Provide the structures of the fragments that result when the molecular ion of 2-heptanone undergoes fragmentation via McLafferty rearrangement. Include charges and single electrons.

Answer:

= 0.2 mL.

Explanation:

Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:

[tex]C_{1} V_{1} = C_{2} V_{2}[/tex]

where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.

Let C1 = 0.5M, V2 = ?

C2 = 0.010M; V2 = 10mL

⇒Volume of stock solution to be diluted, V2

= [tex]\frac{10}{0.5}[/tex] × 0.010

= 0.2 mL.

Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.

I hope this was a detailed explanation given the missing details of "Trial 1" in the question.

Answer:

a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.

Answer:

[tex]\triangle G = -911.296 \ kJ[/tex]

Explanation:

ΔG = ΔH-TΔS

Where ΔH = -176 kJ = -176000 J , T = 25°C + 273 = 298 K , ΔS = -284.8 JK⁻¹

=> [tex]\triangle G =-176000 - (298)(-284.8)[/tex]

=> [tex]\triangle G = -176000+84870.4[/tex]

=> [tex]\triangle G = -91129.6 \ J[/tex]

=> [tex]\triangle G = -911.296\ kJ[/tex]

Since the value is negative, the reaction is spontaneous under standard conditions at 298 K and the reactants have more free energy than the products.

Answer:

The given statement is false.

Explanation:

So that the given is incorrect.

Answer:

It will take 6486.92 minutes  for [H2O2] to fall from 0.95 M to 0.33 M

Explanation:

The order of reaction is defined as the sum of the powers of the concentration terms in the equation. Order of a reaction is given by the number of atoms or molecule whose concentration change during the reaction and determine the rate of reaction.

In first order reaction;

[tex]In \dfrac{a}{a_o-x}= k_1 t[/tex]

where;

a = concentration at time t

[tex]a_o[/tex] = initial concentration

and k = constant.

[tex]In (\dfrac{0.33}{0.95})= -1.63 \times 10^{-4} \times t[/tex]

[tex]-1.05736933 = -1.63 \times 10^{-4} \times t[/tex]

[tex]t = \dfrac{-1.05736933}{ -1.63 \times 10^{-4} }[/tex]

t = 6486.92 minutes

Answer:

CH3CH=NH2+>CH3CH2NH3+

Explanation:

If we look at the both species under review, we will realize that they are both amines hence they possess the polar N-H bond.

Electrons are ordinarily attracted towards the nitrogen atom hence making both compounds acidic. It is worthy of note that certain features of a compound may make it more acidic than another of close structural proximity. 'More acidic' simply means that the proton is more easily lost.

CH3CH=NH2+ contains an sp2 hybridized carbon atom which is highly electronegative and further withdraws electron density from the N-H bond thereby leading to a greater acidity of CH3CH=NH2+ compared to CH3CH2NH3+

Answer:

[tex]\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl[/tex].

One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.

Explanation:

Start by finding the formula for each of the compound.

Therefore, the ratio between [tex]\rm K[/tex] atoms and [tex]\rm Br[/tex] atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula [tex]\rm KBr[/tex]. Similarly, the ratio between

The formula for chlorine gas is [tex]\rm Cl_2[/tex], while the formula for bromine gas is [tex]\rm Br_2[/tex].

Write down the equation using these chemical formulas.

[tex]\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl[/tex].

Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both [tex]\rm KBr[/tex] and [tex]\rm KCl[/tex] features two elements each.

Assume that the coefficient of [tex]\rm KCl[/tex] is one. Hence:

[tex]\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl[/tex].

Note that [tex]\rm KBr[/tex] is the only source of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms among the reactants of this reaction.

There would thus be one [tex]\rm K[/tex] atom and one [tex]\rm Br[/tex] atom on the reactant side of the equation.

Because atoms are conserved in a chemical equation, there should be the same number of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms on the product side of the equation.

In this reaction, [tex]\rm Br_2[/tex] is the only product with [tex]\rm Br[/tex] atoms.

One [tex]\rm Br[/tex] atom would correspond to [tex]0.5[/tex] units of [tex]\rm Br_2[/tex].

Similarly, in this reaction, [tex]\rm KCl[/tex] is the only product with [tex]\rm K[/tex] atoms.

One [tex]\rm K[/tex] atom would correspond to one formula unit of [tex]\rm KCl[/tex].

Hence:

[tex]\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].

Similarly, there should be exactly one [tex]\rm Cl[/tex] atom on either side of this equation. The coefficient of [tex]\rm Cl_2[/tex] should thus be [tex]0.5[/tex]. Hence:

[tex]\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].

That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:

[tex]\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl[/tex].

Answer:

[tex]2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)[/tex]

[tex]Entalpy=-2861.9~KJ[/tex]

Explanation:

In this case, we have to start with the reagents:

[tex]Al~+~NH_4NO_3[/tex]

The compounds given by the problem are:

-) Nitrogen gas =  [tex]N_2[/tex]

-) Water vapor  =  [tex]H_2O[/tex]

-) Aluminum oxide =  [tex]Al_2O_3[/tex]

Now, we can put the products in the reaction:

[tex]Al_(_S_) ~+~NH_4NO_3_(_aq_) ->N_2_(_g_) ~+~H_2O_(_g_) ~+~Al_2O_3_(_S_) [/tex]

When we balance the reaction we will obtain:

[tex]2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)[/tex]

Now, for the enthalpy change, we have to find the standard enthalpy values:

[tex]Al_(_S_)=0~KJ/mol[/tex]

[tex]NH_4NO_3_(_a_q_)=-132.0~KJ/mol[/tex]

[tex]N_2_(_g_)=0~KJ/mol[/tex]

[tex]H_2O_(_g_)=~-~241.8~KJ/mol[/tex]

[tex]Al_2O_3_(_S_)=~-~1675.7~KJ/mol[/tex]

With this in mind, if we multiply the number of moles (in the balanced reaction) by the standard enthalpy value,  we can calculate the energy of the reagents:

[tex](0*2)~+~(-132*3)=~-396~KJ[/tex]

And the products:

[tex](0*3)~+~(-241.8*6)~+~(-1675.7*1)=-3125.9~KJ[/tex]

Finally, for the total enthalpy we have to subtract products by reagents :

[tex](-3125.9~KJ)-(-396~KJ)=-2729.9~KJ[/tex]

I hope it helps!

Answer:

[tex]N_2~+~3H_2~+~4O_2~->~2HNO_3~+~2H_2O[/tex]

Explanation:

We can start with the reaction of hydrogen and nitrogen to produce ammonia, so:

[tex]N_2~+~H_2~->~NH_3[/tex]

When we balance the reaction we will obtain:

[tex]N_2~+~3H_2~->~2NH_3[/tex]

Now, the production of nitric acid with oxygen would be:

[tex]NH_3~+~O_2~->~HNO_3~+~H_2O[/tex]

If we balance the reaction we will obtain:

[tex]NH_3~+~2O_2~->~HNO_3~+~H_2O[/tex]

Now, if we put the reactions together we will obtain:

[tex]N_2~+~3H_2~->~2NH_3[/tex]

[tex]NH_3~+~2O_2~->~HNO_3~+~H_2O[/tex]

We can multiply the second reaction by "2":

[tex]N_2~+~3H_2~->~2NH_3[/tex]

[tex]2NH_3~+~4O_2~->~2HNO_3~+~2H_2O[/tex]

We have "[tex]2NH_3[/tex]" on both sides. In the first reaction is in the right in the second reaction is on the left. Therefore we can cancel out this compound and we will obtain:

[tex]N_2~+~3H_2~+~4O_2~->~2HNO_3~+~2H_2O[/tex]

On this reaction, we will have 2 nitrogen atoms on both sides, 6 hydrogen atoms on both sides, and 8 oxygen atoms on both sides. So, this would be the net reaction for the production of nitric acid.

Answer:

1.00x10^-9

Explanation:

Answer:

a) Specific Rotation of (R)-carvone = -61°

b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%

c) The percentage of (S)-carvone in the mixture = 4.9%

Explanation:

a) The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.

Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.

Specific Rotation of (R)-carvone = -(61°) = -61°

b) Enantiometic excess is used to measure the optical purity of an enantiomeric mixture.

The enantiomeric excess is given mathematically as

ee% = (Observed rotation × 100)/(Specific rotation)

Hence, to calculate the enantiomeric excess of (R)-carvone,

Observed rotation of the mixture = -55°

Specific Rotation of (R)-carvone = -61°

ee% = (-55×100)/(-61) = 90.16% = 90.2%

c) An enantiomeric excess of 90.2% for (R)-carvone indicates that it's actual percentage is 90.2% more than the percentage of its enantiomeric partner, (S)-carvone, in the mixture.

Let the percentage of (R)-carvone in the mixture be x

Let the percentage of (S)-carvone in the mixture be y

x + y = 100

x - y = 90.2

2x = 190.2

x = (190.2/2) = 95.1%

y = 100 - x = 100 - 95.1 = 4.9%

Hence, the percentage of (R)-carvone in the mixture = 95.1%

The percentage of (S)-carvone in the mixture = 4.9%

Hope this Helps!!!

a) Specific Rotation of (R)-carvone = -61°

b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%

c) The percentage of (S)-carvone in the mixture = 4.9%

The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.

Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.

Specific Rotation of (R)-carvone = -(61°) = -61°

b) Calculation for Enantiomeric excess:

The enantiomeric excess is given mathematically as

ee% = (Observed rotation × 100)/(Specific rotation)

Hence, to calculate the enantiomeric excess of (R)-carvone,

Observed rotation of the mixture = -55°

Specific Rotation of (R)-carvone = -61°

ee% = (-55×100)/(-61) = 90.16% = 90.2%

c) Calculation of percentage:

Let the percentage of (R)-carvone in the mixture be x

Let the percentage of (S)-carvone in the mixture be y

x + y = 100

x - y = 90.2

2x = 190.2

x = (190.2/2) = 95.1%

y = 100 - x = 100 - 95.1 = 4.9%

Hence, the percentage of (R)-carvone in the mixture = 95.1%

The percentage of (S)-carvone in the mixture = 4.9%

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Answer:

760 mmHg

Explanation:

Step 1: Given data

Step 2: Calculate the atmospheric pressure

Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.

P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg

Answer:

c iodine

Explanation:

fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride

Answer:

a. 1

b. 0.465M NaOH

Explanation:

KHP reacts with NaOH as follows:

KHP + NaOH → KP⁻ + Na⁺ + H₂O

a. Molar ratio represents how many moles of NaOH reacts per mole of KHP. As you can see in the reaction, 1 mole of NaOH reacts with 1 mole of KHP. Molar ratio is:

1/1 = 1

b. With volume and molar concentration of the KHP solution you can find how many moles of KHP were added until equivalence point, thus:

15.5mL = 0.0155L ₓ (0.600 moles KHP / L) = 0.0093 moles of KHP

In equivalence point, moles of NaOH = Moles KHP. That means moles of NaOH titrated are 0.0093 moles NaOH.

The volume of the NaOH solution was 20mL = 0.020L. Molarity of the solution is:

0.0093 moles NaOH / 0.020L =

a. The balanced equation shows a 1:1 molar ratio between NaOH and KHC₈H₄O₄. This means that for every 1 mole of NaOH, we require 1 mole of KHC₈H₄O₄. Therefore, the molar ratio of NaOH:KHC₈H₄O₄ is 1:1.

The balanced equation for the reaction:

NaOH + KHC₈H₄O₄ → NaKC₈H₄O₄ + H₂O

b. Molarity of KHP solution × volume of KHP solution = Molarity of NaOH solution × volume of NaOH solution at the equivalence point

Molarity of KHP solution = 0.600 M

Volume of KHP solution = 15.5 mL = 0.0155 L

Volume of NaOH solution at the equivalence point = 20 mL = 0.0200 L

Molarity of NaOH solution = (Molarity of KHP solution × volume of KHP solution) / volume of NaOH solution at the equivalence point

Molarity of NaOH solution = (0.600 M × 0.0155 L) / 0.0200 L

Molarity of NaOH solution ≈ 0.465 M

To learn more about the balanced equation, follow the link:

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Answer:

Explanation:

Let the monoprotic acid be HX

HX ⇄ H⁺ + X⁻

pH = 2.53

Hydrogen ion concentration

[tex][ H^+]=10^{-2.53}[/tex]

[tex][ X^-]=10^{-2.53}[/tex]

Concentration of undissociated acid will remain almost the same as it is a weak acid

So

Ka = concentration of H⁺ x concentration of Cl⁻ / concentration of acid

=  [ H⁺] x [Cl⁻ ] / [ HX]

[tex]k_a=\frac{10^{-2.53}\times 10^{-2.53}}{.0166}[/tex]

[tex]k_a=\frac{.00295^2}{.0166}[/tex]

= 5.24 x 10⁻⁴ M .

Answer:

D.

Explanation:

If a compound is composed of oppositely charged ions, it has to be formed by metal and non-metal.

Li2O

Li - metal

O - non-metal

Answer:

91.29 g of Na₃PO₄ are needed to produce 1.67 moles of sodium ions.

Explanation:

The formula for sodium phosphate is Na₃PO₄

Molar mass of sodium phosphate = 164 g/mol

The dissociation of one mole of sodium phosphate produces 3 moles of sodium ions;

Na₃PO₄ ------> 3Na⁺ + PO₄³

Number of moles of Na₃PO₄ that will produce 1.67 moles of Na⁺ = 1/3 * 1.67 = 0.556 moles  of Na₃PO₄

Mass of 0.556 moles of Na₃PO₄ = 0.556 moles * 164 g/mol = 91.29 g

Therefore, 91.29 g of Na₃PO₄ are needed to produce 1.67 moles of sodium ions.

Answer:

P SO₂ = 0.06atm

P O₂ = 2.635atm

P SO₃ = 0.53atm

Kp = 29.6

Explanation:

The reaction of Sulfur dioxide and oxygen react to form sulfur trioxide is as follows:

2SO₂(g) + O₂(g) ⇄ 2SO₃(g)

And Kp is defined as:

[tex]Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}[/tex]

Where P represents the pressure at equilibrium of each reactant.

If you add, in the first, 0.59atm of SO₂ and 2.9atm of O₂, the equilibrium pressures will be:

P SO₂ = 0.59atm - 2X

P O₂ = 2.9atm - X

P SO₃ = 2X

Where X represents the reaction coordiante.

As equilibrium pressure of SO₃ is 0.53atm:

0.53atm = 2X

0.265atm = X

Replacing, equilibrium pressures of each species will be:

P SO₂ = 0.59atm - 2×0.265atm

P O₂ = 2.9atm - 0.265atm

P SO₃ = 2×0.265atm

And Kp will be:

[tex]Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}[/tex]

[tex]Kp = \frac{0.53^2}{{0.06}^2*{2.635}}[/tex]

Answer:

See explanation

Explanation:

In molecules with the same number of electron groups but different molecular geometries, the bond angles differ markedly owing to the presence of lone pairs on the central atom. Recall that lone pairs of electrons take up more space around the central atom and causes more repulsion thus squeezing the bond angle and making it less than the value expected on the basis of the Valence Shell Electron Pair Repulsion Theory.

As the number of bonding groups increases, the bond angle increases since the repulsion due to lone pairs of electrons is being progressively removed by increase in the number of bonding groups.

For 5 electron group molecules, the axial groups are oriented at a bond angle of 90° while the equatorial groups are oriented at a bond angle of 120°. In the presence of lone pairs, the equatorial bonds are removed because the equatorial bonds often have a greater bond length than the axial bonds.

In the tetrahedral geometry, four groups are bonded to the central atom while in a bent molecular geometry, only two groups are bonded to the central atom with two lone pairs present in the molecule.

Answer:

Explanation:

Calcium bicarbonate  dissolved in hard water can easily be removed by heating the hard water . On heating , it decomposes to give calcium carbonate which is insoluble and therefore can be filtered out .

Ca( HCO₃)₂  =  CaCO₃ + CO₂ + H₂O.

In this way hardness of water is removed .

Answer:

H2<Ne<O2<NH3

Explanation:

Intermolecular forces refer to the force of attraction between molecules of a substance in any given state of matter whether solid, liquid or gas. Molecules in a substance must be held together by intermolecular forces of attraction. The magnitude of these intermolecular forces of attraction depends on many factors.

For H2, He and O2, the intermolecular force present in these gases are London forces. As the relative molecular mass of individual gas molecules becomes greater, London forces increases significantly with molecular mass. This explains the sequence shown in the answer.

NH3 has the strongest intermolecular interaction because it contains hydrogen bonds since nitrogen is an electronegative element. This a greater intermolecular interaction than dispersion forces.

Intermolecular forces are those forces that bind the molecules of the substance and the polarity of molecules. These forces range from the strongest to the weakest in ion-dipole, hydrogen bonding, and dipole to dipole.

Hence the form the strongest to the weakest is NH3, 02, Ne and H2.

Learn more about the substances in order of increasing the strength of intermolecular forces.

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Answer:

The answer to your question is given below.

Explanation:

Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:

Al (13) —› 1s² 2s²2p⁶ 3s²3p¹

The orbital diagram is shown on the attached photo.

Answer: screen shot

Explanation:

Answer:

[tex][H^+]=0.000285[/tex]

[tex]pH=3.55[/tex]

Explanation:

In this, we can with the ionization equation for the hydrazoic acid ([tex]HN_3[/tex]). So:

[tex]HN_3~<->~H^+~+~N_3^-[/tex]

Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:

[tex]Ka=\frac{[H^+][N_3^-]}{[HN_3]}[/tex]

For each mol of [tex]H^+[/tex] produced we will have 1 mol of [tex]N_3^-[/tex]. So, we can use "X" for the unknown values and replace in the Ka equation:

[tex]Ka=\frac{X*X}{[HN_3]}[/tex]

Additionally, we have to keep in mind that [tex]HN_3[/tex] is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:

[tex]Ka=\frac{X*X}{0.004-X}[/tex]

Finally, we can put the ka value and solve for "X":

[tex]2.2X10^-^5=\frac{X*X}{0.004-X}[/tex]

[tex]2.2X10^-^5=\frac{X^2}{0.004-X}[/tex]

[tex]X= 0.000285[/tex]

So, we have a concentration of 0.000285 for [tex]H^+[/tex]. With this in mind, we can calculate the pH value:

[tex]pH=-Log[H^+]=-Log[0.000285]=3.55[/tex]

I hope it helps!

The [H+] and pH of a 0.0040 M hydrazoic acid solution is 0.000296648 and  3.527759

What information do we have?

Hydrazoic acid solution = 0.0040 M

Ka of hydrazoic acid = 2.20 × 10⁻⁵

We know that weak acids

[H+] = √( Ka × C)

[H+] = √( 2.2 × 10⁻⁵ × 0.0040)

[H+] = 0.000296648

So,

pH = -log [H+]

pH = -log [0.000296648]

Using log calculator

pH = 3.527759

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Answer:

filtration, drying, and weighing

Explanation:

The procedures that would facilitate the measurement of the actual yield of the solid would be filtration of the precipitate, drying of the precipitate, and weighing of the precipitate respectively.

The liquid/solid mixture resulting from the reaction can be separated by the process of filtration using a filter paper. The residue in the filter paper would be the solid while the filtrate would be the liquid portion of the reaction's product.

The residue can then be allowed to dry, and then weighed using a laboratory-grade weighing balance. The weight of the solid represents the actual yield of the solid.

The molar mass of stearic acid is 18*AC+36*AH+2*AO=18*12+36*1+16*2=284g/mol.

n=m/M=2.831/284=0.01 moles

C18H36O2+27O2-->18CO2+18H2O

we have 18*0.01=0.18 moles of CO2

18*0.01=0.18 moles of H2O

0.01*948=9.48kJ from stearic acid

0.18*393.5=70.83kJ from CO2

0.18*241.826=43.52kJ from H2O

9.48+70.83+43.52=123.83kJ

123.83*4.184=518.10kcal

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄  4CO₂ + 10H₂O + 2N₂    

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄  4CO₂ + 10H₂O + 2N₂              

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.  

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

[tex]n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles[/tex]      

[tex]n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles[/tex]

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

[tex]n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles[/tex]

[tex]m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g[/tex]

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

[tex] \% = \frac{R_{r}}{R_{T}}*100 [/tex]

Donde:

[tex]R_{r}[/tex]: es el rendimiento real

[tex]R_{T}[/tex]: es el rendimiento teórico

[tex]\% = \frac{3,5}{5,043}*100 = 69,4[/tex]

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!        

Answer:

2 moles of HNO3

Explanation:

The equation seems to be balanced correctly. The problem is we done know what you started with. We will assume it is 3 moles of NO2.

If that is the case then 2 moles of HNO3 will be produced.

Answer:

[tex]\boxed{From \ +6 \ to \ 0}[/tex]

Explanation:

2 Fe2O3(s) + 3 C(s) → 4 Fe(s) + 3 CO2(g)

In the given reaction, Iron in the reactants side have the oxidation number of +6. This is because [tex]O_{3}[/tex] with [tex]Fe_{2}O_{3}[/tex] has oxidation state -6, So any atom with it would have an oxidation state of +6 to give the resultant of zero.

In the products side, Iron acts as a free element reacting with no other atom. So, as per the rule of oxidation states, the oxidation state of Iron in the products side will be zero.

So, the oxidation number changes from +6 to 0 .

Extra Info: Decrease in oxidation state is Reduction , So Iron is being reduced here.

The change in the oxidation number of the iron atom in the reaction is from +3 to 0

Oxidation is simply defined as the the loss of electron. However, Oxidation number simply talks about the number of electrons that is either gained or lossed during bond formation.

The change in the oxidation number of iron in the reaction can be obtained as follow:

2Fe₂O₃(s) + 3C(s) → 4Fe(s) + 3CO₂(g)

Oxidation number of Fe₂O₃ = 0 (ground state)

Oxidation number of oxygen = –2

Fe₂O₃ = 0

2Fe + 3O = 0

2Fe + 3(–2) = 0

2Fe – 6 = 0

Collect like term

2Fe = 6

Divide both side by 2

Fe = 6/2

Thus, the oxidation number of Fe in Fe₂O₃ is +3

Therefore, the change in the oxidation number of the iron, Fe, atom in the reaction is from +3 to 0

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Answer:

(a) The diode voltage,  [tex]V_D =[/tex]  0.776 V

(b) The diode current, [tex]I_D =[/tex] 3.81 x 10⁻²⁰ A

Explanation:

Given;

saturation current in diode, [tex]I_s[/tex] = 10⁻¹⁷ A

nonideality factor, n = 1.05

(a) the diode voltage

Given diode current, [tex]I_D[/tex] = 70 μA = 7 x 10⁻⁶ A

Diode voltage is calculated as;

[tex]V_D = nV_Tln(1+ \frac{I_D}{I_S} )[/tex]

Where;

[tex]V_T[/tex] is thermal voltage at 25°C = 0.025

[tex]V_D = 1.05 * 0.025 ln(1+ \frac{70*10^{-6}}{1*10^{-17}})\\\\V_D = 0.02625ln(1+ 7*10^{12})\\\\V_D = 0.776 \ V[/tex]

b) the diode current for VD = 0.1 mV

[tex]V_D = nV_Tln(1 +\frac{I_D}{I_S} )\\\\ln(1 +\frac{I_D}{I_S} ) = \frac{V_D}{nV_T} \\\\ln(1 +\frac{I_D}{I_S} ) = \frac{0.1*10^{-3}}{1.05*0.025} \\\\ln(1 +\frac{I_D}{I_S} ) = 0.00381\\\\1 +\frac{I_D}{I_S} = e^{0.00381}\\\\1+ \frac{I_D}{I_S}= 1.00381\\\\ \frac{I_D}{I_S}=1.00381 - 1\\\\ \frac{I_D}{I_S}= 0.00381\\\\I_D = 0.00381(I_S)\\\\I_D = 0.00381(10^{-17})\\\\I_D = 3.81*10^{-20} \ A[/tex]

Answer: Mass of methyl 4-fluoro-3-nitrobenzoate = 199 mg;

Mass of 4-chlorobenzylamine = 282 mg

Explanation: The mass and mol of a molecule is related by its molar mass, which is given in g/mol.

The molar mass of methyl 4-fluoro-3-nitrobenzoate, which has molecular formula: [tex]C_{8}H_{6}FNO_{4}[/tex] is:

[tex]C_{8}H_{6}FNO_{4}[/tex]  = 12.8 + 6.1 + 19 + 14 + 16.4 = 199 g/mol

Since it is asking in mg: MM = 199.10³mg/mol

For 4-chlorobenzylamine, with molecular formula [tex]C_{7}H_{8}ClN[/tex]:

[tex]C_{7}H_{8}ClN[/tex] = 12.7 + 8.1 + 35 + 14 = 141 g/mol

In mg: MM = 141.10³mg/mol

The reaction requires 1 mmol of [tex]C_{8}H_{6}FNO_{4}[/tex] , then its mass is:

m = 1.10⁻³ mol * 199.10³mg/mol = 199 mg

For [tex]C_{7}H_{8}ClN[/tex], it requires 2mmol:

m = 2.10⁻³ mol * 141.10³ mg/mol = 282 mg

For the SNAr reaction, it is necessary 199 mg of methyl 4-fluoro-3-nitrobenzoate and 282 mg of 4-chlorobenzylamine

The mass of 1 mmol methyl 4-fluoro-3-nitrobenzoate is 199.14 mg and the mass of 2 mmol of 4-chlorobenzyl amine is 283.2 mg

To determine the masses of both reagents,

First we will determine their molar masses

Molar mass = 199.14 g/mol

Now, for the mass of 1 mmol of methyl 4-fluoro-3-nitrobenzoate

Using the formula

Mass = Number of moles × Molar mass

Mass = 1 mmol × 199.14 g/mol

Mass = 199.14 mg

Molar mass = 141.6 g/mol

Now, for the mass of 2 mmol of 4-chlorobenzyl amine

Mass = 2 mmol × 141.6 g/mol

Mass = 283.2 mg

Hence, the mass of 1 mmol methyl 4-fluoro-3-nitrobenzoate is 199.14 mg and the mass of 2 mmol of 4-chlorobenzyl amine is 283.2 mg

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