Draw the structure of the organic product(s) of the Grignard reaction between methyl benzoate and excess phenylmagnesium bromide, followed by aqueous workup. You do not have to consider stereochemistry. If a compound is formed more than once, add another sketcher and draw it again. Alternatively, you may use the square brackets tool to add stoichiometries greater than one. Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate multiple products using the sign from the drop-down menu.

Answer: Hello your question is incomplete attached below is the missing image

answer:

Reduction half-reaction: Zn(s) ⇒ Zn⁺² (aq)  + 2e⁻ ( occurs at anode )

Oxidation half-reaction; CO²⁺(aq) + 2e⁻ ⇒ CO (s)  ( occurs at the cathode )

Overall cell reaction ; Zn(s)  + CO²⁺(aq)  ⇒ Zn⁺² (aq) +  CO (s)

Explanation:

stating the standard reduction potentials

E° zn²⁺/zn = -0.76 v

E°Co²⁺ / Co = -0.28 v

since ; -0.76 v  < -0.28 v.    Zn will be oxidized while Co²⁺ will be reduced .

Reduction half-reaction: Zn(s) ⇒ Zn⁺² (aq)  + 2e⁻ ( occurs at anode )

Oxidation half-reaction;  CO²⁺(aq) + 2e⁻ ⇒ CO (s)  ( occurs at the cathode )

hence the

Overall cell reaction ;

Zn(s)  + CO²⁺(aq)  ⇒ Zn⁺² (aq) +  CO (s)

Answer:

Biphenyl

Explanation:

The reaction of bromo benzene with magnesium-ether solution yields a Grignard reagent.

The byproduct of this reaction is biphenyl. It is formed when two unreacted bromobenzene molecules are coupled together.

Hence, It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of biphenyl by-product formed.

Answer:

Two molecules of hydrogen combine with two molecules of oxygen to form hydrogen peroxide. Hence, its chemical formula is H2O2. It is the simplest peroxide (since it is a compound with an oxygen-oxygen single bond). Hydrogen peroxide has basic uses as an oxidizer, bleaching agent and antiseptic

Answer:

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Answer:

A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.

Silver atoms are oxidized while oxygen atoms are reduced by a loss of electrons and a gain of electrons respectively.

Explanation:

A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.

In a redox reaction,the two reactions occurring simultaneously can be divided into two half reactions; an oxidation half-reaction and a reduction half-reaction.

The oxidation half-reaction involves losing electrons and thus an increase in oxidation number of the species being oxidized. Whereas, the reduction half reaction involves gaining electrons and thus, a reduction innthe oxidation number of the species being reduced.

The species which oxidizes another species is known as an oxidizing agent and isnitself reduced due to its accepting electrons from the species being oxidized. Th reducing agent reduces another species and is itself oxidized as it loses electrons to the oxidized agent.

In the given reaction as shown below:

4 Ag (s) + 2 H₂S (g) + O₂ (g) ---> 2 Ag₂S (s) + 2 H₂0 (g)

The reaction is a redox reaction as a change innthe oxidation number of the reacting species; both oxidation and reduction occurs simultaneously and to the same extent.

The metallic silver atoms, have an oxidation number of zero initially. However, each of the four moles of atoms give up one mole of a electrons each to become oxidized to silver (i) ions, Ag+.

On the other hand, molecular oxygen gas also having oxidation number of zero becomes reduced to oxygen ion, O²-. Each of the two moles of atom in the oxygen gas molecule accept two electrons each donated by the metallic silver atoms to become reduced to oxygen ion, O²-.

The oxidation numbers of hydrogen ion and sulfide ion do not change.

Answer:

A

Explanation:

helium (alpha particle)

Answer:

-401.06 atm

Explanation:

Applying,

P = (nRT/V-nb)-(an²/V²)............... Equation 1

Where P = Pressure, R = Universal gas constant, V = molar Volume, T = Temperature in Kelvin, a = gas constant a , b = gas constant b, n = numbers of mole

From the question,

Given: T = 299.8 K, V = 0.8006 L, a = 2.318 L.atm/mol, b = 3.978×10²L/mol

Constant: R = 0.0082 atm.dm³/K.mol

Substitute these values into equation 1

P = [(0.0082×299.8×10.53)/(0.8006-(10.53×397.8)]-[(10.53²×2.318/0.8006²)]

P = (25.89/-4188.0334)-(400.995)

P = -0.0618-400.995

P = -401.06 atm

Explanation:

Ethanol is made by the hydration of ethylene in the presence of a catalyst such as sulfuric acid (H 2SO 4).

Answer:

"[tex]N^+(g) \rightarrow N^{2+}(g) + e^-[/tex]" is the appropriate answer.

Explanation:

Thus the above is the correct answer.

Answer:

Vien, sometimes we make problems like this harder than they need to be. Suppose I asked you, “How many dozen wheels are on four dozen automobiles?” You would have no trouble answering 4 x 4 = 16 dozen.

A mole is just a quantity like a dozen. And in this case, instead of 4 wheels, each C2F6 molecule bears 6 fluorine atoms, right?

So 4 x 6 = 24 moles of fluorine atoms. You’ve got this, Vien!

Answer:

1:Part A.

[tex]\bold{42.2 g C_{12}H_{22}O_{11} \:in \:528 g H₂O}[/tex]

Mass Percent=[tex]\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}[/tex]

=[tex]\frac{42.2}{528}*100\%=\bold{\underline{7.99\: or \:8\%}}[/tex]

Part B.

[tex]\bold{198\:m g\: C_{6}H_{12}O_{6} \:in\:4.71 g\: H₂O}[/tex]

mass of solute: 198mg

mass of solvent :4.71g=4710g

Mass Percent=[tex]\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}[/tex]

=[tex]\frac{198}{4710}*100\%=\bold{\underline{4.20\%}} [/tex]

Part C.

[tex]\bold{8.85 g NaCl \:in \:190 g\: H₂O}[/tex]

Mass Percent=[tex]\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}[/tex]

=[tex]\frac{8.85}{190}*100\%=\bold{\underline{4.66\%}}[/tex]

Answer:

It will help you !!!!!!!!!!

Answer:

Tapeworms and roundworms both belong to the same phylum however, their families are different from one another.

Tapeworms are flat, segmented intestinal parasites of the cat and dog and humans sometimes. They are present in the intestines of pets and depend on them, therefore, are parasites. These parasites look like tape which gives it its name.

Roundworms can also infect humans and the most common cases are among children. When not treated immediately, they can cause severe damage to a human host and can even cause blindness. Tapeworms are white in color with a long, segmented body.

Answer:

[tex]\rho_t=1025000 gmL^{-1}[/tex]

Explanation:

From the question we are told that:

Density of acetic acid [tex]\rho_a=205 gL^{-1}[/tex]

Density of Water [tex]\rho_w=820 gL^{-1}[/tex]

Generally the equation for Solution Density is mathematically given by

[tex]\rho_t= \rho_w+\rho_a[/tex]

[tex]\rho_t=205+820[/tex]

[tex]\rho_t=1025 gL^{-1}[/tex]

[tex]\rho_t=1025000 gmL^{-1}[/tex]

Answer:

One compound, called n-butane, where the prefix n- represents normal, has its four carbon atoms bonded in a continuous chain. The other, called isobutane, has a branched chain. Different compounds that have the same molecular formula are called isomers.

Answer:

d. constitutional isomers​

Explanation:

i hope it will help

Copper tube is used in EBRT.

Copper tube is used in EBRT.

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Answer:

The water ligands surrounding the cobalt metal center are being replaced by ethylenediamine and chloride ligands which results in a different crystal field splitting. Thus, the energy associated with electron transitions between the do-orbitals will differ for the two compounds showing a color change.

Explanation:

The five d-orbitals are usually degenerate. Upon approach of a ligand, the d-orbitals split into two sets of orbitals depending in the nature of the crystal field.

The magnitude of crystal field splitting is affected by the nature of the ligand. Ligands having filled p-π orbitals such as ethylenediamine lead to greater crystal field splitting.

The change in the colour that takes place when ethylenediamine is added to the solution of cobalt(II) chloride occurs due to a different crystal field splitting pattern. Thus, the energy associated with electron transitions between the d-orbitals now differ for the two compounds showing a color change.

Answer:

n Al=  10/27( mol)- >n Al2O 3 =5/27(mol)

Explanation:

Answer:

The correct answer is "654.54 nm".

Explanation:

According to the question,

⇒ [tex]\frac{1}{\lambda}= Rh(\frac{1}{n1^2} -\frac{1}{n2^2} )[/tex]

By substituting the values, we get

       [tex]=1.1\times 10^7(\frac{1}{4} -\frac{1}{9} )[/tex]

       [tex]=1.1\times 10^7(\frac{9-4}{36} )[/tex]

       [tex]=1.1\times 10^7(\frac{5}{36} )[/tex]

       [tex]=654.54\ nm[/tex]

Thus the above is the right solution.

Answer:

Average specific heat capacity of metal = 0.57 J/g°C

Explanation:

Heat lost = Heat gained

Heat energy gained or lost, H = mcΔT

where m = mass of substance, c = specific heat capacity, ΔT = temperature change

Trial 1:

Heat lost by metal = -[2.746 g × c × ΔT]

ΔT = (26.3 - 72.1) °C = -45.8 °C

Heat lost by metal = -[2.746 g × c × (-45.8 °C)] = c × (125.7688)g°C

Heat gained by water = 15.200 × 4.18 × ΔT

ΔT = (26.3 - 24.7) = 1.6 °C

Heat gained by water = 15.200 × 4.18 × 1.6 = 101.6576 J

From Heat lost = Heat gained

c × (125.7688)g°C = 101.6576 J

c = 101.6576 J / 125.7688 g°C

c = 0.8083 J/g°C

Trial 2:

Heat lost by metal = -[2.750 g × c × ΔT]

ΔT = (26.2 - 72.2)°C] = - 46 °C

Heat lost by metal = -[2.750 g × c × (-46 °C)

Heat lost by metal = c × (126.5) g°C

Heat gained by water = 15.206 × 4.18 × ΔT

ΔT = (26.2 - 24.6) = 1.6 °C

Heat gained by water = 15.206 × 4.18 × 1.6 = 101.697728 J

From Heat lost = Heat gained

c × (126.5)g°C = 101.6977 J

c = 101.697728 J / 126.5 g°C

c = 0.8039 J/g°C

Trial 3:

Heat lost by metal = -[2.900 g × c × ΔT]

ΔT = (24.7 - 71.9)°C] = - 47.2 °C

Heat lost by metal = -[2.900 g × c × (- 47.2 °C)

Heat lost by metal = -[2.900 g × c × (- 47.2)°C] = c × (136.88)g°C

Heat gained by water = 15.201 × 4.18 × ΔT

ΔT = (24.7 - 24.5) = 0.2 °C

Heat gained by water = 15.201 × 4.18 × 0.2 = 12.708036 J

From Heat lost = Heat gained

c × (136.88)g°C = 12.708036 J

c = 12.708036 J / 136.88 g°C

c = 0.0928 J/g°C

Average specific heat capacity of metal = (0.8083 + 0.8039 + 0.0928) J/g°C / 3

Average specific heat capacity of metal = 0.57 J/g°C

Answer:

89°c

Explanation:

i think this is the answer cause the temperature changed from 11 to 100 and so the atmospheric temperature would be the change in temperature

100-11=89°c

I hope this helps and sorry if it's wrong

Answer:

B.bonds are broken and new bonds are formed

Answer:

The answer is "17200 years".

Explanation:

Given:

[tex]A = 20 \ \frac{counts}{minute}\\\\A_{o} = 160\ \frac{counts}{minute}[/tex]

Let the half-life of carbon-14, is beta emitter, is [tex]T = 5730\ years[/tex]

Constant decay [tex]\ w = \frac{0.693}{ T}[/tex]

[tex]= 1.209 \times 10^{-4} \ \frac{1}{year}\\[/tex]

The artifact age [tex]t= ?[/tex]

[tex]A = A_{o} e^{-wt} \\\\e^{-wt} = \frac{A}{A_{o}}\\\\-wt = \ln \frac{A}{A_{o}}\\\\= -2.079\\\\t = 1.7199 \times 10^{4} \ years\\\\\sim \ 17200\ years\\[/tex]

Suppose you have a material in it's liquid phase. As you give energy to that liquid, the temperature of the liquid will increase gradually, and the relation between the increase of temperature and the given energy is the specific heat.

Now, there is a point, a critical point, where the temperature stops to increase, which means that we are near a change of phase. So from this point on, the energy is not used to increase the kinetic energy of the particles (which would increase the temperature), the energy is used to break the bonds and allow a change of phase, for example, from liquid to gas.

So, we know that if you have a mol of water at 100°C, then you need to add 40.7 kJ of energy to change the phase of the water from liquid to gas phase.

This means that if you have a mol of water and you give that exact energy, the temperature will not change, instead, you now will have a mol of water at the temperature of 100°C.

Similarly with the case at 25°C (which happens for a particular pressure only)

So the heat of vaporization can not really be related to increases in temperature as you thought.

For changes in temperature, you need to use the specific heat.

We know that for water it is:

c = 4.184 J/g*°C = 76.15 J/mol*°C

So, if you want to increase the temperature from 25° to 100°

This means an increase of 75°C of one mol of water.

We just need to multiply the above number by:

1mol*(75°C)

Energy needed = (76.15 J/mol*°C)*1mol*(75°C) = 5,711.25 J

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Explanation:

Amount of water required in each case:

(a)The mass% of the solution is:9.95

Mass of solute that is urea is 6.80 g

To determine the mass of solvent water use the formula:

[tex]mass percent=\frac{mass of solute}{mass of solution} x 100\\\\9.95=(6.80g/mass of solution )x100\\mass of solution =(6.80 /9.95)x100\\=68.3 g[/tex]

Hence the mass of solvent = mass of solution - the mass of solute

=68.3 g - 6.80g

=61.5 g

Hence, the answer is mass of solvent water required is 61.5 g.

(b) Given mass%=1.70

mass of solute MgBr2 = 29.3 g

The mass of solvent water required can be calculated as shown below:

[tex]mass percent=\frac{mass of solute}{mass of solution} x 100\\\\1.70=(29.3g/mass of solution )x100\\mass of solution =(29.3 g /1.70)x100\\=1720 g[/tex]

The mass of the solution is 1720 g.

Mass of solvent water = mass of solution - mass of solute

=1720 g - 29.3 g

=1690.7 g

Answer: The mass of water required is 1690.7 g.

Answer:

Molarity = Moles of solute/Volume of solution(in Litres)

Solute = C10H12

moles of solute = Mass/Molar Mass

Molar Mass of C10H12 = 10(12) + 12(1)

= 132gmol-¹

Mole = 45.7/132 = 0.346mol

Molarity = 0.346/5.23

=0.066M

Answer:

A fire class is a system of categorising fire with regard to the type of material and fuel for combustion. Class letters are often assigned to the different types of fire, but these differ between territories. There are separate standards for the United States, Europe, and Australia

The correct order of the substances in order of decreasing boiling point is,

KF , CH30OH , CO2 , Ne

Boiling point, the temperature at which the force exerted by the surroundings upon a liquid exists equaled by the pressure exerted by the vapor of the liquid; under this situation, the addition of heat affects the transformation of the liquid into its vapor without increasing the temperature.

The main difference between the boiling point and the melting point stands that the melting point is determined as the temperature at which solid and liquid phases exist in equilibrium, whereas the boiling point stands as the temperature at which the vapor pressure of a liquid stands equal to the external pressure.

Hence, The correct order of the substances in order of decreasing boiling point is,

KF , CH30OH , CO2 , Ne

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Answer:

0.1 mol

Explanation:

6/(15*3+15)

0.1 mol moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+

The mole, symbol mol, exists as the SI base unit of the amount of substance. The quantity amount of substance exists as a measure of how many elementary entities of a provided substance exist in an object or sample.A mole corresponds to the mass of a substance that includes 6.023 x 1023 particles of the substance. The mole exists the SI unit for the amount of a substance. Its symbol stands mol.

The compound trimethylamine, (CH3 )3N, exists as a  weak base when dissolved in water.

A mole exist expressed as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole exists as a convenient unit to utilize because of the great number of atoms, molecules, or others in any substance.

To find the amount of the substance (CH3)3NH+ to calculate its molar mass:

M((CH3)3NH+) = (12+3)*3 + 14+1 = 60 g/mol

n((CH3)3NH+) = m/M

m((CH3)3NH+) = 6g

Thus,

n((CH3)3NH+) = 6g/60 g/mol = 0.1 mol

Hence,

n((CH3)3NH+) = 0.1 mol

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