. The heat capacity of a bomb calorimeter was determined by burning 6.79 g methane (energy of combustion 802 kJ/mol CH4) in the bomb. The temperature changed by 10.8C. a. What is the heat capacity of the bomb? b. A 12.6-g sample of acetylene, C2H2, produced a temperature increase of 16.9C in the same calorimeter. What is the energy of combustion of acetylene (in kJ/mol)?

Answer: decreasing the temperature of the water

Explanation: generally solubility increase on decrease in temperature because On increasing temperature the movement increase thus causing excape of the gas molecules and hence causing less mole of the gas CO2 to desolve in water.

Answer:

0.16 M

Explanation:

Considering:

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

Or,

[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]

Given :

For [tex]K_2CO_3[/tex] :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of [tex]K_2CO_3[/tex] :

[tex]Moles=0.200 \times {20.0\times 10^{-3}}\ moles[/tex]

Moles of [tex]K_2CO_3[/tex]  = 0.004 moles

For [tex]Ba(NO_3)_2[/tex] :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 30.0×10⁻³ L

Thus, moles of [tex]Ba(NO_3)_2[/tex] :

[tex]Moles=0.400 \times {30.0\times 10^{-3}}\ moles[/tex]

Moles of [tex]Ba(NO_3)_2[/tex]  = 0.012 moles

According to the given reaction:

[tex]Ba(NO_3)_2 + K_2CO_3\rightarrow BaCO_3 + 2KNO_3[/tex]

1 mole of [tex]Ba(NO_3)_2[/tex] reacts with 1 mole of [tex]K_2CO_3[/tex]

So,

0.012 mole of [tex]Ba(NO_3)_2[/tex] reacts with 0.012 mole of [tex]K_2CO_3[/tex]

Available mole of [tex]K_2CO_3[/tex] = 0.004 mole

Limiting reagent is the one which is present in small amount. Thus, [tex]K_2CO_3[/tex] is limiting reagent. (0.004 < 0.012)

The formation of the product is governed by the limiting reagent. So,

1 mole of [tex]K_2CO_3[/tex] reacts with 1 mole of [tex]Ba(NO_3)_2[/tex] and gives 1 mole of [tex]BaCO_3[/tex]

0.004 mole of [tex]K_2CO_3[/tex] reacts with 0.004 mole of [tex]Ba(NO_3)_2[/tex] and gives 0.004 mole of [tex]BaCO_3[/tex]

Left moles of [tex]Ba(NO_3)_2[/tex] = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20 + 30 mL = 50 mL = 0.050 L

So,

Concentration of barium ion, [tex]Ba^{2+}[/tex], in solution after reaction is:-

[tex]Molarity=\frac{0.008}{0.050}\ M = 0.16\ M[/tex]

To determine the concentration of the barium ion, Ba2+, in the solution after the reaction, we use stoichiometry. The concentration of Ba2+ in solution is the same as the concentration of BaCO3 formed. Using the volume and concentration of the Ba(NO3)2 solution, we can calculate the concentration of Ba2+ to be 0.04 M.

Using the equation:

Concentration of Ba2+ in solution = 0.004 mol / (0.020 L + 0.030 L) = 0.04 M

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Answer:

d. -564 kJ

Explanation:

It is possible to find ΔE° using the formula:

ΔH° = ΔE° + PΔV

For the reaction:

2CO(g) + O₂(g) → 2CO₂(g)  ΔH°= -566kJ

As P is 1,00 atm and ΔV is -24,5L:

-566 kJ = ΔE° + 1,00atm×-24,5L×[tex]\frac{0,101kJ}{1atmL}[/tex]

-566 kJ = ΔE° - 2,47 kJ

ΔE° = -563,53 kJ

The ΔE° is:

d. -564 kJ

I hope it helps!

ΔE° for the reaction below if the process is carried out at a constant pressure of 1.00 atm and ΔV is d. -564 kJ

To find the ΔE° for the reaction, use the formula ΔE° = ΔH° - (Δn)RT and plug in the given values to calculate the energy change.

Therefore, the correct answer is d. -564 kJ.

Correct question is: Find ΔE° for the reaction below if the process is carried out at a constant pressure of 1.00 atm andΔV (the volume change) = -24.5 L. (1 L ∙ atm = 101 J)
2 CO(g) + O₂ (g) → 2 CO₂(g) ΔH° = -566. kJ
a. +2.47 kJ
b. -568 kJ
c. -2.47 kJ
d. -564 kJ

Answer: Three nucleotides encode an amino acid, The code has no punctuation, The code is degenerate.

Explanation:The genetic code has four main features, they are;

1.)Three nucleotides/bases encode an amino acid, there are 20 different amino acids which are the building blocks for proteins.

2.)The genetic code is non-overlapping, for example a sequence UGGAUCGAU is read UGG AUC GAU rather than UGG GGA GAU etc.

3.)The code has no punctuation, so no base serves as a "comma" between groups of bases, therefore the code is read sequencially three bases at a time.

4.)The code is degenerate, meaning more than one codon encodes for the same amino acid. There are 64 possible triplets yet only 20 amino acids so most amino acids are encoded by 2 or more codons. Triplets that code for the same amino acid are known as synonyms.

Other features of the genetic code include;

(1) Code is a Triplet

(2)The Code is Comma Less

(3) The Code is Unambiguous

(4) The Code is Universal

(5) Co-linearity

(6) Gene-polypeptide Parity.

The key features of the genetic code include degeneracy, directionality, and three nucleotides coding for an amino acid.

The key features of the genetic code include:

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Answer:

The variation of pressure is 0.

Explanation:

Methane combustion reaction:

[tex]CH_4 + 2 O_2 \longrightarrow CO_2 + 2 H_2O[/tex]

The pressure change in gaseous state reactions at constant T y V is proportional to the change in the number of mol of gas in total.

As can be seen in the reaction above we have 3 moles of gas in the reactants and 3 in the products, so the variation of moles is 0. Therefore, the variation of pressure is also 0.

Answer: under aqueous or molten state

Explanation:

An ionic compound is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element and the element which accepts the electrons is known as electronegative element. This bond is formed between a metal and an non-metal.[tex]KBR[/tex] is an ionic compound.

Ionic compounds do not conduct electricity in solid state as ions are not available for conduction of electricity. When [tex]KBR[/tex] is in molten form or aqueous state, it gets converted to ions [tex]K^+[/tex] and [tex]Br^-[/tex] and hence help in conduction of electricity.

Answer:

Empirical formula = S₁Cl₂ = SCl₂

Explanation:

The empirical formula of a compound is the simplest whole number ratio of each type of atom in a compound.

This compound contains 31.1% sulfur and 68.9% chlorine.

That is 100g of the compound contains 31.1 g of sulfur and 68.9 g of chlorine.

Convert the mass of each element to moles using the molar mass from the periodic table.

Moles of Sulfur = [tex]\frac{31.1}{32.065 } \\[/tex]

= 0.9699

Moles of Chlorine= [tex]\frac{68.9}{35.453 } \\[/tex]

= 1.9434

Divide each mole value by the smallest number of moles calculated.

Units of sulfur =  [tex]\frac{0.9699}{0.9699} \\[/tex]

= 1

Units of Chlorine =  [tex]\frac{1.9434}{0.9699} \\[/tex]

= 2

Empirical formula = S₁Cl₂ = SCl₂

Answer: 9.56

Explanation:

First we need to know the moles of each species involved:

NH3: 50.00* 10^-3* 0.10 M =5.0* 10^-3

NH4Cl: 25.00* 10^-3* 0.10 M = 2.5 * 10^-3

Then we calculate the molarities:

5.0* 10^-3 moles/75 * 10^-3 L =0.67

2.5 * 10^-3 moles/75 * 10^-3 L =0.33

Ka and Kb are related by:

pKa = 14 - pKb

If Kb is 1.8 *10^-5

then pKb is - log (1.8 *10^-5)=4.74

Therefore

pKa =14 -4.74 =9.26

Using the Henderson-Hasselbalch (H-H) equation,

pH = pKa + log (0.67)/(0.33) = 9.26+ log 2 =9.26 + 0.30 = 9.56

Final answer:

To calculate the pH of a buffer solution prepared with NH₃ and NH₄Cl, use the Henderson-Hasselbalch equation after determining moles and volume. The pH calculated for the given condition is 8.96, which would be presented as 8.96 E0 in scientific notation.

Explanation:

The pH of a solution prepared by mixing 50.00 mL of 0.10 M NH₃ with 25.00 mL of 0.10 M NH₄Cl can be found using a buffer solution calculation. First, calculate the moles of NH₃ and NH₄⁺ introduced:

NH₃: 0.10 M × 0.0500 L = 0.0050 moles

NH₄⁺ (from NH₄Cl): 0.10 M × 0.0250 L = 0.00250 moles

Since the volume is additive, the total volume is 0.0750 L. The NH₄⁺/NH₃ ratio remains the same, and we can use the Henderson-Hasselbalch equation to find pH:

pH = pKa + log({[NH₄⁺]}/{[NH₃]})

Ka for NH₃ is the ionization constant of NH₄⁺, which is Kw/Kb (Kw = 1.0 × 10⁻¹⁴), giving us:

Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻µ) = 5.56 × 10⁻¹³

Hence, pKa = -log(Ka) = -log(5.56 × 10⁻¹³) = 9.26.

Using the Henderson-Hasselbalch equation:

pH = 9.26 + log(0.00250/0.0050) = 9.26 + log(0.5) = 9.26 - 0.301 = 8.959

Thus, the pH of the solution is 8.96 when rounded to two decimal places, which is typically presented in scientific notation as 8.96 E0.

Answer:

The quantity of chlorine gas left, in grams, after the reaction has occurred is 156.2 g

Explanation:

2Al (s)  +  3Cl₂ (g)  →  2AlCl₃ (s)

First step: We should know the moles we have, of each reactant.

Mass / Molar weight = Moles

Moles Cl₂ : 235g / 70.9 g/m = 3.31 moles

Moles Al : 19.9 g/ 26.98 g/m = 0.73 moles

Now the equation:

2 moles of Al ___ react ___ 3 moles Cl₂

0.73 moles of Al ___ react ___ 1.10 moles Cl₂

(0.73 .3) / 2 = 1.10

I have 3.31 moles of Cl₂ and I only need 1.10 moles to complet the total reaction of Al.

3.31 moles - 1.10 moles = 2.21

These are the moles that remain to react.

Moles . molar weight = mass

2.21 moles . 70.9 g/m = 156.2 g

Answer:

The answer is A. Newton's First law

Explanation:

Newton's First Law states that an object will stay at rest if it's at rest and an object in motion will stay in motion unless another object comes.

Hope this helps : )

The relations among the forces acting on a body and the motion of the body is first formulated by physicist Isaac Newton. An object which is at rest resume at rest whereas an object in motion carry on with motion is given by Newton's First Law. The correct option is A.

The Newton's First Law is also called the Law of inertia which states that if a body at rest or moving at a constant speed, it will remain at rest or keep its motion in a straight line unless it is acted upon by a force.

As long as the forces are not unbalanced, which means as long as the forces are balanced, the first law of motion is obeyed. When we shake the branch of a mango tree, the mangoes fall, which is an example of the inertia of rest.

During the breaking of a bus or train immediately, the passengers who are sitting lean forward. This denotes the inertia of motion.

Thus the correct option is A.

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To calculate the mass of the product formed in each reaction, use stoichiometry and the given mass of the reactant. Balance the chemical equation, calculate the molar mass of the reactant and product, and use the mole ratio to convert the given mass of the reactant to the mass of the product.

To calculate the mass of the product formed in each reaction, we need to use stoichiometry and the given mass of the reactant. The first step is to balance the chemical equation, and then use the balanced equation to calculate the molar mass of the reactant and the product. Finally, we can use the mole ratio to convert the given mass of the reactant to the mass of the product.

For example, let's consider reaction (a). The balanced equation is 2K(s) + Cl2(g) → 2KCl(s). The molar mass of KCl is 74.55 g/mol. We can calculate the number of moles of KCl using the given mass of K, and then use the mole ratio to determine the mass of KCl formed.

Keywords: calculate, reactions, mass, grams, product formed, reactant, balanced equation, stoichiometry, molar mass, mole ratio

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The mass of products formed from 15.16 g of the underlined reactant was calculated using molar masses and stoichiometry. For KCl, 28.91 g is formed; for KBr, 46.13 g is formed; for Cr₂O₃, 22.14 g is formed; and for SrO, 17.93 g is formed.

Calculating Mass of Products from Reactions

We need to calculate the mass of the products formed when 15.16 g of the underlined reactant completely reacts. First, we'll convert the mass of the underlined reactant to moles. Then, use the stoichiometry of the balanced reactions to find the moles of the products and finally convert this to mass.

Part a)

Reaction:

2K(s) + Cl₂(g) → 2KCl(s)

Stoichiometry:

Part b)

Reaction:

Stoichiometry:

Part c)

Reaction:

4Cr(s) + 3O₂(g) → 2Cr₂O₃(s)

Stoichiometry:

Part d)

Reaction: 2Sr(s) + O₂(g) → 2SrO(s)

Stoichiometry:

The empirical formula of the hydrocarbon is C2H3. The molar mass of the hydrocarbon is about nine times the molar mass of the empirical formula, thus the molecular formula of the hydrocarbon is C18H27.

First off, we need to find the empirical formula using the given data. The molar mass for carbon (C) is 12 g/mol and for oxygen (O) is 16 g/mol. Since the combustion of the hydrocarbon produced 440 mg of CO2, which contains 12/44 of carbon, we multiply 440 mg by 12/44 to get 120 mg of carbon.

The molar mass for hydrogen (H) is 1 g/mol. The combustion of the hydrocarbon also produced 135 mg of H2O, which contains 2/18 of hydrogen, so we multiply 135 mg by 2/18 to get 15 mg of hydrogen.

The empirical formula mass of carbon (120 mg/12 g/mol = 10 mmol) and hydrogen (15 mg/1 g/mol= 15 mmol) gives us CH1.5. But because we can't have fractional moles of atoms in a molecule, we multiply by 2 to get C2H3.

Therefore, the empirical formula is C2H3.

The molecular formula is a multiple of the empirical formula. The molar mass of C2H3 is 29 g/mol. As the molar mass of the hydrocarbon is given as 270 g/mol, we find that this is approximately 270/29 = around 9 times the empirical formula mass. So we multiply C2H3 by this to give the molecular formula: C18H27.

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The molecular formula is 18 times the empirical formula:

[tex]\[ \text{Molecular formula} = (\text{CH}_3)_{18} = \text{C}_{18}\text{H}_{54}[/tex]

To determine the molecular formula of the hydrocarbon, we need to first find the empirical formula, which is the simplest whole-number ratio of atoms in the compound. Then, using the molar mass, we can find the molecular formula.

 Step 1: Calculate the moles of carbon and hydrogen in the sample.

 From the given information, we have:

- Mass of CO2 produced = 440 mg = 0.440 g

- Mass of H2O produced = 135 mg = 0.135 g

Using the molar masses of CO2 (44.01 g/mol) and H2O (18.015 g/mol), we can calculate the moles of carbon and hydrogen:

Moles of carbon (from CO2):

[tex]\[ n_C = \frac{\text{mass of CO2}}{\text{molar mass of CO2}} = \frac{0.440 \text{ g}}{44.01 \text{ g/mol}} \approx 0.0100 \text{ mol} \][/tex]

Moles of hydrogen (from H2O):

[tex]\[ n_H = \frac{\text{mass of H2O} \times 2}{\text{molar mass of H2O}} = \frac{0.135 \text{ g} \times 2}{18.015 \text{ g/mol}} \approx 0.0150 \text{ mol} \][/tex]

 Step 2: Determine the simplest whole-number ratio of moles of carbon to hydrogen.

 The ratio of moles of hydrogen to moles of carbon is:

[tex]\[ \frac{n_H}{n_C} = \frac{0.0150}{0.0100} = \frac{3}{2} \][/tex]

This means there are 3 moles of hydrogen for every 2 moles of carbon. Therefore, the empirical formula of the hydrocarbon is CH3 (methane).

Step 3: Calculate the molar mass of the empirical formula.

 The molar mass of CH3 is:

[tex]\[ \text{Molar mass of CH}_3 = \text{molar mass of C} + 3 \times \text{molar mass of H} \][/tex]

[tex]\[ \text{Molar mass of CH}_3 = 12.01 \text{ g/mol} + 3 \times 1.008 \text{ g/mol} \approx 15.024 \text{ g/mol} \][/tex]

Step 4: Determine the molecular formula.

The given molar mass of the hydrocarbon sample is 270 g/mol. We can find the molecular formula by comparing the empirical formula molar mass to the given molar mass:

[tex]\[ \text{Molecular formula molar mass} = n \times \text{Empirical formula molar mass} \][/tex]

[tex]\[ 270 \text{ g/mol} = n \times 15.024 \text{ g/mol} \][/tex]

[tex]\[ n = \frac{270}{15.024} \approx 18 \][/tex]

 Therefore, the molecular formula is 18 times the empirical formula:

[tex]\[ \text{Molecular formula} = (\text{CH}_3)_{18} = \text{C}_{18}\text{H}_{54}[/tex]

The final answer is the molecular formula of the hydrocarbon, which is [tex]\(\boxed{\text{C}_{18}\text{H}_{54}}\).[/tex]

Answer:

a. Yes

b. 143.5 mmHg

Explanation:

The vapor pressure is the pressure of the vapor that is in equilibrium with the liquid. At a constant temperature, some molecules of the liquid will vaporize, and then will do pressure at the surface of the liquid.

If the pressure at the container is higher then the vapor pressure, the liquid will evaporate.

a. Let's calculate the pressure at the container by the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (62,364 mmHg.mL/mol.K), and T is the temperature.

The molar mass of CH₃I is 142 g/mol

n = mass/ molar mass

n = 0.453/142

n = 0.0032 mol

P*370 = 0.0032*62,364*266

370P = 53,084.24

P = 143.5 mmHg

So, all the liquid will evaporate.

b. Because all liquid evaporates, when the equilibrium is reached, the pressure is the gas pressure: 143.5 mmHg.

Final answer:

All of the liquid iodomethane will evaporate in the closed, evacuated container set at a constant temperature of 266 K, and the pressure in the container when equilibrium is reached will be 100 mm Hg, corresponding to the vapor pressure of iodomethane at this temperature.

Explanation:

To determine if all of the liquid will evaporate and the pressure at equilibrium, we can apply knowledge of vapor pressure and the ideal gas law.

a. Yes, all of the liquid iodomethane will evaporate. Given that the temperature remains constant at 266 K and the container is sealed and evacuated, the liquid will evaporate until the vapor pressure reaches 100 mm Hg, the equilibrium vapor pressure of iodomethane at this temperature.

b. Pressure at Equilibrium: The pressure in the container when equilibrium is reached will be 100 mm Hg. This is because the vapor pressure of a liquid is independent of the amount of liquid as long as some liquid is present and the system has reached equilibrium. The vapor pressure is solely dependent on temperature, and since the container's temperature is fixed at 266 K, the vapor pressure inside it will stabilize at the given vapor pressure of iodomethane which is 100 mm Hg.

Answer:

[tex]\DeltaG=158235.4 J[/tex]

Explanation:

The free-energy equation for a redox reaction is:

[tex]\DeltaG=-n*F*E[/tex]

Where:

In this case:

[tex]n=2 mol[/tex]

[tex]F=96485 C/mol[/tex]

[tex]E=0.82 V[/tex]

So:

[tex]\DeltaG=-2mol*96485 C/mol*0.82 V[/tex]

[tex]\DeltaG=158235.4 J[/tex]

Taking into account the reaction stoichiometry, ideal gas law and the definition of STP conditions, 13.70 L of O₂ (at STP) is formed when 50.0 g of KClO₃ decomposes according to the following reaction.

The balanced reaction is:

2 KClO₃ → 2 KCl + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass for KClO₃ is 122.55 g/mol. Then, by reaction stoichiometry, the following mass quantity of the compound participates in the reaction:

[tex]2 molesx\frac{122.55 g}{1 mole} =[/tex]245.1 grams of KClO₃

Then it is possible to apply the following rule of three: if by reaction stoichiometry 245.1 grams of KClO₃ form 3 moles of O₂, 50 grams of KClO₃ form how many moles of O₂?

[tex]amount of moles of O_{2} =\frac{50 grams of KClO_{3} x3 moles of O_{2} }{245.1grams of KClO_{3}}[/tex]

amount of moles of O₂= 0.612 moles

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

On the other side, the STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases.

Then, in this case you know:

Replacing in the ideal gas:

1 atm× V= 0.612 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 273 K

Solving:

[tex]V=\frac{0.612 molesx 0.082\frac{atmL}{molK} x 273 K}{1 atm}[/tex]

V=13.70 L

Finally, 13.70 L of O₂ (at STP) is formed when 50.0 g of KClO₃ decomposes according to the following reaction.

Learn more:

STP conditions refer to the standard temperature and pressure. The volume of the oxygen at STP will be 13.70 L when 50 grams of potassium chlorate decomposes.

The chemical equation of decomposition if potassium chlorate can be written as:

[tex]\text{KClO}_3\;\rightarrow\; 2\;\text{KCl + 3\; O}_2[/tex]

From the stoichiometry of the reaction, it can be observed as:

Now, calculating the mass of potassium chlorate participating in the reaction:

Now, substituting the values to calculate the moles of oxygen:

[tex]\text{amount of moles oxygen}&=\dfrac{50\;\text g} {245.1 \;\text g} \times 3\;\text{moles of oxygen}[/tex]

From the ideal gas equation:

P is pressure, Tis  the temperature, and the volume, V. R is ideal gas constant.

Substituting the values:

Therefore, the 13.70 L of oxygen at STP will be formed when 50 grams of potassium chlorate decomposes into potassium and oxygen.

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Answer:

2.75g

Explanation:

Firstly, we write a balanced equation of the reaction.

Fe2O3(s)+3CO(g)-->2Fe(s)+3CO2(g).

Now, we write the formula for the percentage yield.

Percentage yield = Actual yield/theoretical yield * 100%

From the equation, we can see that we have one unknown

We convert the 5kg to g. It must be known that 5kg is 5000g as 1kg equals 1000g.

Now, we are asked to calculate the mass of the iron iii oxide required.

From the balanced equation we can see that 1 mole of iron trioxide yielded 2 moles of Fe.

Now, let’s calculate the mass required. We calculate the actual number of moles of iron produced. This is the mass of iron divided by the molar mass. The atomic mass of iron is 56 amu. The number of moles of iron produced is 5000/56 moles

Since the mole ratio is 1 to 2, the number of moles of iron trioxide thus used is 5000/56 divided by 2 which equals 5000/112 moles.

Now, we proceed to calculate the mass of iron trioxide used. The mass is equal the molar mass multiplied by the number of moles. The molar mass of iron trioxide is 2(56) + 3(16) = 112 + 48 = 160g/mol

The mass thus required to produce 5kg of iron is 5000/112 * 160 which is 3,500g or 3.5kg. We know this to be the theoretical mass, the actual mass is calculated using the formula given above.

78.5 = actual mass/3.5 * 100

Actual mass = (3.5 * 78.5) /100

= 2.75g

Final answer:

To determine the enthalpy of neutralization, we can use the equation q = mcΔT to calculate the heat absorbed or released by the solution. Then, we can use the molar ratios from the balanced equation to calculate the enthalpy of neutralization.

Explanation:

From the given information, we can calculate the enthalpy of neutralization using the equation q = mcΔT. The heat absorbed or released by the reaction is equal to the heat absorbed or released by the solution. First, we need to calculate the heat absorbed or released by the solution using the formula q = mcΔT, where q is the heat, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature. Then, we can use the molar ratios from the balanced equation to calculate the enthalpy of neutralization.

Answer:

p1/T1=p2/T2

760mmHg/212°F=731mmHg/T2

T2= 203.91°F

760mmHg/100°C=731mmHg/T2

T2= 96.18°C

Explanation:

You'd have to choose in which units you want to express the temperature.

Answer:

Explanation:

The gaseous diffusion process utilizes uranium hexafluoride, UF₆, because although it is a solid at room temperature it is easily vaporized. [1] UF6 is not only convenient for its volatility, but also due to the fact that fluorine only consists of the isotope ¹⁹F, meaning the difference in molecular weights for UF6 are purely reliant on 235U and 238U.Here arises another problem however, for the masses of the two uranium isotopes are so nearly equal there is very little separation of 235UF6 and 238UF6 with one pass through a diffuser.Therefore a cascade process is needed to obtain any measurable amount of enrichment. In a cascade the feed stream at diffuser 1 is the UF6 prior to enrichment (meaning it will contain 0.711% 235U and 99.289% 238U) and marks the start of the cascade. There will be hundreds to thousands of diffusers on the upward or enriching side as well as on the downward or depleted side. The slightly enriched UF6 is sent up the cascade process to the next diffuser where it will be enriched again. The slightly depleted UF6 will be sent downward through the cascade where it will also be enriched again. In this way, the enriched uranium keeps getting enriched and sent onward, and the depleted uranium also gets enriched and sent onward. The depleted uranium always gets sent downward where it will eventually be ejected from the downward stream as depleted uranium.

Answer:

a. 7,24g of CsBr

b. 4,63g of CaSO₄

c. 5,57g of  Na₃PO₄

d. 7,82g of Li₂Cr₂O₇

e. 5,65g of K₂C₂O₄

Explanation:

To make 3,40x10² mL of a 0,100M solution you need:

3,40x10² mL × 0,100 mmol/mL = 34mmol ≡ 0,034moles of solute.

a. 0,034 moles of CsBr are:

0,034 mol CsBr×[tex]\frac{212,81g}{1mol}[/tex] = 7,24g of CsBr -Where 212,81g/mol is molar mass of CsBr-

b. 0,034 moles of CaSO₄ are:

0,034 mol CaSO₄×[tex]\frac{136,14g}{1mol}[/tex] = 4,63g of CaSO₄

c. 0,034 moles of Na₃PO₄ are:

0,034 mol Na₃PO₄×[tex]\frac{163,94g}{1mol}[/tex] = 5,57g of  Na₃PO₄

d. 0,034 moles of Li₂Cr₂O₇ are:

0,034 mol Li₂Cr₂O₇×[tex]\frac{229,87g}{1mol}[/tex] = 7,82g of Li₂Cr₂O₇

e. 0,034 moles of K₂C₂O₄ are:

0,034 mol K₂C₂O₄×[tex]\frac{166,2g}{1mol}[/tex] = 5,65g of K₂C₂O₄

I hope it helps!

To prepare a 0.100 M solution, the weight of each solute needed varies with its molar mass. For instance, you need 72.35 g of Cesium Bromide, 46.29 g of Calcium Sulfate, 55.74 g of Sodium Phosphate, 98.94 g of Lithium Dichromate, and 56.51 g of Potassium Oxalate.

To determine how many grams of each solute is needed, you need to know the molar mass of each solute. The formula to use is M=n/V, where M is molarity, n is moles and V is volume in liters.

(a) The molar mass of cesium bromide (CsBr) is 212.81 g/mol. Using the equation, you would require 0.100 M x 212.81 g/mol x 3.40 L = 72.35 g of CsBr.

(b) The molar mass of calcium sulfate (CaSO4) is 136.14 g/mol. Therefore, you would need 0.100 M x 136.14 g/mol x 3.40 L = 46.29 g of CaSO4.

(c) The molar mass of Sodium Phosphate (Na3PO4) is 163.94 g/mol. Thus you require 0.100 M x 163.94 g/mol x 3.40 L = 55.74 g of Na3PO4.

(d) The molar mass of Lithium Dichromate (Li2Cr2O7) is 291.0 g/mol. Hence, you would need 0.100 M x 291.0 g/mol x 3.40 L = 98.94 g of Li2Cr2O7.

(e) The molar mass of Potassium Oxalate (K2C2O4) is 166.22 g/mol. Hence, you need 0.100 M x 166.22 g/mol x 3.40 L = 56.51 g of K2C2O4.

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The values of ΔH and ΔE for the combustion of one mole of butane are 2658 kJ and 2655 kJ, respectively.

Part A:

The values of ΔH and ΔE for the combustion of one mole of butane can be determined using the first law of thermodynamics:

ΔH = q + w

where ΔH is the change in enthalpy, q is the heat released or absorbed, and w is the work done during the reaction.

From the given information, we know that 1 mole of butane produces 2658 kJ of heat and does 3 kJ of work. Therefore, ΔH = 2658 kJ and ΔE = ΔH - w = 2658 kJ - 3 kJ = 2655 kJ.

Part B:

Expressing the answers using four significant figures, we have ΔH = 2658 kJ and ΔE = 2655 kJ.

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The change in enthalpy (∆H) for the combustion of one mole of butane is -2658 kJ and the change in internal energy (∆E) is -2655 kJ, both values are expressed with four significant figures.

In thermodynamics, the heat transfer at constant pressure is defined as the change in enthalpy (∆H), while the change in internal energy (∆E) is given by heat transfer minus work done. For the combustion of one mole of butane, given that the heat produced is 2658 kJ (which is released so it is -2658 kJ), and the work done is 3 kJ (the system does work so it's -3kJ).

For Part A, we can calculate these values as follows: ∆H = q_p = -2658 kJ and ∆E = ∆H - work = -2658 kJ - (-3 kJ) = -2655 kJ

For part B, using four significant figures, the values will be ∆H = -2658 kJ and ∆E = -2655 kJ

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Answer: The percent yield of silver is 86.96 %

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     .....(1)

Given mass of copper = 40.0 g

Molar mass of copper = 63.55 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of copper}=\frac{40.0g}{63.55g/mol}=0.629mol[/tex]

The given chemical equation follows:

[tex]Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag[/tex]

By Stoichiometry of the reaction:

1 mole of copper produces 2 moles of silver

So, 0.629 moles of copper will produce = [tex]\frac{2}{1}\times 0.629=1.258mol[/tex] of silver

Now, calculating the mass of silver from equation 1, we get:

Molar mass of silver = 107.9 g/mol

Moles of silver = 1.258 moles

Putting values in equation 1, we get:

[tex]1.258mol=\frac{\text{Mass of silver}}{107.9g/mol}\\\\\text{Mass of silver}=(1.285mol\times 107.9g/mol)=135.7g[/tex]

To calculate the percentage yield of silver, we use the equation:

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield of silver = 118 g

Theoretical yield of silver = 135.7 g

Putting values in above equation, we get:

[tex]\%\text{ yield of silver}=\frac{118g}{135.7g}\times 100\\\\\% \text{yield of silver}=86.96\%[/tex]

Hence, the percent yield of silver is 86.96 %

The percent yield of silver when 40.0 g of copper react with silver nitrate, producing 118 g of silver, is calculated to be 86.9%.

When 40.0 g of copper (Cu) are reacted with silver nitrate (AgNO3) to produce 118 g of silver (Ag), the percent yield of silver can be calculated using the actual yield and the theoretical yield. The balanced chemical equation for the reaction is: Cu (s) + 2 AgNO3(aq) → Cu(NO3)2 (aq) + 2 Ag (s).

First, we calculate the theoretical yield of silver that should have been produced from 40.0 g of copper. The molar mass of copper is 63.55 g/mol, so 40.0 g of copper is equal to 40.0 g / 63.55 g/mol = 0.629 mol of copper. According to the balanced equation, 1 mol of copper will produce 2 mol of silver. Hence, 0.629 mol of copper will produce 0.629 × 2 = 1.258 mol of silver. With the molar mass of silver being 107.9 g/mol, the theoretical yield of silver is 1.258 mol × 107.9 g/mol = 135.8 g.

To find the percent yield, divide the actual yield (118 g of Ag) by the theoretical yield and multiply by 100:
Percent yield = (118 g / 135.8 g) × 100 = 86.9%

Answer: The process is endothermic and the heat of fusion is positive

Explanation:

All solids absorb heat as they melt to become liquids. The gain of heat in this endothermic process goes into changing the state rather than changing the temperature

Answer:

The electron domain and molecular geometry of NF₃ are option b

Explanation:

First of all, think the total valence electrons of each element.

N : 5

F: 7, but we have 3, so 7.3 = 21

Total: 26 e⁻

Now let's draw a scheme

             F   ----  N  ----  F

                          |

                         F

We have to put 26 e⁻ around the elements, so 6 e⁻ are been used for the bonds N-F, so now, there are 20 e⁻  to add. As the F, have used 1 e⁻ to form the bond, there are 6 more, around. Each fluorine which has 7 e⁻,  shares 1e⁻ with the N, to complete the octet rule.

So, there are still 2e⁻ to add. They will be above the N and they are free. That's why the molecular geometric is trigonal, and the electronic geeometry is tetrahedral because the non-binding electron pair of N

NF₃ has a tetrahedral electron domain geometry and trigonal pyramidal molecular geometry, due to the presence of one lone pair on the central nitrogen atom.

The electron domain and molecular geometry of NF₃ can be described as tetrahedral, trigonal. The electron domain refers to the number of electron pairs around the central atom, while the molecular geometry describes the arrangement of the bonded atoms. In the case of NF₃, the central atom is N, which has four electron domain pairs, giving it a tetrahedral electron domain geometry. However, due to the presence of one lone pair on N, the molecular geometry becomes trigonal pyramidal, with bond angles less than 109.5 degrees.

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Answer:

1) Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.

2) Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.

Explanation:

[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g) ,\Delta H_{rxn} =-802.3 kJ[/tex]

1) Minimum mass of  methane required to raise the temperature of water by 21.0°C.

Mass of water = m = 45.0 g

Specific heat capacity of water = c = 4.18 J/g°C

Change in temperature of water = ΔT = 21.0°C.

Heat required to raise the temperature of water by 21.0°C = Q

[tex]Q=mc\Delta T= 45.0 g\times 4.18 J/g^oC\times 21.0^oC[/tex]

Q = 3,950.1 J = 3.9501 kJ

According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.

Then 3.950.1 kJ of heat will be given by:

[tex]=\frac{3.950.1 kJ}{802.3 kJ}=0.004923 mol[/tex]

Mass of 0.004923 moles of methane :

0.004923 mol × 16 g/mol=0.0788 g

Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.

2) Minimum mass of  methane required to raise the temperature of water by 26.0°C.

Mass of water = m = 50.0 g

Specific heat capacity of water = c = 4.18 J/g°C

Change in temperature of water = ΔT = 26.0°C.

Heat required to raise the temperature of water by 21.0°C = Q

[tex]Q=mc\Delta T= 50.0 g\times 4.18 J/g^oC\times 26.0^oC[/tex]

Q = 5,434 J= 5.434 kJ

According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.

Then 5.434 kJ of heat will be given by:

[tex]=\frac{5.434 kJ}{802.3 kJ}=0.006773 mol[/tex]

Mass of 0.006773 moles of methane :

0.006773 mol × 16 g/mol= 0.108 g

Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.

The minimum mass of methane required to heat up 45g by 21°C is 0.079g

Data;

From the chemical equation, we can say that each gram of water requires 4.18g of heat to increase by 1°C

[tex]45* 21 = 945J[/tex]

945J of heat is required to raise 45g of methane

One mole of methane weighs 16.04g/mol and burning one mole supplied 802.3kJ of heat.

The specific heat Q is given as

[tex]Q = _mC_pdt\\[/tex]

The heat added is the product of the specific heat, the mass and the change in temperature.

The heat added is

[tex]\delta H=4.18*45*(21-0)\\\delta H = 3950J = 3.950kJ[/tex]

The number of moles of methane needed is

[tex]\frac{\delta H}{\delta H_r} = \frac{3.950}{802.3} = 4.9234*10^-^3 mols[/tex]

The mass of methane needed will be

[tex]mass = 4.9234*10^-^3 * 16.04 = 0.079g[/tex]

The minimum mass of methane required to heat up 45g by 21°C is 0.079g

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Answer:

1. lead

2. tropospheric ozone

Explanation:

1. A gasoline engine mixes the air with a small amount of gasoline, is compressed, and detonated with a small spark. The more the mixture is compressed, the more power is achieved, but if the mixture is compressed too much it detonates alone, the piston trips and effectiveness is lost.

Therefore, it was decided to incorporate an antidetonating additive to gasoline, thus allowing greater compression and preventing the mixture from detonating alone. These additives (Pb (CH3) 4) originally contained lead.

With the passage of time it was discovered that this solution was not the most appropriate. On the one hand there was the problem that this lead containing gasoline is expelled through the exhaust pipe with the rest of the gases, and this pollutes the environment, and on the other hand the lead also "poisoned" and rendered catalysts useless, when these began to be used.

These emissions generated the most important source of lead contamination.

To resolve this situation (eliminating the emission of lead, and being able to continue applying catalysts to reduce the impact of the other substances expelled) around the 1980s, it was necessary to develop new types of fuels, using other types of lead-free anti-knock additives, such as methyl t-butyl ether (MTBE). This was the origin of unleaded gasoline.

2.Tropospheric ozone (not to be confused with the stratospheric, whose layer protects the Earth from solar radiation) is a secondary pollutant, that is, it is produced from other pollutants emitted by cars or industry and, in addition, several kilometers where they occur

Its effects on health depends on your level of concentration. From 180 micrograms per cubic meter (the level of information), certain people - especially asthmatics and those with respiratory problems - could see their ailments increased.

Answer:

The molarity of Z in this solution is 1.80 mol/L

Explanation:

29% (by mass) means that in 100 g of solution, you have 29 grams of solute.

So, by the solution density we can find our volume.

Density = mass / volume

Solution density = mass solution / volume solution

1.5 g/ml = 100 g / volume solution

volume solution = 100 g/ 1..5 g/ml

volume solution = 66.6 mL

So we have just found out that in 66.6 mL, we have 29 g of solute Z, so if we want to know the molarity, we have to find out the mass in 1 L, and afterwards, calculate the moles, we have

1000 mL = 1L

66.6 mL ____ 29 g of solute Z

1000 mL ____ (29 g . 1000 mL)/ 66.6 mL = 435.4 g

With the molar mass (formula weight), we can find out the moles

Mass /Molar mass = Mol

435.4 g /242 g/m = 1.80 mol

Molarity is 1.80 mol/L

The molarity of solute Z in the solution is 1.797 M.

To calculate the molarity of solute Z in the solution, we need to use the given information. The solution contains 29% (by mass) of solute Z, which has a formula weight of 242 g/mol. The density of the solution is 1.5 g/mL.

First, we need to calculate the mass of solute Z in the solution. Assuming we have 100 g of the solution, the mass of solute Z would be 29 g (29% of 100 g).

Next, we can calculate the volume of the solution using the density. Since the density is 1.5 g/mL, the mass of the solution would be 100 g / 1.5 g/mL = 66.67 mL.

Finally, we can calculate the molarity of solute Z by converting the mass of solute Z to moles and dividing by the volume of the solution in liters. The molar mass of solute Z is 242 g/mol, so the moles of solute Z would be 29 g / 242 g/mol = 0.1198 mol. The volume of the solution is 66.67 mL = 0.06667 L. Therefore, the molarity of solute Z in the solution is 0.1198 mol / 0.06667 L = 1.797 M.

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The reason and what will happen if one atom in the given type of covalent bond reaction can win the tug of war is explained below.

      In chemistry, tug of war is usually used in polar covalent bond. Polar covalent bond is the bond that occurs as a result of unequal sharing of electrons between 2 atoms.

    Now, the reason why polar covalent bond is referred to as tug of war is because in tug of war games, the person that is the strongest usually wins. Likewise in polar covalent bonds, between the 2 sharing atoms, the atom that has higher electronegativity will have a stronger pull for electrons and as a result of this, it will have the shared electrons closer to it.

     Therefore, the molecule of the atom with higher electronegativity will become  negative because it has drawn the electrons closer to itself while the molecule of the atom that has the lesser electronegativity will become positive.

      In conclusion, Yes one atom in polar covalent bond reaction can win the ''tug of war''.

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For the given reaction, the enthalpy change calculated with bond energies is 36 kJ/mol.

Given the following balanced reaction:

H₂(g) + CO₂(g) → H₂O(g) + CO(g)     (1)

The enthalpy change ([tex]\Delta H_{rxn}[/tex]) can be calculated with the next equation:

[tex] \Delta H_{rxn} = \Sigma (n*E_{r}) - \Sigma (m*E_{p}) [/tex]

Where:

n and m are the stoichiometric number of moles of the reactants and the products, respectively          

[tex]E_{r}[/tex] and [tex]E_{p}[/tex] are the bond energies of the reactants and the products, respectively

For reaction (1), the enthalpy change is given by:

[tex]\Delta H_{rxn} = \Sigma (n_{H_{2}}*E_{H-H} + n_{CO_{2}}*2*E_{C=O}) - \Sigma (m_{H_{2}O}*2*E_{H-O} + m_{CO}*E_{C\equiv O})[/tex]

[tex] \Delta H_{rxn} = \Sigma (1*436 kJ/mol + 1*2*799 kJ/mol) - \Sigma (1*2*463 kJ/mol} + 1*1072 kJ/mol) = 36 kJ/mol [/tex]                        

Therefore, the enthalpy change for the given reaction is 36 kJ/mol.

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The enthalpy change for the reaction H2(g) + CO2(g) ⟶ H2O(g) + CO(g) can be calculated by subtracting the sum of bond energies of bonds formed from the sum of bond energies of bonds broken in the reactants and products. Specific bond energies are needed from a bond enthalpy table to complete the calculation.

To calculate the enthalpy change for the reaction H2(g) + CO2(g) ⟶ H2O(g) + CO(g), you have to consider the bonds being broken in the reactants and the bonds being formed in the products. This involves using bond energies for each type of bond involved in the reaction:

The general approach to calculate the enthalpy change (ΔH) is:

ΔH = Sum of bond energies of bonds broken - Sum of bond energies of bonds formed

However, to provide a precise answer, the specific bond energies for each bond must be known, which are not provided in the question. Typically, you can find these values in a table of average bond enthalpies. For the sake of this example, if we had the bond energies for H-H, O=O, O-H, and C=O, we would use the following formula:

ΔH = (Bond energy of H-H + Bond energy of O=O) - (2 x Bond energy of O-H + Bond energy of C=O)

Since we don't have the actual values, you would need to look up the specific bond energies and plug them into the formula to get the enthalpy change for this reaction.

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The rate of appearance of Br₂ at the same instant is 6.0 x 10⁻³ mol/L • s.

The given information, involving the rate of disappearance of Br⁻ and its relationship to the formation of Br₂, provides crucial insights into the reaction kinetics. The rate of disappearance, specified as 2.0 x 10⁻³ mol/L • s, plays a significant role in determining the rate of appearance of Br₂. This correlation is derived from the balanced chemical equation, which illustrates that for every mole of Br⁻ consumed, three moles of Br₂ are generated.

Hence, the rate of appearance of Br₂ at the same instant is logically calculated as three times the rate of Br⁻ disappearance, resulting in a rate of 6.0 x 10⁻³ mol/L • s. These quantitative relationships offer valuable insights into reaction mechanisms and enable the precise monitoring of reactant and product concentrations in chemical reactions, aiding in the study and application of kinetics.

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The rate of appearance of Br is 1.0 x 10 mol/L

To determine the rate of appearance of Br we need to look at the stoichiometry of the balanced chemical equation:

[tex]\[ BrO^- + 5Br^- + 6H^+ \rightarrow 3Br_2 + 3H_2O \][/tex]

 From the stoichiometry, we can see that for every 5 moles of Brâ» that disappear, 3 moles of Brâ‚‚ appear. The rate of disappearance of Brâ» is given as 2.0 x 10â»Â³ mol/L • s. To find the rate of appearance of Brâ‚‚, we use the stoichiometric ratio of the products to the reactants:

[tex]\[ \text{Rate of appearance of Br}_2 = \left( \frac{3 \text{ moles of Br}_2}{5 \text{ moles of Br}^-} \right) \times \text{Rate of disappearance of Br}^- \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = \left( \frac{3}{5} \right) \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = \frac{3}{5} \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 0.6 \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 1.2 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

 [tex]\[ \text{Rate of appearance of Br}_2 = \frac{3}{5} \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 0.6 \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \[/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 1.2 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

 This is the correct rate of appearance of Br. However, upon reviewing the initial question and the balanced equation, it is clear that the rate of disappearance of B 2.0 x 10³ mol/L  and the stoichiometry dictates that for every 5 moles of Br that react, 3 moles of Br are produced. Therefore, the rate of appearance of Br should be calculated as:

[tex]\[ \text{Rate of appearance of Br}_2 = \frac{3}{5} \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 0.6 \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 1.2 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

 This calculation is still incorrect because the stoichiometric ratio was not correctly simplified. The correct simplification of the stoichiometric ratio is 3/5, not 0.6. Therefore, the correct rate of appearance of Br‚ is:

[tex]\[ \text{Rate of appearance of Br}_2 = \frac{3}{5} \times 2.0 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = \frac{3 \times 2.0}{5} \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = \frac{6.0}{5} \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \text{Rate of appearance of Br}_2 = 1.2 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

 This is the correct rate of appearance of Br. However, the final answer should be simplified to one decimal place, as that is the precision given in the rate of disappearance of Br. Therefore, the final answer is:

[tex]\[ \text{Rate of appearance of Br}_2 = 1.2 \times 10^{-3} \text{ mol/L} \cdot \text{s} \][/tex]

[tex]\[ \boxed{1.2 \times 10^{-3} \text{ mol/L} \cdot \text{s}} \][/tex]