Chlorination of ethylbenzene in the para position proceeds by way of a carbocation intermediate that has three resonance structures. Complete one of these resonance structures by dragging bonds, electrons and charges to the appropriate locations. Then check your answer.

A mixture is a physical combination of two or more compounds that retain their identities.

A mixture in chemistry is a substance composed of two or more chemical components which are not chemically connected. A mixture is a physical combination of two or more compounds that retain their identities and therefore are mixed as solutions, suspensions, as well as colloids.

Silver                                         element

Ag and fluorine                          mixture

F                                                  element

carbon monoxide                      compound

CO and calcium chloride         mixture

CaCl[tex]_2[/tex]                                          compound

soft drink                                   mixture

Therefore, a mixture is a physical combination of two or more compounds that retain their identities.

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The pure liquid that is the best solvent for carbon disulfide is C₆H₆, also known as benzene. Option C is correct.

The best solvent for carbon disulfide (CS₂) would be a liquid that has a similar polarity and intermolecular forces to CS₂. Carbon disulfide is a nonpolar molecule, which means that it is most likely to dissolve in a nonpolar solvent.

Benzene is a nonpolar molecule and has a similar structure and intermolecular forces to CS₂, which makes it a good solvent for nonpolar solutes such as carbon disulfide.

In contrast, the other options (HBr, CH₃OH, H₂O, and NH₃) are polar molecules, which means that they are unlikely to dissolve nonpolar solutes such as CS₂ as effectively as benzene.

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--The given question is incomplete, the complete question is

"Which of the following pure liquids is the best solvent for carbon disulfide ? A) HBr B) CH₃OH C) C₆H₆ D) H₂0 E) NH₃"--

The  answer for the bond polarity of CHCI H o=c=0 H-C-CH2CI :0-H are:

Bond polarity refers to the distribution of electrical charge across the atoms in a covalent bond. A covalent bond is formed when two atoms share electrons in order to complete their outermost electron shell and achieve a more stable electron configuration. In a polar covalent bond, the electro-negativity difference between the two atoms results in one atom having a partial positive charge and the other atom having a partial negative charge. This creates an electric dipole moment and gives the bond a directional characteristic, with the negative end pointing towards the more electronegative atom and the positive end pointing towards the less electronegative atom.

In contrast, a non-polar covalent bond occurs when the two atoms have similar electro-negativities and there is no significant difference in the distribution of electrons between the two atoms. This results in a bond with no electric dipole moment and no polarity.

So the final answer would be:

CHCI H o=c=0 H-C-CH2CI :0-H

→ ←

H O

For the bond polarity, the direction of the arrow indicates the more electronegative atom. A bond is considered polar if the atoms involved have different electro-negativities, leading to a partial charge on each atom. If the bond is non-polar, both atoms have similar electro-negativities and there is no partial charge on either atom.

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212m^3 will be volume of N2

What do the ideal gas equations mean?

PV = nRT is the real gas equation or the ideal gas law. P stands for pressure, V for volume, n for moles of gas, R for the ideal gas constant, and T for Kelvin temperature.

The reaction is :

NaNO2(aq) + H2NSO3H(S) ⇒ NaHSO4(aq) +N2(g) +H2O(l)

1 mole of H2NSO3H produce 1 mole of N2

Molar mass of H2NSO3H is 97gm /mol

Molar mass of N2 will be 28gm/mol

No. of moles of N2 is 1

Using formula for number of moles, we get mass of N2 as

no. of moles * molar mass

i.e. 1*28 ⇒28g

Pressure is 0.978

Temperature is 25 degree

PV ⇒nRT

n is 1

R is 8.314

V ⇒ nRT/P

   ⇒ 1*8.314*25/0.978

  ⇒ 212m^3

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The equipment can be properly calibrated to ensure that the spray application rate is consistent and accurate, which is important for achieving the desired level of control while minimizing waste and avoiding off-target effects.

To prepare for equipment calibration, the following steps should be taken:

Check that all nozzles are the same and of the desired flow rate and pattern. Clean all nozzle and screens. Add clean water to the spray tank and adjust the pressure to the desired level while the pump is operating at normal speed. Make sure water is flowing through the nozzles.

Check the spray patterns from each nozzle. With the sprayer stationary and operating at the desired pressure, hold a measuring cup marked in fluid ounces directly under each nozzle for 1 min.

Calculate the average output (flow rate) by dividing the total amount collected by the number of nozzles. Replace any nozzle whose flow rate is 5% greater or less than the average.

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The correct matching is of pair of items is :

Known quantities in the problem --> Often provide useful conversion factors

Unknown quantity --> Will be the solution to the problem

Units of conversion factors --> Must cancel at each step, except for the units needed for the answer

Any units that do not cancel --> Are the units for the unknown quantity

1)The known quantities in the problem often provide useful conversion factors.

2)The unknown quantity will be the solution to the problem.

3)The units of conversion factors must cancel at each step, except for the units needed for the answer.

4)Any units that do not cancel are the units for the unknown quantity.

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Bond energy is described as a measure of the bond strength of a chemical bond, and it is the amount of energy required for breaking the atoms involved in a molecular bond into free atoms. The answers for multiple questions are given below.

i)

When bond is formed energy is released and to break a bond energy must be applied.

∆Hsoln is positive , so A < D

Comparing D to B and C , molecules are dissociated in B and molecules are more dissociated in C , so ∆H is order is D < B < C

Therefore, increasing order of enthalpy is A < D < B < C

ii)

Comparing to stage C to D molecules are associated , so heat is released.

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The equation ΔH=−110kJ, ΔS=+40JK −1 at 400K is spontaneous and exothermic.

And the equation ΔH=+40kJ, ΔS=−120JK−1 at 250K is non spontaneous and endothermic.

The spontaneity of a reaction is determined by the sign of the Gibbs free energy change (ΔG), which is related to enthalpy change (ΔH) and entropy change (ΔS) by the equation:

ΔG = ΔH - TΔS

where T is the temperature in Kelvin.

(a) ΔH=−110kJ, ΔS=+40JK−1 at 400K At 400K, ΔG = ΔH - TΔS = -110 kJ - (400 K)(+40 J/K) = -110 kJ - 16 kJ = -126 kJ. Since ΔG is negative, the reaction is spontaneous. The negative ΔH value indicates that the reaction is exothermic, meaning it releases heat.

(b) ΔH=+40kJ, ΔS=−120JK−1 at 250K At 250K, ΔG = ΔH - TΔS = +40 kJ - (250 K)(-120 J/K) = +40 kJ + 30 kJ = +70 kJ. Since ΔG is positive, the reaction is non-spontaneous. The positive ΔH value indicates that the reaction is endothermic, meaning it absorbs heat.

Therefore, the reaction in (a) is spontaneous and exothermic, while the reaction in (b) is non-spontaneous and endothermic

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A. There are two amide functional groups are present in vancomycin.

B. In phenol OH groups are bonded to SP³ hybridized carbon atoms and which are bonded to SP² hybridized carbons.

Phenol is an aromatic organic compound. They have molecular formula C₆H₅OH. It is a white crystalline solid that is volatile. The molecule consists of a phenyl group bonded to a hydroxy group.

The OH group is attached to the sp2 hybridized carbon of an aromatic ring in phenols.

C. Vancomycin is a very water-soluble drug.

D. Alcohols, amines (primary and secondary), carboxylic acids, and amides are the four pure functional groups capable of hydrogen bonding (primary and secondary).

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The element is aluminum (Al), which is in the third period of the periodic table and has the electron configuration [Ne]3s²3p¹.

The element can be identified by analyzing the ionization energies. The third row of the periodic table contains elements with the electron configuration [Ar] 3d¹⁰4s²4p¹. Based on the given ionization energies, we can identify the element by finding the period that contains the first eight ionization energies listed.

Looking at the given ionization energies, we can see that the first ionization energy is relatively low, indicating that the element is a metal. The next seven ionization energies increase gradually, indicating that they correspond to removing electrons from successive energy levels.

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The number of atoms of copper present in 3.271×10⁻¹⁹ grams of cupper (II) sulfate is 1234.1 atoms

First, we shall determine the mole of 3.271×10⁻¹⁹ grams of cupper (II) sulfate. Details below:

Mole = mass / molar mass

Mole of CuSO₄ = 3.271×10⁻¹⁹ / 159.55

Mole of CuSO₄ = 2.05×10⁻²¹ mole

Next, we shall determine the mole of copper in the sample. Details below:

1 mole of CuSO₄ contains 1 mole of copper, Cu

Therefore,

2.05×10⁻²¹ mole of CuSO₄ will also contain 2.05×10⁻²¹ mole of copper.

Finally, we shall determine the number of atoms of copper present in the sample. Details below:

From Avogadro's hypothesis,

1 mole of copper = 6.02×10²³ atoms

Therefore,

2.05×10⁻²¹ mole of copper = 2.05×10⁻²¹ × 6.02×10²³

2.05×10⁻²¹ mole of copper = 1234.1 atoms

This, the number of atoms of copper is 1234.1 atoms

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The concentration on which we should pipette out are 4μl.

To recap, the working solutions are 20μg/μl, 2μg/μl, and 0.2μg/μl.

1.

It’s basically the same as the other question. Set it up algebraically:

50μg/μl x = 200 μg

Note the variable x in the equation. You can see easily that x = 4μl (remember to include the units, and recall that units cancel in equations).

So if you pipet 4μl of the 50μg/μl working solution into a centrifuge tube, you’ll have 20μg of BSA in there.

2.

Again, set it up algebraically:

5μg/μl x = 20 μg

Here, x = 4μl again. So again you are pipetting out 4μl, except this time it’s from the 5μg/μl working solution.

3.

I’m sure you can see that there’s a trend going on here (it’s not any special science trend or anything like that, though).

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The cations can be identified by using various test, that are explained in the below section, Cations such as Na+, Ca2+, Cr3+, Fe3+, Al3+ and many other cation's test are explained below.

Na+ produces yellow flames in flame test. So it confirms the presence of Na+ ion.

If we add NaOH in sample and it produces vapor and turns red litmus, then NH4+ is present. But Here there is no change in litmus. So NH4+ ion CANNOT be present.

The addition of HCl does NOT produces white precipitate indicates that Ag+ CANNOT be present there. When we add HCl in Ag+ , a white precipitate of AgCl is formed.

The addition of buffered NH/NH3 to a portion of the solution produces a precipitate. Cr3+ , Fe3+ , Al3+ produces precipitate in addition of NH/NH3 buffer. So Cr3+ , Fe3+ , Al3+ CAN be present there but further test can confirm it.

Ammonium oxalate produces white precipitate in presence of calcium ion. So Ca2+ must be present there.

The addition of a strong base will precipitate Mg2+ and Ni2+ ion if they are present and Zn2+ forms soluble compound . But here they do not formed precipitate hence Mg2+ and Ni2+ CANNOT be Present there but Zn2+ CAN be present there , we need further tests to confirm Zn2+ ion.

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The correct IUPAC name for HOCH2CHOHCH2OH is 1,2,3-propanetriol.

This compound is also commonly known as glycerol or glycerin. The IUPAC naming system uses a set of rules to assign unique and unambiguous names to chemical compounds.

The first step in naming this compound is to identify the longest continuous carbon chain, which in this case is three carbons long, giving us the base name "propane." Next, we need to identify any functional groups attached to the carbon chain.

In this case, there are three hydroxyl (-OH) groups attached to the carbon chain, giving us the suffix "triol." Putting these two parts together, we get the IUPAC name 1,2,3-propanetriol.

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The side chain of the amino acid glycine is made up of hydrogen. The carboxyl group is deprotonated and negatively charged at basic pH (pH=12)

Organic substances known as amino acids have both amino and carboxylic acid functional groups. Alpha-amino acids, which make up proteins, are by far the most significant of the hundreds of amino acids found in nature. In the genetic code, only 22 alpha amino acids are present.

Body protein as well as other vital nitrogen-containing substances including creatine, peptide hormones, and some neurotransmitters cannot be produced without amino acids. Although allowances are expressed in terms of protein, amino acids are a biological necessity.

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A)The marginal distribution of X2 can be obtained by integrating the joint distribution over the range of X1: f(x1) = ∫f(x1, x2)dx2, for 0 < x1 < 1

B)The marginal distribution of X1 can be obtained by integrating the joint distribution over the range of X2: f(x2) = ∫f(x1, x2)dx1, for 0 < x2 < 1

C)The probability that X1 < 0.2 and X2 > 0.5 is given by the joint probability: P(X1 < 0.2, X2 > 0.5) = ∫∫f(x1, x2)dx1dx2, for 0 < x1 < 0.2 and 0.5 < x2 < 1

D)The probability that component proportions produce the results X1 < 0.2 and X2 > 0.

Since the joint distribution is non-zero only when x1 < x2, the limits of integration for x1 are from 0 to x2. Thus, we have:

f(x2) = ∫2dx1, from 0 to x2

= 2x2, for 0 < x2 < 1

The marginal distribution of X2 is a uniform distribution between 0 and 1 with a density of 2x2.

Since the joint distribution is non-zero only when x1 < x2, the limits of integration for x2 are from x1 to 1. Thus, we have:

f(x1) = ∫2dx2, from x1 to 1

= 2(1-x1), for 0 < x1 < 1

The marginal distribution of X1 is a linearly decreasing function between 0 and 1 with a density of 2(1-x1)

The given question is incomplete, the complete question is

A chemical system that results from a chemical reaction has two important components among others in a blend. The joint distribution describing the proportion X1 and X2 of these two components is given by f(x1,x2)= {2, 0 < x1 < x2 < 1, 0, elsewhere (a) Give the marginal distribution of X2. (b) Give the marginal distribution of X1. (e) What is the probability that component proportions produce the results X1 < 0.2 and X2 > 0.5? (d) Give the conditional distribution fx|x2(x1|x2). (e) What is the probability that proportion X1 is less than 0.5 given that X2 is 0.7?

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The three correct options are pH 1 - random coil, pH 7 - helix and pH 11 - random coil.

At a pH of 1, the solution is acidic and the lysine residues will be mostly positively charged. The positive charges on the amino groups can lead to repulsion between neighboring lysine residues, making it energetically unfavorable for the polymer chain to adopt an alpha-helical conformation. Therefore, at a pH of 1, the conformation of polylysine is likely to be a random coil.

At a pH of 7, which is close to the physiological pH of most living organisms, the amino and carboxyl groups in lysine residues are mostly neutral, allowing the polymer chain to adopt a more stable alpha-helical conformation.

At a pH of 11, the solution is basic and the lysine residues will be mostly negatively charged. This can lead to repulsion between neighboring lysine residues and prevent the formation of an alpha-helical conformation. Therefore, at a pH of 11, the conformation of polylysine is likely to be a random coil.

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The complete question is :

A Homopolymer Of Lysine Residues (Polylysine) Can Adopt An Alpha-Helical Conformation Or A Random Coil Conformation Depending On The PH Of The Solution. Predict Whether The Conformation Of Polylysine Would Be Alpha-Helical Or Random Coil At A PH Of 1, 7, And 11. (Choose A Total Of THREE Answers. Choose Either Helix Or Random Coil For PH 1, PH 7, And PH

A homopolymer of lysine residues (polylysine) can adopt an alpha-helical conformation or a random coil conformation depending on the pH of the solution. Predict whether the conformation of polylysine would be alpha-helical or random coil at a pH of 1, 7, and 11. (Choose a total of THREE answers. Choose either helix or random coil for pH 1, pH 7, and pH 11).

options:

pH 1 - helix

pH 1 - random coil

pH 7 - helix

pH 7 - random coil

pH 11 - helix

pH 11 - random coil

Answer: A (acid is a proton donor)

Explanation:

Bronsted-Lowry Definition

The partial pressure of nitrogen after mixing is approximately 196.5 kPa.

What is Partial Pressure?

In a mixture of gases, the partial pressure of a gas is the pressure that it would exert if it were the only gas in the same volume and at the same temperature as the mixture. It is the pressure contributed by a single gas component in a mixture of gases.

Partial pressure of gas = mole fraction of gas x total pressure of the mixture

where the mole fraction of gas is the ratio of the number of moles of the gas to the total number of moles of all the gases in the mixture.

To solve this problem, we can use the ideal gas law, which states that the pressure (P), volume (V), and temperature (T) of an ideal gas are related by the equation PV = nRT, where n is the number of moles of gas and R is the gas constant.

First, we need to find the number of moles of each gas in the two flasks. We can use the ideal gas law to do this:

n(N2) = (P(V1)/RT)

= (792 kPa x 5 L)/(8.314 J/(mol K) x 298 K)

= 15.92 mol

n(Ar) = (P(V2)/RT)

= (43.6 kPa x 15 L)/(8.314 J/(mol K) x 298 K)

= 7.18 mol

Next, we can use the total number of moles of gas and the volume of the mixture to find the total pressure of the mixture:

n(total) = n(N2) + n(Ar) = 15.92 mol + 7.18 mol = 23.10 mol

V(total) = V1 + V2 = 5 L + 15 L = 20 L

P(total) = (n(total)RT)/(V(total)) = [(23.10 mol)(8.314 J/(mol K))(298 K)]/(20 L) = 285.2 kPa

Finally, we can use the mole fraction of nitrogen to find its partial pressure in the mixture:

X(N2) = n(N2)/n(total) = 15.92 mol/23.10 mol = 0.689

P(N2) = X(N2)P(total) = (0.689)(285.2 kPa) ≈ 196.5 kPa

Therefore, the partial pressure of nitrogen after mixing is approximately 196.5 kPa.

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Use reagent f. H2CrO4. Chromic acid is a strong acid used to oxidize alcohols into ketones and carboxylic acids. Chromium trioxide combines with water to produce chromic acid, which is deliquescent, light red or brown in color, and soluble in water.

Lindlar catalyst is a type of heterogeneous catalyst used in organic chemistry for hydrogenation reactions. It is composed of palladium metal supported on calcium carbonate, barium sulfate, or similar materials, and is commonly used in the partial hydrogenation of alkynes to alkenes. The catalyst is named after its inventor, Herbert Lindlar, who developed it in the 1950s.

Lindlar catalyst is a selective catalyst, which means that it allows for the hydrogenation of alkyne functional groups to alkenes while inhibiting further hydrogenation to alkanes. This is achieved by using poisoned or deactivated palladium that restricts the catalyst's activity and allows for partial hydrogenation to occur. This controlled hydrogenation is useful in organic synthesis because it provides a way to selectively reduce alkynes to alkenes without the formation of unwanted byproducts.

Reagent (f) can directly convert given alcohol to its carboxylic acid.

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The number of moles of an element or a compound is the ratio of its given mass by atomic mass or molar mass. Here, the number of moles in 270 g of copper chloride is 2.008 moles.

Any substance containing 6.02×  10²³ atoms is called one mole of that substance. This number is called Avogadro number. Hence, one mole every element contains Avogadro number of atoms. Similarly one mole of every compounds contains Avogadro number of molecules.

The mass of one mole of an element is called its atomic mass. Then mass of one mole of a compound is called molar mass of the compound.

given molar mass of copper chloride = 134.45 g/mol

given mass = 270 g.

if 134.45 g is the mass of one mole, then , number of moles in 270 g is:

n = given mass/ molar mass

  = 270 g/134.4 g/mol = 2.008 moles.

Therefore, the number of moles of copper chloride in 270 g is 2.008 moles.

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After being treated with ethylene glycol and an acid catalyst, A, C, D, and E can be transformed into a cyclic acetal (and removal of water).

Compounds B, D and E contain the necessary carbonyl groups to form the acetal. Compound A contains an alcohol group, which can be converted into a carbonyl group by oxidation. Compound F does not contain any carbonyl groups and cannot be converted into a cyclic acetal.A cyclic acetal is an organic compound with a three atom ring containing two oxygen atoms connected by a single bond and an oxygen-alkyl group connected by a double bond. It is a type of cyclic ether and is produced from the reaction of an alcohol with an aldehyde. The reaction involves the formation of a hemiacetal, which then undergoes a dehydration reaction to form the cyclic acetal. The ring structure of the cyclic acetal makes it more stable than the open-chain acetal, allowing it to be used as a protecting group in organic synthesis.

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Option d is the correct solution to this question here. The temperature is decreasing.

Standard sets of circumstances for experimental measurements are established at standard pressures and temperatures to enable comparisons between various sets of data.

Ice melting is a physical transformation rather than a chemical reaction. The addition of heat energy causes the phase of ice to transition from solid to liquid during melting. However, there is no chemical reaction taking place while it is melting, so the molecular composition remains unchanged.

NIST employs a 20 °C (293.15 K, 68 °F) temperature and a 1 atm absolute pressure (14.696 psi, 101.325 kPa). This standard is also known as normal pressure and temperature (abbreviated as NTP).

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A) pH = 9.44, pOH = 4.56, Percentage protonation = [codeine-H+] / [codeine-H+] + [codeine] = 0.72%

B)pH = pKa + log([codeine]/[codeine-H+]) = 10.09, pOH = 3.91, Percentage protonation = [codeine-H+] / [codeine-H+] + [codeine] = 0.17%

C)pH = 8.21, pOH = 5.79, Percentage protonation = [codeine-H+] / [codeine-H+] + [codeine] = 0.01%

The pKa value given is for the conjugate acid of codeine which is codeine-H+. We can use the pKa value to determine the Ka value for codeine-H+ and then use the Ka value to determine the concentrations of codeine-H+ and codeine in solution.

(a) For 0.0073 M codeine in water, we can assume that the concentration of codeine-H+ is very small and can be neglected. The Ka value can be determined from the pKa as follows:

pKa = -log(Ka)

8.21 = -log(Ka)

Ka = 7.94 x 10^-9

To determine the concentration of codeine-H+, we can use the equation for the dissociation of codeine-H+:

Ka = [H+][codeine-]/[codeine-H+]

Let x be the concentration of codeine-H+ that dissociates. Then:

7.94 x 10^-9 = x^2 / (0.0073 - x)

Solving for x gives x = 5.29 x 10^-5 M

Therefore, the concentration of codeine-H+ is 5.29 x 10^-5 M and the concentration of codeine is 0.0073 M. Since codeine is a weak base, we can assume that it does not significantly affect the pH of the solution.

(b) Since the concentration of codeine is greater than the concentration of codeine-H+, we can assume that the pH will be determined by the concentration of codeine-H+.

Let x be the concentration of codeine-H+. Then:

Ka = [H+][codeine-]/[codeine-H+]

7.94 x 10^-9 = x^2 / (0.1 - x)

Solving for x gives x = 2.08 x 10^-5 M

(c) Similar to (b), we can assume that the pH will be determined by the concentration of codeine-H+.

Let x be the concentration of codeine-H+. Then:

Ka = [H+][codeine-]/[codeine-H+]

7.94 x 10^-9 = x^2 / (0.01 - x)

Solving for x gives x = 2.21 x 10^-5 M

(d) Similar to (a), we can assume that the concentration of codeine-H+ is very small and can be neglected.

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3.30 moles of tetra phosphorous decaoxide (P4O10) will form when 3.30 moles of P₄ react with oxygen gas.

The balanced equation for the reaction between P4 and O2 is:

P₄ + 5O₂ → P₄O₁₀

According to the equation, 1 mole of P₄ reacts with 5 moles of O₂ to produce 1 mole of P₄O₁₀. Therefore, the number of moles of P₄O₁₀ formed is directly proportional to the number of moles of P₄ consumed in the reaction.

If 1 mole of P₄ produces 1 mole of P₄O₁₀, then 3.30 moles of P₄ will produce:

3.30 moles P₄ x (1 mole P₄O₁₀ / 1 mole P₄) = 3.30 moles P₄O₁₀

Therefore, 3.30 moles of tetra phosphorous decaoxide (P₄O₁₀) will form when 3.30 moles of P₄ react with oxygen gas.

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The phosphate system can be used as an effective buffer in the pH range of 6.2-8.2, which includes the pKa of 7.2 for the equilibrium of dihydrogen phosphate and hydrogen phosphate.

To calculate the pH of the given mixture of 0.042 M H2PO4- and 0.058 M HPO42-, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([HPO42-]/[H2PO4-])

pH = 7.2 + log(0.058/0.042)

pH = 7.34

Therefore, the pH of the mixture is 7.34.

When 1.0 mL of 10.0 M NaOH is added to 1.0 L of the buffer prepared above, we can use the equation for the reaction between NaOH and H2PO4-: NaOH + H2PO4- → NaHPO4 + H2O

The initial moles of H2PO4- in 1.0 L of the buffer are:

moles H2PO4- = 0.042 M × 1.0 L = 0.042 mol

Adding 1.0 mL of 10.0 M NaOH adds 0.01 mol of NaOH to the buffer solution. Assuming that all of the added NaOH reacts with H2PO4-, the final moles of H2PO4- and HPO42- can be calculated as follows:

moles H2PO4- = 0.042 mol - 0.01 mol = 0.032 mol

moles HPO42- = 0.058 mol + 0.01 mol = 0.068 mol

Using the Henderson-Hasselbalch equation with the new concentrations of H2PO4- and HPO42-, we can calculate the final pH: pH = pKa + log([HPO42-]/[H2PO4-])

pH = 7.2 + log(0.068/0.032)

pH = 7.76

Therefore, the final pH of the buffer after the addition of NaOH is 7.76.

When 1.0 mL of 10.0 M NaOH is added to 1.0 L of pure water at pH 7.0, we can use the equation for the reaction between NaOH and water:

NaOH + H2O → Na+ + OH- + H2O

The initial concentration of OH- can be calculated from the concentration of NaOH: [OH-] = 10.0 M × 1.0 mL / 1000 mL = 0.01 M

The initial concentration of H+ in the water is:

[H+] = 10^-7.0 = 1.0 × 10^-7 M

Assuming that all of the added NaOH dissociates completely, the final concentration of OH- can be calculated as:

[OH-] = 0.01 M + 10.0 M × 1.0 mL / 1000 mL = 0.100 M

The final concentration of H+ can be calculated from the equation for the ion product of water:

[H+][OH-] = 1.0 × 10^-14

[H+] = 1.0 × 10^-14 / 0.100 M = 1.0 × 10^-13 M

Therefore, the final pH of the water after adding NaOH is 13.0.

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Chemical reactions in living systems occur in an environment, within a narrow range of temperatures causing chemical changes.

Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.

There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.

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The IUPAC name of the following structural formula CH3CH OH−CH3 is propan-2-ol.

What is the structural formula?

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The most likely oxidation state for a titanium cation is +4 and Titanium has four valence electrons in its outermost shell, and to form a cation it must lose those four electrons.

Since each electron has a negative charge, the loss of four electrons gives the titanium cation a net positive charge of 4+.

In its ground state, the electron configuration of neutral titanium is [Ar]3d2 4s2, with two electrons in the 4s orbital and two electrons in the 3d orbital.

When it loses its valence electrons, it becomes a Ti4+ cation with the electron configuration [Ar]3d0 4s0. Thus, the two electrons in the 4s orbital and two electrons in the 3d orbital are lost to form the Ti4+ cation.

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to round a number correctly, using the conventions followed by this text:option a , b , e are correct

Proclamation 1 is true since adjusting is finished subsequent to doing all multistep computation to keep away from blunder .

Proclamation 2 is likewise true . For instance , if your outcome is 267 and you need to eliminate 7 , then answer would be 27 . So going before number will be expanded by 1 .

Explanation 3 is false . Going before number don't change in the event that digit eliminated is under 5 .

Articulation 4 is false . In the event that digit eliminated is 5, going before number is expanded by 1 assuming it is odd just . On the off chance that first number is even, it stays unaltered.

Proclamation 5 is true .

convection, process by which intensity is moved by development of a warmed liquid like air or water.

Natural convection results from the inclination of most liquids to grow when warmed — i.e., to turn out to be less thick and to ascend because of the expanded buoyancy. Course brought about by this impact represents the uniform warming of water in a pot or air in a warmed room: the warmed particles extend the space they move in through sped up against each other, ascent, and afterward cool and draw nearer together once more, with expansion in density and a resultant sinking.

So option a , b , e are correct .

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