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Horner-Wadsworth-Emmons (HWE) reaction is an important synthetic reaction in organic chemistry. It is widely used for synthesizing various compounds. The reaction is between an aldehyde or ketone and a phosphonate or phosphonate ester in the presence of a strong base.

The Horner-Wadsworth-Emmons reaction is one of the most convenient and well-known methods of constructing carbon-carbon double bonds. The reaction proceeds via the formation of an ylide intermediate. The HWE reaction is particularly useful for the synthesis of compounds with a Z-configuration.

The mechanism for the reaction of diethyl benzylphosphonate and 3,4-methylenedioxybenzaldehyde in the presence of aqueous NaOH, forming 3,4-methylenedioxystilbene as the product, can be explained in the following steps:

Step 1: Formation of the ylide intermediate

The reaction starts with the formation of an ylide intermediate. This is achieved by the reaction of diethyl benzylphosphonate and 3,4-methylenedioxybenzaldehyde in the presence of a strong base like NaOH or KOH. In this reaction, a deprotonated species called an ylide intermediate is generated.

Step 2: Addition of the ylide intermediate to the aldehyde

The ylide intermediate then attacks the aldehyde, leading to the formation of a betaine intermediate.

Step 3: Formation of the phosphonate ester

The betaine intermediate undergoes elimination to form the final product, 3,4-methylenedioxystilbene, and the by-product phosphonate ester.

The mechanism of the Horner-Wadsworth-Emmons (HWE) reaction between diethyl benzylphosphonate and 3,4-methylenedioxybenzaldehyde in the presence of aqueous NaOH, forming 3,4-methylenedioxystilbene as the product, is complete. This reaction is significant in organic chemistry and finds applications in the pharmaceutical industry.

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Yes, the qs for the formation and decomposition of 1 mol of NO have opposite signs.

The heat of formation, or the heat absorbed when 1 mol of a substance is formed from its elements, represents the symbol qf. The heat of decomposition or the heat is given off when 1 mol of a substance breaks down into its elements, is represented by the symbol qd. The relationship between the two qs is that qf and qd are equal in magnitude but opposite in sign; qf is negative and qd is positive.

The relationship between qf and qd is that they are equal in magnitude but opposite in sign. For example, if qf for the formation of NO is -100 kJ, then qd for the decomposition of NO will be +100 kJ.

In summary, qf and qd is the heat absorbed or released during the formation and decomposition of a substance respectively. They are equal in magnitude but opposite in sign. The heat absorbed during formation is negative and the heat released during decomposition is positive.

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When NH₄OH and CaS are mixed the following reaction occurs:

NH₄OH + CaS ---> Ca(OH)₂ + (NH₄)₂S

Ca(OH)₂ is a poorly soluble base in water. So the formula of the precipitade formed is Ca(OH)₂ and its name is calcium hydroxide.

Bases with cations of the group II of the periodic table are generally poorly soluble in water.

The formula [Cu(C2O4)2]2- represents the specific complex formed between Cu2+ and oxalate with a coordination number of 4.

The formula for the complex formed between Cu2+ and oxalate (C2O2−4) with a coordination number of 4 is [Cu(C2O4)2]2-. In this complex, Cu2+ acts as the central metal ion, while the oxalate ligand (C2O2−4) coordinates around it.

The coordination number of 4 suggests that four oxalate ligands are involved in binding to the central Cu2+ ion. Each oxalate ligand has two negatively charged oxygen atoms (O2-) and can form two coordination bonds with the metal ion, resulting in a total of four coordination bonds.

The resulting complex, [Cu(C2O4)2]2-, consists of the Cu2+ ion at the center surrounded by two oxalate ligands, with each ligand providing two oxygen atoms for coordination. The overall charge of the complex is 2-, balancing the charge of the Cu2+ ion.

Therefore, the formula [Cu(C2O4)2]2- represents the specific complex formed between Cu2+ and oxalate with a coordination number of 4.

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Answer:

Explanation:

Decomposers get nutrients and energy by breaking down dead organisms and animal wastes. Through this process, decomposers release nutrients, such as carbon and nitrogen, back into the environment. These nutrients are recycled back into the ecosystem so that the producers can use them.

Answer:

False

Explanation:

Chemical reaction equation:

                   C₅H₁₂ + 8O₂ → 2CO₂ + 6H₂O

Most chemical reactions obey the law of conservation of mass. By so doing, the number of chemical elements on both sides of the expression must be balanced.

                 Reactants                                          Products

C                    5                                                           4

H                  12                                                           12

O                 16                                                             10

We see that for C and O, the number of atoms on both sides of the expression differs and so, it is not balanced.

The reaction depicted in the figure would take place in a fission reactor. Option 3.

A fission reactor, also known as a nuclear reactor, is a device that generates electricity by using the heat produced by a controlled nuclear chain reaction.

This reaction involves splitting atomic nuclei, a process known as nuclear fission, and releasing large amounts of energy in the form of heat.

The kind of reaction depicted by the image is a fission reaction - a reaction involving the fission of heavy nuclei to smaller ones. Thus, the correct place for it to take place is a fission reactor.

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Gay-Lussac's law: \(\frac{V1}{T1}\)=\(\frac{V2}{T2}\)

We need to convert temperature from Celsius to Kelvin

-55 C = 218 K

30 C = 303 K

V2 = V1*T2 / T1 = 0.550 *303 / 218 = 0.764 l

Answer:

67.8%

Explanation:

La reacción de descomposición del CaCO₃ es:

CaCO₃ → CO₂ + CaO

Donde 1 mol de CaCO₃ al descomponerse produce 1 mol de CO₂ y 1 mol de CaO.

Usando la ley general de los gases, las moles de dioxido de carbono son:

PV = nRT.

Donde P es presión (1atm), V es volumen (20L), n son moles de gas, R es la constante de los gases (0.082atmL/molK) y T es temperatura absoluta (15 + 273.15 = 288.15K). Reemplazando los valores en la ecuación:

PV / RT = n

1atmₓ20L / 0.082atmL/molKₓ288.15K = 0.846 moles

Como 1 mol de CO₂ es producido desde 1 mol de CaCO₃, las moles iniciales de CaCO₃ son 0.846moles.

La masa molar de CaCO₃ es 100.087g/mol. Así, la masa de 0.846moles de CaCO₃ es:

0.846moles ₓ (100.087g / mol) = 84.7g de CaCO₃

Así, la pureza del marmol es:

(84.7g de CaCO₃ / 125g) ₓ 100 =

If 10 moles of P4S3 were used, the mass of P4O6 produced would be 2838.8 grams.

To determine the number of grams of P4O6 produced from 10 moles of P4S3, we need to use the balanced chemical equation and the molar masses of the compounds involved.The balanced equation for the reaction between P4S3 and oxygen to produce P4O6 is:

P4S3 + 8 O2 → P4O6 + 6 SO2

From the balanced equation, we can see that the molar ratio between P4S3 and P4O6 is 1:1. This means that for every 1 mole of P4S3 consumed, 1 mole of P4O6 is produced.The molar mass of P4S3 is 220.25 g/mol, and the molar mass of P4O6 is 283.88 g/mol.

To calculate the mass of P4O6 produced, we can use the following equation:

Mass of P4O6 = Moles of P4O6 × Molar mass of P4O6

Since the molar ratio between P4S3 and P4O6 is 1:1, the number of moles of P4O6 produced is also 10 moles.

Mass of P4O6 = 10 moles × 283.88 g/mol = 2838.8 grams

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Answer: Radhika loves to eat sweets and chocolates. She asks her mother for the same before and after every meal. This is an example of bad eating habit.

Explanation:

Both sweets and chocolates contain high amount of fat and sugar. When Radhika is eating the same before and after every meal then it means she is consuming it in excess.

When we eat too much of sweets and chocolates then its remains get deposited in between our tooth and when they remain over there for hours then more amount of bacteria is formed.

Due to this bacteria our tooth starts to decay.

Hence, it is advised even by the doctors to eat sweets and chocolates occasionally and not regularly.  

Thus, we can conclude that Radhika loves to eat sweets and chocolates. She asks her mother for the same before and after every meal. This is an example of bad eating habit.

Answer:0

Explanation: zero because it is the most stable form of oxygen in its standard state

Explanation:

The Sun is Earth’s major source of energy, yet the planet only receives a small portion of its energy and the Sun is just an ordinary star. Many stars produce much more energy than the Sun

Fresh water pollutants  are substances which pollute fresh water and  industrial waste, is most harmful fresh water pollutant to man and aquatic organisms.

Pollutants are substances which cause harm when they are present in the environment.

Pollutants include chemicals such as petroleum and material such as sewage.

The presence of pollutants in freshwater results in water pollution and make the water unfit for drinking purposes and also harms aquatic life in freshwaters.

Some freshwater pollutants include:

Of these pollutants, the most dangerous fresh water pollutant is industrial wastes as they kill aquatic organisms most due to the presence of harmful chemicals in them.

Therefore, fresh water pollutants such as industrial waste is most harmful to man and aquatic organisms.

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The mass of CH2Cl2 that can be melted by applying 7.80 kJ of energy at the melting point can be calculated using the equation of q = m * c * ΔT, where q is the energy applied, m is the mass, c is the heat capacity, and ΔT is the difference between the final and initial temperatures. In this case, the mass can be calculated as m = q / (c * ΔT). Plugging in the given values yields a mass of 0.126 g, rounded to three significant figures.

Therefore, 7.80 kJ of energy can melt 0.126 g of solid CH2Cl2 at the melting point. The equation used for this calculation assumes that the heat capacity and melting point of CH2Cl2 remain constant throughout the process, and thus the calculated value is only an estimate.

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Answer:

3) 333.75 Grams

Explanation:

2Al + 6HCl --- 2AlCl3 + 3H2

Work out and label what we know and workout Ar and Mr

Al = 67.15grams         AlCl3

Al = Ar=27                  Mr = 133.5

Mass/Mr == Moles

Al = 2.5 Moles          AlCl3 = 333.75

Moles * Mr = Mass

2.5 * 133.5 = 333.75 Grams

4) Apply the same concept (numbers in front of elements are ratios, in this case 12 and 8) (when working out MR/AR ignore numbers in front)

1) Workout Mr/Ar (The relative atomic masses added up)

2) Do 36.72/(Mr of O2) = Your Moles

3) Do workout what the moles would be in the ratio 12:8 (Times/divide moles to find out the correct moles for 12:8

4) Times the new moles by the Mr of SO3

5) = Answer

In this reaction when heated, elemental phosphorus, p4, produces phosphine, ph3, and phosphoric acid, h3po4.The  grams of phosphine are produced if 56 g p4 have reacted are 38.4 g.

Write and stability the equation.

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Answer:

Hey mate...

Explanation:

This is ur answer....

Effervescence is the escape of gas from an aqueous solution and the foaming or fizzing that results from that release.

Hope it helps!

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Explanation:

Your answer is effervensecence

Answer:

magnesium nitride

Explanation:

To solve this problem, we need to use the principles of stoichiometry and combustion thermodynamics. The A/F ratio is the air-to-fuel ratio by mass, and it is defined as the mass of air required to completely combust one unit mass of fuel. The dew point temperature is the temperature at which the water vapor in the combustion products begins to condense, and it depends on the composition and temperature of the products.

The A/F ratio can be calculated as follows:

First, we need to write the balanced chemical equation for the combustion of ethane:

C2H6 + 3.5 O2 + (0.2 x 3.5) O2 -> 2 CO2 + 3 H2O + (0.2 x 3.5) N2

Here, the 3.5 O2 represents the stoichiometric amount of oxygen required to completely combust one unit mass of ethane, and the 0.2 x 3.5 O2 represents the excess oxygen supplied (20% excess air). The (0.2 x 3.5) N2 represents the corresponding amount of nitrogen in the excess air.

The molecular weights of the reactants and products are:

C2H6: 2 x 12.01 + 6 x 1.01 = 30.07 g/mol

O2: 2 x 16.00 = 32.00 g/mol

CO2: 1 x 12.01 + 2 x 16.00 = 44.01 g/mol

H2O: 2 x 1.01 + 16.00 = 18.02 g/mol

N2: 2 x 14.01 = 28.02 g/mol

Using these values, we can calculate the mass of air required to combust one unit mass of ethane:

A/F = (mass of O2 + mass of N2) / mass of C2H6

= [(3.5 x 32.00) + (0.2 x 3.5 x 28.02)] / 30.07

= 12.53

Therefore, the A/F ratio is 12.53 by mass.

To determine the dew point temperature of the products, we need to calculate the mole fractions of the water vapor and nitrogen in the products, and then use a psychrometric chart or equations to find the dew point temperature. Alternatively, we can use the following simplified equation:

Tdp = (243.12 x ln(RH/100) + 17.62 x T) / (17.62 - ln(RH/100) - T)

Here, Tdp is the dew point temperature in °C, RH is the relative humidity of the products, and T is the temperature of the products in °C.

Assuming the combustion products are initially at a temperature of 25°C and a relative humidity of 100% (i.e., the products are fully saturated with water vapor), we can calculate the mole fractions of the water vapor and nitrogen as follows:

Mole fraction of H2O = (n H2O) / (n H2O + n N2)

= (3 x 0.2) / [(3 x 0.2) + (3.5 x 0.2)]

= 0.3

Mole fraction of N2 = (n N2) / (n H2O + n N2)

= (3.5 x 0.2) / [(3 x 0.2) + (3.5 x 0.2)]

= 0.35

Using these values

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The metallic ion that has 27 electrons and that when united with chlorine forms CoCl2 is cobalt.

In a neutral atom, the number of electrons will be equal to the number of protons, so the number of electrons can be easily determined from the atomic number.

In this case, the metal ion has 27 electrons.

And since electrons = protons (P+)

And since Z (atomic number) is equal to P+

Then

z=27

And in the periodic table the element with Z equal to 27 is cobalt

And the compound that would be forming would be CoCl2

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Answer:

500mL.

Explanation:

Answer:

6 Energy levels

Explanation:

First Energy Level: 2

Second Energy Level: 8

Third Energy Level: 18

Fourth Energy Level: 25

Fifth Energy Level: 8

Sixth Energy Level: 2

Hope this helps.

This law states that matter can neither be created nor destroyed in an ordinary chemical reaction, but can change from one form to another.

What this implies is that when two elements or compound combine as a reactant to form a product, there's no loss of matter and the mass of the reactants must be equal to the mass of the product.

Given that

AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃

If we look closely at the above reaction, we would see that aluminuim is not balanced, potassium is also not balanced, bromine is not balance as well as sulphur and oxygen.

Let's put two moles attached to the AlBr₂ and three moles of K₂SO₄

2AlBr₃ + 3K₂SO₄ →

This would give us

6KBr + Al₂(SO₄)₃

If we add the two equations together,

2AlBr₃ + 3K₂SO₄ → 6KBr + Al₂(SO₄)₃

From the above, we have

2 atoms of Al on the reactant side and 2 atoms of Al on the product side

6 atoms of Br on the reactant side and 6 atoms of Br on the product side

6 atoms of K on the reactant side and 6 atoms of K on the product side

3 atoms of S on the reactant side and 3 atoms of S on the product side

12 atoms of O on the reactant side and 12 atoms of O on the product side

The reactions for which ΔH∘rxn is equal to the ΔH∘f of the product(s) are: 2Na(s) + Cl₂(g) → 2NaCl(s), C(s, graphite) + O₂(g) → CO₂(g), and CO(g) + 1/2O₂(g) → CO₂(g).

The enthalpy change of a reaction, ΔH∘rxn, is defined as the difference between the enthalpy of the products and the enthalpy of the reactants. The enthalpy of the products can be expressed as the sum of the enthalpies of formation of each product, ΔH∘f, while the enthalpy of the reactants can be expressed as the sum of the enthalpies of formation of each reactant, also ΔH∘f. Therefore, if ΔH∘rxn is equal to the sum of the ΔH∘f values of the products, it means that the enthalpy change of the reaction is equal to the enthalpy of formation of the products.

1. CaCO₃(g) → CaO + CO₂(g)
The enthalpy change of this reaction is the difference between the enthalpy of the products (ΔH∘f of CaO and CO₂) and the enthalpy of the reactant (ΔH∘f of CaCO₃). However, since the products are two different substances, we cannot simply add their enthalpies of formation. Therefore, ΔH∘rxn is not equal to the sum of the ΔH∘f values of the products.

2. Na(s) + 1/2Cl₂(l) → NaCl(s)
The enthalpy change of this reaction is the difference between the enthalpy of the product (ΔH∘f of NaCl) and the enthalpy of the reactants (ΔH∘f of Na and Cl₂). Since there is only one product, we can equate ΔH∘rxn to the ΔH∘f of NaCl. Therefore, ΔH∘rxn is equal to the sum of the ΔH∘f values of the products.

3. 2Na(s) + Cl₂(g) → 2NaCl(s)
Enthalpy change of this reaction is the difference between the enthalpy of the product (ΔH∘f of NaCl) and the enthalpy of the reactants (ΔH∘f of Na and Cl₂). However, there are two moles of product, so we need to multiply ΔH∘f of NaCl by 2. Therefore, ΔH∘rxn is equal to the sum of the ΔH∘f values of the products.

4. C(s, graphite) + O₂(g) → CO₂(g)
The enthalpy change of this reaction is the difference between the enthalpy of the product (ΔH∘f of CO₂) and the enthalpy of the reactants (ΔH∘f of C and O₂). Therefore, ΔH∘rxn is equal to the sum of the ΔH∘f values of the products.

5. CO(g) + 1/2O₂(g) → CO₂(g)
The enthalpy change of this reaction is the difference between the enthalpy of the product (ΔH∘f of CO₂) and the enthalpy of the reactants (ΔH∘f of CO and O₂). Therefore, ΔH∘rxn is equal to the sum of the ΔH∘f values of the products.

6. Na(s) + 1/2Cl₂(g) → NaCl(s)
Similar to reaction 2, the enthalpy change of this reaction is the difference between the enthalpy of the product (ΔH∘f of NaCl) and the enthalpy of the reactants (ΔH∘f of Na and Cl₂). Therefore, ΔH∘rxn is equal to the sum of the ΔH∘f values of the products.

Therefore, the reactions for which ΔH∘rxn is equal to the ΔH∘f of the product(s) are: 2Na(s) + Cl₂(g) → 2NaCl(s), C(s, graphite) + O₂(g) → CO₂(g), and CO(g) + 1/2O₂(g) → CO₂(g).

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A symmetric confidence interval is used to determine the degree of certainty for a specific estimate, and it is critical when it comes to hypothesis testing. When constructing symmetric confidence intervals, a precise estimate of the standard error is required.

A symmetric confidence interval is used to determine the degree of certainty for a specific estimate, and it is critical when it comes to hypothesis testing. When constructing symmetric confidence intervals, a precise estimate of the standard error is required. A symmetric confidence interval has the following characteristics: The lower boundary is equidistant from the estimate and the upper boundary is equidistant from the estimate. The sample distribution is symmetric, and the estimator is equal to the mean.
When determining whether a hypothesis test is two-tailed, we use symmetric confidence intervals. A two-tailed hypothesis test is when the null hypothesis is rejected or the alternate hypothesis is accepted when the result is either in the tail or in the central part of the distribution. Symmetric confidence intervals are particularly useful when testing the variance of a population. This is because the symmetric confidence interval contains the same percentage of the distribution as the central area of the distribution, which is the area containing the most likely values. The distribution of a symmetric confidence interval is particularly useful when it comes to two-sided hypothesis tests, and it provides more reliable results than an asymmetrical confidence interval would. Therefore, symmetric confidence intervals are frequently used to draw conclusions about two-sided hypothesis tests.

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Stirring is important for a Wittig reaction because it ensures that the reactants are thoroughly mixed and in contact with each other.

In a Wittig reaction, an aldehyde or ketone reacts with a phosphorus ylide to form an alkene. The reaction is usually carried out in a solvent such as tetrahydrofuran (THF) or dimethyl sulfoxide (DMSO), and typically requires heating to initiate the reaction.In order for the reaction to proceed efficiently, it is important to ensure that the reactants are well mixed and in contact with each other. Stirring the reaction mixture helps to achieve this by promoting mixing and dispersion of the reactants throughout the solvent. This helps to increase the reaction rate and ensure that the maximum amount of product is formed.

Stirring also helps to prevent localized heating or cooling, which can affect the reaction rate and yield.Stirring is particularly important when carrying out a Wittig reaction using a solid phosphorus ylide. In this case, it is important to ensure that the ylide is well dispersed throughout the solvent to ensure maximum contact with the aldehyde or ketone. If the ylide is not well mixed, the reaction may proceed slowly or not at all, leading to poor yields or incomplete conversion.

In conclusion, stirring is an important factor in achieving a successful Wittig reaction. It helps to ensure that the reactants are well mixed and in contact with each other, which promotes efficient and rapid reaction.

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Answer:

Tungsten(VI) fluoride, also known as tungsten hexafluoride, is an inorganic compound with the formula WF6. It is a toxic, corrosive, colorless gas, with a density of about 13 g/L (roughly 11 times heavier than air.[

Explanation:

The WF6 molecule is octahedral with the symmetry point group of Oh. The W–F bond distances are 183.2 pm. Between 2.3 and 17 °C, tungsten hexafluoride condenses into a pale yellow liquid having a density of 3.44 g/cm^3 at 15 °C. At 2.3 °C it freezes into a white solid having a cubic crystalline structure, the lattice constant of 628 pm, and a calculated density of 3.99 g/cm^3. At −9 °C this structure transforms into an orthorhombic solid with the lattice constants of a = 960.3 pm, b = 871.3 pm, and c = 504.4 pm, and the density of 4.56 g/cm3. In this phase, the W–F distance is 181 pm, and the mean closest intermolecular contacts are 312 pm. Whereas WF6 gas is one of the densest gases, with the density exceeding that of the heaviest elemental gas radon (9.73 g/L), the density of WF6 in the liquid and solid-state is rather moderate. The vapor pressure of WF6 between −70 and 17 °C can be described by the equation

log10 P = 4.55569 −

1021.208

T + 208.45,

where the P = vapor pressure (bar), T = temperature (°C).