Predict the products of the following elimination reaction, and draw the major product formed. Make sure to consider the stereochemistry of the reaction. 3-chloro-3-methylpentane reacts with sodium tertbutoxide in tertbutanol.
Predict the products of the following elimination reaction, and draw the major product formed. Make sure to consider the stereochemistry of the reaction. 3-chloro-3-methylpentane reacts with sodium ethoxide in ethanol.

Answer:

[tex]M=0.0637M[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)[/tex]

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

[tex]n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2} *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}[/tex]

Finally, the resulting molarity in 30.8 mL (0.0308 L):

[tex]M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M[/tex]

Regards.

Answer:

I think that the fire will continue burning, because the oil and water don't mix and the water is heavier (denser) than oil, so the oil will go up and the fire with it. That's why because the gas station have sand instead of water

Water is heavier than oil. Because oil is lighter and immiscible with water, it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire. As a result, the fire won't be put out.

A small amount of water will instantly sink to the bottom of a pan or deep fryer filled with hot, burning oil and explode there. The Scientific American claims that the characteristics of oils explain why they do not mix with water.

Oil or petroleum-related fires cannot be put out with water. Water sinks below the oil because it is heavier than oil and does not float, allowing the fire to continue to burn. Oil and petroleum fires can be put out with fire extinguishers or sand.

The temperature of the burning substance is lowered by water. The fire goes out when the temperature drops below the burning substance's ignition temperature. Here, the water serves as an acclimatizer.

Thus,  it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire.

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Answer:

The correct answer is 15.80 grams.

Explanation:

The reaction taking place in the given question,  

Pb(CH₃COO)₂ + Na₂SO₄ ⇒ PbSO₄ + 2NaCH₃COO

The number of moles can be calculated by using the formula,  

n = weight / molecular mass

Based on the given question, the weight of lead (II) acetate is 23.81 grams and the weight of sodium sulfate is 7.410 grams.  

The number of moles of Pb(CH₃COO)₂ is,  

n = 23.81 g / 325.29 g/mol = 0.0732 moles

The number of moles of Na₂SO₄ is,  

n = 7.410 g / 142.04 g/mol = 0.0521 moles

As one mole of lead (II) acetate needs one mole of sodium sulfate. Therefore, 0.0732 moles of lead (II) acetate needs 0.0732 moles of sodium sulfate.  

However, as sodium sulfate is less, that is, 0.0521, therefore, Na₂SO₄ is a limiting reactant.  

One mole of sodium sulfate produces one mole of PbSO₄. So, 0.0521 moles of Na₂SO₄ produces 0.0521 moles of PbSO₄.  

Now the mass of PbSO₄ is,  

mass = moles × molecular mass

mass = 0.0521 × 303.26 g/mol

mass = 15.80 grams.  

Answer:

The falling standards of Ghanaian youths can be minimized by proper upbringing of the children by their parents. The youths should be taught about what is wrong or right and there should be a corresponding reward for those who do good and exceptional in order to encourage others in towing that line and punishment should also be meted out to those who break the law. Mediocrity shouldn’t be celebrated and the elders should lead by example.

These will make the falling standards of Ghanaian youth get reduced.

Complete Question

The complete question is shown on the first uploaded image

Answer:

Part A

    activation barrier for the reaction [tex]E_a = 84 .0 \ KJ/mol[/tex]

Part B

    The frequency plot is  [tex]A = 2.4*10^{13} s^{-1}[/tex]    

Explanation:

From the question we are told that

     at  [tex]T_1 = 300 \ K[/tex]   [tex]k_1 = 5.70 *10^{-2}[/tex]

and  at  [tex]T_2 = 310 \ K[/tex]   [tex]k_2 = 0.169[/tex]

The  Arrhenius plot is mathematically represented as

      [tex]ln [\frac{k_2}{k_1} ] = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2} ][/tex]

Where [tex]E_a[/tex] is the activation barrier for the reaction

         R is the gas constant with a value of  [tex]R = 8.314*10^{-3} KJ/mol \cdot K[/tex]

Substituting values

          [tex]ln [\frac{0.169}{6*10^-2{}} ] = \frac{E_a}{8.314*10^{-3}} [\frac{1}{300} - \frac{1}{310} ][/tex]

=>       [tex]E_a = 84 .0 \ KJ/mol[/tex]

The  Arrhenius plot can also be  mathematically represented as

      [tex]k = A * e^{-\frac{E_a}{RT} }[/tex]

Here we can use any value of k from the data table with there corresponding temperature let take  [tex]k_2 \ and \ T_2[/tex]

So substituting values

        [tex]0.169 = A e ^{- \frac{84.0}{8.314*10^{-3} * 310} }[/tex]

=>       [tex]A = 2.4*10^{13} s^{-1}[/tex]    

Answer:

The correct IUPAC name for the organic reactant is :

d) 3-methylbutene

Explanation:

Firstly the  missing diagram is attached in the diagram below.

The objective of this question is to draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.

From the image attached below; we would see the reaction that occurs between the alkene and the HBr (hydrobromic acid). What really occur in the reaction is that; in the presence of HBr with an alkene compound a secondary 2° carbocation is usually formed. This secondary 2° carbocation formed is usually unstable, so what we called an hydride shift occurs (Markovnikov's product) here to form a stable tertiary 3° carbocation.

The correct IUPAC name for the organic reactant is : 3-methylbutene

Answer:

a) ortho-para isomers predominates

b) 3-nitrobenzoic acid ( meta isomer predominates)

c) 3-bromo nitrobenzene ( meta isomer predominates)

d) the ortho- para isomers predominates

Explanation:

a) Toluene contains -CH3 which is an ortho- para- director hence the major product of the sulphonation of toluene should be the ortho- para isomers.

b) The major product of the nitration of benzoic acid is 3-nitrobenzoic acid. This is an electrophilic substitution in which the meta isomer predominates.

c) The meta isomer predominates giving 3-bromo nitrobenzene as the major product.

d) The isopropyl group is an ortho- para director hence the ortho- para isomers predominates .

Answer:

The correct answer is 66.35 kilograms.

Explanation:

Based on the data given in the question, the energy consumed by the body of a human being is 50%. Based on the given data, the energy consumed in a day is 8000 kJ, 50 percent is the energy transfer efficiency. Thus, the consumption of total energy is 4000 kJ, and for the transformation of ADP to ATP, the energy involved is 30.6 kJ per mole.  

Hence, the total ATP produced in the process is,  

ATP = 4000 kJ / 30.6 kJ/mol

= 130.7189 mol.  

Thus, with the energy transfer efficiency of 50 percent, the total moles of ATP produced is 130.7 mol.  

The mass of ATP can be calculated by using the formula,  

moles = mass/molecular mass

The molecular mass of ATP is 507.18 g per mol

Now by putting the values we get,  

mass of ATP = 130.7189 mol * 507.18 g/mol

= 66298.011 g or 66.298 kg

It is mentioned that human comprise 50 g of ATP or 0.05 kg of ATP. Therefore, the sum of the available ATP will be.  

= Total production of ATP + Total ATP available

= 66.298 kg + 0.05 kg

= 66.348 kg

Hence, the sum of the ATP that is turned over by a resting human in a day is 66.35 kg.  

Answer:

pH = - log [concentration]

pH = - log (0.0043M)

pH = 2.37

Answer: D

Explanation: The elements in two compunds switch places

Answer:

0.031 m

HgO(s) ⇒ Hg(l) + 1/2 O₂(g)

Chemical change

Element

Explanation:

A 75 gram solid cube of mercury (II) oxide has a density of 2.4 × 10³ kg/m³.  What is the length of one side of the cube in cm?

Step 1: Convert the mass to kilograms

We will use the relationship 1 kg = 1,000 g.

[tex]75g \times \frac{1kg}{1,000g} = 0.075kg[/tex]

Step 2: Calculate the volume (V) of the cube

[tex]0.075kg \times \frac{1m^{3} }{2.4 \times 10^{3} kg} = 3.1 \times 10^{-5} m^{3}[/tex]

Step 3: Calculate the length (l) of one side of the cube

We will use the following expression.

[tex]V = l^{3} \\l = \sqrt[3]{V} = \sqrt[3]{3.1 \times 10^{-5} m^{3} }=0.031m[/tex]

The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?

The balanced chemical equation is:

HgO(s) ⇒ Hg(l) + 1/2 O₂(g)

This is a chemical change because new substances are formed.

The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?

The liquid gray substance is Hg(l), which is an element because it is formed by just one kind of atoms.

Answer: Fossil fuels

Explanation:

Fossil fuels such as petroleum, oil,  and natural gas, are non-renewable energy resources which are formed from the remains of  prehistoric ancient  plants and animals beneath layers of rock of the earth surface.

By analyzing and studying fossil fuels using Radiocarbon analyses by archaeologists, earth scientists  etc, Information about the history of the earth can be obtained from the decomposition of dead organisms present in fossil fuels.

Answer:

6-metyl-2-heptyne

Explanation:

C-C-C-C-C-C-C hept

   2

C-C≡C-C-C-C-C  2-heptyne

                   C

                    | 6

C-C≡C-C-C-C-C

6-metyl-2-heptyne

The IUPAC name for the above structure is 6 methyl, hept-2-yne.

IUPAC stands for international Union of pure and applied chemistry. It is the body in charge of naming organic chemical compounds.

The naming is is based on a molecule's longest chain of carbons connected by single/double/triple bonds, whether in a continuous chain or in a ring etc.

According to this question, a structure is given. The following applies;

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Answer:

  1703 mmHg

Explanation:

Volume and pressure are presumed to be inversely proportional. Hence a change in volume by a factor of 125/325 = 5/13 is expected to change the pressure by a factor of 13/5:

  (13/5)(655 mmHg) = 1703 mmHg

Answer:

10 g of CO2

Explanation:

Equation of the reaction:

CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2

Fom the above balanced equation,

1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2

Molar mass of Octane = 114 g/mol

Molar mass of oxygen gas = 32 g/mol

Molar mass of CO2 = 44 g/mol

Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.

From the given mass of reactants;

3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.

Therefore oxygen is the limiting reactant.

15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.

Mass of CO2 produced will be

(352 * 15.6)/544 = 10 g of CO2

Answer: The oxygen's oxidation number is -2.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

In [tex]Fe_2O_3[/tex], Fe is having an oxidation state of +3 called as  cation and oxygen  is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]Fe_2O_3[/tex]

The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.

Answer:

[CH₂Cl₂] = 7.07x10⁻² M

[CH₄] = 0.319 M

[CCl₄] = 0.164 M  

Explanation:

The equilibrium reaction is the following:

2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)  

The equilibrium constant of the above reaction is:

[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]

When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:

[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]

[tex]C_{CH_{4}} = 0.328 M[/tex]      

Now, the concentrations at the equilibrium are:

2CH₂Cl₂(g)   ⇄   CH₄(g)  +  CCl₄(g)

5.35x10⁻² - 2x   0.328 + x   0.173 + x    

[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]

[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]

Solving the above equation for x:  

x₁ = 0.076 and x₂ = -0.0086

Hence, the concentration of the three gases once equilibrium has been reestablished is:

[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M

[CH₄] = 0.328 + (-0.0086) = 0.319 M

[CCl₄] = 0.173 + (-0.0086) = 0.164 M  

We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.

I hope it helps you!

Answer:

[tex]\Delta H=1962.3J[/tex]

Explanation:

Hello,

In this case, we can compute the change in the solution enthalpy by using the following formula:

[tex]\Delta H=mC\Delta T[/tex]

Whereas the mass of the solution is 350 g, the specific heat capacity is 4.184 J/g °C and the change in the temperature is 1.34 °C, therefore, we obtain:

[tex]\Delta H=350g*4.184\frac{J}{g\°C} *1.34\°C\\\\\Delta H=1962.3J[/tex]

It is important to notice that the mass is just 350 g that is the reacting amount and by means of the law of the conservation of mass, the total mass will remain constant, for that reason we compute the change in the enthalpy as shown above, which is positive due to the temperature raise.

Best regards.

Answer:

Explanation:

In cobalt - 65 ,

no of protons is 27 ( p )

no of neutron = 65 - 27 ( n )

= 38

n / p ratio

=  38 / 27

= 1.41

If case of emission of alpha particle

no of proton p = 27 - 2 = 25

no of neutrons = 38 - 2 = 36

n / p ratio = 36 / 25

= 1.44

So it increases

In case of emission of beta particle

No of neutron n = 38 - 1 = 37

No of proton = 27 + 1 = 28

n / p ratio = 37 / 28

= 1.32

Hence ratio decreases.

Hence beta ray decay will result in decrease in n / p ratio.

Answer:

D

Explanation:

trust me its correct i think

Answer:

C3H6O

Explanation:

Step 1:

Data obtained from the question include the following:

Mass of the compound = 0.225g

Mass of CO2 = 0.512g

Mass of H2O = 0.209g

Step 2:

Determination of the masses of carbon, hydrogen and oxygen present in the compound.

This is illustrated below:

For Carbon, C:

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

Mass of C in CO2 = 12/44 x 0.512 = 0.1396g

For Hydrogen, H:

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

Mass of H in H2O = 2/18 x 0.209 = 0.0232g

For Oxygen, O:

Mass of O = 0.225 – (0.1396 + 0.0232)

Mass of O = 0.0622g

Step 3:

Determination of the empirical formula for caprioc acid.

This can be obtain as follow:

C = 0.1396g

H = 0.0232g

O = 0.0622g

Divide by their molar mass

C = 0.1396/12 = 0.0116

H = 0.0232/1 = 0.0232

O = 0.0622/16 = 0.0039

Divide by the smallest

C = 0.0116/0.0039 = 3

H = 0.0232/0.0039 = 6

O = 0.0039/0.0039 = 1

Therefore, the empirical formula for caprioc acid is C3H6O

Answer:

Mass of Ga = 0.73694 gram

Explanation:

Given:

Current = 0.850 A

Time = 60 minutes

Find:

Amount of gas deposit.

Computation:

Total charge = Current × Time in second

Total charge = 0.850 × 60 × 60

Total charge = 3,060 C

Mole of electron = Total charge / Faraday constant         [Faraday constant = 96,485.3329]

Mole of electron = 3,060 / 96,485.3329

Mole of electron = 0.0317146

Moles of Ga = 1/3 [Mole of electron]

Moles of Ga = 1/3 [0.0317146]

Moles of Ga = 0.01057

Mass of Ga = molar mass × Moles of Ga

Mass of Ga = 69.72 × 0.01057

Mass of Ga = 0.73694 gram

Answer:

[tex]2.11\ * 10^{-2}[/tex]  is the correct answer to the given question.

Explanation:

Given k=6.40 x 10-3 min-1.

According to the first order reaction .

The concentration of time can be written as

[tex][\ A\ ]\ = \ [\ A_{0}\ ] * e \ ^\ {-kt}[/tex]

Here [tex][\ A\ ]_{0}[/tex] = Initial concentration.

So  [tex][\ A\ ]_{0}= 0.0314 M[/tex]

Putting this value into the above equation.

[tex]0.0314 \ *\ e^{6.40 x 10^{-3} \ * \ 62.0 }[/tex]

=0.211 M

This can be written as

[tex]=\ 2.11 *\ 10^{-2}[/tex]

Answer:

The answer is Hg.

Explanation:

Symbol for Mercury is Hg.