Why is carbon used to extract metals from their oxides
1-cheap
2- high_____

The Diels-Alder mechanism between a diene and a dienophile is concerted, which means that bond breaking happens simultaneously as bond forming. To help the mechanism succeed, the diene should have an electron-donating group and the dienophile should have an electron-withdrawing group.

Diels-Alder reaction is a [1,4]-electrocyclic reaction. It includes the formation of a cyclic molecule with six carbons. It's a concerted reaction where the reaction occurs simultaneously in a single step. The dienophile and diene are the two reactants.

The diene is a hydrocarbon with two double bonds, while the dienophile is a substance with one double bond that is rich in electrons. The concerted mechanism results from the electrocyclic addition of the dienophile to the diene and a single molecule are produced.

The diene is the nucleophile in this mechanism, whereas the dienophile is the electrophile in this mechanism. The reaction is best achieved when the dienophile has an electron-withdrawing group, and the diene has an electron-donating group. As a result, the Diels-Alder reaction is an excellent example of a stereoselective reaction.

The stereochemistry of the diene and dienophile determines the stereochemistry of the final product. The endo-product is generated when the dienophile and diene groups are near to one another, and the exo-product is generated when they are far apart.

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Answer to the question : D.                                                            

Answer:

my answer is C. having mass and taking space

Answer:

checking g down to jajrsw

Explanation:

Answer:

Explanation: Hello Sonic, the explanation and answer to your question is that When vinegar (acetic acid) is mixed with dolomite (a mineral composed of calcium magnesium carbonate), a chemical reaction occurs. The acetic acid reacts with the calcium and magnesium ions in the dolomite, causing them to dissolve into the vinegar solution. This results in the formation of calcium acetate and magnesium acetate, which are soluble in water. The chemical equation for this reaction is: CaMg(CO3)2 + 2CH3COOH → Ca(CH3COO)2 + Mg(CH3COO)2 + H2O + CO2 In this equation, CaMg(CO3)2 represents dolomite, CH3COOH represents acetic acid, Ca(CH3COO)2 represents calcium acetate, Mg(CH3COO)2 represents magnesium acetate, H2O represents water, and CO2 represents carbon dioxide gas. The reaction also produces bubbles of carbon dioxide gas, which can be seen as the mixture fizzes and bubbles. Overall, the reaction between vinegar and dolomite is an example of an acid-base reaction, where the acetic acid acts as the acid and the dolomite acts as the base. This reaction can be used to dissolve dolomite rocks or to create calcium and magnesium acetate solutions for various applications. Hope this helps :)

If the combustion reaction is reversible, the likely magnitude of the equilibrium constant, Kc would depend on the stoichiometric coefficients of the reactants and products involved in the reversible reaction, as well as the temperature and pressure conditions of the reaction.

If a combustion reaction was a reversible reaction, the likely magnitude of the equilibrium constant, Kc would be determined by the equation below:

Kc = [C]^c [D]^d / [A]^a [B]^b

Where A, B, C, and D are the reactants and products of the reversible reaction, and a, b, c, and d are their corresponding stoichiometric coefficients. Since hydrogen is used as a rocket fuel because it is very light and reacts explosively and completely with oxygen, its combustion reaction with oxygen is an exothermic reaction, and the combustion of hydrogen and oxygen is not a reversible reaction. It can be represented by the following equation:

2H2 + O2 → 2H2O

However, if the combustion reaction is reversible, the likely magnitude of the equilibrium constant, Kc would depend on the stoichiometric coefficients of the reactants and products involved in the reversible reaction, as well as the temperature and pressure conditions of the reaction.

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The percent mass of hydrogen in glucose is approximately 6.73%. The molar mass of glucose (C₆H₁₂O₆) can be calculated as follows:

Molar mass of C₆H₁₂O₆ = (6 x atomic mass of carbon) + (12 x atomic mass of hydrogen) + (6 x atomic mass of oxygen)

= (6 x 12.01 g/mol) + (12 x 1.01 g/mol) + (6 x 16.00 g/mol)

= 72.06 g/mol + 12.12 g/mol + 96.00 g/mol

= 180.18 g/mol

The mass of hydrogen in one molecule of glucose is 12 x 1.01 g/mol = 12.12 g/mol.

To calculate the percent mass of hydrogen in glucose, we can use the following formula:

Percent mass of hydrogen = (mass of hydrogen / molar mass of glucose) x 100%

Substituting the values, we get:

Percent mass of hydrogen = (12.12 g/mol / 180.18 g/mol) x 100%

= 6.73%

Therefore, the percent mass of hydrogen in glucose is approximately 6.73%.

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Explanation:

First, heat it from 55-100 C

11.8 g   ( 100 - 55) C  4.18 J / g C = 2219.6 J   = 2.22 kJ

Your heat of vaporization is in units of moles

   so 11.8 g of H2O =   11.8 gm / 18 gm /mole = .656 moles

Then

.656 moles * 40.6 kJ / mole = 26.6 kJ

Total kJ = 2.22 + 26.6 = 28.8 kJ

The formal charges ought to add up to the molecule's overall charge, which for HCNO is zero. The genuine electronic circulation in HCNO is a cross breed of the two reverberation structures displayed here.

Isocyanic acid, which is a small, unstable molecule made up of hydrogen, carbon, nitrogen, and oxygen atoms, has the chemical formula HCNO. It has the construction HNCO or H-O=C=N, where the nitrogen particle is clung to both the carbon and oxygen iotas through twofold bonds. At room temperature, isocyanic acid is a colorless gas that is very reactive, making it difficult to isolate and study. Numerous organic compounds, including ureas, carbamates, and biurets, are synthesized using this important intermediate. The thermal decomposition of urea also has the potential to produce it.

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To calculate Laila's average running speed, the following data should be recorded:

The distance she ran - This can be measured using a GPS device or by using a running track.

The time taken to run that distance - This can be measured using a stopwatch or a timer.

By dividing the distance by the time taken, we can calculate Laila's average running speed.

Average Speed = Distance / Time

For example, if Laila runs a distance of 5 kilometers in 30 minutes, her average running speed would be:

Average Speed = 5 km / (30/60) hr = 10 km/hr

Therefore, to calculate Laila's average running speed, we need to record the distance she ran and the time taken to run that distance.

The two conformations of the same disaccharide shown on the right likely differ in their free energy levels due to the arrangement and interactions of their constituent molecules. The top conformation, with a lower free energy, is likely more stable because of the formation of favorable interactions, such as hydrogen bonding, between the mono saccharide units, and a more optimal orientation of their functional groups.

In addition, the lower free energy conformation could be attributed to a reduced level of steric hindrance, where the spatial arrangement of the atoms in the disaccharide allows for minimal clashes and more efficient packing of the molecule. This can lead to a more stable and energetically favorable structure.

Moreover, the stability of the top conformation may be further enhanced by the presence of intramolecular forces, such as van der Waals interactions and hydrophobic effects. These interactions can significantly contribute to the overall stability of the molecule and subsequently result in a lower free energy state.

In summary, the top conformation of the disaccharide likely has a lower free energy due to a combination of factors, including favorable interactions between the monosaccharide units, optimal orientation of functional groups, reduced steric hindrance, and the presence of intramolecular forces. These factors collectively contribute to a more stable and energetically favorable molecular structure.

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Metallic behavior correlates with large atomic size and low ionization energy. This is true. In metallic behavior, a metal atom tends to lose electrons easily and form positive ions. When these atoms are packed together in a solid metal, they form a lattice that consists of positively charged ions and free electrons, which are free to move throughout the lattice.

When atoms have low ionization energies, it is easier for them to lose electrons and become positive ions. So, large atomic size and low ionization energy can lead to metallic behavior. A larger atomic size means that there is more distance between the nucleus and the outermost electrons. This means that the valence electrons are not as strongly attracted to the nucleus, and it is easier for them to be removed. Low ionization energy means that the energy required to remove an electron from an atom is low. So, when an atom with low ionization energy is in contact with another atom, it tends to lose electrons easily and become a positive ion. This positive ion then attracts the electrons from the other atom, forming a metallic bond. Thus, metallic behavior correlates with large atomic size and low ionization energy. This is because these properties make it easier for metal atoms to lose electrons and become positive ions, leading to metallic bonding.

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The intermolecular forces present in each of the substances are:

1. CH4 (Methane) - Dispersion forces only
2. CH3OH (Methanol) - Dispersion forces, dipole-dipole forces, and hydrogen bonding
3. HOCH2CH2OH (Ethylene glycol) - Dispersion forces, dipole-dipole forces, and hydrogen bonding
4. HCN (Hydrogen cyanide) - Dispersion forces and dipole-dipole forces. Dipole-dipole interactions are attractive forces among polar molecules. Polar molecules have permanent dipoles that are formed due to differences in the electronegativities of the atoms that are associated with a covalent bond.

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The presence of waters of hydration can affect the enthalpy of a substance when it is dissolved. Hydration occurs when a substance is surrounded by or interacts with water molecules.

When a substance is dissolved, the attraction between the solute molecules and the water molecules can cause an increase in the enthalpy of the reaction. This is because more energy is needed to separate the molecules and allow them to interact with the solvent. When hydrated, how does water affect the enthalpy for dissolving?The addition of waters of hydration increases the enthalpy of dissolving.The enthalpy change of dissolving a compound refers to the heat energy absorbed or released when the compound dissolves in water

. Hydration water is present in most solids. This water exists as coordinated water molecules that are bound to the ions present in the solid. As a result, when a compound dissolves in water, hydration water is also present. As a result, when a substance dissolves in water, it consumes energy to break the water's hydrogen bonds and dissolve the compound's particles. The enthalpy change of dissolving a compound is affected by the number of water molecules consumed in the hydration process. The enthalpy of dissolving a compound rises as the number of hydration water molecules increases.

Hence, when water is used to hydrate a substance, it raises the enthalpy of dissolution.The enthalpy of hydration is the energy required to hydrate a mole of ions. The hydration energy is released when the ions and water molecules interact, and the hydration shells form around the ions. The enthalpy change of dissolving a compound, like the enthalpy of hydration, is influenced by the hydration process. The enthalpy of hydration, on the other hand, is negative because it is a release of energy. The hydration energy offsets some of the energy needed to dissolve the compound's particles, resulting in a lower enthalpy of dissolving.

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When answering questions on the Brainly platform, it is important to always be factually accurate, professional, and friendly. It is also important to be concise and not provide extraneous amounts of detail, while using the relevant terms from the student question.Here is the answer to the student's question:

To calculate the partial pressure of argon gas in a mixture of gases, we can use the following formula:Partial pressure of gas = (Moles of gas / Total moles of gas) x Total pressure of mixture Given that the mixture of gases contains 0.50 moles of helium, 1.5 moles of argon, and 2.5 moles of neon and the total pressure of the mixture is 1.6 atm, we can find the partial pressure of argon as follows:Total moles of gas = 0.50 + 1.5 + 2.5 = 4.5 moles Moles of argon gas = 1.5 moles Partial pressure of argon gas = (1.5 / 4.5) x 1.6 atm= 0.5 x 1.6 atm= 0.8 atm Therefore, the partial pressure of argon gas in the given mixture of gases is 0.8 atm.

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The rotational constant, B, of a molecule depends on its deuterated molecule mass and bond length. Isotopic exchange can change both of these parameters and therefore alter the rotational spectra of a molecule.

For the deuterated molecule, D79Br, the reduced mass will increase because the mass of the hydrogen atom is replaced with the heavier deuterium atom. The bond length is not expected to change significantly. The increased reduced mass will result in a decrease in the rotational constant, B, compared to H79Br.

For the bromine exchanged molecule, H81Br, the reduced mass is expected to be similar to that of H79Br, but the bond length will be longer due to the larger size of the 81Br isotope. The longer bond length will result in a decrease in the rotational constant, B, compared to H79Br.

Isotopic exchange generally decreases the rotational constant, B, because it increases the reduced mass or changes the bond length. The exchange of 1H or 79Br has a greater effect on the rotational spectra of a molecule depends on their relative contribution to the reduced mass and bond length of the molecule. In the case of H79Br, the bond length is largely determined by the H-Br bond, so the exchange of 79Br has a greater effect on the rotational spectra than the exchange of 1H. However, the effect of isotopic exchange on the rotational spectra can be complex and may also depend on other factors, such as the electronic structure of the molecule.

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The pH of the solution in case of titration after the addition of 40.0 ml KOH is 21.07.

The given balanced chemical equation is HClO (aq) + KOH (aq) → KClO (aq) + H2O (l).Calculate the pH for each case in the titration of 50.0 ml of 0.200 M HClO (aq) with 0.200 M KOH (aq). The ionization constant for HClO is given asKa = 3.0 × 10-8. To calculate the pH of the solution, we use the following formula:

pH = - log10[H+], The number of moles of HClO is: Moles = Molarity × Volume, Moles = 0.200 mol/L × 50.0 mL × 1 L/1000 mL = 0.010 mol. The number of moles of HClO decreases by the number of moles of KOH. Therefore, the number of moles of HClO left in the solution is0.010 mol - 0.008 mol = 0.002 mol. The number of moles of KClO formed is equal to the number of moles of KOH added, which is0.008 mol.

The number of moles of HClO that has reacted is also equal to the number of moles of KClO formed, which is0.008 mol. The number of moles of HClO remaining in the solution is 0.010 mol - 0.008 mol = 0.002 mol. The concentration of HClO is 0.002 mol/60.0 mL = 0.033 M. The concentration of KOH is0.008 mol/90.0 mL = 0.089 M. Using the ionization constant, we can calculate the pH for the titration of HClO with KOH, pH = pKa + log([A-]/[HA])

The H+ concentration is calculated as follows:[H+] = Ka × [HA] / [A-][H+] = (3.0 × 10-8) × 0.002 mol / 0.008 mol[H+] = 7.5 × 10-9 mol/L. The pH of the solution after the addition of KOH is pH

= 14.00 - pOHpOH

= - log10[OH-]pOH

= - log10(0.089)pOH = 1.05pH = 14.00 - 1.05 - log10[H+]pH = 12.95 - log10(7.5 × 10-9)pH = 12.95 + 8.12pH = 21.07

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Lignin decomposition played a crucial role in the rise and fall of oxygen levels during the Carboniferous period.

During the Carboniferous period, forests covered vast areas of the Earth, and the growth of trees and plants led to the accumulation of lignin in soils and sediments. Lignin is a complex organic polymer that is resistant to decomposition, and its buildup in the soil prevented the complete decay of organic matter.

As a result, oxygen was continually produced and released into the atmosphere through photosynthesis, while the carbon that was not decomposed was stored in the form of coal.

However, as the amount of lignin increased over time, the rate of oxygen production eventually reached a plateau, and the oxygen levels in the atmosphere began to decline. As lignin-rich organic matter accumulated in the soil and was buried over time, the oxygen levels continued to fall, and the carbon in the soil was converted into coal through heat and pressure.

Overall, lignin decomposition played a critical role in shaping the atmospheric conditions during the Carboniferous period, and its buildup and eventual burial had a significant impact on the rise and fall of oxygen levels on Earth.

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The classification of radicals are :

(CH₃)CHCH₂CH₂ = The Secondary radical

(CH₃)CHCHCH₃ = The Tertiary radical

H₂C = The Primary radical

The radical is the atom or the molecule with the unpaired electron. The unpaired electron will gives the radical the chemical reactivity, as it is the highly reactive because of the unpaired electron.

The Primary free radicals: When the radical is attached to the two H-atoms.

The Secondary free radicals: When the radical is attached to the one H-atoms.

The Tertiary free radicals: When the radical is not attached to the any Hydrogen atoms.

The unpaired electron in the free radical is the highly reactive, the leading to the radical reactions that can be the important in the many chemical processes.

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This question is incomplete, the complete question is :

classify the radicals into the appropriate categories. Primary. Secondary. Tertiary. Allylic.

(CH₃)CHCH₂CH₂, (CH₃)CHCHCH₃, H₂C

molar heat of vaporization for liquid water is 40.6 kJ/mole. 15.54 kJ of energy is required to change 6.9 g of liquid water to steam if the water is already at 100°C.

To calculate the energy required to change 6.9 g of liquid water to steam, we need to use the formula:

q = n × ΔH_vap

where q is the energy required, n is the number of moles of water being vaporized, and ΔH_vap is the molar heat of vaporization for water, which is 40.6 kJ/mole.

First, we need to calculate the number of moles of water in 6.9 g of liquid water. We can do this using the molar mass of water, which is 18.015 g/mole:

n = m / M

n = 6.9 g / 18.015 g/mol

n = 0.383 moles

Now we can use this value of n and the molar heat of vaporization to calculate the energy required to vaporize the water:

q = n × ΔH_vap

q = 0.383 moles × 40.6 kJ/mole

q = 15.54 kJ

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The number of moles of MgCO₃ is also 0.200 moles and the mass of magnesium carbonate at the start was 16.86 g.

The balanced chemical equation for the reaction between sulfuric acid (H₂SO₄) and magnesium carbonate (MgCO₃) is:

H₂SO₄ + MgCO₃ → MgSO₄ + CO₂ + H₂O

From the balanced equation, we can see that one mole of sulfuric acid reacts with one mole of magnesium carbonate to produce one mole of magnesium sulfate, one mole of carbon dioxide, and one mole of water.

The moles of magnesium carbonate at the start can be calculated from the number of moles of sulfuric acid that reacted with it, using the mole ratio from the balanced equation:

1 mole of MgCO₃ will reacts with 1 mole of H₂SO₄

Therefore, the number of moles of MgCO₃ is also 0.200 moles.

The mass of magnesium carbonate at the start can be calculated using its molar mass and the number of moles:

molar mass of MgCO₃ = 24.31 g/mol (for Mg) + 12.01 g/mol (for C) + 3(16.00 g/mol) (for 3 O) = 84.31 g/mol

mass of MgCO₃ = number of moles × molar mass

mass of MgCO₃ = 0.200 moles × 84.31 g/mol

mass of MgCO₃ = 16.86 g

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Being the best fit for the situation at hand can aid in supremacy. Plant A therefore dominance the other plants.

The most plentiful producers in the environment are dominant primary producers. Savanna grass is a prime illustration of a dominant primary producer. Keystone species, on the other hand, are by definition those that play a significant role in maintaining the ecosystem's balance despite not being particularly numerous.

According to Mendel's law of dominance, only one version of a trait will manifest in the offspring of two parents who have distinct, opposing traits. The only trait in the phenotype that the hybrid progeny will display is the dominant trait.

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The limiting reagent is TMSCl. This is because it is present in the smallest amount compared to all other reactants.

Amount is a noun used to describe the total number or quantity of something. It is used to refer to an aggregate of items, materials, or money. Amount can also be used figuratively to refer to an indefinite quantity that is not easily measured or counted. For example, one might say "there was an amount of people at the party".

Limiting reactants are the reactants that are used up first in a chemical reaction, and they limit the amount of product that can be formed. In this case, the amount of carvone and NaHCO3 (aq) is greater than the amount of TMSCl, so TMSCl is the limiting reactant.

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Rutherford's gold foil exploration showed that the particle is generally unfilled space with a small, thick, emphatically charged core. In view of these outcomes, Rutherford proposed the nuclear model of the particle.

The following pivotal examination throughout the entire existence of the iota was performed by Ernest Rutherford, a physicist from New Zealand who burned through a large portion of his vocation in Britain and Canada. In his popular gold foil try, Rutherford terminated a meager light emission

αalpha particles (articulated alpha particles) at an extremely slight sheet of unadulterated gold. Alpha particles are helium cores

[tex](24He)^2( 24He)^2[/tex], begin addendum, 2, end addendum, begin superscript, 4, end superscript, begin a text, H, e, end text, begin superscript, 2, or more, end superscript, right enclosure, and they are radiated in different radioactive rot processes. For this situation, Rutherford set an example of radium (a radioactive metal) inside a lead confine with a little pinhole it. The majority of the radiation was consumed by the lead, however, slim light emission particles got away from out of the pinhole toward the gold foil. The gold foil was encircled by an indicator screen that would streak when hit with an αalpha molecule.

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With a pressure rise to 11.1 atm, the hydrogen gas volume is 2.03 L.

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas. The equation for the combined gas law is:

P₁V₁/T₁ = P₂V₂/T₂

Where P₁, V₁, and T₁ are the initial pressure, volume, and temperature, respectively, and P₂, V₂, and T₂ are the final pressure, volume, and temperature, respectively.

We can rearrange this equation to solve for V₂:

V₂ = (P₁V₁T₂)/(P₂T₁)

We are given that the initial volume (V₁) is 4.4 L, the initial pressure (P₁) is 4.9 atm, and the final pressure (P₂) is 11.1 atm. We are asked to find the final volume (V₂) when the pressure is increased to 11.1 atm.

We can assume that the temperature (T₁) is constant, since it is not specified otherwise. Therefore, we can substitute the values into the equation and solve for V₂:

V₂ = (P₁V₁T₂)/(P₂T₁)

V₂ = (4.9 atm)(4.4 L)(273 K)/(11.1 atm)(273 K)

V₂ = 2.03 L

Therefore, the volume of the hydrogen gas is 2.03 L when the pressure is increased to 11.1 atm.

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These next statements are true because of the Bohr model of the hydrogen atom:

These statements are true because of the Bohr model of the hydrogen atom. The Bohr model describes the way the electrons move in a hydrogen atom. This model shows that electrons are arranged in shells around the nucleus. Each shell corresponds to a specific energy level, and each energy level has a fixed circular orbit of the electron around the nucleus. When the electron is in the orbit closest to the nucleus, it has its lowest energy state.

When the electron absorbs a photon with energy equal to the new energy state, it moves to a higher energy state. Only certain energy levels are allowed within the hydrogen atom because electrons can only exist in specific energy levels.

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The amount of water that crystallises is 305.1%.

Crystalization is a process of separating solid particles from a solution. It occurs when a solvent, such as water, is mixed with a solute, such as salt or sugar.The percentage of water of crystallization in 35.7g Na₂CO₃ 10 H₂O can be calculated using the following equation:

Percentage of water of crystallization =[tex](\frac{Mass of water of crystallization}{Mass of anhydrous salt}) * 100[/tex]

Mass of water of crystallization = 10 * (18.015 g/mol) = 180.15 g

Mass of anhydrous salt = (2 * (22.99 g/mol) + 1 * (12.01 g/mol)) = 58.97 g

The percentage of water of crystallization =[tex](\frac{180.15 g}{58.97 g}) * 100 = 305.1 \%.[/tex]

Therefore,The percentage of water of crystallization is 305.1%.

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