What is the difference between elimination and substitution reaction
Identify the key factors that will determine if a reaction undergoes elimination or substitution mechanism.
Use the following reagents to determine the type of reaction pathway expected and determine the products in each reaction.

a. Tert BuO- in tertbutanol and chlorobutane
b. KOH in water and bromobutane
c. NaI in acetone and bromobutane

Write a conclusion of no more than two paragraphs to summarize your results

North America and south africa

Answer:

B) ethers

Explanation:

The functional group of an organic compound defines its specificity. The functional group is responsible for the chemical behavior of an organic compound. For example, alkenes are known to have a carbon-carbon double bond (C=C) functional group.

Likewise, organic compounds known as ETHERS are known to possess an ethoxy functional group i.e. oxygen atom bonded to two alkyl groups (R- OR; where R is an alkyl group). Members of ether functional group includes dimethyl ether (CH3-O-CH3), diethyl ether (C2H5-O-C2H5).

Given concentration of NaCl=15%

Means ,

In every 100g of Solution 15g of NaCl is present .

Now

So ,

[tex]\\ \Large\sf\longmapsto 250\times 15\%[/tex]

[tex]\\ \Large\sf\longmapsto 250\times \dfrac{15}{100}[/tex]

[tex]\\ \Large\sf\longmapsto 37.5g[/tex]

37.5g of NaCl present in 250g of solution.

Answer:

Given concentration of NaCl=15%

Means ,

In every 100g of Solution 15g of NaCl is present .

Now

Given mass=250g

So ,

➡250 × 15%

➡250×15/100

➡37.5g

37.5g of NaCl present in 250g of solution.

Answer:

16

Explanation:

Because the sum of all electron in that compound should be 41 and as it has one electron extra ,total no. of electrons are 42 .

So if we add 26 +16 we get 42

Hence it's correct answer

QUESTION:- Manganese-55 has _____neutrons.

OPTIONS :-

A. 55

B. 30

C. 25

ANSWER:- NUMBER OF NEUTRONS IS EQUAL TO THE DIFFERENCE BETWEEN THE MASS IF THE ATOM AND ATOMIC NUMBER

SO DIFFERENCE IS EQUAL TO :- 55-25 = 30 NEUTRONS.

SO THERE IS 30 NEUTRONS IN SINGLE ATOM OF THE MANGANESE-55 ATOM.

Answer:

the mass of an atom is the sum of proton and neutron which are both concentrated in nocleus of an atom. from the question the mass is given as 55 and the proton is 25.

Una sustancia homogénea es una sustancia que se compone de una sola fase.

Recordemos que definimos una fase en química como "cantidad química y físicamente uniforme u homogénea de materia que se puede separar mecánicamente de una mezcla no homogénea y que puede consistir en una sola sustancia o una mezcla de sustancias" según Ecyclopedia Britiannica.

El hecho de que un sistema esté compuesto por una sola sustancia no lo hace es autóctono. A veces, un sistema puede estar compuesto por partículas sólidas de una sustancia en equilibrio con su líquido. El sistema contiene solo una sustancia pero en diferentes fases, por lo tanto, el sistema contiene una sustancia pero no es homogéneo.

Por tanto, el hecho de que un sistema contenga una sola sustancia no significa necesariamente que sea homogéneo.

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Explanation:

here's the answer to your question

The pressure of the isoflurane gas at the temperature of 27.4 °C is 0.31 atm.

Gay-Lussac's law states that the pressure of a gas is directly proportional to the absolute temperature when the volume of the gas is kept constant.

Mathematically, Gay Lussca's law can be written as follows:

P/T = k

It is also expressed as the pressure of the gas being directly proportional to temperature.

P ∝ T                    (where V is constant)

or

[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex]                                       ................(1)

Given, the initial pressure and initial temperature of the gas:

P₁ = 0.30 atm

and T₁ =  17.9 °C. = 17.9 + 273 = 290.9 K

The final temperature of gas T₂ = 27.4°C = 27.4 + 273 = 300.4 K

Now,  from the equation (1): [tex]P_2=\frac{P_1\times T_2}{T_1}[/tex]

P₂ = (0.30) × (300.4)/290.0

P₂ = 0.31 atm

Therefore, the pressure of the isoflurane at  27.4°C is equal to 0.31 atm.

Learn more about Gay Lussac's law, here:

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Answer:

True

Explanation:

When Methyl Pentene is introduced in a chemical reaction with dichloromethane then the major product will be bromomethylpentene. There can be small amount of bromo methyl pentene than the amount of methyl pentene introduced for reaction.

Answer:

2 m/s

Explanation:

Applying the formulae of velocity,

V = d/t............. Equation 1

Where V = Velocity of the body, d = distance, t = time

From the question,

Given: d = 600 m, t = 5 minutes = (5×60) = 300 seconds.

Substitute these values into equation 1

V = 600/300

V = 2 m/s.

Hence the velocity of the body when it travels is 2 m/s

The sum of the squares of two sides of a right angle is equal to the square of the hypotenuse

Explanation:

The given balanced chemical equation is:

[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]

The value of Kc at 445oC is 0.020.

[HI]=1.5M

[H2]=2.50M

[I2]=0.05M

The value of Qc(reaction quotient ) is calculated as shown below:

Qc has the same expression as the equilibrium constant.

[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]

Qc>Kc,

Hence, the backward reaction is favored and the formation of Hi is favored.

Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.

Answer:

0.886 J/g.°C

Explanation:

Step 1: Calculate the heat absorbed by the water

We will use the following expression

Q = c × m × ΔT

where,

Q(water) = c(water) × m(water) × ΔT(water)

Q(water) = 4.184 J/g.°C × 50.0 g × (34.4 °C - 25.36 °C) = 1.89 × 10³ J

According to the law of conservation of energy, the sum of the energy lost by the solid and the energy absorbed by the water is zero.

Q(water) + Q(solid) = 0

Q(solid) = -Q(water) =  -1.89 × 10³ J

Step 2: Calculate the specific heat capacity of the solid

We will use the following expression.

Q(solid) = c(solid) × m(solid) × ΔT(solid)

c(solid) = Q(solid) / m(solid) × ΔT(solid)

c(solid) = (-1.89 × 10³ J) / 32.53 g × (34.4 °C - 100. °C) = 0.886 J/g.°C

Answer:

A. Cation

B. Ne

Explanation:

The ion is positively charged by 2, making it a cation.

The electron configuration of the nearest noble gas Neon is 1s22s22p6

1. A monatomic ion with a charge of 2 has an electronic configuration of 1s22s22p6 is Neon.

2. chemical symbol of the noble gas is Kr (krypton).

Isoelectronic atom or ion has the same number of valence electrons. Krypton has 36 electrons and 36 protons (atomic number 36).

Neon is a chemical element with the symbol Ne and atomic number 10. It is a noble gas. Neon is a colorless, odorless, inert monatomic gas under standard conditions, with about two-thirds the density of air. It was discovered (along with krypton and xenon) in 1898 as one of the three residual rare inert elements remaining in dry air after nitrogen, oxygen, argon, and carbon dioxide were removed. Neon was the second of these three rare gases to be discovered and was immediately recognized as a new element of its bright red emission spectrum. The name neon is derived from the Greek word, the neuter singular form of  (neos), meaning 'new'. Neon is chemically inert, and no uncharged neon compounds are known. The compounds of neon currently known include ionic molecules, molecules held together by van der Waals forces, and clathrates.

Learn more about Neon

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Answer:

74%

Explanation:

Step 1: Write the balanced equation

2 C₆H₁₀(l) + 17 O₂(g) ⇒ 12 CO₂(g) + 10 H₂O(g)

Step 2: Determine the limiting reactant

The theoretical mass ratio (TMR) of C₆H₁₀ to O₂ is 164:544 = 0.301:1.

The experimental mass ratio (EMR) of C₆H₁₀ to O₂ is 115:199 = 0.578:1.

Since EMR > TMR, the limiting reactant is O₂.

Step 3: Calculate the theoretical yield of H₂O

The theoretical mass ratio of O₂ to H₂O 544:180.

199 g O₂ × 180 g H₂O/544 g O₂ = 65.8 g H₂O

Step 4: Calculate the percent yield of H₂O

%yield = (experimental yield/theoretical yield) × 100%

%yield = (49 g/65.8 g) × 100% = 74%

Answer:

Percentage yield of H₂O = 74.24%

Explanation:

The balanced equation for the reaction is given below:

2C₆H₁₀ + 17O₂ —> 12CO₂ + 10H₂O

Next, we shall determine the masses of C₆H₁₀ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This is can be obtained as follow:

Molar mass of C₆H₁₀ = 82 g/mol

Mass of C₆H₁₀ from the balanced equation = 2 × 82 = 164 g

Molar mass of O₂ = 32 g/mol

Mass of O₂ from the balanced equation = 17 × 32 = 544 g

Molar mass of H₂O = 18 g/mol

Mass of H₂O from the balanced equation = 10 × 18 = 180 g

SUMMARY:

From the balanced equation above,

164 g of C₆H₁₀ reacted with 544 g of O₂ to produce 180 g of H₂O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

164 g of C₆H₁₀ reacted with 544 g of O₂.

Therefore, 115 g of C₆H₁₀ will react to produce = (115 × 544)/164 = 381 g of O₂.

From the calculations made above, we can see that a higher mass (i.e 381 g) of O₂ than what was given (i.e 199 g) is needed to react with 115 g of C₆H₁₀.

Therefore, O₂ is the limiting reactant and C₆H₁₀ is the excess reactant.

Next, we shall determine the theoretical yield of H₂O. This can be obtained by using the limiting reactant as shown below:

From the balanced equation above,

544 g of O₂ reacted to produce 180 g of H₂O.

Therefore, 199 g of O₂ will react to produce = (199 × 180)/544 = 66 g of H₂O.

Thus, the theoretical yield of H₂O is 66 g.

Finally, we shall determine the percentage yield. This can be obtained as follow:

Actual yield of H₂O = 49 g

Theoretical yield of H₂O = 66 g

Percentage yield of H₂O =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of H₂O = 49/66 × 100

Percentage yield of H₂O = 74.24%

Answer:

a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. i. NO₂⁻ is the oxidizing agent

ii. NO₃⁻ is the reducing agent.

Explanation:

a. Balance the following skeleton reaction

The reaction is

NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)

The half reactions are

NO₂ (g) → NO₃⁻ (aq)  (1) and

NO₂ (g) → NO₂⁻  (aq) (2)

We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)

We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms

NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)

The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.

So, NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

We now add equation (4) and (5)

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

+  NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + e⁻  (4)

2NO₂ (g) + H₂O (l)  → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq)  

We now add two hydroxide ions to both sides of the equation.

So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + 2OH⁻ (aq)

The hydrogen ion and the hydroxide ion become a water molecule

2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H₂O (l)

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

So, the required reaction is

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. Identify the oxidizing agent and reducing agent

Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = 0

x - 4 = 0

x = 4

Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = -1

x - 4 = -1

x = 4 - 1

x = 3

Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = -1

x + 3(-2) = -1

x - 6 = -1

x = 6 - 1

x = 5

i. The oxidizing agent

The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus  NO₂⁻ is the oxidizing agent

ii. The reducing agent

The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and  NO₃⁻ is the reducing agent.

Answer:

There are 8Si atoms and 16 O atoms per unit cell

Explanation:

From the question we are told that:

Edge length [tex]l=0.700nm=>0.7*10^9nm[/tex]

Density [tex]\rho=2.32g/cm^3[/tex]

Generally the equation for Volume is mathematically given by

[tex]V=l^3[/tex]

[tex]V=(0.7*10^9)m^3[/tex]

[tex]V=3,43*10^-{22}cm[/tex]

Where

Molar mass of  (SiO2) for one formula unit

[tex]M=28+32[/tex]

[tex]M=60g/mol[/tex]

Therefore

Density of Si per unit length is

[tex]\rho_{si}=\frac{9.96*10^{23}}{3.43*10^22}[/tex]

[tex]\rho=0.29[/tex]

Molar mass of  (SiO2) for one formula unit

[tex]M=28+32[/tex]

[tex]M=60g/mol[/tex]

Therefore

There are 8Si atoms and 16 O atoms per unit cell

Answer:

b) 0.75 mol, 0.9375 mol

Explanation:

According to this question, ammonia reacts with oxygen to produce nitrogen monoxide and water. The chemical equation is as follows:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Based on this balanced equation, 4 moles of ammonia (NH3) reacts with 5 moles of oxygen (O2).

A stoichiometric amount of the two reactants (NH3 and O2) must represent the ratio 4:5.

Given the provided options;

0.75 mol of ammonia (NH3) will react with (0.75 × 5/4) = 0.935 mol of O2 for them to be in stoichiometry.

N.B: 1 mol of NH3 will react with 1.25mol of O2 and not 1g, 1.25g.

Answer:

0.034 M

Explanation:

1/[A] = kt + 1/[A]o

[A] = ?

k= 0.04556

t= 10.0 minutes or 600 seconds

[A]o = 0.50 M

1/[A] = (0.04556 × 600) + 1/0.50

[A] = 0.034 M

Answer:

A) Val-Gly-Arg-Ala-Val-Lys-Ala-Gly-Phe

B) Ala-Val-Lys-Val-Gly-Arg-Ala-Gly-Phe

Explanation:

The possible sequences that could be obtained from the available dat provided are :

A) Val-Gly-Arg-Ala-Val-Lys-Ala-Gly-Phe

B) Ala-Val-Lys-Val-Gly-Arg-Ala-Gly-Phe

Trypsin is generally a catalyst  for the breakdown of proteins into smaller peptides ( during the hydrolysis of peptide bonds )  

Answer:

0.0922 g

Explanation:

Number of moles of acid present = 25/1000 × 0.0981

= 0.00245 moles

Number of moles of base = 5.83/1000 × 0.104

= 0.000606 moles

Since the reaction of HCl and NaOH is 1:1

Number of moles of HCl that reacted with antacid = 0.00245 moles - 0.000606 moles

= 0.001844 moles

From the reaction;

CaCO3 + 2HCl ----> CaCl2 + H2O + CO2

1 mole of CaCO3 reacts with 2 moles of HCl

x moles of CaCO3 reacts with 0.001844 moles ofHCl

x = 1 × 0.001844/2

= 0.000922 moles

Mass of CaCO3 = 0.000922 moles × 100 g/mol

= 0.0922 g

Answer:

X

anode

electrons in the wire flow away

anions from salt bridge flow toward

loses mass

Y

cathode

electrons in the wire flow toward

cations from salt bridge flow toward

gains mass

Explanation:

In a galvanic cell, oxidation occurs at the anode while reduction occurs at the cathode. The metal that is more reactive functions as the anode while the less reactive metal functions as the cathode.

Electrons leave the anode and travel via a wire to the cathode. At the anode cations give up electrons and enter into the solution.

At the cathode, cations pick up electrons and are deposited on the cathode leading to a gain in mass at the cathode.

Positive ions from the salt bridge flow towards the cathode while negative ions from the salt bridge flow towards the anode.

Answer:

There is 13.48 grams of P2O5 formed

Explanation:

Step 1: Data given

A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen.

Mass of P = 5.89 grams

Molar mass of O2 = 32.0 g/mol

atomic mass of P = 30.97 g/mol

molar mass of P2O5 = 141.94 g/mol

Step 2: The balanced equation

4P(s)+5O2(g)⇔ 2P2O5(s)

Step 3: Calculate moles of P

Moles P = Mass P / atomic mass P

Moles P = 5.89 grams / 30.97 g/mol

Moles P = 0.190 moles

Step 4: Calculate moles of P2O5

For 4 moles P we need 5 moles O2 to produce 2 moles P2O5

For 0.190 moles of P we'll have 0.190/2 = 0.095 moles P2O5

Step 5: Calculate mass of P2O5

Mass P2O5 = moles P2O5 * molar mass P2O5

Mass P2O5 = 0.095 moles * 141.94 g/mol

Mass P2O5 = 13.48 grams

There is 13.48 grams of P2O5 formed

Answer:

Boiling T° of solution  → 83.6°C

Explanation:

To solve this, we apply Elevation of boiling point, property

ΔT = Kb . m . i

As we talk about organic solute, i = 1. No ions are formed.

m = molality (moles of solute in 1kg of solvent)

We determine mass of solvent by density

D = m /V so D . V = m

950 mL . 0.877 g/mL = 833.15 g

We convert to kg → 833.15 g . 1 kg/ 1000g = 0.833 kg

Moles of solute (naphtalene): 150 g . 1 mol/ 128.16g = 1.17 mol

m = 1.17mol / 0.833 kg = 1.41 mol/kg

We replace data:

Boiling T° of solution - 80.1°C = 2.53°C/m . 1.41 m . 1

Boiling T° of solution = 2.53°C/m . 1.41 m . 1  +  80.1°C → 83.6°C

Answer:

The answer is c or 17.1 g

Answer:

Kp = 1.10.

Explanation:

Hello there!

In this case, according to the given information about the chemical reaction at equilibrium, it turns out possible for us to find the partial pressures-based equilibrium expression for the decomposition of phosphorous pentachloride by applying the law of mass action whereas the pressure of products is divided by that of the reactants as shown below:

[tex]Kp=\frac{p_{PCl_3}p_{Cl_2}}{p_{PCl_5}}[/tex]

Now, we plug in the given pressures to obtain:

[tex]Kp=\frac{0.870}{0.688} \\\\Kp=1.10[/tex]

Regards!

Answer:

As stored fuels, triacylglycerols have two significant advantages over polysaccharides such as glycogen and starch. The carbon atoms of fatty acids are more reduced than those of sugars, and oxidation of triacylglycerols yields more than twice as much energy, gram for gram, as that of carbohydrates.

Explanation:

Answer:

velocity = distance/time

velocity = wavelength × frequency

Both of these are commonly known equations to calculate velocity with different variables.

Answer:

c. 29 J

Explanation:

Step 1: Given data

Step 2: Calculate the temperature change

ΔT = 37 °C - 22 °C = 15 °C

Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece

We will use the following expression.

Q = c × m × ΔT

Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J

Solution :

Molecular      Molar Mass       Volume      Density       Mass      Moles      nmoles

formula            (g/mol)               (mL)          (g/mL)           (g)

[tex]$C_6H_8N_2$[/tex]            108.14                                                    0.108      0.001          1

HCOOH           46.02                0.064          1.22     0.07808     0.0017       1.7

mmoles of o-phenylenediamine = 1 mmoles

mmoles of formic acid = 1.7 [tex]\approx[/tex] 2 mmoles

From the reaction of o-phenylenediamine and formic acid, we see,

1 mmole of o-phenylenediamine reacts with 1 mmole of formic acid.

But here, 2 mmoles of the formic acid , this means that the formic acid is an excess reagent and the o-phenylenediamine is the limiting reagent here.

The amount of product depends on the limiting reagent that is o-phenylenediamine. So, 1mmole of o-phenylenediamine will give 1mmole of product.

molar mass of Benzimidazole = [tex]118.14[/tex] g/mol

mmoles of Benzimidazole formed = [tex]1[/tex] mmol

Mass of benzimidazole formed = molar mass x [tex]\frac{nmoles}{1000}[/tex]

                                                    [tex]$=\frac{118.14 \times 1}{1000}$[/tex]

                                                     = 0.11814 g

So the theoretical yield of Benzimidazole is = 0.118 g = 118mg