In principle, the equilibrium in the dehydration of an alcohol could be shifted to the right be removal of water. Why is this tactic not a good option for the dehydration of 4-methyl-2-pentanol and cyclohexanol

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:

the electron groups around the central nitrogen atom is trigonal planar

Explanation:

An electron group is defined as a bond (single or multiple) or/and a non-bonding electron pair around the central atom. The central nitrogen (N) atom in NO3- has three (3) electron groups (three bonds to three oxygen, O atoms).

Three electron groups around a central atom are best arranged in a trigonal planer arrangement to minimize repulsions between the bonding and/or non-bonding electrons. Therefore, the arrangement of the electron groups is trigonal planer or planer trigonal.

Answer:

The required hybridization for a central atom that have:

a tetrahedral arrangement of electron pairs        is  [tex]sp^3[/tex]

a trigonal planar arrangement of electron pairs   is  [tex]sp^2[/tex]

a linear arrangement of electron pairs                 is   [tex]sp[/tex]

Explanation:

From the given information:

The required hybridization for a central atom that have:

a tetrahedral arrangement of electron pairs        is  [tex]sp^3[/tex]

a trigonal planar arrangement of electron pairs   is  [tex]sp^2[/tex]

a linear arrangement of electron pairs                 is   [tex]sp[/tex]

Hybridization is the mixing and blending of two or more pure atomic orbitals ( s, p and d) to forma two or more hybrid atomic orbitals that are identical in shape and energy e. sp ,sp² , sp³, sp³d hybrid orbitals.

Examples:

Tetrahedral

In CH₄ , carbon C is the central atom.

A 2s electron is excited from the ground state of boron 1s²2s²2p² to one of the empty orbitals to 2p to give the excited state 1s²2s²2p³.

In the excited state of carbon, the 2-s orbital can be mixed with the 2p orbitals in three ways : sp³, sp² and sp hybridization. For the formation of four sp³ hybrid orbitals, the 2s orbital are mixed with all the three p orbitals. The four sp³ hybrid orbitals are tetrahedrally arranged with a bond angle of 109.5⁰

Trigonal Planar

In BF₃ , Boron B is the central atom

A 2s electron is excited from the ground state of boron 1s²2s²2p¹ to one of the empty orbitals to 2p. The 2s orbital is then mixed with two orbitals of 2p to form three sp² hybrid orbitals that are trigonal planar arranged in the plane in order to minimize repulsion. They bonds between them form an equal strength and with bond angles of 120⁰

Linear arrangement:

In BeCl₂, Be is the central atom

To provide the unpaired electrons for covalent bonds a 2s electron is excited to a 2p orbital. Thereafter, the two atomic orbitals hybridize to give two identical orbitals called sp hybrid orbitals. The  sigma bond for,ed are equal in bond lengths and form a bond angle of 180°

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:

On the 2 hydrogen atoms.

Explanation:

δ+ indicates the atom has a lower electronegativity than the other atom it is bonded with. This only exist in polar covalent bonds, where the 2 atoms have different electronegativity values. When they have different electronegativity values, the one with higher electronegativity has a higher tendency to "pull" the shared electrons towards itself, they have a δ- symbol.

Back to H2O, since the electronegativity of elements increases from left to right horizontally and upwards vertically in the periodic table (except for noble gases, they are unreactive. Note that fluorine has the highest electronegativity), O atom has a higher electronegativity than hydrogen (hydrogen sits at the centre top of the table). hence, we can find δ+ on the hydrogen atoms.

Answer:

0.109 mol/tablespoon

Explanation:

6.37 g/ 58.5 mol = 0.10888888 mol (0.109 significantly)

Answer:

A: 0.109

Explanation:

Edge 2020

Answer:

i. n = 5

ii. ΔE = 7.61 × [tex]10^{-46}[/tex] KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × [tex]10^{-18}[/tex]([tex]\frac{1}{n^{2}_{final} }[/tex] - [tex]\frac{1}{n^{2}_{initial} }[/tex])

    (1/434 × [tex]10^{-9}[/tex]) = -2.178 × [tex]10^{-18}[/tex] ([tex]\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }[/tex])

⇒ 434 × [tex]10^{-9}[/tex] = (1/-2.178 × [tex]10^{-18}[/tex])[tex]\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }[/tex]

But, [tex]n_{final}[/tex] = 2

434 × [tex]10^{-9}[/tex] = (1/2.178 × [tex]10^{-18}[/tex])[tex]\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }[/tex]

434 × [tex]10^{-9}[/tex]  × 2.178 × [tex]10^{-18}[/tex] = [tex](\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })[/tex]

⇒ [tex]n_{initial}[/tex] = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

     = (hc) ÷ λ

    = (6.626 × 10−34 × 3.0 × [tex]10^{8}[/tex] ) ÷ (434 × [tex]10^{-9}[/tex])

    = (1.9878 × [tex]10^{-25}[/tex]) ÷ (434 × [tex]10^{-9}[/tex])

    = 4.58 × [tex]10^{-19}[/tex] J

    = 4.58 × [tex]10^{-22}[/tex] KJ

But 1 mole = 6.02×[tex]10^{23}[/tex], then;

energy in KJ/mole = (4.58 × [tex]10^{-22}[/tex] KJ) ÷ (6.02×[tex]10^{23}[/tex])

         = 7.61 × [tex]10^{-46}[/tex] KJ/mole

The initial energy level is 5  and the energy of this transition in units of kJ/mole is 7.57 * 10^-43 kJ/mole

We must first calculate ΔE as follows;

ΔE = hc/λ

h = Plank's constant = 6.6 * 10^-34 Js

c = speed of light = 3 * 10^8 m/s

λ = wavelength = 434 * 10^-9

ΔE =  6.6 * 10^-34 * 3 * 10^8/434 * 10^-9

ΔE = 0.0456 * 10^-17 J

ΔE = [tex]ΔE = -2.178 x10^-18 (\frac{1}{n^2final} - \frac{1}{n^2initial}) \\ΔE = -2.178 x10^-18 (\frac{1}{2^2} - \frac{1}{n^2initial} )\\\\4.56 * 10^-19/2.178 x10^-18 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\0.210 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\\frac{1}{n^2initial} = 0.25 - 0.210\\\frac{1}{n^2final} = 0.04\\n = (\sqrt{(0.04)^-1} \\n = 5[/tex]

Energy of this transition in units of kJ/mole = 4.56 * 10^-19/ 6.02 * 10^23

= 7.57 * 10^-43 kJ/mole

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Answer: materials and design Techniques that reduce the negative environmental impact of a structure

Explanation:

Answer:

Q = 1.5 kJ

Explanation:

It is given that,

The specific heat for aluminum is 0.900 J/g°C

Mass of sample, m = 3.8 g

Initial temperature, [tex]T_i=450^{\circ} C[/tex]

Final temperature, [tex]T_f=25^{\circ} C[/tex]

We need to find the heat released. The amount of heat released is given by the formula:

[tex]Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=3.8\times 0.9\times (25-450)\\\\Q=1453.5\ J\\\\Q=1.45\ kJ[/tex]

or

[tex]Q=1.5\ kJ[/tex]

So, the correct option is (A) i.e. 1.5 kJ.

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

Answer:

[tex]\Delta _cH=-1328.3kJ/mol[/tex]

Explanation:

Helllo,

In this case, for the given chemical reaction in gaseous state:

[tex]CH_3OCH_3+3O_2\rightarrow 2CO_2+3H_2O[/tex]

We comoute the combustion enthalpy as the reaction enthalpy for one mole of fuel (dimethyl ether) considering the formation enthalpy of each given substance and whether they are reactants (subtracting) or products (adding), therefore we write:

[tex]\Delta _cH=2*\Delta _fH_{CO_2}+3*\Delta _fH_{H_2O}-\Delta _fH_{CH_3OCH_3}-3*\Delta _fH_{O_2}[/tex]

Whereas the formation enthalpies for carbon dioxide, water, dimethyl ether and oxygen are -393.5, -241.8, -184.1 and 0 kJ/mol respectively, thereby, the combustion enthalpy turns out:

[tex]\Delta _cH=2(-393.5)+3*(-241.8)-(-184.1)-3(0)\\\\\Delta _cH=-1328.3kJ/mol[/tex]

Notice that enthalpy of formation of oxygen is zero since forming an element has no chemical sense, it just exists as it has been early demonstrated.

Regards.

Answer:

The answer is "−847 J/K".

Explanation:

The given expression is:

2Al(s)+ Fe2O3(s) → Al2O3(s)+ 2Fe(s)

Δ[tex]H^{\circ}_{rxn}=[/tex] ∑(Δ[tex]H^{\circ}_{products}-H^{\circ}_{reactants}[/tex])

by the above definition Δ[tex]H^{\circ}_{element}= 0\cdot KJ \cdot Mol^{-1}[/tex] For Such a Component under standard conditions from its standard state, that also applies here. But, we start taking the overview and follow the conventions of signing:

[tex]\to (-1669)-(-822) \frac{KJ}{mol}\\\\\to (-1669+822) \frac{KJ}{mol}\\\\\to -847\frac{KJ}{mol}\\\\[/tex]

Δ[tex]H^{\circ}_{rxn}=[/tex] -847 [tex]\frac{KJ}{mol} \ mol^{-1} \texttt{ we mean \mole of Reaction as written....}\\[/tex]

Answer:

Ba(NO₃)₂(s) → Ba²⁺ + 2NO₃⁻

Explanation:

A strong electrolyte is a salt (A compound that has an anion and a cation and are neutral) that, in water, dissociates completely in its ions.

In Barium nitrate, Ba(NO₃)₂, the cation is Ba²⁺ (Alkaline earth metal), and the anion is the nitrate ion, NO₃⁻.

Thus, when Ba(NO₃)₂ (s) is dissolved in water, its transformation is:

When solid barium nitrate (Ba(NO₃)₂) dissolves in water, it undergoes a dissociation process where the compound breaks apart into its constituent ions.

Dissociation refers to the process in which a compound breaks apart into its constituent ions when dissolved in a solvent, typically water. In this process, the chemical bonds within the compound are disrupted, resulting in the separation of positive and negative ions.

The dissociation occurs due to the interaction between the solute particles and the solvent molecules, leading to the formation of hydrated ions.

The transformation can be represented as follows:

Ba(NO₃)₂(s) → Ba²⁺(aq) + 2NO₃⁻(aq)

In this process, the barium nitrate compound dissociates into barium ions (Ba²⁺) and nitrate ions (NO₃⁻) in the aqueous solution. The resulting ions are free to move and conduct electricity, indicating that barium nitrate is a strong electrolyte when dissolved in water.

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Answer:

The difference between an ideal and a nonideal solution is given below:-

Explanation:

Ideal Solution:  

Non-ideal Solution:  

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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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:

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:

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!

Answer:

Following are the answer to this question:

Explanation:

The answer are:

1) ketopentose

2) Triose

3) Aldose

4) Ketose

5) Glucose

6) Aldohexose

Answer:

pH is a measure of the concentration of H+ ions in a solution of an acid or base. The pH plots the concentration of solutions in a range from 0–14.

Explanation:

The pH is a measure of the hydrogen ion(H^+) concentration in an acid or base. It can be obtained mathematically by the formula:

pH = —Log [H^+]

The pH scale ranges from 0 to 14

Answer:

it really is A

Explanation:

just got wrong answer because i put 16 and clearly b and c makes no sence : )

Answer:

An example of rancidity is when a chips pack is exposed to atmospheric air which results in a change in taste and odor.

Explanation:

Rancidity is a condition in which the substance with oil and fats get oxidized when they are exposed to air. A substance is said to be rancid when there is a change in smell, taste, and colour.Oil becomes rancid due to decomposition of fats it contains or sometimes milk becomes rancid due to not heating it in humid atmosphere, etc.

Answer:

Change in the taste and smell of food is the answer.

I'm leaving this app.

you can meet on instagram :- nikhil.pandey_1728.

Answer:

here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.

hope it helps...

The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.

Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.

There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.

The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.

Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

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Answer:

[tex]m=0.512m[/tex]

Explanation:

Hello,

In this case, we can consider the n-propanol as the solute (lower amount) and the t-butanol as the solvent (higher amount), for which, initially, we must compute the moles of n-propanol (molar mass = 60.1 g/mol) as shown below:

[tex]n_{solute}=0.400g*\frac{1mol}{60.1g0}=6.656x10^{-3}mol[/tex]

Since the molality is computed via:

[tex]m=\frac{n_{solute}}{m_{solvent}}[/tex]

Whereas the mass of the solvent is used in kilograms (0.0130g for the given one). Thus, we compute the resulting molality of the solution:

[tex]m=\frac{6.656x10^{-3}mol}{0.0130kg}\\ \\m=0.512\frac{mol}{kg}[/tex]

Or just:

[tex]m=0.512m[/tex]

Best regards.

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:

1105.6 K

Explanation:

The following data were obtained from the question:

CaCO3 —> CaO + CO2

Enthalpy (H) data:

CaCO3 = -1207 kJ/mol

CaO = -635 kJ/mol

CO2 = -394 kJ/mol

Entropy (S) data:

CaCO3 = +93 J/K mol

CaO = +40 J/K mol

CO2 = +214 J/K mol

Next, we shall determine the enthalphy change (ΔH). This can be obtained as follow:

CaCO3 —> CaO + CO2

CaCO3 = -1207 kJ/mol

CaO = -635 kJ/mol

CO2 = -394 kJ/mol

Heat of product (Hp) = -635 + (-394) = - 1029 KJ/mol

Heat of reactant (Hr) = -1207 kJ/mol

Enthalphy change (ΔH) = Hp – Hr

Enthalphy change (ΔH) = -1029 – (-1207)

Enthalphy change (ΔH) = 178 KJ/mol

Next, we shall determine the entropy change (ΔS). This can be obtained as follow:

CaCO3 —> CaO + CO2

CaCO3 = +93 J/K mol

CaO = +40 J/K mol

CO2 = +214 J/K mol

Entropy of product (Sp) = 40 + 214 = +254 J/Kmol

Entropy of reactant (Sr) = +93 J/Kmol

Entropy change (ΔS) = Sp – Sr

Entropy change (ΔS) = 254 – 93

Entropy change (ΔS) = + 161 J/Kmol

Finally, we can obtain the temperature at which the reaction is feasible as follow:

Enthalphy change (ΔH) = 178 KJ/mol = 178000 J/mol

Entropy change (ΔS) = + 161 J/Kmol

Temperature (T) =..?

Entropy change (ΔS) = Enthalphy change (ΔH) / Temperature (T)

ΔS = ΔH/T

161 = 178000/T

Cross multiply

161 x T = 178000

Divide both side by 161

T = 178000/161

T = 1105.6 K

Therefore, the temperature at which the reaction is feasible is 1105.6 K

Answer:

the mass of HBr that would react is 25.41 g of HBr

Explanation:

attached is the calculations.

Answer:

The answer is

Explanation:

Density = mass / volume

mass = density × volume

volume = 2cm³

density = 520g/cm³

mass = 2 × 520

= 1040g

Hope this helps you

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

[tex]\frac{dm}{dt} = -\frac{m}{\tau}[/tex]

Where:

[tex]m[/tex] - Current isotope mass, measured in kilograms.

[tex]t[/tex] - Time, measured in years.

[tex]\tau[/tex] - Time constant, measured in years.

The solution of this differential equation is:

[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]

Where [tex]m_{o}[/tex] is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

[tex]\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%[/tex]

[tex]1 - \frac{m(t+1)}{m(t)} = 0.002[/tex]

[tex]1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002[/tex]

[tex]1 - e^{-\frac{1}{\tau} } = 0.002[/tex]

[tex]e^{-\frac{1}{\tau} } = 0.998[/tex]

[tex]-\frac{1}{\tau} = \ln 0.998[/tex]

The time constant associated to the decay is:

[tex]\tau = -\frac{1}{\ln 0.998}[/tex]

[tex]\tau \approx 499.500\,years[/tex]

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

[tex]t_{1/2} = \tau \cdot \ln 2[/tex]

If [tex]\tau \approx 499.500\,years[/tex], the half-life of the isotope is:

[tex]t_{1/2} = (499.500\,years)\cdot \ln 2[/tex]

[tex]t_{1/2}\approx 346.227\,years[/tex]

The half-life of the radioactive isotope is 346 years.

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