(a) Your TA will give you a 1H NMR spectrum of the 3-nitroaniline product. Using your NMR knowledge and the special NMR section in the lab book (especially pages 48 and 49), assign the 4 peaks in the spectrum to each of the protons in the product - you will need to use coupling pattern and chemical shift to complete the assignment. (b) Explain your assignment, specifically how you differentiated between HA and Hc

Answer:

At pH0 the more thermodynamically stable species is lower and the most strongly oxidizing species is higher. At pH14 the more thermodynamically stable species is higher and the most strongly oxidizing species is lower . At pH0 convex curve will tend to disproportionate and pH14 concave curce will tend to disproportionate. Under acidic conditions mixing Mn3+(aq) and MnO42- (aq) will result in the formation of redox oxidation-reduction reaction.

Answer: The final temperature in °C is 167

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay lussac's Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant volume.

Mathematically,

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

where,

[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.

[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.

We are given:

[tex]P_1=4.40atm\\T_1=25^oC=(25+273)K=298K\\P_2=6.50atm\\T_2=?[/tex]

Putting values in above equation, we get:

[tex]\frac{4.40}{298}=\frac{6.50}{T_2}\\\\T_2=440K=(440-273)^0C=167^0C[/tex]

Thus the final temperature in °C is 167

To determine at what temperature a gas cylinder will reach 6.50 atm pressure from 4.40 atm at 25°C, we use Gay-Lussac's Law and find that the temperature is 167°C.

We can solve this problem using Gay-Lussac's Law, which states that for a given mass and constant volume of an ideal gas, the pressure is directly proportional to its temperature.

The formula for Gay-Lussac's Law is:

P1/T1 = P2/T2

Where:

First, we need to convert the initial temperature into Kelvin:

T1 (in Kelvin) = 25°C + 273 = 298 K

Now, we rearrange the equation to solve for T2:

T2 = (P2 × T1) / P1

Substitute the known values into the formula:

T2 = (6.50 atm × 298 K) / 4.40 atm

T2 ≈ 440 K

Finally, convert the temperature back to Celsius:

T2 (in °C) = 440 K - 273 = 167°C

Therefore, the temperature at which the gas will reach a pressure of 6.50 atm is 167°C.

Answer:

The mass of the gold sample is 28.59 g

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system. The amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without changing its physical state (solid, liquid or gaseous) is calculated by:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

Replacing:

[tex]1564 J=0.129\frac{J}{g*C}*m*424 C[/tex]

Solving:

[tex]m=\frac{1564 J}{0.129\frac{J}{g*C}*424 C}[/tex]

m=28.59 g

The mass of the gold sample is 28.59 g

Answer:

produces carbon dioxide when reacting with carbonates

Answer:

% is 51.6% of Oxygen

Explanation:

According to the working in the photo

51.6% is the percent of oxygen in sodium oxide. It represents the amount of solute in a given amount of solution.

A concentration is expressed as a mass percent. Furthermore, it defines the component of a particular blend. The solution component can be expressed in terms of mass percentage. It represents the amount of solute in a given amount of solution.

The amount of solute is measured in mass or moles. We shall study the mass percent equation using numerous solved numerical examples in this post. Mass is often represented in grams, although any unit of measurement is acceptable as long as an equivalent unit is used both for the component or solute mass and thus the total or solution mass.

mass percentage= (mass of oxygen / mass of  sodium oxide)×100

                           =(32/62)×100

                           = 51.6%

Therefore, 51.6% is the percent of oxygen in sodium oxide.

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Answer:  The carbon-14 left after 10 years is 25 g

Explanation:

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for rate constant

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]

[tex]k=\frac{0.693}{5years}=0.139years^{-1}[/tex]

b) for amount left after 10 years

[tex]t=\frac{2.303}{k}\log\frac{a}{(a-x)}[/tex]

[tex]10=\frac{2.303}{0.139}\log\frac{100}{(a-x)}[/tex]

[tex](a-x)=25g[/tex]

Thus carbon-14 left after 10 years is 25 g

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Final answer:

A call option is a financial contract that gives the buyer the right to purchase shares of a stock at a predetermined price. The net profit from this call option transaction can be calculated by determining the difference between the market price and the strike price, multiplied by the number of shares.

Explanation:

A call option is a type of financial contract that gives the buyer the right, but not the obligation, to purchase a specified number of shares of a stock at a predetermined price within a certain time frame. In this case, Nash Company purchased a call option to buy 1,050 shares of Merchant stock at a strike price of $51 per share. On March 31, 2020, the market price for Merchant stock is $54 per share, and the time value of the option is $210.

The net profit from this stock transaction can be calculated by determining the difference between the market price of the stock and the strike price, multiplied by the number of shares:

Net Profit = (Market Price - Strike Price) x Number of Shares

Using the given information, the net profit from this call option transaction would be:

Net Profit = ($54 - $51) x 1,050 = $3 x 1,050 = $3,150

Answer : The new volume will be, 238.9 mL

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles of gas.

Mathematically,

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where,

[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.

[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.

We are given:

[tex]V_1=222mL\\T_1=30.0^oC=(30.0+273)K=303K\\V_2=?\\T_2=53.1^oC=(53.1+273)K=326.1K[/tex]

Putting values in above equation, we get:

[tex]\frac{222mL}{303K}=\frac{V_2}{326.1K}\\\\V_2=238.9mL[/tex]

Therefore, the new volume will be, 238.9 mL

Answer:

D. The experiment includes a good plan for what measurements to take.

Explanation:

Answer:

1. NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂

2. The reactant that is reduced is Q

3. The charge on iron on the right side is +2, Fe²⁺

Explanation:

NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂

The reaction above is catalysed by NADH:ubiquinone oxidoreductase (complex 1), which transfers a hydride ion from NADH to FMN, from which two electrons pass through a series of of Fe-S centers to the iron-sulfur protein N-2. Electron transfer from N-2 to Ubiquinone forms QH₂

The species in a reaction which gains hydrogen irons is reduced,   Therefore, the reactant that is reduced is Q, ubiquinone to form QH₂, ubiquinol.

To determine the oxidation number of iron on the right side of the reaction below,

QH2 + 2cyt c ( Fe3+) ⟶ Q + 2cyt c(Fex) + 2H^+

Sum of charges on the left side = Sum of charges on the right side

Sum of charges on the left side = 2 *+3 = +6

Therefore 2 * x + 2= 6

2x = 6 -2 = 4

x = 4/

x = 2

Therefore the charge on iron on the right side is +2, Fe²⁺

Answer:

42050 J.

Explanation:

Data obtained from the question:

Mass (M) of ethanol = 50g

Heat of vaporisation (ΔHv) of ethanol = 841 J/g

Heat (Q) =.?

The heat required to boil 50g of ethanol can be obtained as follow:

Q = MΔHv

Q = 50 x 841

Q = 42050 J.

Therefore, the heat required to boil 50g of ethanol is 42050 J.

Answer:

A weakly acidic solution.

Explanation:

Final answer:

A solution with a Ka value much less than 1 refers to a weak acid, where only a small portion of the acid is ionized, resulting in relatively low concentrations of hydronium ions and the conjugate base.

Explanation:

The type of solution that would have a Ka value much less than 1 is a weak acid solution. This is because the Ka (acid dissociation constant) value indicates the strength of the acid in solution; the lower the Ka, the weaker the acid. For a Ka value significantly less than 1, this means that only a small fraction of the acid molecules donate protons (H+) to the water, resulting in few hydronium ions (H₋₃O⁺) and the conjugate base (A-).

In contrast, strong acids have a Ka value that approaches infinity since they are almost completely ionized in solution. Acetic acid, with a Ka value of 1.75 x 10⁻⁵, is an example of a weak acid, indicating that in equilibrium, the concentration of the undissociated acetic acid is much greater than the concentrations of the acetate and hydronium ions.

Answer: 0.28 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to the neutralization law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of stock [tex]LiOH[/tex] solution = 4.5 ml

[tex]V_1[/tex] = volume of stock [tex]LiOH[/tex] solution = 27 ml

[tex]M_2[/tex] = molarity of dilute c solution = ?

[tex]V_2[/tex] = volume of dilute [tex]LiOH[/tex] solution = 435 ml

[tex]4.5\times 27=M_2\times 435[/tex]

[tex]M_2=0.28M[/tex]

Thus the concentration of LiOH be if 27 ml of a 4.5 M LiOH solution is diluted to 435 ml is 0.28 M

Answer:

increases

Explanation:

Now look at the matter closely. Given the values of electrode potential stated in the question, one can see that lead will function as the anode and copper as the cathode since lead has a more negative electrode potential.

This implies that at the anode, the half reaction going on is this;

Pb(s) -------> Pb^2+(aq) + 2e.

There will be a build up of Pb^2+ in the anode compartment. The addition of H2SO4 and formation of PbSO4 favours the removal of the Pb^2+ ions;

Pb^2+(aq) + SO4^2-(aq) -------> PbSO4(s)

As this continues, Pb^2+ concentration begins to decrease, in order to maintain equilibrium, more Pb^2+ is formed thereby increasing the current and voltage flowing in the cell as more electrons are transferred from anode to cathode(more current flows).

In the case when the sulfuric acid is added to the Pb(NO3)2  solution increases so the cell potential is increased.

Here the anode and copper treated as the cathode because it contains more negative electrode potential. Due to this, it should be half reaction

Pb(s) -------> Pb^2+(aq) + 2e.

Also, there should be the addition of H2SO4 and creation of the PbSO4 that eliminates the Pb^2+ ions.

So,

Pb^2+(aq) + SO4^2-(aq) -------> PbSO4(s)

Here the concentration of Pb^2+ should reduced for maintaining the equilibrium.

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Answer:

0.981atm

Explanation:

According ot Dalton's law total pressure of a mixture of non-reactive gas is equal to sum of partial pressures of individual gases.

total pressure= 1.01at

Number of gases=2

Gases: water vapor and hydrogen

partial pressure of water vapor= 0.029atm

1.01= partial pressure of water vapor+ partial pressure of hydrogen

1.01= 0.029 + partial pressure of hydrogen

partial pressure of hydrogen = 0.981atm

Answer:

Solvent

Explanation:

Many people use the solute to describe the solid being dissolved and the solvent to describe the thing doing the dissolving, but really solvent means the part of the solution with a greater amount/concentration.

For example, if you have 1 gram of salt in 10 liters of water, the water is the solvent.

Final answer:

A solution is a homogeneous mixture where the component with the largest quantity is known as the solvent, and the substance in the smaller quantity is the solute.

Explanation:

A solution is a homogeneous mixture of two or more pure substances. When we talk about a solution, we refer to its components as the solvent and the solute. The solvent is the part of the solution that is present in the greater amount and acts as the medium in which the other substance, or substances, are dispersed.

The solute, on the other hand, is any component of a solution that is not the solvent. In other words, it is the material present in the smaller amounts within the solution. When a solute is dissolved in a solvent, the resulting mixture is uniform on a microscopic level, ensuring that the composition is consistent throughout.

Solutions can come in various forms and are not limited to liquids; for instance, air is a solution that is in a gaseous state. It is important to note that the largest component by mole fraction is always considered the solvent, and this remains the same whether the components are in solid, liquid, or gaseous form.

Answer:

Explanation:

check below for explanation in the attached files.

Answer:

0.122 moles

Explanation:

Final answer:

0.122 moles of KCl will be produced from the reaction of 15.0 g of KClO3, using stoichiometry and the molar mass of KClO3 which is 122.55 g/mol.

Explanation:

To determine the amount of KCl produced from 15.0 g of KClO3, we first need to know the molar mass of KClO3, which is 122.55 g/mol according to LibreTexts. By using stoichiometry, we can find out how many moles of KClO3 we have and, subsequently, how many moles of KCl will be produced according to the balanced chemical equation.

First, calculate the number of moles of KClO3:

15.0 g KClO3 × (1 mol KClO3 / 122.55 g) = 0.122 mol KClO3

From the balanced equation Î{2KClO3 -> 2KCl + 3O2}, we can see that 2 moles of KClO3 produce 2 moles of KCl. Therefore, the number of moles of KCl produced is equal to the number of moles of KClO3 available:

0.122 mol KClO3 × (2 mol KCl / 2 mol KClO3) = 0.122 mol KCl

Thus, 0.122 mol of KCl are produced from the reaction of 15.0 g of KClO3.

Answer:

False

False

True

False

False

False

Explanation:

The colour emitted by an excited atom depends on the emission Spectra of the atom rather than on the number of free electrons passing through the lamp.

When a free electrons hits an atom, the atom can be excited to various intermediate states other than the highest energy level depending on the amount of energy it absorbed.

Increase in voltage of the battery also increases the kinetic energy of the electron

The kinetic energy of the electron depends on the voltage difference not on the distance from the atom.

The Paschen, Brackett, and Pfund series of lines are due to transitions from higher-energy orbits to orbits with n = 3, 4, and 5, respectively; these transitions release substantially less energy, corresponding to infrared radiation.

Excited atoms can jump from a higher level to the ground state in a series of steps or directly to the ground state

Answer:

= -356KJ

therefore, the reaction where heat is released is exothermic reaction since theΔH is negative

Explanation:

given that enthalpy of gaseous reactants decreases by 162KJ and workdone is -194KJ

then,

change in enthalpy (ΔH) = -162( released energy)

work(w) = -194KJ

change in enthalpy is said to be negative if the heat is evolved during the reaction while heat change(ΔH) is said to be positive if the heat required for the reaction occurs.

At constant pressure the change in enthalpy is given as

ΔH = ΔU + PΔV

ΔU = change in energy

ΔV = change in volume

P = pressure

w =  -pΔV

therefore,

ΔH = ΔU -W

to evaluate  energy change we have,

ΔU =ΔH + W

ΔU = -162+ (-194KJ)

= -356KJ

therefore, the reaction where heat is released is exothermic reaction since theΔH is negative

Final answer:

The change in energy of the gas mixture during the reaction is 32 kJ. The reaction is exothermic because it releases heat to the surroundings as indicated by the negative enthalpy change.

Explanation:

To calculate the change in energy of the gas mixture during the reaction, we use the formula for change in internal energy (ΔU) at constant pressure:

ΔU = ΔH - PΔV

where ΔH is the change in enthalpy and PΔV is the work done on or by the system. Given that the enthalpy of the system decreases by 162 kJ and the work done on the system is -194 kJ, we can substitute these values into the equation:

ΔU = -162 kJ - (-194 kJ)

ΔU = -162 kJ + 194 kJ

ΔU = 32 kJ

Therefore, the change in energy of the gas mixture during the reaction is 32 kJ. The reaction can be classified as exothermic since the enthalpy decreases (ΔH is negative), which means heat is released by the system to the surroundings.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

The direction of heat flow in each scenario will be:

It should be noted that heat transfer means the transfer of heat from a location to another.

It is the process in which the molecules are moved from a region of higher temperature to that of a lower temperature.

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Answer:

It Is Scenario D.

Explanation:

Answer:

2Na(s) + C(s) + 3/2 O2(g) -----> Na2CO3(s).

Explanation:

Balancing and writring chemical reactions in Chemistry is a very important aspect of chemistry and,

writting and balancing of chemical reaction is often regarded as the heart of chemistry. A balanced Chemical reaction shows the number of each species in a reaction.

Hence, the balanced chemical equation for the standard formation reaction of solid sodium carbonate is given below;

(1). The reaction(unbalanced Reaction):

Na(s) + C(s) + O2(g) -----> Na2CO3(s).

Where solid sodium carbonate = Na2CO3(s).

(2). Balanced Reaction;

2Na(s) + C(s) + 3/2 O2(g) -----> Na2CO3(s)

The balanced chemical equation for the standard formation reaction of

solid sodium carbonate is shown below:

2Na(s) + C(s) + 3/2 O2(g) -----> Na2CO3(s).

The reaction involves elements which include Sodium(Na) , Carbon(C) and

Oxygen(O2). They react under certain conditions to form solid sodium

carbonate.

The reaction involves specific number of atoms involved in the reaction

which is why the equation must be balanced. The number of atoms of an

element on the left hand side should always be equal to those on the right

hand side.

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Answer:

ionizing

I hope I helped :)

Answer:

the minimum [OH−] triggers precipitation = 2.3x10^-6 M

Explanation:

Mg2+ + 2OH- --> Mg(OH)2(s)

Ksp = [Mg2+][OH-]^2

(2.06x10^-13) = (0.039*x^2)

x = [OH-] = 2.3*10^-6 M

the minimum [OH−] triggers precipitation = 2.3x10^-6 M

Final answer:

The minimum [OH-] that triggers the precipitation of Mg2+ ion can be calculated using the Ksp value.

Explanation:

The minimum [OH-] that triggers the precipitation of Mg2+ ion can be calculated using the Ksp value. Given that the concentration of Mg2+ in the solution is 0.039 M, and the Ksp value is 2.06×10⁻¹³, the minimum [OH-] can be determined to initiate precipitation.

Answer:

Which of the following

properties distinguishes a solution

oversaturated with a dilute?

Answer:

THE NEW PRESSURE IS 15.5 L.

Explanation:

Using Boyle's law which states that the pressure of a given mass of gas is inversely proportional to the volume of the gas at constant temperature.

Mathematically,

P1V1 =P2V2

P1 = 345 torr = 345/760 atm = 0.45 atm

P2 = 760 torr = 760/760 atm = 1 atm

V1 = 34.5 L

V2 = ?

Rearranging the variables, making V2 the subject of the formula, we obtain:

V2 = P2 V1 / P1

V2 = 1 * 34.5 / 0.45

V2 = 15.525 Litres

The volume of the gas when the pressure becomes 760 torr or 1 atm is 15.5 Litres

Answer:

(a)

Rate of appearance of sucrose = - d[C12H22O11] / dt = - ( 0.808 - 1.002 ) / ( 60.0 - 0.0) = 0.00323 M/s

(b)

Rate of appearance of fructose = d[C6H12O6] / dt = (1.002 - 0.808) / (60.0 - 0.0) = 0.00323 M/s

(c)

k = (1 / t ) * ln[A]/[A]t

k = ( 1 / 60.0 ) * ln[1.002 / 0.808]

k = 0.00359 min-1

(d)

0.00359 = ( 1 / t ) * ln[1.002 / 0.212]

t = 432.6 min

(e)

Half life time = 0.693 / k = 0.693 / 0.00359 = 193 min

Explanation:

First-order reactions are defined as the chemical reactions in which rate of the reaction is linearly dependent on the concentration of only one reactant.

The answers can be explained as:

(a) Rate of appearance of the sucrose from the chemical reaction is:

Rate = [tex]\dfrac{\text d [\text C_{12}\text H_{22}\text O_{11}]}{\text {dt}}[/tex] = [tex]\dfrac{0.808 - 1.002}{60.0 -0.0}[/tex]

Rate = 0.00323 m/s

(b) Rate of appearance of Fructose from the given chemical reaction is:

Rate = [tex]\dfrac{\text d [\text C_{6}\text H_{12}\text O_{6}]}{\text {dt}}[/tex] = [tex]\dfrac{1.002 - 0.808 }{60.0 -0.0}[/tex]

Rate = 0.00323 m/s

(C) Rate constant for the reaction is:

[tex]\text k &= \dfrac{1}{\text t}\times \dfrac {\text{ln [A]}}{\text {[A]} \text t}[/tex]

[tex]\text k &= \dfrac{1}{60}\times \dfrac {\text{ln} (1.002)}{(0.808)}[/tex]

k = 0.00359 minute⁻¹

(d) Time required for the concentration of sucrose to drop from 1.002 to 0.212 M is:

[tex]0.00359 &= \dfrac{1}{t} \times {\text{ln}\dfrac{[1.002]}{[0.212]}[/tex]

t = 432.6 minutes

(e) The half-life of the decomposition of sucrose at 25°C is:

Half-life = [tex]\dfrac{0.693}{\text k} = \dfrac{0.693}{0.00359}[/tex]

Half-life = 193 minutes.

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