10. Show as many ways as you can think of: a) to make benzamide from benzene; b) to make cyclohexyl methyl ketone from bromocyclohexane;

Answer:

Explanation:

C₈H₁₆ + 12O₂ = 8 CO₂ + 8H₂O.

a )

Heat of formation of C₈H₁₆

[tex]\triangle H_f (C_6H_{16})=-174.5 kJ[/tex]

[tex]\triangle H_f (CO_2)=-393.5 kJ[/tex]

[tex]\triangle H_f (O_2)= 0[/tex]

[tex]\triangle H_f (H_2O)=-285.82 kJ[/tex]

[tex]\triangle H_{reaction} =[/tex] 8 x - 393.5 - 8 x 285.82 + 174.5x 1

= - 5260.06 kJ

b ) Energy required = 2.905 x 10¹⁵kJ

moles of C₈H₁₆ require to be burnt

= 2.905 x 10¹⁵ / 5260.06

= 55.23 x 10¹⁰ moles

= 55.23 x 10¹⁰ x mol weight of C₈H₁₆ g

= 55.23 x 10¹⁰ x 112 g

= 6185.5 x 10¹⁰ g

= 6185.5 x 10⁷ kg

c )

No of litres of CO₂ produced at NTP = 8 x 22.4 x 55.23 x 10¹⁰ L

=  9897.22 x 10¹⁰ L

At 1520 mm of Hg pressure and 250°C

volume of CO₂

= 9897.22  x 10¹⁰ x 760 x ( 273 + 250) / ( 1520 x 273 )

= 9480.3 x 10¹⁰ L .

Answer:

[tex]V=27992L=28.00m^3[/tex]

Explanation:

Hello,

In this case, the combustion of methane is shown below:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

[tex]n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4[/tex]

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3[/tex]

Best regards.

Answer:

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10.

Explanation:

The element nitrogen is in the Group 15 in the Periodic Table, so it tends to gain 3 electrons (3 negative charges) to fill its valance shell with 8 electrons.

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10 (the original 7 plus the 3 gained). It is isoelectronic with the gas Neon, which accounts for its stability.

Answer:

483.27 minutes

Explanation:

using second faradays law of electrolysis

D. The name for the set of independent and dependent variables that will be controlled by the scientist.

The statement, that describes the ‘control’ in an experiment is "the name for the set of independent and dependent variables that will be controlled by the scientist."

A control is an element in an experiment that remains intact or unaffected by other variables. An experiment or observation aiming to minimise the influence of variables other than the independent variable is referred to as a scientific control. It serves as a standard or point of reference against which other test findings are measured.

In a scientific experiment, an independent variable is the variable that is modified or manipulated in order to assess the effects on the dependent variable. In a scientific experiment, the dependent variable is the variable that is being tested and measured. The designation given to the set of independent and dependent variables that the scientist will regulate.

Hence the correct option is D.

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Answer:

An ionic compound is formed between a metal and a nonmetal (or polyatomic ions), and is held together through the attraction of opposite charges. An example is KCl. A molecular compound is usually formed between two or more nonmetals, and is held together through the sharing of electrons between atoms. An example is SO2.

Explanation:

When we talk about ionic bonds, the first thing that must come to mind is the electrostatic attraction between oppositely charged ions. Hence, we know that metals form cations and nonmetals form anions, thus metals could transfer electrons to nonmetals to facilitate the formation of ionic bonds. Ionic bonds could also be formed by the combination of metals with polyatomic ions such as CaCO3. Always keep it in mind that ionic bonds are characterized by electrostatic attraction between any pair of oppositely charged ions.

Molecular compounds are formed by sharing of electrons between nonmetals. We find covalent or polar covalent bonds in molecular compounds such as SO2.

Answer:

The answer to your question is given below

Explanation:

The balanced equation for the reaction is given below:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

1. Writing an expression for the equilibrium constant, K.

The equilibrium constant, K for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Thus, we can write the equilibrium constant, K for the reaction as follow:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

K = [CaCO3] [H2]⁴ / [CaO] [CH4] [H2O]²

2. Based on the value of K, more products will be in the equilibrium mixture since the value of K is a positive large number.

Answer:

H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)

Explanation:

In a redox reaction, one half-reaction is the oxidation (where the atom loss electrons) whereas the other reaction is the reduction (Where the atom is gaining electrons.

In the reactions:

H₂(g) → 2H⁺(aq) + 2e⁻ oxidation

Here, the reaction is written as the oxidation because the hydrogen H₂ is in oxidation state 0 and H⁺ in +1. That means each atom is loosing one electron.

Cu²⁺(aq) + 2e⁻ → Cu(s) reduction

And here, the Cu²⁺ is in +2 oxidation state and after the reaction is in Cu(s) 0 state. Thus, each atom is gaining 2 electrons.

The sum of both reactions is:

H₂(g) + Cu²⁺(aq) + 2e⁻ → 2H⁺(aq) + 2e⁻ + Cu(s)

Subtracting the electrons in both sides of the reaction:

Answer:

double replacement

Explanation:

Double replacement reactions are where both cations and anions of reactants switch.

For eg: AB + CD -----> AD + BC (Here we see both things are replaced)

Synthesis reactions is where two simple things make one complex thing.

So A + B --- > AB

Combustion is usually reaction where Oxygen is reactant but here we don't have that.

Single replacement only replaces either cation or anion.

So AB + C .------> AC + B

Answer:

Replacement

Explanation:

Answer:

[tex]n_S=1.076molS[/tex]

Explanation:

Hello,

In this case, given the undergoing chemical reaction, we can see a 4:3 mole ratio between the consumed moles of gallium and sulfur respectively, therefore, the consumed moles of sulfur, from the 100.0 g of gallium (use its atomic mass) turn out:

[tex]n_{S}=100.0gGa*\frac{1molGa}{69.72gGa}*\frac{3molS}{4molS} \\\\n_S=1.076molS[/tex]

Best regards.

Answer:

[tex]n_{CO_2}=1.93 gCO_2[/tex]

Explanation:

Hello,

In this case, considering the given chemical reaction, we can use the molar mass of octane (114.23 g/mol) and the 2:16 molar ratio with carbon dioxide to compute the emitted moles of CO2 to the atmosphere via the following stoichiometric procedure:

[tex]n_{CO2}=27.6gC_8H_{18}*\frac{1molC_8H_{18}}{114.23gC_8H_{18}} *\frac{16molCO_2}{2molC_8H_{18}} \\\\n_{CO_2}=1.93 gCO_2[/tex]

Which also corresponds to the following mass:

[tex]m_{CO_2}=1.93molCO_2*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=85.0gCO_2[/tex]

Best regards.

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

[tex]n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC[/tex]

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

[tex]n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH[/tex]

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

[tex]m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO[/tex]

4. Compute the moles of oxygen by using its molar mass:

[tex]n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO[/tex]

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

[tex]C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1[/tex]

6. Search for the closest whole number (in this case multiply by 2):

[tex]C_3H_6O_2[/tex]

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

[tex]M=12*3+1*6+16*2=74g/mol[/tex]

Which is about three times in the molecular formula, for that reason, the actual formula is:

[tex]C_9H_{18}O_6[/tex]

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

Answer:

the answer is monerans

Explanation:

When Carl Woese developed the modern system of classification, he broke the previous kingdom of Monera into the two kingdoms of Bacteria and Archaea.

Some biologists believed it made sense to classify prokaryotes as belonging to their own kingdom, the Monera. That served as the foundation for Richard Whittaker  and Lynn Margulis's five-kingdom proposal, which enhanced the Haeckel plan by include a kingdom of fungus.

Protists, protozoa, monera, fungi, and viruses have long been proposed as belonging to different kingdoms, but traditional evolutionists during the majority of the 20th century had given none of them any thought.

Later, the Monera kingdom was split into Eubacteria and Archaebacteria by Carl Woese .  Moreover, he divided the five kingdoms into three domains: Eukaryotes, Archaea, and Bacteria.

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Your question is incomplete. But your complete question is as follows:

When Carl Woese developed the modern system of classification, he broke the previous kingdom of into the two kingdoms of  _____  into  Bacteria and Archaea.

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density  * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

Answer:

Hope this helps...

Good luck on your assignment...

Answer:

[tex]C_{base}=0.302M[/tex]

Explanation:

Hello,

In this case, we can evidence that when calcium hydroxide solution reacts with hydrochloric acid solution, the balanced neutralization reaction turns out:

[tex]2HCl(aq)+Ca(OH)_2\rightarrow CaCl_2(aq)+2H_2O(l)[/tex]

Moreover, the concentration of neutralized calcium hydroxide can be computed by using the 2:1 mole ratio between the base and the acid:

[tex]C_{acid}V_{acid}=2*C_{base}V_{base}\\\\C_{base}=\frac{C_{acid}V_{acid}}{2*V_{base}} =\frac{0.225M*26.85mL}{2*10.0mL}\\ \\C_{base}=0.302M[/tex]

Regards.

Answer:

[tex]m_{KAl(SO_4)_2\dot \ 12H_2O}^{actual}=32.23gKAl(SO_4)_2\dot \ 12H_2O[/tex]

Explanation:

Hello,

In this case, we balance the given equations as shown below:

[tex]Al(s)+KOH(aq)+3H_2O(l)\rightarrow KAl(OH)_4(aq)+\frac{3}{2} H_2(g)\\\\KAl(OH)_4(aq)+2H_2SO_4(aq)\rightarrow KAl(SO_4)_2(aq)+4H_2O(l)\\\\KAl(SO_4)_2(aq)+12H_2O\rightarrow KAl(SO_4)_2\dot\ 12H_2O(aq)[/tex]

Now, with 3.00 grams of aluminium, 50.00 mL of water and 10.00 mL of 8.00M potassium hydroxide, the first step is to identify the limiting reactant by firstly computing the moles of all of them:

[tex]n_{Al}=3.00 gAl*\frac{1molAl}{27gAl}=0.111molAl\\ \\n_{KOH}=0.010L*8.00mol/L=0.08molKOH\\\\n_{H_2O}=50.00mL*\frac{1g}{1mL} *\frac{1mol}{18g}=2.78molH_2O[/tex]

Thus, we can notice that 0.111 mol of aluminium will consume 0.11. moles of potassium hydroxide and 2.78 moles of water will consume 0.927 moles of potassium hydroxide, for that reason, we can infer that since there are only 0.08 moles of potassium hydroxide, it is the limiting reactant, therefore, we compute the yielded moles of potassium aluminium hydroxide in the first reaction:

[tex]n_{KAl(OH)_4}=0.08molKOH*\frac{1molKAl(OH)_4}{1molKOH} =0.08molKAl(OH)_4[/tex]

Next, we compute the yielded moles of potassium aluminium sulfate in the second reaction assuming sulfuric acid is in excess:

[tex]n_{KAl(SO_4)_2}=0.08molKAl(OH)_4*\frac{1molKAl(SO_4)_2}{1molKAl(OH)_4}=0.08molKAl(SO_4)_2[/tex]

Finally, in the third reaction, we compute the yielded grams of potassium aluminum sulphate dodecahydrate by using its molar mass and its mole ratio with potassium aluminium sulfate:

[tex]m_{KAl(SO_4)_2\dot \ 12H_2O}=0.08molKAl(SO_4)_2*\frac{1molKAl(SO_4)_2\dot \ 12H_2O}{1molKAl(SO_4)_2} *\frac{474.00gKAl(SO_4)_2\dot \ 12H_2O}{1molKAl(SO_4)_2\dot \ 12H_2O} \\\\m_{KAl(SO_4)_2\dot \ 12H_2O}=37.92gKAl(SO_4)_2\dot \ 12H_2O[/tex]

Which is the theoretical yield, thus, by using the percent yield the actual yielded mass turns out:

[tex]m_{KAl(SO_4)_2\dot \ 12H_2O}^{actual}=0.85*37.92gKAl(SO_4)_2\dot \ 12H_2O\\\\m_{KAl(SO_4)_2\dot \ 12H_2O}^{actual}=32.23gKAl(SO_4)_2\dot \ 12H_2O[/tex]

Best regards.

Answer:

pH = 7.29

Explanation:

Ka of butanoic acid is 1.54x10⁻⁵

To obtain the pH of the solution you must use H-H equation for butanoic acid:

pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]

Where pKa is defined as -log Ka = 4.81

Now, you need to find [C₃H₇COO⁻] and [C₃H₇COOH] concentrations (Also, you can find moles of each substance and replace them in the equation.

Butanoic acid reacts with LiOH, producing C₃H₇COO⁻, thus:

C₃H₇COOH + LiOH → C₃H₇COO⁻ + H₂O + Li⁺

Moles of both reactants, C₃H₇COOH and LiOH are:

C₃H₇COOH = 0.0300L ₓ (0.1000mol / L) = 0.003000moles of C₃H₇COOH

LiOH = 0.0299L ₓ (0.1000mol / L) = 0.00299 moles of LiOH.

That means moles of C₃H₇COO⁻ produced are 0.00299 moles.

And moles of C₃H₇COOH that remains in solution are:

0.00300 - 0.00299 = 0.00001 moles of C₃H₇COOH

Replacing in H-H equation:

pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]

pH = 4.81 + log₁₀ [0.00299moles] / [0.00001moles]

Answer:

The colour of the orange solution becomes yellow.  

Explanation:

1. Before adding NaOH

Assume the picture showed a beaker of potassium chromate and one of potassium dichromate.

Both solutions are involved in the same equilibrium:

[tex]\rm\underbrace{\hbox{2CrO$_{4}^{2-}$(aq)}}_{\text{yellow}} +2H^{+}(aq) \rightleftharpoons \, \underbrace{\hbox{Cr$_{2}$O$_{7}^{2-}$}}_{\text{orange}} + H_{2}O[/tex]

The first beaker contains mostly chromate ions with a few dichromate ions.

The position of equilibrium lies to the left and the solution is yellow.

The second beaker contains mostly dichromate ions with a few chromate ions.

The position of equilibrium lies to the right and the solution is orange.

2. After adding NaOH

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

Beaker 1

If you add OH⁻ to the equilibrium solution, it removes the H⁺ (by forming water).

The system responds by having the dichromate react with water to replace the H⁺.  

At the same time, the system forms more of the yellow chromate ion.

The position of equilibrium shifts to the left.

However, the solution is already yellow, so you see no change in colour.

Beaker 2

The reaction is the same as in Beaker 1.

This time, however, as the dichromate ion disappears, do does its orange colour.

Also, the yellow chromate is being formed and its yellow colour appears .

The colour changes from orange to yellow.

Answer:

(a)

Explanation:

Hello,

In this case, the temperature required to boil argon, it means, transform it from liquid to gas is -197 °C. In such a way, since the temperature inside the steel sphere is -190 °C, which is greater than the boiling point, we realize argon is gaseous, therefore, the molecules will be spread inside the sphere as they will be moving based on the kinetic theory of gases.

For that reason, answer is scheme (a).

Best regards.

Answer:

[tex][Ar]4s^23d^5[/tex]

Explanation:

Hello,

In this case, we write the electron configuration of the manganese atom by noticing its atomic number is 25, so we fill the orbitals and levels up-to 25 electrons as shown below:

[tex]1s^22s^22p^63s^23p^64s^23d^5[/tex]

Moreover, for the condensed electron configuration, we consider the previous noble gas, that is argon, electron configuration which is:

[tex]1s^22s^22p^63s^23p^6[/tex]

By cause of its atomic number that is 18. In such a way, we combine argon's electron configuration with manganese's to obtain its condensed version:

[tex][Ar]4s^23d^5[/tex]

Regards.

Answer:

[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, we can write the law of mass action for the undergoing chemical reaction, based on the rates and the stoichiometric coefficients:

[tex]\frac{1}{-2}r_{NH_3} =\frac{1}{1} r_{N_2}=\frac{1}{3}r_{H_2}[/tex]

In such a way, knowing the rate of formation hydrogen (H₂), we can know the  rate of change of ammonia, that must be negative for consumption:

[tex]r_{NH_3} =\frac{-2}{3}r_{H_2}=\frac{-2}{3}*9.00\frac{M}{s} \\\\r_{NH_3} =-6.00\frac{M}{s}[/tex]

Nevertheless, the absolute magnitude will be positive:

[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]

Best regards.

Answer:

See figure 1

Explanation:

In the structure of nylon 6,6 we have amide groups. In this functional group, We have a nitrogen bond to hydrogen, so in this bond, we will have a dipole, due to the electronegativity difference. Nitrogen has more electronegativity than hydrogen, therefore a positive dipole would be generated in the hydrogen atom. Additionally, in the carbonyl group (C=O) due to the oxygen, we will have also a dipole, in this case, a negative dipole because the oxygen atom has more electronegativity (compare with carbon).

When we put two strings of nylon 6,6 the positive dipole will interact with the negative dipole and vice-versa and we will obtain the "hydrogen bonds".

See figure 1

I hope it helps!

Answer:

Copper (II) chloride.

Explanation:

Hello,

In this case, considering the described reaction which is also given as:

[tex]2Al + 3CuCl_2 \rightarrow 2AlCl_3 + 3Cu[/tex]

For us to identify the limiting reactant we first compute the available moles of aluminium:

[tex]n_{Al}=512gAl*\frac{1molAl}{27gAl}=19.0molAl[/tex]

Next, we compute the consumed moles of aluminium by the 1147 grams of copper (II) chloride by using their 2:3 molar ratio:

[tex]n_{Al}^{consumed}=1147gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2}*\frac{2molAl}{3molCuCl_2} =5.69molAl[/tex]

Thereby, we can infer aluminium is in excess since less moles are consumed than available whereas the copper (II) chloride is the limiting reactant.

Best regards.

Answer: The Answer should be B. 0.887 (:

Explanation:

The mole fraction of the solution is 0.887.

Mass percent of LiCl present =  23.0%

Mass of the LiCl can be obtained from;

23 = x/100 × 100/1

x = 23 g

Number of moles of LiCl = 23g/42 g/mol = 0.55 moles

Number of moles of water = (100 - 23) g/18 g/mol = 4.3 moles

Total number of moles = 0.55 + 4.3 = 4.85 moles

Mole fraction of water = 4.3 moles/4.85 moles = 0.887

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Answer:

the density if vinegar will also be needed

Explanation:

Because this is an experiment of volumetric analysis

Answer:

2. Due to the possession of resonance

Explanation:

In the benzene ring, the electrons that results in the bonds between the carbon atoms are delocalized. That is, they do not belong to a specific carbon atom. It is this unique feature that enables them to have a bond length between single and double bond lengths.

This feature is as a result of resonance.

The correct option is 2.

Answer:

Option C. Will always.

Explanation:

A spontaneous reaction is a reaction that occurs without an external supply of heat.

This implies that spontaneous reaction will always occur as no external supply of heat is needed.

Answer:

a

Explanation:

Matter can be volume or density. So, this concludes that it is when it takes up space.

Answer: A.

Explanation:

it takes up space