Conceptual Checkpoint 18.13 When 1,3-dinitrobenzene is treated with nitric acid and sulfuric acid at elevated temperature, the product is 1,3,5-trinitrobenzene. Draw the sigma complex for each possible pathway to compare the relative stability of each sigma complex. For the mechanism, draw the curved arrows as needed. Include lone pairs and charges in your answer. Do not draw out any hydrogen explicitly in your products. Do not use abbreviations such as Me or Ph.

Answer:

The value of the equilibrium constant Kc for this reaction is 3.34*10^-5

Explanation:

Step 1: Data given

Volume of the reaction vessel = 9.3 L

Mass of CH3OH at equilibrium = 1.47 grams

Mass of O2 at equilibrium = 1.56 grams

Mass of CO2 at equilibrium = 2.28 grams

Mass of H2O at equilibrium = 3.33 grams

Molar mass CH3OH = 32.04 g/mol

Molar mass of O2 = 32.0 g/mol

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

2CH3OH (l) + 3O2 (g) → 2CO2 (g) + 4H2O (g)

Step 3: Calulate moles

Moles = mass / molar mass

Moles CH3OH = 1.47 grams / 32.04 g/mol

Moles CH3OH = 0.0459 moles

Moles O2 = 1.56 grams / 32.0 g/mol

Moles O2 = 0.0489 moles

Moles CO2 = 2.28 grams / 44.01 g/mol

Moles CO2 = 0.0518 moles

Moles H2O = 3.33 grams / 18.02 g/mol

Moles H2O = 0.185 moles

Step 4: Calculate molarity

Molarity = moles / volume

[CH3OH] = 0.0459 moles / 9.3 L

[CH3OH] = 0.00494 M

[O2] = 0.0489 moles / 9.3 L

[O2] = 0.00526 M

[CO2] = 0.0518 moles / 9.3 L

[CO2] = 0.00557 M

[H2O] = 0.185 moles / 9.3 L

[H2O] = 0.0199 M

Step 5: Calculate Kc

Since liquids and solids do not affect the equilibrium, CH3OH will not take part

Kc = [CO2]²[H2O]^4 / [O2]³

Kc = (0.00557² * 0.0199^4) / 0.00526³

Kc = 3.34*10^-5

The value of the equilibrium constant Kc for this reaction is 3.34*10^-5

The value of the equilibrium constant [tex]\( K_c \)[/tex] for this reaction is approximately 1.4.

To calculate the equilibrium constant [tex]\( K_c \)[/tex], we need to determine the molar concentrations of each species involved in the reaction at equilibrium. The balanced chemical equation is:

[tex]\[ 2CH_3OH (l) + 3O_2 (g) \rightleftharpoons 2CO_2 (g) + 4H_2O (g) \][/tex]

Given the masses of the compounds and the volume of the reaction vessel, we can calculate the molar concentrations as follows:

1. Calculate the moles of each compound:

[tex]- For \( CH_3OH \): \( n_{CH_3OH} = \frac{1.47 \text{ g}}{32.04 \text{ g/mol}} \)\\ - For \( O_2 \): \( n_{O_2} = \frac{1.56 \text{ g}}{32.00 \text{ g/mol}} \)\\ - For \( CO_2 \): \( n_{CO_2} = \frac{2.28 \text{ g}}{44.01 \text{ g/mol}} \)\\ - For \( H_2O \): \( n_{H_2O} = \frac{3.33 \text{ g}}{18.015 \text{ g/mol}} \)[/tex]

2. Calculate the molar concentrations (using the volume of the reaction vessel, 9.3 L):

[tex]- For \( CH_3OH \): \( [CH_3OH] = \frac{n_{CH_3OH}}{9.3 \text{ L}} \)\\ - For \( O_2 \): \( [O_2] = \frac{n_{O_2}}{9.3 \text{ L}} \)\\ - For \( CO_2 \): \( [CO_2] = \frac{n_{CO_2}}{9.3 \text{ L}} \)\\ - For \( H_2O \): \( [H_2O] = \frac{n_{H_2O}}{9.3 \text{ L}} \)[/tex]

3. Write the expression for [tex]\( K_c \)[/tex] and substitute the concentrations:

[tex]- \( K_c = \frac{[CO_2]^2 [H_2O]^4}{[CH_3OH]^2 [O_2]^3} \)[/tex]

4. Substitute the calculated molar concentrations into the expression for [tex]\( K_c \)[/tex] and solve.

5. Round the result to two significant digits.

After performing these calculations, we find that [tex]\( K_c \)[/tex] is approximately 1.4 when rounded to two significant digits. This value indicates the position of equilibrium in this reaction mixture, with higher values of [tex]\( K_c \)[/tex]corresponding to a greater extent of the reaction proceeding to the right (towards the products).

Answer:

1 < 3 < 2.

Explanation:

Entropy is the measurement of the degree of disorderliness in a system. The term ''entropy" is one of the important aspect in kinetics and thermodynamics. Entropy is designated with the alphabet symbol "S" and a change in entropy is ∆S.

The main idea in order to solve this question is very simple. Let us make reference to your clothes that you wear. At first, your clothes when you wear them, they are neat and clean but as time goes by the clothes will get dirty, this is a layman example of entropy.

Okay, let me explain further if you look at the first reaction you will notice that we have 2 moles in the reactant sides and 4 moles in the products sides, which means that the entropy has increased.

CH4(g) + H2O(g) --> CO(g) + 3 H2(g).

For the second equation, the entropy is increasing because the the solid reactant produces a gaseous product.

But in this reaction the number of moles of the products has decreased.

C(s) + O2(g) --> CO2(g)

For the third Equation, the the state changes from liquid to gas.

H2O2(l) --> H2O(l) + 1/2 O2(g)

Note that the entropy increases from solid to liquid to gas. That is, gas has the highest degree of entropy and solid has the lowest degree of entropy.

Look at reaction (1) gas to gas.

Look at Reaction (2) solid to gas( great improvement/Increament).

Look at Reaction (3) liquid to gas( increases).

1. The concentration of[tex]\( \text{Q}^- \)[/tex] at equilibrium is[tex]\( 2.10 \times 10^{-3} \)[/tex] M.

2. The[tex]\( K_{sp} \)[/tex] value for [tex]\( \text{XQ}_3 \) is \( 6.48 \times 10^{-12} \).[/tex]

To find the concentration of [tex]\( \text{Q} \)[/tex] at equilibrium and the solubility product constant [tex](\( K_{sp} \))[/tex]for the ionic compound [tex]\( \text{XQ}_3 \)[/tex], follow these steps:

The compound [tex]\( \text{XQ}_3 \)[/tex] dissociates in water as follows:

[tex]\[ \text{XQ}_3(s) \rightleftharpoons \text{X}^{3+}(aq) + 3\text{Q}^-(aq) \][/tex]

Let ( s ) be the solubility of [tex]\( \text{XQ}_3 \)[/tex] in mol/L, which is given as \[tex]7.00 \times 10^{-4} \) M.[/tex]

When [tex]\( \text{XQ}_3 \)[/tex] dissociates:

- The concentration of[tex]\( \text{X}^{3+} \)[/tex] ions will be ( s ).

- The concentration of [tex]\( \text{Q}^- \)[/tex] ions will be ( 3s ) (since three[tex]\( \text{Q}^- \)[/tex] ions are produced for each formula unit of[tex]\( \text{XQ}_3 \)).[/tex]

Given [tex]\( s = 7.00 \times 10^{-4} \) M:[/tex]

[tex]{Q}^-] = 3s = 3 \times 7.00 \times 10^{-4} = 2.10 \times 10^{-3} \text{ M} \][/tex]

Thus, the concentration of [tex]\( \text{Q}^- \)[/tex] at equilibrium is[tex]\( 2.10 \times 10^{-3} \)[/tex].

The solubility product constant [tex]\( K_{sp} \) for \( \text{XQ}_3 \)[/tex]   can be calculated using the equilibrium concentrations of the ions:

[tex][ K_{sp} = [\text{X}^{3+}][\text{Q}^-]^3 \][/tex]

Substitute the equilibrium concentrations:

[tex][ [\text{X}^{3+}] = s = 7.00 \times 10^{-4} \text{ M} \][/tex]

[tex][ [\text{Q}^-] = 3s = 2.10 \times 10^{-3} \text{ M} \][/tex]

Now, calculate[tex]K_{sp} \):[/tex]

[tex]\[ K_{sp} = (7.00 \times 10^{-4}) \times (2.10 \times 10^{-3})^3 \[/tex]

First, compute[tex]( (2.10 \times 10^{-3})^3 \)[/tex]

[tex]\[ (2.10 \times 10^{-3})^3 = 2.10^3 \times (10^{-3})^3 = 9.261 \times 10^{-9} \][/tex]

Then, multiply by[tex]7.00 \times 10^{-4} \):[/tex]

[tex][ K_{sp} = (7.00 \times 10^{-4}) \times (9.261 \times 10^{-9}) \][/tex]

[tex][ K_{sp} = 6.4827 \times 10^{-12} \][/tex]

1. The concentration of[tex]{Q}^- \)[/tex] at equilibrium is [tex]\( 2.10 \times 10^{-3} \) M.[/tex]

2. The [tex]K_{sp} \) value for {XQ}_3 \) is \( 6.4827 \times 10^{-12} \).[/tex]

To calculate the atoms of an element in a given molecule, we need to multiply stoichiometry by the number that is written on the foot of that element. Therefore, the balanced equation is

1B+4St+1Y+3Ic[tex]\rightarrow[/tex]1 BSt[tex]_4[/tex]YIc[tex]_3[/tex]

Balanced equation is the one in which the total number of atoms of a species on reactant side is equal to the total number of atoms on product side. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, displacement reaction.

The other characteristic of balanced reaction is that physical state should be written with each compound or molecule on reactant and product side. Physical state should be written in brackets. s means solid, l means liquid, g means gas.

The word equation is

1 banana+ 4 strawberries+1 container of yogurt+3 ice cubes[tex]\rightarrow[/tex] 1 smoothie

The balanced chemical equation is

1B+4St+1Y+3Ic[tex]\rightarrow[/tex]1 BSt[tex]_4[/tex]YIc[tex]_3[/tex]

Therefore, the balanced equation is

1B+4St+1Y+3Ic[tex]\rightarrow[/tex]1 BSt[tex]_4[/tex]YIc[tex]_3[/tex]

Learn more about the balanced equation, here:

https://brainly.com/question/7181548

#SPJ6

Final answer:

To calculate the pH of a hydroiodic acid solution, you need to determine the initial concentration of hydronium ions. For a 0.100 M solution of hydroiodic acid, the pH is 1.00.

Explanation:

The concentration of hydronium ion in a solution of hydroiodic acid can be expressed as the pH of the solution. To calculate the pH of the solution, you need to determine the initial concentration of hydronium ions. Since hydroiodic acid is a strong acid, you can assume that all of it dissociates. If you dissolve 0.100 mol of hydroiodic acid in enough water to make up 1.00 L of solution, the initial concentration of hydronium ions is 0.100 M. The pH of the solution can be calculated using the formula pH = -log[H3O+]. Therefore, the pH of the solution is -log(0.100) = 1.00.

Answer:

fraction of the compound will be in the unprotonated, amine (amino) form is evaluated to be 83.37%

Explanation:

check the attachment for explicit explanations.

Using the Henderson-Hasselbalch equation, we determine that approximately 86% of the compound will be in the unprotonated, amine form after the pH change from 11 to 10.6.

We can use the Henderson-Hasselbalch equation to find the ratio of the protonated form to the unprotonated form:

Henderson-Hasselbalch equation:

Insert the known values:

Rearrange to solve for the ratio of [tex][R-NH_2][/tex] to [tex][R-NH_3^+][/tex]:

Raise 10 to the power of both sides to remove the logarithm: [tex]10^{0.8} \approx 6.31[/tex]

To find the fraction of the compound in the unprotonated form, we can use:

Convert this fraction to a percentage:

Therefore, approximately 86% of the compound will be in the unprotonated, amine form.

Question: The question is incomplete and can't be comprehended. See the complete question below and the answer.

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

a) Adding Cu2+ ions to the copper half reaction (assuming no volume change).

b) Adding equal amounts of water to both half reactions.

c) Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

d) Adding Ag+ ions to the silver half reaction (assume no volume change)

Explanation:

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q    

where                            

             Ecell = actual cell potential

             E°cell­ ­​ = standard cell potential

             R = the universal gas constant = 8.314472(15) J K−1 mol−1

             T = the temperature in kelvins

              n = the number of moles of electrons transferred                                    

 F = the Faraday constant, the number of coulombs per mole of electrons:

  (F = 9.64853399(24)×104 C mol−1)

 Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from  the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.

Answer:

The difference in energy between the [tex]S_1[/tex] and [tex]T_1[/tex] states is 35.7 kJ/mol.

Explanation:

The Planck's equation :

[tex]E=\frac{hc}{\lambda} [/tex]

Where:

E = Energy of the electromagnetic radiations.

h = Planck's constant = [tex]6.626\times 10^{-34} Js[/tex]

c = speed of light = [tex]3\times 10^8 m/s[/tex]

[tex]\lambda [/tex] = Wavelength of the electromagnetic radiations.

Energy associated with [tex]T_1[/tex] transition : [tex]E_1[/tex]

Wavelength associated  with [tex]T_1[/tex] transition :

[tex]\lambda _1=397 nm=397\times 10^{-9} m[/tex]

Energy associated with [tex]S_1[/tex] transition : [tex]E_2[/tex]

Wavelength associated  with [tex]S_1[/tex] transition : [tex]\lambda _2=355 nm=355\times 10^{-9} m[/tex]

The difference in energy between the [tex]S_1[/tex] and [tex]T_1[/tex] states:

[tex]E=E_1-E_2=\frac{hc}{\lambda _2}-\frac{hc}{\lambda _1}[/tex]

[tex]E=hc\times (\frac{1}{\lambda _2}-\frac{1}{\lambda _1})[/tex]

[tex]E=6.626\times 10^{-34} Js \times 3\times 10^8 m/s(\frac{1}{355\times 10^{-9}m}-\frac{1}{397\times 10^{-9} m})[/tex]

[tex]E=5.92\times 10^{-20} J[/tex]

1 J = 0.001 kJ

[tex]E=5.92\times 10^{-20}\times 10^{-3} kJ=5.92\times 10^{-23} kJ[/tex]

1 mole = [tex]6.022\times 10^{-23}[/tex]

The difference in energy(kJ/mol) between the [tex]S_1[/tex] and [tex]T_1[/tex] states:

[tex]=5.92\times 10^{-23} kJ\times 6.022\times 10^{23} mol^{-1}=35.7 kJ/mol[/tex]

Final answer:

The energy difference between the S1 and T1 states in formaldehyde can be calculated using the wavelengths given and Planck's equation, with the resulting value representing the difference in energy due to electronic transitions between these two states.

Explanation:

The student is asking about the difference in energy between the singlet state (S1) and triplet state (T1) of formaldehyde, which corresponds to n → π* electronic transitions occurring at different wavelengths. The wavelengths given are 397 nm (T1) and 355 nm (S1). We can calculate the energy difference using the equation E = hv (where h is Planck constant and v is frequency), and converting wavelengths to frequencies using c = λv (where c is the speed of light and λ is the wavelength). To find the difference in energy (ΔE) in kJ/mol, we first convert the energy of each transition from Joules to kJ/mol and then subtract the energy of T1 from S1.

To calculate the energy for each wavelength:
E(397nm) = (6.626 x 10^-34 J·s) (3 x 10^8 m/s) / (397nm x 10^-9 m/nm) x (1/1000) kJ/J x (6.022 x 10^23 mol^-1)
E(355nm) = (6.626 x 10^-34 J·s) (3 x 10^8 m/s) / (355nm x 10^-9 m/nm) x (1/1000) kJ/J x (6.022 x 10^23 mol^-1)
ΔE = E(355nm) - E(397nm)

The difference in energy between the S_1 and T_1 states:

E=E_1-E_2=(hc)/(\lambda _2)-(hc)/(\lambda _1)

E=hc* ((1)/(\lambda _2)-(1)/(\lambda _1))

E=6.626* 10^(-34) Js * 3* 10^8 m/s((1)/(355* 10^(-9)m)-(1)/(397* 10^(-9) m))

E=5.92* 10^(-20) J

1 J = 0.001 kJ

E=5.92* 10^(-20)* 10^(-3) kJ=5.92* 10^(-23) kJ

1 mole = 6.022* 10^(-23)

The difference in energy(kJ/mol) between the S_1 and T_1 states:

=5.92* 10^(-23) kJ* 6.022* 10^(23) mol^(-1)=35.7 kJ/mol

Answer:

Explanation:

Find attached the solution

Complete Question (in order)

The growth of baker's yeast (S. cerevisiae) on glucose may be simply described by following equation:

C₆H₁₂0₆ +  30₂ + 0.48 NH₃ => 0.48C₆H₁₂NO₃ + 4.32H₂O + 3.12CO₂ yeast

In a batch reactor of volume 10 1, the final desired yeast concentration is 50 gdw/1.

a. Determine the concentration and total amount of glucose and (NH4)₂SO4 in the nutrient medium.

b. Determine the yield coefficients [tex]Y_{x/s}[/tex] (biomass/glucose) and [tex]Y_{x/o}[/tex] (biomass/oxygen).

c. Determine the total amount of oxygen required.

d. If the rate of growth at exponential phase is [tex]r_x[/tex] = 0.7 gdw/l-h, determine the rate of oxygen consumption (g O₂/1-h)

e. Calculate the heat-removal requirements for the reactor (recall equation 6.26).

Answer:

a. 2.3 × 10⁴ kg

b. 0.384 and 0.72

c. 6.94 × 10³ kg

d. 0.973

Explanation:

The given equation that describes the growth of baker's yeast on glucose

C₆H₁₂0₆ +  30₂ + 0.48 NH₃ => 0.48C₆H₁₂NO₃ + 4.32H₂O + 3.12CO₂ yeast

The volume of batch reactor is given as 10⁵ Liters

The final desired yeast concentration is 50 gdw/l

The following attached documents gives additional solution for all other answers

Answer: The molar concentration of sulfuric acid in the original sample is 1.943 M

Explanation:

To calculate the molarity of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=2\\M_1=?\\V_1=56.5mL\\n_2=1\\M_2=0.5824M\\V_2=43.37mL[/tex]

Putting values in above equation, we get:

[tex]2\times M_1\times 56.5=1\times 0.5824\times 43.37[/tex]

[tex]M_1=0.2235[/tex]

Now to calculate the molarity of original solution:

[tex]M_1\times 6.5=0.2235\times 56.5[/tex]

[tex]M_1=1.943[/tex]

Thus the molar concentration of sulfuric acid in the original sample is 1.943 M

Final answer:

The molar concentration of sulfuric acid in the original 6.5 mL sample is calculated to be approximately 1.9436 M, based on titration with a 0.5824 M NaOH solution.

Explanation:

Determining the Concentration of Sulfuric Acid in a Sample Using Titration:

To determine the molar concentration of sulfuric acid, H2SO4, we will use the data that 43.37 mL of a 0.5824 M NaOH solution was needed to titrate a 6.5 mL sample of the acid.

Firstly, we calculate the moles of NaOH used in the titration:

Moles of NaOH = Volume (in Liters)  imes Molarity

Moles of NaOH = 0.04337 L  imes 0.5824 mol/L

Moles of NaOH = 0.02526888 mol

According to the balanced chemical equation, H2SO4 + 2NaOH
ightarrow Na2SO4 + 2H2O, the stoichiometry of the reaction is 1:2. This means one mole of sulfuric acid reacts with two moles of sodium hydroxide. Thus, the moles of H2SO4 will be half of the moles of NaOH used.

Moles of H2SO4 = 0.02526888 mol / 2

Moles of H2SO4 = 0.01263444 mol

Now, we calculate the molarity of the sulfuric acid in the original sample:

Molarity of H2SO4 = Moles of H2SO4 / Volume of sample in liters

Molarity of H2SO4 = 0.01263444 mol / 0.0065 L

Molarity of H2SO4 = 1.9436 M

Therefore, the molar concentration of the sulfuric acid in the original sample is approximately 1.9436 M.

Answer:

1. 3.29mol

2. 0.125molH2o/mol

3. 52.5'C

Explanation: The step by step explanation are attached to the answer.

Answer:

Mole fraction of propane = 0.74 mol C₃H₈/mol

Mole fraction of water = 0.29 mol H₂0 /mol

Dew point temperature = 52.5°C

Explanation:

See the attached file for the calculation.

Answer:

Molar mass= 297.1gmol-1

Explanation:

BaBr2 = 137.3+ (79.9*2)= 297.1gmol-1

Answer:

c = 4

Explanation:

In general, for the reaction

       a A     +    b B     ⇒ c C   +    d D

the rate is given by:

rate = - 1/a ΔA/Δt = - 1/b ΔB/Δt = + 1/c ΔC/Δt = + 1/d ΔD/Δt

this is done so as to express the rate in a standarized way which is the same to all the reactants and products irrespective of their stoichiometric coefficients.

For this question in particular we know the coefficient of A and need to determine the coefficient c.

- 1/2 ΔA/Δt = + 1/c ΔC/Δt

- 1/2 (-0.0080 ) = + 1/c ( 0.0160 mol L⁻¹s⁻¹ )

0.0040 mol L⁻¹s⁻¹  c = 0.0160 mol L⁻¹s⁻¹

∴ c = 0.0160 / 0.0040 = 4

Answer:

During a phase change, the temperature of a substance remains constant

Explanation:

Step 1:

During phase change, the energy that is required, will be used to separate the molecules ( to change phase).

This energy is not used to change the temperature.

Since the kinetic energy of the molecules remains the same, the temperature of the substance remains constant.

Final answer:

Internal standards and standard addition are techniques used in analytical chemistry to ensure accuracy. The incorrect statement is about the use of internal standards.

Explanation:

Internal standards and standard addition are both techniques used in analytical chemistry to ensure the accuracy and reliability of quantitative measurements. In internal standard method, a known amount of a compound is added to all samples and standards, which allows for compensation of errors that may occur during sample preparation and analysis.

Standard addition, on the other hand, involves adding known amounts of a standard solution to an unknown solution to determine the concentration of the analyte of interest. It can be a single standard addition or multiple standard additions, depending on the requirements of the analysis.

Based on the given options, the incorrect description is:

d) Internal standard is used when instrument response varies from run to run.

The correct answer is a) Internal standard is useful when the matrix in the unknown is complicated.

Let's analyze each option to understand why option (a) is incorrect:

a) Internal standard is useful when the matrix in the unknown is complicated.

This statement is incorrect because it confuses the roles of internal standards and standard addition.

An internal standard is a substance that is added in a constant amount to all samples, including the calibration standards and the unknowns. It is used to correct for any variations in the analytical procedure, such as changes in instrument response, sample preparation, and matrix effects that affect the analyte's signal.

However, it is not specifically used because the matrix is complicated; rather, it is used to account for variations in the analytical process. The complexity of the matrix is typically addressed by the standard addition method, which involves adding known quantities of the analyte to the sample to overcome matrix effects.

b) Standard addition could be a single standard addition or multiple standard additions to an unknown solution.

This statement is correct. Standard addition can involve adding a known amount of analyte to the sample once (single standard addition) or several times (multiple standard additions) to construct a calibration curve.

This method is used to compensate for matrix effects that might not be accounted for by using external calibration alone.

c) Standard addition is useful when the matrix in the unknown is complicated.

This statement is correct. The standard addition method is particularly useful for samples with complex matrices that can interfere with the analysis.

By adding known amounts of the analyte directly to the sample, the method allows for the determination of the analyte's concentration while accounting for the matrix effects.

d) Internal standard is used when instrument response varies from run to run.

This statement is correct. An internal standard is used to normalize the response of the analyte and to correct for any variations in the analytical procedure, including changes in instrument response over time.

It helps to ensure the accuracy and precision of the analytical results, regardless of the variations that occur between different runs of the analysis.

Therefore, the statement in option (a) is the one that is not correct, as it misrepresents the use of internal standards in the context of complex matrices.

Answer:

The daughter nuclide is selenium; 76:34 Se

Explanation:

The complete question is as follows;

is desired to determine the concentration of arsenic in a lake sediment sample by means of neutron activation analysis. The nuclide 75:33 As captures a neutron to form 76:33 As, which in turn undergoes beta decay. The daughter nuclide produces the characteristic gamma rays used for the analysis. What is the daughter nuclide?

solution:

Please check attachment for decay equations and explanations

Answer:

13

Explanation:

1M MOH solution dissociates into 0.1M OH- ions.

pOH= -log(OH-)

       =    1

pH=14-1

    =13

The pH of this solution is 13.

We can solve this question knowing that a hydroxide as MOH, dissociates in water as follows:

MOH(aq) ⇄ M⁺(aq) + OH⁻(aq)

Based on the reaction, 1 mole of MOH produces 1 mole of OH⁻

If 10% of the 1M MOH dissociates, a 0,1M of M⁺ and a 0.1M of OH⁻ are produced.

Now, with molarity of OH⁻ we can find pOH (pOH = -log[OH⁻]) and pH (pH = 14-pOH) as follows:

pOH:

pOH = -log [OH⁻] = -log [0.1] =1

And pH:

pH:

pH = 14 - pOH

pH = 13

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Answer:

Nitrogen

Explanation: nitrogen is around 70% of the air hope this helps god bless

Answer:

The Keq value for the reaction is 0.13

The reactants are favored in the reaction.

Explanation:

The equilibrium expression:  N[tex]^{2}[/tex]O4(g) ⇌ 2 [NO[tex]^{2}[/tex]](g)  

Keq = squared of  [NO[tex]^{2}[/tex]] divided by [N[tex]^{2}[/tex]O4]

Keq = squared of [0.10] divided by [0.075]

Keq = 0.13

Equilibrium constant (Keq) indicates what will be favoured at equilibrium, as it is a ratio that quantifies the position of a chemical equilibrium at a given temperature. It measures the extent to which reactants are converted to products.

If equilibrium constant is less than 1, the mixture contains mostly reactants. Which indicates more reactants are present at equilibrium. If equilibrium constant is equal to 1, then, the amount of products is roughly equal to the amount of reactants at equilibrium. If equilibrium constant is much greater than 1, more products are present at equilibrium.

In the question treated above, the reactants are N[tex]^{2}[/tex]O4(g). While, the products are: 2 [NO[tex]^{2}[/tex]](g).

The value of [tex]\( K_{eq} \)[/tex] is 0.1333. Since [tex]\( K_{eq} \)[/tex] is less than 1, the reactants are favored at equilibrium is 0.1333.

The value of [tex]\( K_{eq} \)[/tex] for the decomposition of dinitrogen tetroxide [tex](\( N_2O_4 \))[/tex] into nitrogen dioxide [tex](\( NO_2 \))[/tex] is calculated by the following equilibrium expression:

[tex]\[ K_{eq} = \frac{[NO_2]^2}{[N_2O_4]} \][/tex]

Given that the initial concentration of [tex]\( N_2O_4 \)[/tex] is 0.125 mol/L and the equilibrium concentration is 0.0750 mol/L, we can determine the concentration of [tex]\( NO_2 \)[/tex] at equilibrium by using the stoichiometry of the reaction:

[tex]\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \][/tex]

For every mole of [tex]\( N_2O_4 \)[/tex] that decomposes, 2 moles of [tex]\( NO_2 \)[/tex] are produced. The change in concentration of [tex]\( N_2O_4 \) (\( \Delta[N_2O_4] \))[/tex] is the initial concentration minus the equilibrium concentration:

[tex]\[ \Delta[N_2O_4] = [N_2O_4]_{initial} - [N_2O_4]_{equilibrium} \][/tex]

[tex]\[ \Delta[N_2O_4] = 0.125 \text{ mol/L} - 0.0750 \text{ mol/L} \][/tex]

[tex]\[ \Delta[N_2O_4] = 0.0500 \text{ mol/L} \][/tex]

Since 2 moles of [tex]\( NO_2 \)[/tex] are produced for every mole of [tex]\( N_2O_4 \)[/tex] that decomposes, the change in concentration of [tex]\( NO_2 \) (\( \Delta[NO_2] \))[/tex] is twice the change in concentration of [tex]\( N_2O_4 \)[/tex]:

[tex]\[ \Delta[NO_2] = 2 \times \Delta[N_2O_4] \][/tex]

[tex]\[ \Delta[NO_2] = 2 \times 0.0500 \text{ mol/L} \][/tex]

[tex]\[ \Delta[NO_2] = 0.1000 \text{ mol/L} \][/tex]

The equilibrium concentration of [tex]\( NO_2 \) (\( [NO_2]_{equilibrium} \))[/tex] is the initial concentration of [tex]\( NO_2 \)[/tex] (which is 0, since none was present initially) plus the change in concentration:

[tex]\[ [NO_2]_{equilibrium} = [NO_2]_{initial} + \Delta[NO_2] \][/tex]

[tex]\[ [NO_2]_{equilibrium} = 0 + 0.1000 \text{ mol/L} \][/tex]

[tex]\[ [NO_2]_{equilibrium} = 0.1000 \text{ mol/L} \][/tex]

Now we can calculate [tex]\( K_{eq} \)[/tex]:

[tex]\[ K_{eq} = \frac{[NO_2]_{equilibrium}^2}{[N_2O_4]_{equilibrium}} \][/tex]

[tex]\[ K_{eq} = \frac{(0.1000)^2}{0.0750} \][/tex]

[tex]\[ K_{eq} = \frac{0.01000}{0.0750} \][/tex]

[tex]\[ K_{eq} = 0.1333 \][/tex]

Answer:

see explanation below

Explanation:

First, let's remember the principle of a friedel crafts alkylation. This kind of reaction are often used to alkylate an aromatic ring like bencene, or a mono substitued bencene. This reactions often involves reactions with alkyl halides or alkenes and aluminum or iron halides.

Now, in this case we have Iodobenzene, bromobenzene, fluorobenzene and benzoic acid.

First, the reactivity order always raise with the polarization of the C - X bond. Therefore, the one with the most polar bond, will react first. In this case, with the halides, the reactivity order would be RI > RBr > RCl >> RF

In the case of the benzoic acid, the COOH is a very strong deactivating group by resonance, therefore, the benzene it's really weak to do a reaction of alkylation, so this will most likely to not react.

In conclude, we have the following order:

Compound A: Iodobencene (1)

Compound B: Bromobencene (2)

Compound C: Fluorobencene (3)

Compound D: Benzoic acid (non)

In Friedel-Crafts alkylation, compounds are ordered by reactivity based on the strength of their carbon-halogen bonds and ability to form stable carbocations. The ranking is Bromobenzene, Iodobenzene, Fluorobenzene, and Benzoic Acid, with the latter being 'non' reactive due to the presence of a deactivating carboxylic acid group.

The reactivity of these compounds in a Friedel-Crafts alkylation scenario depends on their ability to form stable carbocation intermediates and the strength of the carbon-halogen bond. The stronger the bond, the less likely it is to break and form the carbocation necessary for the reaction. Therefore, the reactivity order is as follows:

It's worth noting that Benzoic Acid (Compound D) would be not considered in Friedel-Crafts alkylation due to the presence of the deactivating carboxylic acid group, which hinders this reaction. Therefore, in effect, it would be ranked as 'non' reactive for this specific reaction.

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Answer:

523°C

Explanation:

Step 1:

Data obtained from the question.

Diameter (d) = 31.0 cm

Height (h) = 37.2 cm

Pressure (P) = 5.90 MPa

Mass of ClF5 = 3.27 kg

Temperature (T) =?

Step 2:

Determination of the volume of the stainless-steel cylinder.

Volume of cylinder = πr^2h = π/4d^2h

V = π/4 x (31)^2 x 37.2

V = 28077.36 cm3

Converting the volume to L, we have:

1 cm3 = 0.001 L

Therefore, 28077.36 cm3 = 28077.36 x 0.001 = 28.07736 L

Step 3:

Conversion of 5.90 MPa to atm.

1 MPa = 9.869 atm

Therefore, 5.90 MPa = 5.90 x 9.869

= 58.2271 atm

Step 4:

Determination of the number of mole of chiorine pentafluoride gas (ClF5). This is illustrated below:

Molar Mass of ClF5 = 35.5 + (19x5) = 35.5 + 95 = 130.5g/mol

Mass of ClF5 = 3.27kg =3.27x10000

Mass of ClF5 = 3270g

Number of mole = Mass /Molar Mass

Number of mole of ClF5 = 3270/130.5

Number of mole of ClF5 = 25.057 moles

Step 5:

Determination of the temperature.

Applying the ideal gas equation

PV = nRT, the temperature T, can be obtained as follow:

P = 58.2271 atm

V = 28.07736 L

n = 25.057 moles

R (gas constant) = 0.082atm.L/Kmol

T=?

PV = nRT

Divide both side by nR

T = PV /nR

T = (58.2271x28.07736)/(25.057x0.082)

T = 795.68 K

Step 6:

Conversion of Kelvin temperature to celsius temperature.

°C = K - 273

K = 795.68 K

°C = 795.68 - 273

°C = 523°C

Therefore, the maximum safe operating temperature the engineer should recommend is 523°C

Answer:

A,C,F/Air,water,orginisams in water

Explanation:

The resources and organisms are affected by oil spillages are Air, Water, and Organisms in water

Oil spillage can cause to the general environment and the organisms in the environment.

Based on the passage, the resources and organisms are affected by oil spillages include the following;

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Answer:

The standard free energy of combustion of 1 mole of methane = -801.11 kJ

The negative sign shows that this reaction is spontaneous under standard conditions.

The negative sign on the standard free energy also means this combustion reaction is product-favoured at equilibrium.

Explanation:

The chemical reaction for the combustion of methane is given by

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

The standard free energy of formation for the reactants and products as obtained from literature include

For CH₄, ΔG⁰ = -50.50 kJ/mol

For O₂, ΔG⁰ = 0 kJ/mol

For CO₂, ΔG⁰ = -394.39 kJ/mol

For H₂O(g), ΔG⁰ = -228.61 kJ/mol

ΔG(combustion) = ΔG(products) - ΔG(reactants)

ΔG(products) = (1×-394.39) + (2×-228.61) = -851.61 kJ/mol

ΔG(reactants) = (1×-50.50) + (2×0) = -50.50 kJ/mol

ΔG(combustion) = ΔG(products) - ΔG(reactants)

ΔG(combustion) = -851.61 - (-50.5) = -801.11 kJ/mol

Since we're calculating for 1 mole of methane, ΔG(combustion) = -801.11 kJ

- A negative sign on the standard free energy means that the reaction is spontaneous under standard conditions.

- A positive sign indicates a non-spontaneous reaction.

- A Gibb's free energy of 0 indicates that the reaction is at equilibrium.

- A negative sign on the standard free energy also means that if the reaction reaches equilibrium, it will be product favoured.

- A positive sign on the standard free energy means that the reaction is reactant-favoured at equilibrium.

Hope this Helps!!!

The enthalpy of combustion of ethanol is approximately -118.45 kJ/mol.

To determine the enthalpy of combustion (ΔH) of ethanol, we can use the heat transfer equation:

q = mcΔT

where q is the heat absorbed by the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

First, calculate the heat absorbed by the water:

q = (730 g)(4.184 J/g ⋅ K)(20.5 °C - 16.8 °C)

q = 730 × 4.184 × 3.7 J

q = 11,300.984 J

Next, convert the mass of ethanol burned (4.40 g) to moles using the molar mass of ethanol (C2H5OH):

Molar mass of ethanol = 2 × molar mass of C + 6 × molar mass of H + molar mass of O

Molar mass of ethanol = 2 × 12.01 + 6 × 1.01 + 16.00 g/mol

Molar mass of ethanol = 46.07 g/mol

Moles of ethanol = 4.40 g / 46.08 g/mol

Moles of ethanol ≈ 0.0954 mol

Now, calculate ΔH using the formula:

ΔH = -q / moles of ethanol

ΔH = -11,300.984 J / 0.0954 mol

ΔH ≈ -118,458.952 J/mol

Convert to kJ/mol:

ΔH ≈ -118.45 kJ/mol

Answer:

Mix 11.73 mL of CHCl₃ and 8.27 mL of CHBr₃

Explanation:

To begin, it seems your question lacks the name of the pure samples that will be mixed to prepare the liquid mixture. The names are not necessary, as the numerical values are there. However, an internet search tells me they are CHCl₃ (d=1.492g/mL) and CHBr₃ (d=2.890 g/mL).

The mixture has a density of 2.07 g/mL, so 20 mL of the mixture would weigh:

20 mL * 2.07 g/mL = 41.4 g

Let X be the volume of CHCl₃ and Y the volume of CHBr₃:

X + Y = 20 mL

The mass of CHCl₃ and CHBr₃ combined have to be equal to the mass of the mixture. We can write that equation using the volume of the samples and their density:

X * 1.492 + Y * 2.890 = 41.4 g

So now we have a system of two equations and two unknowns, we use algebra to solve it:

   1. Express Y in terms of X:

X + Y = 20

   2. Replace Y in the second equation:

X * 1.492 + Y * 2.890 = 41.4

   3. Solve for X:

   4. Using the now known value of X, solve for Y:

X + Y = 20

So, to prepare the liquid mixture we would mix 8.27 mL of CHBr₃ and 11.73 mL of  CHCl₃.

Answer: The final temperature of the water is [tex]33.85^{o}C[/tex].

Explanation:

We know that molar mass of [tex]C_{6}H_{6}[/tex] is 78 g/mol. And, the amount of heat produced when 2 mol of [tex]C_{2}H_{6}[/tex] burns is 6542 KJ.

This means that,

   [tex]78 \times 2[/tex] = 156 g of [tex]C_{2}H_{6}[/tex] burns, heat produced is 6542 kJ.

Therefore, heat produced (Q) by burning  7.3 g of [tex]C_{6}H_{6}[/tex] is as follows.

               [tex]\frac{6542 \times 7.3 g}{156 g}[/tex]

              = 306.13 kJ

or,           = 306130 J      (as 1 KJ = 1000 J)

For water, mass is given as 5691 g and specific heat capacity of water is 4.186 [tex]J/g^{o}C[/tex].

So, we will calculate the value of final temperature as follows.

            Q = [tex]m \times C \times (T_{f} - T_{i})[/tex]

  306130 J = [tex]5691 g \times 4.186 J/g^{o}C \times (T_{f} - 21)^{o}C[/tex]

       [tex](T_{f} - 21)^{o}C = \frac{306130 J}{23822.53 J/^{o}C}[/tex]

          [tex]T_{f}[/tex] = 12.85 + 21

                      = [tex]33.85^{o}C[/tex]

Thus, we can conclude that the final temperature of the water is [tex]33.85^{o}C[/tex].

Final answer:

To determine the final temperature of water when 7.300 g of benzene is burned, use the enthalpy of combustion for benzene and apply stoichiometry to find the heat transferred to the water, then use the specific heat capacity formula to find the change in water's temperature.

Explanation:

The question asks about finding the final temperature of water after burning 7.300 g of benzene (C6H6) and adding the heat produced to the water at an initial temperature of 21 °C. To answer this, we need to use thermochemical principles and calculate the energy involved in the combustion of benzene which will then be absorbed by the water, causing a change in its temperature.

The heat absorbed by the water (q_water) can be calculated using the formula q = m * c *
igtriangleup T, where m is the mass of the water, c is the specific heat capacity of water (4.184 J/g°C), and
igtriangleup T is the change in temperature. Since we're given the enthalpy of combustion of benzene (6,542 kJ per 2 moles of C6H6), we can calculate the heat produced by burning 7.300 g of benzene using stoichiometry and the molar mass of C6H6 (78.11 g/mol).

Using this information allows us to solve for the final temperature of the water. Remember to convert the energy units appropriately when performing these calculations.

Answer:

I think it is D.

Explanation:

Answer: i know it d but i think it is c to

Explanation:

Answer: 1. 1-propanol can form hydrogen bonds with water, but butane cannot.

2. 1-propanol is more soluble than ethyl methyl ether because it can form more hydrogen bonds with water.

3. Ethanol is more soluble in water than 1-hexanol because it has a shorter carbon chain.

Explanation:

It is know that like dissolves like. That is polar substance will be soluble in polar solvent and non-polar substance will be soluble in non-polar solvent.

As 1-propanol is polar in nature due to the alcoholic group present in it so, it is able to form hydrogen bonds with water. Hence, it readily dissolves in water. Whereas butane is non-polar in nature and it will not dissolve in water (polar solvent).

Also, smaller is the number of carbon chain present in a compound smaller will be its surface area. Hence, more readily it will dissolve in water.

For example, ethanol has less number of carbon atoms than 1-hexanol. So, ethanol will be more soluble in water than 1-hexanol.

Thus, we can conclude that given blanks are filled as follows.

1. 1-propanol can form hydrogen bonds with water, but butane cannot.

2. 1-propanol is more soluble than ethyl methyl ether because it can form more hydrogen bonds with water.

3. Ethanol is more soluble in water than 1-hexanol because it has a shorter carbon chain.

Answer : The rate constant at 525 K is, [tex]0.0606M^{-1}s^{-1}[/tex]

Explanation :

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

or,

[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]

where,

[tex]K_1[/tex] = rate constant at [tex]701K[/tex] = [tex]2.57M^{-1}s^{-1}[/tex]

[tex]K_2[/tex] = rate constant at [tex]525K[/tex] = ?

[tex]Ea[/tex] = activation energy for the reaction = [tex]1.5\times 10^2kJ/mol=1.5\times 10^5J/mol[/tex]

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = 701 K

[tex]T_2[/tex] = final temperature = 525 K

Now put all the given values in this formula, we get:

[tex]\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}][/tex]

[tex]K_2=0.0606M^{-1}s^{-1}[/tex]

Therefore, the rate constant at 525 K is, [tex]0.0606M^{-1}s^{-1}[/tex]

Answer:

MnO2 is the limiting reactant

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

4Al(s) + 3MnO2(s) —> 3Mn(l) + 2Al2O3(s)

Step 2:

Determination of the masses of Al and MnO2 that reacted from the balanced equation. This is illustrated below:

Molar Mass of Al = 26.98g/mol

Mass of Al from the balanced equation = 4 x 26.98 = 107.92g

Molar Mass of MnO2 = 86.94g/mol

Mass of MnO2 from the balanced equation = 3 x 86.94 = 260.82g

From the balanced equation above,

107.92g of Al reacted with 260.82g of MnO2 reacted.

Step 3:

Determination of the limiting reactant. This is illustrated below:

Let us consider using all the 203g of Al given to verify if there will be any leftover for MnO2. This is illustrated below:

From the balanced equation above,

107.92g of Al reacted with 260.82g of MnO2,

Therefore, 203g of Al will react with = (203 x 260.82)/107.92 = 490.61g of MnO2.

We can see that the mass of MnO2 that will react with 203g of Al is 490.61g which far greater than the 472g that was given. Therefore, it is not acceptable.

Now let us consider using all 472g of MnO2 given to verify if there will be any leftover for Al. This is shown below:

From the balanced equation above,

107.92g of Al reacted with 260.82g of MnO2,

Therefore, Xg of Al will react with 472g of MnO2 i.e

Xg of Al = (107.92 x 472)/260.82

Xg of Al = 195.30g

From the calculations above, we can see clearly that there are leftover for Al as 195.30g out of 203g reacted.

Therefore, MnO2 is the limiting reactant