuterium nuclei, so their presence cannot be confirmed by this method. The exchange rate for OH and NH2 protons is generally much higher than that of CH protons, which makes them easier to detect.

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Based on the Laws of Thermodynamics, when a beaker of isobutyl alcohol is put inside a sealed tank containing isobutyl alcohol gas at 103.

Degree c and 619. torr, the pressure inside the tank will be higher than the pressure in the beaker, causing some of the liquid to evaporate and increase the liquid volume in the beaker. Therefore, after ten minutes, there will be more liquid in the beaker than before.

When a beaker of isobutyl alcohol is put inside a sealed tank containing isobutyl alcohol gas at 103°C and 619 torr, the molecules of isobutyl alcohol gas in the tank will be more energetic due to the increased temperature. As a result, they will move faster and collide with each other more frequently, raising the pressure inside the tank. The higher pressure in the tank will push molecules of isobutyl alcohol gas into the beaker, where they will condense into liquid due to the decrease in pressure.

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It is favorable rather than the many oil to form the one large oil droplet droplets in an aqueous solution , the correct option is A) because there is less ordered water with large oil droplets, so higher entropy overall.

The entropy is the measure of the randomness or the disorder in the solution. The entropy increases the randomness increases and when the entropy decreases the arrangement will be the more ordered.

Thus, Because of the less ordered water with large oil droplets, so higher entropy overall in the energetically favorable rather than the many oil to form the one large oil droplet droplets that is protein monomers in that is protein complex in an aqueous solution.

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The minimum concentration of dissolved oxygen required for a complete degradation of the organic matter in this water body is 4.30 mg/L.

To determine the minimum concentration of dissolved oxygen in the water body, we need to consider the stoichiometry of the reaction between the organic matter (CH2O) and oxygen (O2) to form carbon dioxide (CO2) and water (H2O):

CH2O + O2 → CO2 + H2O

The stoichiometric ratio between CH2O and O2 is 1:1, meaning that one molecule of CH2O reacts with one molecule of O2. Therefore, the minimum concentration of dissolved oxygen in the water body can be calculated as follows:

O2 (mg/L) = CH2O (mg/L) = 4.30 mg/L

So, the minimum concentration of dissolved oxygen required for a complete degradation of the organic matter in this water body is 4.30 mg/L.

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A sample of 0.53 g of carbon dioxide will be obtained by heating 1.31 g of calcium carbonate. The percent yield for this reaction will be 92%.

1CaCO₃(s) → 1Cao(s) + 1CO₂(g)

1.31 g                                  ?

% yield will be = actual yield / theoretical yield × 100

actual yield will be 0.53 gram and theoretical yield will be 0.576 gram

CaCO₃ will be 1.31 gram

1.31 gram of CaCO₃ → moles of CaCO₃ → moles of CO₂ → grams

1.31 g CaCO₃ × 1 mol CaCO₃ / 100.09 g CaCO₃ × 1 mole CO₂ / 1 mole CaCO₃ × 44.01g CO₂ / 1 mole CO₂ = 0.576 g of CO₂

% yield = 0.53 / 0.576 × 100

% yield = 92%

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In 22.4 gm of SnO₂ 8.933 * 10²²  molecules are present which is .148 moles.

The molar mass of a chemical compound is defined as the ratio between the mass and the amount of that substance in any sample. A material's molar mass is a bulk characteristic rather than a molecular one.

The molar mass of a substance is its mass expressed in grams per mole. G/mol, or grams per mole, is the sign for molar mass. The ratio between the mass of an isotope and the mass of the isotope carbon is known as the isotopic atomic mass, or mass of a single isotope of any particular element. -12

The Avogadro constant, also known as NA or L, is the proportionality factor that links the amount of material in a sample to the number of component particles in that sample. It is a SI defining constant with an exact value of 6.02214076*1023 reciprocal moles.

Molar mass of Sn = 119

Molar mass of O₂ = 2* 16 = 32

Molar mass of SnO₂ = 119 + 32 = 151 gm

In 151 gm of SnO₂ = 6.022*10²³ molecules are present

In 1 gm of SnO₂ = 6.022*10²³/151  molecules are present

In 22.4 gm of SnO₂ = 6.022*10²³* 22.4/151 = 8.933 * 10²² molecules are present

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Answer:

Explanation:

Reactivity of halogens basically decreases. Imagine this halogens are these people who need love, they need to attract metals who are hot.

The number of shells increases as you go down the group, so they become fatter. Fluorine is the hottest , chlorine is the second hottest, the rest are fat-fattest. The plus size elements (Bromine, Iodine) find it easier to lose an electron, because they are "insecure", the attraction between protons and valence electrons are not high, because they are so far away. Metals need their partners do be stiff so that they can give away their electrons. Halogens are not metals, they need to attract them, so it is ideal here to not lose electrons.

Hope that explanation helps, phew!

The pH of the given solution is 6.25 when it contains 6 x 10-5 mg/l of OH-

pH is measure if acidity or basicity of a chemical depending on the concentration of OH- ions or H+ ions present in it. Highly acidic and basic molecule have high concentration of H+ ions and OH- ions in the respectively. Highly acidic molecule have pH near 0 and basic molecule have it near 14. molecule which are neutral have pH of 7.

Consider volume of the solution is 1 L.

mass of OH=3x10-4 mg=3x10-7g

Molar mass of OH=17g/mole

Number of mol of OH- =mass /molar mass

=3×10-7 g / 17g mol-1

= -1.765 × 10-mol

Calculate concentration of OH.

[OH-] = number of moles / volume =  1.765×10¯8mol / 1L

=1.765x10-8 mol

pOH = -log[OH¯]

=-log(1.765×10-8)=7.75

Now calculate pH of the solution as shown below. pH = 14 - pOH

= 14 -7.75 = 6.25

Therefore, pH of the solution is 6.25.

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According to Rutherford's theory, an atom's nucleus is surrounded by electrons that are negatively charged.

He also asserted that the electrons that surround the nucleus travel in a circular pattern at extremely high speeds. He gave these elliptical routes the name orbits. A positively charged, compact, and tiny nucleus with protons and neutrons makes up an atom. The nucleus of an atom contains all of its mass. (2) Negatively charged electrons surround the nucleus. The electrons are moving extremely quickly in a circular pattern around the nucleus. According to Rutherford's theory, an atom's nucleus is surrounded by electrons that are negatively charged. He also asserted that the electrons that surround the nucleus travel in a circular pattern at extremely high speeds.

The complete question is- according to rutherford’s model of the atom, how many electrons would be found in each of the following atoms?

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Answer:

As you go up the fractionating column, the hydrocarbons have: lower boiling points lower viscosity (they flow more easily) higher flammability (they ignite more easily) This means that in general hydrocarbons with small molecules make better fuels than hydrocarbons with large molecules.

Energy from the sun promotes condensation, which in turn creates winds.

Water vapor turns into liquid through a process called condensation. Either the air cools to its air temperature or it gets so dissolves in water that it is no longer hold any more water, which leads to condensation. Dry Point

Condensation happens when warm air bumps into cold surfaces or when your environment is very humid. When this moist warm air strikes a chilly area, it swiftly cools down and sheds the water, which condenses into liquid on the chilly surface.

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By adding the specified compounds to 18 liters of deionized water, the chemical analysis of the resulting simulated groundwater would show the following concentrations:

Chemical analysis is a process of determining the chemical composition of a given sample. It can be used to identify the presence of particular elements and compounds, as well as to measure the concentrations of those components. Chemical analysis is typically done through a variety of techniques, such as spectroscopy, chromatography, and mass spectrometry.

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It goes as follows: Zn(NO₃)² + NaCl = ZnCl₂ + NaNO₃. Double displacement reaction is the name given to this kind of reaction.

In a double displacement reaction, two chemicals react and the affirmative (cation) and negative (anion) ions of the two reactants switch positions, creating two new compounds or outputs.

Although technically the bonds created between the different molecules may be either ionic or covalent in nature, the reactions most commonly occurs between ionic compounds. Double ejection processes also involve foundations or acids.

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The gradual color change observed when the sodium thiosulfate ([tex]Na_{2}S_{2}O_{3}[/tex]) crystal was added to the aqueous solution of KI/[tex]I_{2}[/tex] in Station C evidence of a chemical change.

Let's mix а solution of sodium thiosulfаte, [tex]Na_{2}S_{2}O_{3}[/tex], with iodine, [tex]I_{2}[/tex], dissolved in аqueous potаssium iodide, KI. The mixture of iodine аnd potаssium iodide mаkes potаssium triiodide. [tex]KI_{3}[/tex]. We will put the triiodide solution in the empty beаker аnd аdd some sodium thiosulfаte.

A balanced equation for the oxidation-reduction reaction between sodium thiosulfate and potassium triiodide:

[tex]S_{2}O_{3} ^{2-} + I_{2}\neq S_{4} O_{6} ^{2-} +I^{-}[/tex]

Thus, this equation shows evidence of a chemical change.

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