In your discussion consider a SN2 reaction mechanism concept. Propose a modification of experimental procedure that would improve reaction yield. Give an example of another method of ether synthesis. Illustrate it with drawn reaction scheme; describe it in few sentences.

Answer:

Sedimentary rocks

Explanation:

My explanation is that when an animal decomposes it body returns to the ground eventually being used in the rock cycle and rocks form this through the rock cycle when broken down by weathering and erosion.

Hope this helps you

Answer:

sedimentary rocks

They form from deposits that accumulate on the Earth's surface.

Answer:

1 g

Explanation:

From the formula;

N/No = (1/2)^t/t1/2

Where;

N= mass of radioactive element left after a time t = the unknown

No= mass of radioactive element originally present in the sample = 8g

t= time taken for N mass of the sample to remain = 1500

t1/2= half-life of the radioactive element = 500 years

Substituting values, we have;

N/8 = (1/2)^1500/500

N/8 = (1/2)^3

N/8 = 1/8

N= 1/8 ×8

N= 1 g

Therefore; mass of radioactive element left after 1500 years is 1 g

Answer:

0.025 M

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 5 mL

Initial concentration (C1) = 0.5 M

Final volume (V2) = 100 mL

Final concentration (C2) =..?

Using the dilution formula, we can obtain the final concentration of the diluted solution as follow:

C1V1 = C2V2

0.5 x 5 = C2 x 100

Divide both side by 100

C2 = (0.5 x 5)/100

C2 = 0.025 M

Therefore, the final concentration of the diluted solution is 0.025 M

The concentration of the final diluted solution is 0.025M

The dilution formula is expressed according to the formula:

[tex]C_1V_1=C_2V_2[/tex]

Given the following parameters

[tex]C_1=0.5M\\V_1=5.00mL\\V_2=100.0mL\\C_2=?[/tex]

Substitute the given parameters into the formula:

[tex]C_1V_1=C_2V_2\\0.5(5)=100C_2\\2.5=100C_2\\C_2=\frac{2.5}{100}\\C_2= 0.025M[/tex]

Hence the concentration of the final diluted solution is 0.025M

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Answer:

14.3mL you require to reach the half-equivalence point

Explanation:

A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:

CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻

As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.

Initial moles of CH₃CH₂NH₂ are:

20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =

0.0106moles CH₃CH₂NH₂

To reach the half-equivalence point you require:

0.0106moles ÷ 2 = 0.005304 moles HClO₄

As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:

0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =

When The Kb of ethylamine is 6.5 x 10-4 is = 14.3mL you require to reach the half-equivalence point.

When A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, Therefore:

Then CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻

So when the reaction is 1:1, to reach the equivalence point then you instruct to that add the moles of HClO₄ equal to the moles CH₃CH₂NH₂ you add originally. Also, When the half-equivalence point requires you to add half-moles of CH₃CH₂NH₂ you add originally.

Then Initial moles of CH₃CH₂NH₂ are:

After that 20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =

Then 0.0106moles CH₃CH₂NH₂To get the half-equivalence point you require is:

Then 0.0106moles ÷ 2 = 0.005304 moles HClO₄

After that As the concentration of HClO₄ is 0.37M, the volume you require to add 0.005304moles is:

Then 0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =

Therefore, 14.3mL you require to reach the half-equivalence point.

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Answer:

2 Al(s) + 3 CuSO₄(aq) ⇒ 3 Cu(s) + Al₂(SO₄)₃(aq)

Explanation:

Let's consider the single displacement reaction of Al(s) with CuSO₄(aq). Copper has a higher reduction potential than aluminum, so aluminum will take the place of copper to form aluminum sulfate and metallic copper. The corresponding balanced chemical equation is:

2 Al(s) + 3 CuSO₄(aq) ⇒ 3 Cu(s) + Al₂(SO₄)₃(aq)

The chemical equation is 2 Al(s) + 3 CuSO₄(aq) ⇒ 3 Cu(s) + Al₂(SO₄)₃(aq)

here we considered the single displacement reaction of Al(s) with CuSO₄(aq). Also, Copper contained a higher reduction potential as compared to aluminum, due to this aluminum will take the place of copper to create aluminum sulfate and metallic copper. So the above should be the balance chemical equation.

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Answer:

Dissolving NH4F in water will form a weak acidic solution.

Explanation:

That it is a weak acid solution means that it has a pH below 7 but close to the value, that is, it does not contain as many acids as those substances that are around a pH of 1 to 4, generally weak acids have a pH approximately 5 to 6

The solution of solid ammonium fluoride in pure water has been slightly acidic in nature.

Ammonium fluoride has been an ionic compound formed by the interaction of cationic ammonia and anionic fluoride ions. The dissolution of ionic compounds will result in the compound in its dissociated ionic state.

The dissociation results in the formation of ammonium cation. The ammonium has been a strong acid.

The resulted anion has been fluoride. It has been a strong base, but slightly weaker than ammonia.

Thus the resultant solution will result in slightly acidic nature.

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Answer:

Product: propan-1-ol

Explanation:

IIn this case, we have to remember that [tex]LiAlH_4[/tex]  is a reduction agent.  So, this is a reduction reaction. The [tex]LiAlH_4[/tex] has the ability to produce hydride ions [tex]H^-[/tex]. This ion can attack the carbonyl group generating a negative charge in the oxygen. In the next step, the negative charge in the oxygen can attack a water molecule to protonate the molecule and produce propan-1-ol.

See figure 1

I hope it helps!

Answer:

D. He showed that most of an atom's mass was located in the atom's

nucleus.

Explanation:

Ernest Rutherford changed the atomic model because of his experiment which was the gold foil experiment. A beam of alpha particles was aimed at a piece of gold foil, most particles passed through but some were scattered backward which showed that the middle of an atom (nucleus) is the where most of the mass is located.

Rutherford's model of atoms is the improved version of Thomson's model. In the model, it is stated that most of an atom's mass is located in the nucleus. Thus, option D is correct.

Ernest Rutherford gave the improved atomic model that postulated the failure of Thomson's model. Rutherford's model described the atom to consist of a sub-atomic particle with a positively charged nucleus.

The nucleus is in the center of the atom and had nearly all mass concentrated in it due to the presence of the protons and neutrons. The electrons were called negatively charged species that were present in the shells around the nucleus like the planets around the Sun.

Therefore, Rutherford's model showed mass concentrated in the center of the nucleus.

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Answer:

[tex]m=368 g[/tex]

Explanation:

Hello,

In this case, in order to compute the required mass, we first must notice that 6.022x10²³ formula units of sodium sulfate contain 1 mole of such compound (Avogadro's relationship). Moreover, one mole of sodium sulfate contains 142.04 g, which is in fact, the molar mass. Thereby, the required mass is computed via the following mole-mass-particles relationship:

[tex]m=1.56x10^{24}f.u*\frac{1mol}{6.022x10^{23}f.u}*\frac{142.04g}{1mol} \\\\m=368 g[/tex]

Regards.

Answer:

A) One that occurs on its own

Answer:

Body-centered cubic

Explanation:

As the edge length of the cubic unit is 350pm, its volume is:

(350x10⁻¹²m)³ = 4.288x10⁻²⁹m³. In cm³

4.288x10⁻²⁹m³ ₓ ([100cm]³ / 1m³) = 4.288x10⁻²³cm³.

As the density of the structure is 0.535g/cm³, the mass of a cubic unir is:

4.288x10⁻²³cm³ ₓ (0.535g / cm³) = 2.29x10⁻²³g,

As atomic mass of Lithium is 6.941g/mol. Mass of 1 atom of Li is:

6.941g/mol ₓ 1mol / 6.022x10²³ atoms = 1.1526x10⁻²³ g / atoms.

Thus, atoms of Li in 2.29x10⁻²³g are:

2.29x10⁻²³g ₓ ( 1 atom / 1.1526x10⁻²³g) = 2 atoms in 1 cubic unit

In body-centered cubic unit cell, there are 2 atoms per cubic unit cell. Thus, structure of Li is:

Answer:

Explanation:

Skeleton equation is opposite of word equation because here you use chemical formulas to write down the components.

Potassium Oxide  =  K2O

Magnesium Oxide = MgO

Sulfur Trioxide = SO3

Sodium Chloride = NaCl

Complete question:

(a) compute the specific heat capacity at constant volume of nitrogen gas. the molar mass of N₂ is 29.0 You warm 1.8 kg of water at a constant volume 1.00 L from  21 C to 30.5 C in a kettle. For the same amount of heat, how many kilograms of 21∘C air would you be able to warm to 30.5∘C ?

(b) What volume (in liters) would this air occupy at 21∘C and a pressure of 1.00 atm? Make the simplifying assumption that air is 100% N₂​

Answer:

(a) The specific heat capacity of N₂ is 715.86 J/kg.K

(b) The volume the air occupy at 21∘C is 8784.29 Liters

Explanation:

Given;

M is the molar mass of N₂ = 29 x 10⁻³ kg/mol

specific heat of N₂ at constant volume, Cv = 20.76 J/mol.K

(a)

The specific heat capacity of N₂ is calculated as;

[tex]C = \frac{C_v}{M} \\\\C = \frac{20.76}{29 *10^{-3}} \\\\C = 715.86 \ J/kg.K[/tex]

(b) heat capacity of water;

Q = mcΔθ

where;

c is the specific heat capacity of water = 4200 J/kg.K

m is mass of water, = 1.8 kg

Δθ is change in temperature, = 30.5 - 21 = 9.5 °C

Q = 1.8 x 4200 x  9.5

Q = 71820 J

Mass of nitrogen gas N₂, at this quantity of heat;

[tex]m_{N_2} = \frac{Q}{C*\delta \theta} \\\\m_{N_2} = \frac{71820}{715.86*9.5}\\\\m_{N_2} = 10.56 \ kg[/tex]

The volume this air occupy at 21∘C

Apply ideal gas law;

[tex]PV = nRT = \frac{m}{M} RT[/tex]

[tex]PV = \frac{mRT}{M} \\\\V = \frac{mRT}{MP}\\\\V = \frac{10.56(kg)*8.314*10^3(L.Pa/mol.K)*294(K)}{29*10^{-3}(kg)1.01325*10^5 (Pa)}\\\\V = 8784.29 \ Liters[/tex]

Answer:

Boiling point of the solution is 100.964°C

Explanation:

In this problem, first, you must use Raoult's law to calculate molality of the solution. When you find the molality you can obtain the boiling point elevation because of the effect of the solute in the solution (Colligative properties).

Using Raoult's law:

Psol = Xwater × P°water.

As vapour pressure of the solution is 23.0torr and for the pure water is 23.78torr:

23.0torr= Xwater × 23.78torr.

0.9672 = Xwater.

The mole fraction of water is:

[tex]0.9672 = \frac{X_{H_2O}}{X_{H_2O}+X_{solute}}[/tex]

Also,

[tex]1 = X_{H_2O}+X_{solute}[/tex]

You can assume moles of water are 0.9672 and moles of solute are 1- 0.9672 = 0.0328 moles

Molality is defined as the ratio between moles of solute (0.0328moles) and kg of solvent. kg of solvent are:

[tex]09672mol *\frac{18.01g}{1mol}* \frac{1kg}{1000g} = 0.01742kg[/tex]

Molality of the solution is:

0.0328mol Solute / 0.01742kg = 1.883m

Boiling point elevation formula is:

ΔT = Kb×m×i

Where ΔT is how many °C increase the boiling point regard to pure solvent, Kb is a constant (0.512°C/m for water), m molality (1.883m) and i is Van't Hoff factor (Assuming a i=1).

Replacing:

ΔT = 0.512°C/m×1.882m×1

ΔT = 0.964°C

As the boiling point of water is 100°C,

Boiling point of the solution is 100.964°C.

It says that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present.

[tex]P_{sol} = X_{water} * P^o_{water}[/tex]

Given:

Substituting the values:

[tex]23.0torr = X_{water} * 23.78torr\\\\0.9672 = X_{water}[/tex]

[tex]0.9762=\frac{X_{water}}{X_{water}+X_{solute}}[/tex]

The sum of the mol fractions of water and solute is 1.

We can consider,

Moles of water = 0.9672

Moles of solute = 1- 0.9672 = 0.0328 moles

It is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.

[tex]\text{Mass of solvent}=0.9672*\frac{18g/mol}{1mol} *\frac{1kg}{1000g}\\\\\text{Mass of solvent} =0.01745kg[/tex]

[tex]\text{Molality}= \frac{0.0328mol}{0.01742kg} \\\\\\text{Molality}= 1.883m[/tex]

[tex]\triangle T = K_b*m*i[/tex]

Substituting the values in the above formula:

[tex]\triangle T = 0.512^oC/m*1.882m*1\\\\\triangle T = 0.964^oC[/tex]

Thus, Boiling point of the solution is 100.964°C, since boiling point of water is 100°C.

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Answer:

3-methylenehexane

Explanation:

In this case, we have two clues.

1) The hydrogenation reaction

2) The ozonolysis reaction

See figure 1.  

With this in mind, lets analyze each clue. In the first reaction, we know that only 1 molecule of [tex]H_2[/tex] is added to the unknown molecule. This indicates that we only have 1 double bond in the molecule. Now, the next question is where is placed the double bond?

To answer this question, we have to use the second clue. In the ozonolysis reaction, a double bond is broken and is replaced with a carbonyl group. If, formaldehyde is formed the double bond is formed with a primary carbon. The primary carbons in the structure (given in the first reaction: 3-methylhexane) are carbons 1, 6, and 7. So, the double bond can be placed between carbons:

a) 6 and 5

b) 7 and 3

c) 1 and 2

To decide which one is the position of the double bond we have to keep in mind the second product of the ozonolysis reaction a ketone. With this in mind, the carbon bonded to the primary one (deduced by the formaldehyde) it has to be a tertiary carbon. The only option that has a primary carbon bonded to tertiary carbon is b). (See figure 2)

Finally, with this in mind the structure is 3-methylenehexane. To be sure, we can check the formula for the compound, [tex]C_7H_1_4[/tex] and the reactions. (See figure 3)

I hope it helps!

Answer:

It is important to separate mixture into pure or relatively pure substances when performing a chemical analysis SO AS TO KNOW THE PROPERTIES COMING FROM EACH PART MIXTURE WHICH MAY INTERFERE WITH THE SEPARATION.

Explanation:

In chemistry, Mixture is the combination of two or more substances which are not combine chemically.

Mixture contain different substances with different physical and chemical properties.

It is important to purify the substances in a mixture so as to identify what properties are coming from each mixture and also some part of the mixture can interfere with the properties of other mixture present for skewing analysis.

Answer:

THE ENTHALPY CHANGE PER MOLE OF KOH IS 8400 Joules/ mole OF HEAT.

Explanation:

Heat = heat capacity * change in temperature

Heat capacity = 36 J/K

Temperature of the mixture before mixing = 21.5 C

Temperature of mixtire after mixing = 23.6 C

Calculate the change in temperature:

Change in temperature = 23.6 C - 21.5 C = 2.1 C

Heat = 36 * 2.1

Heat = 75.6 J of heat

In essence, 45 ml of 0.20 M of KOH produces 75.8 J of heat

The enthalpy change per mole of water:

It is important t obtain the number of moles involved in the reaction of 45 mL of 0.20 M of KOH

n = C V

n = 0.20 M * 45 *10^-3

n = 0.009 moles

Since number of moles = mass / molar mass

The mass of 45 ml of 0.20 M of KOH is then:

Molar mass = ( 39 + 16 + 1) g/mol = 56 g/mol

Mass = number of moles * molar mass

Mass = 0.009 * 56

Mass = 0.504 g

So therefore 0.504 g of KOH produces 75.6 J of heat

1 mole of KOH will produce x J of heat

1 mole of KOH = 56 g of KOH

0.504 g = 75.6 J

56 g = x J

x J = 56 * 75.6 / 0.504

x J = 8400 J / mole of KOH

Answer:

See explanation

Explanation:

In this question, we have to follow the IUPAC rules. Lets analyze each compound:

a. 1-methylbutane

In this compound we have a chain of 5 carbons, so the correct name is Pentane.

b. 1,1,3-trimethylhexane

In this compound, we longest chain is made of 7 carbons, so, we have to use the name "heptane". Carbon one would be the closet one to the methyl group, so the correct name is  2,4-dimethylheptane.

c. 5-octyne

In this case, carbon 1 would be the closet one to the triplet bond. With this in mind, the correct name is oct-3-yne.

d. 2-ethyl-1-propanol

In this compound, we longest chain is made of 4 carbons, so, we have to use the name "butane". Carbon one would be the carbon with the "OH" group, so the correct name is  2-methylbutan-1-ol.

e. 2.2-dimethyl-3-butanol

In this case, carbon 1 would be the closet one to the "OH". With this in mind, the correct name is 3,3-dimethylbutan-2-ol.

See figure 1

I hope it helps!

Answer:

C8H17N

Explanation:

Mass of the unknown compound = 5.024 mg

Mass of CO2 = 13.90 mg

Mass of H2O = 6.048 mg

Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:

For carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C = 12/44 x 13.90 = 3.791 mg

For hydrogen, H:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H = 2/18 x 6.048 = 0.672 mg

For nitrogen, N:

Mass N = mass of unknown – (mass of C + mass of H)

Mass of N = 5.024 – (3.791 + 0.672)

Mass of N = 0.561 mg

Now, we can obtain the empirical formula for the compound as follow:

C = 3.791 mg

H = 0.672 mg

N = 0.561 mg

Divide each by their molar mass

C = 3.791 / 12 = 0.316

H = 0.672 / 1 = 0.672

N = 0.561 / 14 = 0.040

Divide by the smallest

C = 0.316 / 0.04 = 8

H = 0.672 / 0.04 = 17

N = 0.040 / 0.04 = 1

Therefore, the empirical formula for the compound is C8H17N

Answer:

The element potassium forms a cation with the charge +1 . The symbol for this ion is K⁺, and the name is potassium ion. The number of electrons in this ion is 18.

Explanation:

Potassium is a metal. It belongs to the group 1 elements. Metals form cations by losing electrons. Since potassium is a group element, it forms a cation by losing one electron. The charge it has is +1 due to the excess of the protons compared t the electrons by 1.

Potassium has  19 electrons. Potassium io on the other hand has 19-1 = 18 electrons.

Answer:

284 calories

Explanation:

There are 284 calories in 50 grams of peanuts.

Calorie breakdown: 73% fat, 11% carbs, 17% protein.

Answer:

Explanation: M(PCL5)= 31 + 5(35.5)

=208.5g/mol

M(H20)= 18g/mol

n(PCL5) = 75.5÷208.5

= 0.362mol

n(HCl)/n(PCL5)= 5/1

n(HCl)= 5×0.362

=1.81mol of HCl

Answer:

A molecule is the answer.

Answer:

See the explanation and answer below.

Explanation:

In chemistry, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound. The formula gives the proportions of the elements present in a compound but not the actual arrangement of atoms.

[tex]\mathrm{Molecular \:Formula}\quad \quad |\quad \quad \mathrm{Empirical \:Formula}[/tex]

[tex]1.\:\:\:NH_4OH\quad | \quad H_5NO[/tex]                           (Ammonium hydroxide)

[tex]2.\:\:\:Fe(OH)_3\quad |\quad FeH_3O_3[/tex]                       (Iron(III) hydroxide)

[tex]3.\:\:\:NH_4C_2H_3O_2\quad |\quad C_2H_7NO_2[/tex]              (Ammonium acetate)

[tex]4.\:\:\:Fe(C_2H_3O_2)_3\quad |\quad C_6H_9FeO_6[/tex]            (Iron(III) Acetate)

I hate chemistry but best regards!

Answer:

5.42

Explanation:

Step 1: Consider the dissociation of NH₄Br

NH₄Br(aq) ⇒ NH₄⁺(aq) + Br⁻(aq)

Br⁻ is the conjugate base of HBr, a strong acid, so it doesn´t react with water. NH₄⁺ is the conjugate acid of NH₃, so it does react with water.

Step 2: Consider the acid reaction of NH₄⁺

NH₄⁺(aq) + H₂O(l) ⇄ NH₃(aq) + H₃O⁺(aq)

Step 3: calculate the acid dissociation constant for NH₄⁺

We will use the following expression.

[tex]K_a \times K_b = K_w\\K_a = \frac{K_w}{K_b} = \frac{1.00 \times 10^{-14} }{1.76 \times 10^{-5}} = 5.68 \times 10^{-10}[/tex]

Step 4: Calculate the concentration of H₃O⁺

We will use the following expression.

[tex][H_3O^{+} ]= \sqrt{K_a \times C_a } = \sqrt{5.68 \times 10^{-10} \times 0.0255 } = 3.81 \times 10^{-6}M[/tex]

Step 5: Calculate the pH

We will use the following expression.

[tex]pH = -log [H_3O^{+} ] = -log (3.81 \times 10^{-6}) = 5.42[/tex]

The pH of 0.0255 M solution should be 5.42.

Since we know that

ka * kb = kw

So,

ka = kw/kb

= 1.00*10^-14 / 1.76*10^-5

= 5.68*10^-10

Now the concentration of H3O should be

= √ka * Ca

= √5.68*10^-10 * 0.0255

= 3.81*10^-6M

Now the pH value should be

= -log(H3O+)

= -log(3.81*10^-6)

= 5.42

hence, The pH of 0.0255 M solution should be 5.42.

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Answer:

Edge length of the unit cell is 4.07x10⁻¹⁰m

Explanation:

In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:

a² + a² = b² = (4r)²

2a² = 16r²

a = √8 r

That means edge lenght is = √8 r

adius

As radius of Silver is 144pm = 144x10⁻¹²m:

a = √8 r

a = √8 ₓ 144x10⁻¹²m

a = 4.07x10⁻¹⁰m

Answer:

204.7 g

Explanation:

(taking the atomic mass of C, H, O as 12, 1 and 16 respectively).

no. of moles of C3H8 burnt =  56.3 / (12x3 + 1x8)

                                              = 1.27955 mol

From the equation, the mole ratio of C3H8 :  O2 = 1:5

Hence,

the no. of moles of O2 required will be

=1.27955 x 5

= 6.397727 mol

Mass of O2 required = 6.397727 x (16x2)

= 204.7 g

Answer:

See explanation

Explanation:

1H NMR

In the 2-chloro-pentanal we have 4 different types of hydrogens. Therefore, we will have 4 different signals. (See figure 1)

Red hydrogen

For the red hydrogens we have only 1 neighbor. So, if we follow the n+1 rule we can calculate the multiplicity of this hydrogen. In this case a doublet.

Blue hydrogens

In this case, we have 3 neighbors (one in the right, two in the left). Therefore we will have a quartet.

Purple hydrogens

For these hydrogens, we have also will have a quartet, because we have 3 neighbors (one in the right, two in the left).

Green hydrogens

In the green hydrogen,s we have 5 neighbors (2 in the right 3 in the left). Therefore a sextet would be produced.

Orange hydrogens

Finally, in these hydrogens, we have 2 neighbors. Therefore a triplet is expected.

13C NMR

For the 13C NMR, we have again 4 different kinds of carbons. Therefore we will have 4 signals. The most deshielded carbon, in this case, is the red one (see figure 2), so this carbon would be on the left side (around 190). Then the next deshield carbon is the blue one, due to the "Cl" atom placed on this carbon.

I hope it helps!

The approximate age of the cloth is 17190 years.

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

2ⁿ = 100 / 13

2ⁿ = 8

2ⁿ = 2³

n = 3

Finally, we shall determine the age of the cloth.

t = n × t½

t = 3  × 5730

t = 17190 years

Thus, the approximate age of the cloth is 17190 years

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Answer:

Explanation:

Total mass of substance = 7.22 + 2.53 + 5.25 g

= 15 g

percentage of nickel = 7.22 x 100 / 15

= 48.13

= 48.1 %

percentage of phosphorus  = 2.53 x 100 / 15

= 16.87%  

= 16.9%

percentage of oxygen  = 5.25 x 100 / 15

= 35 %  

The percent composition of the pure substance should be 48.1%, 16.9%, and 35%.

Total mass of substance = 7.22 + 2.53 + 5.25 g

= 15 g

Now

percentage of nickel = 7.22 x 100 / 15

= 48.13

= 48.1 %

And,

percentage of phosphorus  = 2.53 x 100 / 15

= 16.87%  

= 16.9%

And, finally

percentage of oxygen  = 5.25 x 100 / 15

= 35 %  

learn more about oxygen here: https://brainly.com/question/11820632