Assume that Aluminum and Silver Sulfide are the starting substances (reactants) in the reaction: a. Write a balanced chemical equation describing the "re-creation" of silver, using the information in the case study. b. State the names of the products that are produced from this reaction. c. What type of reaction(s) is/are being represented by the chemical reaction you wrote in part (a)? d. Is the reaction in part (a) an oxidation-reduction (redox) reaction? e. If this is a redox reaction, then identify the following: What is undergoing oxidation (what is being oxidized)? Support your answer using oxidation numbers. What is undergoing reduction (what is being reduced)? Support your answer using oxidation numbers. What is the reducing agent? What is the oxidizing agent?

Answer:

Pb(NO₃)₂ (aq)  + 2NaCl (aq)  → PbCl₂ (s) ↓  + 2NaNO₃ (aq)

Explanation:

The reactants are:

Lead(II) nitrate → Pb(NO₃)₂ (aq)

Sodium chloride → NaCl (aq)

The products are:

Lead(II) chloride → PbCl₂ (s)

Sodium nitrate → NaNO₃ (aq)

Salts form nitrate are soluble. The chloride makes a precipitate with the Pb²⁺. The chemical equation for this reaction is:

Pb(NO₃)₂ (aq)  + 2NaCl (aq)  → PbCl₂ (s) ↓  + 2NaNO₃ (aq)

The balanced chemical equation for the reaction between aqueous lead(II) nitrate (Pb(NO3)2) and aqueous sodium chloride (NaCl) is:

Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq).

The balanced chemical equation for the reaction between aqueous lead(II) nitrate (Pb(NO3)2) and aqueous sodium chloride (NaCl) is as follows:

Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq)

In this equation:

- Pb(NO3)2(aq) represents aqueous lead(II) nitrate, which is a soluble compound in water.

- 2NaCl(aq) represents aqueous sodium chloride, which is also soluble in water.

- PbCl2(s) represents solid lead(II) chloride, which is insoluble and precipitates out of the solution.

- 2NaNO3(aq) represents aqueous sodium nitrate, another soluble compound.

This chemical equation illustrates a double displacement or metathesis reaction, where the cations (Pb²⁺ and Na⁺) and anions (NO3⁻ and Cl⁻) swap partners. As a result, a solid product, lead(II) chloride (PbCl2), forms as a precipitate because it is sparingly soluble in water. Sodium nitrate (NaNO3) remains in solution because it is highly soluble.

The equation is balanced because the number of atoms of each element on both sides of the reaction arrow is the same, ensuring the law of conservation of mass is obeyed.

This reaction can be used in laboratory settings to demonstrate the formation of a precipitate when solutions of lead(II) nitrate and sodium chloride are mixed. The lead(II) chloride formed can be separated from the remaining solution and collected as a solid.

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Final answer:

The density of the compound is 5.84 x 10^27 g/m3.

Explanation:

The density of a compound can be calculated using the formula      density = mass/volume. NaCl has a molar mass of 58.4 g/mol, which means that 1 mole of NaCl has a mass of 58.4 grams.

Given that the length of an edge of the unit cell is 461 pm = 461 x 10^-12 m, we can calculate the volume of the unit cell.

The volume of a cube (unit cell) is calculated using the formula volume = (length of edge)^3.�

So, the volume of the unit cell is (461 x 10^-12 m)^3 = 0.100 x 10^-27 m^3

Now, we can calculate the density using the formula density = mass/volume.

So, the density of the compound is 58.4 g/mol / 0.100 x 10^-27 m^3 = 5.84 x 10^27 g/m3

Therefore, the density of the compound with a molar mass of 120.0 g/mol and crystallizes with the sodium chloride structure is 5.84 x 10^27 g/m3.

Answer:

Explanation:

Given reaction

N₂H₄ + O₂ ⇒ N₂ + 2H₂O   ΔH₁ = -543 KJ ---------- ( 1 )

2H₂ + O₂ ⇒ 2H₂O              ΔH₂ = -484 KJ ---------- ( 2 )

N₂ + 3 H₂ ⇒ 2NH₃              ΔH₃ = -92 KJ  -----------( 3 )

( 1 ) -  ( 2 ) +( 3 )

N₂H₄ + O₂ - 2H₂ - O₂ +N₂ + 3 H₂ ⇒ N₂ + 2H₂O - 2H₂O +2NH₃  

                                                         ΔH =       -543 + 484 -92 = -151 KJ

     N₂H₄ + H₂ ⇒ 2NH₃     ΔH  = -151 KJ .

2NH₃ ⇒ N₂H₄ + H₂     ΔH  = + 151 KJ

Answer:

Explanation:

The reaction that takes place is:

With a percent yield of 87.5%.

To determine the limiting reactant, first we convert the masses of each reactant to moles, using their molar mass:

Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.

To calculate the mass of precipitate (PbCl₂) formed, we use the moles of the limiting reactant:

- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:

Now we convert that amount to moles of KCl and finally into grams of KCl:

3.86 g of KCl would react, so the amount remaining would be:

For the reaction between 9.82 g of lead(II) nitrate with 5.76 g of potassium chloride, we have:

a. The balanced chemical equation is:

Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

b. The limiting reactant is lead(II) nitrate (Pb(NO₃)₂).

c. The mass of the precipitate (PbCl₂) formed is 7.20 g.

d. The mass of the excess reactant (KCl) that remains in solution after the reaction is 1.35 grams.

a. The chemical reaction between lead(II) nitrate and potassium chloride is the following:

Pb(NO₃)₂(aq) + KCl(aq) → PbCl₂(s) + KNO₃(aq)   (1)

In this reaction, lead (II) chloride precipitates in the solution.

We need to balance equation (1). We can see that on the reactants side, we have 2 molecules of NO₃, and on the products side, we have one, so we need to add a coefficient of 2 before KNO₃  

Pb(NO₃)₂(aq) + KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

Now, we have 2 K atoms on the products, so we need to add a coefficient of 2 before KCl.

Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)   (2)

Reaction (2) is balanced now.

b. To find the limiting reactant we need to calculate the initial number of moles of Pb(NO₃)₂ and KCl

[tex] n_{Pb(NO_{3})_{2}}_{i} = \frac{m_{Pb(NO_{3})_{2}}}{M_{Pb(NO_{3})_{2}}} [/tex]

Where:

m: is the mass

M: is the molar mass

[tex] n_{Pb(NO_{3})_{2}}_{i} = \frac{9.82 g}{331.2 g/mol} = 0.0296 \:moles [/tex]

[tex] n_{KCl}_{i} = \frac{m_{KCl}}{M_{KCl}} = \frac{5.76 g}{74.5513 g/mol} = 0.0773 \:moles [/tex]

From reaction (2), we have that 1 mol of Pb(NO₃)₂ react with 2 moles of KCl so, the number of moles of Pb(NO₃)₂ needed to react with 0.0773 moles of KCl is:

[tex] n_{Pb(NO_{3})_{2}} = \frac{1 \:mol \:Pb(NO_{3})_{2}}{2\: mol \:KCl}*n_{KCl}_{i} = \frac{1 \:mol \:Pb(NO_{3})_{2}}{2\: mol \:KCl}*0.0773 \:moles    = 0.0387 \:moles [/tex]

Since we need 0.0387 moles of Pb(NO₃)₂ to react with KCl, and initially we have 0.0296 moles, the limiting reactant is Pb(NO₃)₂.

c. The mass of PbCl₂ can be calculated as follows.

From reaction (2) we have that 1 mol of Pb(NO₃)₂ produces 1 mol of PbCl₂, so:

[tex] n_{PbCl_{2}} = n_{Pb(NO_{3})_{2}}_{i} = 0.0296 \:moles [/tex]

[tex] m_{PbCl_{2}}_{t} = n_{PbCl_{2}}*M_{PbCl_{2}} = 0.0296 \:moles*278.1 g/mol = 8.23 g [/tex]

Since the percent yield is 87.5%, the mass of PbCl₂ formed (experimental mass) is:

[tex] m_{PbCl_{2}}_{e} = 8.23 g*\frac{87.5}{100} = 7.20 g [/tex]

Therefore, the mass of PbCl₂ formed is 7.20 g.

d. Mass of excess reactant

The mass of the excess reactant that remains in solution after the reaction is given by:

[tex] m_{KCl} = m_{KCl}_{i} - m_{KCl}_{r} [/tex]

The mass of KCl that reacts with Pb(NO₃)₂ ([tex]m_{KCl}_{r} [/tex]) can be calculated from the number of moles:

[tex] n_{KCl} = \frac{2\: mol \:KCl}{1 \:mol \:Pb(NO_{3})_{2}}*n_{Pb(NO_{3})_{2}}_{i} = \frac{2\: mol \:KCl}{1 \:mol \:Pb(NO_{3})_{2}}*0.0296 \:moles = 0.0592 \:moles [/tex]

[tex] m_{KCl}_{r} = n_{KCl}*M_{KCl} = 0.0592\:moles*74.5513 g/mol = 4.41 g [/tex]

Then, the mass of KCl that remains is:

[tex] m_{KCl} = m_{KCl}_{i} - m_{KCl}_{r} = 5.76 g - 4.41 g = 1.35 g [/tex]  

Hence, 1.35 grams of KCl remains in solution after the reaction.

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The balanced combustion reaction of octane (C₈H₁₈) shows that 1 mole of octane can produce 8 moles of  (CO₂). To give an exact amount of  (CO₂) produced globally per year, specific data of the amount of octane used is needed which the question does not provide.

The topic at hand pertains to combustion reactions, specifically focusing on the combustion of octane (C₈H₁₈), a major component of gasoline, and the resultant carbon dioxide (CO₂) produced.

Firstly, we start by writing down the balanced chemical equation for the combustion of octane:

2 C₈H₁₈ + 25 O₂ -> 16 CO₂ +18 H₂O

Through this equation, we can observe that from the combustion of 2 moles of octane, we yield 16 moles of carbon dioxide. Effectively, 1 mole of octane will produce 8 moles of CO₂. Using the molar mass of CO₂ (approximately 44.01 g/mol), we can calculate the mass of CO₂ produced from the combustion of 1 mol of octane.

However, to provide the actual quantity of CO₂ produced annually worldwide from fossil fuels, we would need to know the exact quantity of octane combusted yearly worldwide. This is data that is not provided in the question. It is implied that there's a mix of types of fossil fuels used globally as each specific type produces different quantities of CO₂ when combusted.

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Answer:

Explanation:

a. The reaction is exothermic, since it releases heat (300 KJ that flow out of the system).

b. The water bath will receive the 300 KJ of heat that the system has released, therefore, its temperature will rise.

c. The pressure of the system is constant (1 atm) and its temperature goes down after the reaction, so the volume will decrease (according tho Charles' Law). Therefore, the piston will go in.

d. Since the reaction is exothermic, it releases energy.

e. The reaction releases 300 KJ of energy in the form of heat.

Answer:

that are glucogenic all produce oxaloacetate and then glucose.

Explanation:

The major aim of protein catabolism during a state of starvation is to provide the glucogenic amino acids especially alanine and glutamine, that serve as substrates for endogenous glucose production ie gluconeogenesis in the liver.

Glucogenic amino acid are known for their production of oxaloacetate and then glucose.

Answer:

lignands, the central atom/metal ion

Explanation:

When a complex ion forms, ligands arrange themselves around , the central atom/metal ion creating a new ion.

This is defined as an atom or molecule which possesses an electric charge such as positive or negative.

Complex ion form as a result of ligands arrangement around central atom/metal ion with a charge equal to the sum of the charges of its components.

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CH3COOLi and KF are basic in a 0.10 M solution, while C2H5NHCl, KNO3, and KClO4 are neutral.

The salts CH3COOLi and KF will be basic in a 0.10 M solution. The salt CH3COOLi is basic because the anion CH3COO- slightly reacts with water to form the weak acid CH3COOH and OH-. The salt KF is basic because the K+ cation does not react with water and the F- anion does not react significantly with water, resulting in a solution with a pH greater than 7.

The salts C2H5NHCl, KNO3, and KClO4 will be neutral in a 0.10 M solution. Both C2H5NH+ cation and Cl- anion do not react significantly with water, resulting in a solution with a pH close to 7. K+ cation does not react with water and the NO3- anion does not react significantly with water, resulting in a neutral solution. Similarly, K+ cation does not react with water and the ClO4- anion does not react significantly with water, resulting in a neutral solution.

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Salts derived from strong bases and weak acids form basic solutions, while those from strong acids and weak bases form acidic solutions. Salts from both strong acids and bases form neutral solutions. CH3COOLi and KF are basic; C2H5NHCl is acidic; KNO3 and KClO4 are neutral.

Here we classify several salts as basic, acidic, or neutral based on their behavior in a 0.10 M solution:

Understanding the nature of the conjugate acids and bases helps determine whether the salt solution will be acidic, basic, or neutral.

Answer:

C. attaching an active metal to make the pipe the cathode in an electrochemical cell.

Explanation:

Cathodic protection is a technique which helps in controlling the increased rate of corrosion of a metal surface by making it the cathode of an electrochemical cell. It connects the metal to be protected to a more easily corroded which is usually referred to as the sacrificial metal to act as the anode.

This technique preserves the metal by providing a highly active metal that can act as an anode and provide free electrons. By introducing these free electrons, the active metal sacrifices its ions and keeps the less active steel from corroding.

Answer:

Excess reagent is the ZnS. After the reaction is complete, 1.67 moles of sulfide remain.

Explanation:

This is the reaction to work with:

2ZnS + 3O₂  →  2ZnO + 2SO₂

Limiting reagent is the oxygen. We confirm it by a rule of three

2 moles of sulfide can react to 3 moles of O₂

Therefore 3 moles of ZnS will react to (3 . 3) / 2 = 4.5 moles (we need 4.5 moles of O₂ and we only have 2 moles, that's why the O₂ is the limiting)

Excess reagent is the zinc (II) sulfide

3 moles of oxygen react to 2 moles of ZnS

2 moles of O₂ will react to (2 . 2) / 3 = 1.33 moles of ZnS (it is ok, be cause we have 3 moles, and we need only 1.33)

After the reaction is complete ( 3 - 1.33) = 1.67 moles of sulfide remain.

Answer:

Answer: Zinc(II)sulfide (ZnS) is in excess. there will remain 1.67 moles

Explanation:

Step 1: Data given

Zinc(II)sulfide = ZnS

oxygen = O2

Number of moles ZnS = 3.0 moles

Number of moles O2 = 2.0 moles

Step 2: The balanced equation

2ZnS + 3O2 → 2ZnO + 2SO2

Step 3: Calculate the limiting reactant

For 2 mles zinc(II) sulfide we need 3 moles oxygen to produce 2 moles zinc oxide and 2 moles sulfur dioxide

O2 is the limiting reactant. There willreact 2.0 moles.

ZnS is in excess. There will react 2/3*2.0 = 1.33 moles

There will remain 3.0 - 1.33 = 1.67 moles ZnS

Step 4: Calculate products

For 2 mles zinc(II) sulfide we need 3 moles oxygen to produce 2 moles zinc oxide and 2 moles sulfur dioxide

For 2.0 moles O2 we'll have 1.33 moles ZnO and 1.33 moles SO2

Answer: Zinc(II)sulfide (ZnS) is in excess. there will remain 1.67 moles

Answer:

Gases are easily compressed. We can see evidence of this in Table 1 in Thermal Expansion of Solids and Liquids, where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same β. This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.

The answer lies in the large separation of atoms and molecules in gases, compared to their sizes, as illustrated in Figure 2. Because atoms and molecules have large separations, forces between them can be ignored, except when they collide with each other during collisions. The motion of atoms and molecules (at temperatures well above the boiling temperature) is fast, such that the gas occupies all of the accessible volume and the expansion of gases is rapid. In contrast, in liquids and solids, atoms and molecules are closer together and are quite sensitive to the forces between them.

Answer:

Explanation:

Mass of ammonium chloride  = 11.5 gm

Molar mass  of ammonium chloride  = 53.491 gm

No. of moles

[tex]N = \frac{mass}{molar \ mass}[/tex]

[tex]N = \frac{11.5}{53.491}[/tex]

N = 0.215 moles

[tex]Molarity= \frac{N}{volume}[/tex]

[tex]molarity = \frac{0.215}{0.255}[/tex]

M = 0.843 M

Thus the no. of moles present in one liter solution is 0.843

Answer:

The drawing of the structure is found in diagram 1 of the attached figure.

Explanation:

Diagram 1 shows that three different types of protons are found in the structure. The nine hydrogen atoms have a similar behavior, the six hydrogen atoms also have a similar behavior and finally, the three hydrogen atoms adjacent to oxygen have a similar behavior. The number of peaks are as follows:

9H = singlet peak = between 3 and 4 ppm

6H = singlet peak = 4 ppm

3H = singlet peak = 3 ppm.

The 9 protons are around 3.5 ppm and the 6 hydrogen atoms show a peak at 4 ppm, and finally, the 3 protons have a peak around 3 ppm. Therefore, the corresponding drawing can be seen in diagram 2.

To construct a simulated 1H NMR spectrum for CH3OC(CH2OCH3)3, analyze the structure and determine the chemical shifts and integration values for each proton.

In order to construct a simulated 1H NMR spectrum for CH3OC(CH2OCH3)3, we need to analyze the structure of the molecule and determine the chemical shifts and integration values for each proton. Let's break down the molecule:

By analyzing the structure and applying the appropriate splitting patterns and chemical shifts, we can construct a simulated 1H NMR spectrum for CH3OC(CH2OCH3)3.

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Answer:

The balanced equation is given below:

CH4 + 2O2 —> CO2 + 2H2O

The coefficients are: 1, 2, 1, 2

Explanation:

The equation for the Combustion of methane (CH4) is given below:

CH4 + O2 —> CO2 + H2O

Now we can balance the equation above as follow:

There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

CH4 + O2 —> CO2 + 2H2O

Now, there are a total of 4 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 2 in front of O2 as shown below:

CH4 + 2O2 —> CO2 + 2H2O

Now the equation is balanced.

The coefficients are: 1, 2, 1, 2

Answer:

The coefficients for CH4; O2; CO2; H2O  are 1; 2; 1; 2

Explanation:

Step 1: Data given

The reactants are : CH4 and 02

The products are CO2 and H2O

Step 2: The unbalanced equation

CH4 + O2 → CO2 + H2O

Step 3: Balancing the equation

CH4 + O2 → CO2 + H2O

On the left side we have 4x H (in CH4) on the right side we have 2x H in (H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 2.

CH4 + O2 → CO2 + 2H2O

On the left side we have 2x O (in O2), on the right side we have 4x O (2x in CO2 and 2x in H2O). To multiply the amount of O on both side we have to multiply O2 on the left side by 2. Now the equation is balanced.

CH4 + 2O2 → CO2 + 2H2O

The coefficients for CH4; O2; CO2; H2O  are 1; 2; 1; 2

Answer:

1.131g of MnO2

Explanation:

Step 1:

The balanced equation for the reaction.

MnO2(s) + 4HCl(aq)—> MnCl2(aq) + 2H2O(l) + Cl2(g)

Step 2:

Data obtained from the question. This includes:

Volume (V) of Cl2 = 295mL = 0.295L

Temperature (T) = 25°C = 25°C + 273 = 298K

Pressure (P) = 795 Torr = 795/760 = 1.05 atm

Number of mole (n) of Cl2 =.?

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

Step 3:

Determination of the number of mole of Cl2 produced. The number of mole of Cl2 produced from the reaction can be obtained by applying the ideal gas equation as follow:

PV = nRT

1.05 x 0.295 = n x 0.082 x 298

Divide both side by 0.082 x 298

n = (1.05 x 0.295)/(0.082 x 298)

n = 0.013 mole.

Therefore, the number of mole of Cl2 produced is 0.013 mole.

Step 4:

Determination of number of mole of MnO2 needed to produce 0.013 mole of Cl2.

This is illustrated below:

From the balanced equation above,

1 mole of MnO2 produced 1 mole of Cl2.

Therefore, it will also take 0.013 mole of MnO2 to produce 0.013 mole of Cl2.

Step 5:

Converting 0.013 mole of MnO2 to grams. This is illustrated below:

Number of mole MnO2 = 0.013 mole

Molar Mass of MnO2 = 55 + (2x16) = 87g/mol

Mass of MnO2 =?

Mass = number of mole x molar Mass

Mass of MnO2 = 0.013 x 87

Mass of MnO2 = 1.131g

Therefore 1.131g of MnO2 should be added to excess HCl.

Answer:

10 mmoles

Explanation:

moles = Molarity x Volume in Liters

moles NaCl = (0.20M)(0.050L) = 0.010mol x (1000 mmoles/mole) = 10 mmoles

Answer:

10 millimoles

Explanation:

moles = Molarity x Volume in Liters

moles NaCl = (0.20M)(0.050L) = 0.010mol x (1000 millimoles/mole) = 10 millimoles

Answer: [tex]N_2(g)+3H_2(g)+4O_2(g)\rightarrow 2HNO_3(g)+2H_2O(g)[/tex]

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced chemical reaction for nitrogen and hydrogen react to form ammonia:

[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]     (1)

The balanced chemical reaction for ammonia and oxygen react to form nitric acid and water:

[tex]NH_3(g)+2O_2(g)\rightarrow HNO_3(g)+H_2O(g)[/tex] [tex]\times 2[/tex]

[tex]2NH_3(g)+4O_2(g)\rightarrow 2HNO_3(g)+2H_2O(g)[/tex]    (2)

The net chemical equation for the production of nitric acid from nitrogen, hydrogen and oxygen by adding 1 and 2

[tex]N_2(g)+3H_2(g)+4O_2(g)\rightarrow 2HNO_3(g)+2H_2O(g)[/tex]

Final answer:

The net balanced chemical equation for the manufacturing of nitric acid from nitrogen, hydrogen, and oxygen is N₂(g) + O₂(g) + 3H₂(g) → 2HNO₃(g) + 2H₂O(l). This reflects a combination of the Haber process for ammonia synthesis and the Ostwald process for oxidizing ammonia into nitric acid.

Explanation:

To produce nitric acid (HNO₃) from nitrogen (N₂), hydrogen (H₂), and oxygen (O₂), we start with the Haber process to produce ammonia (NH₃) followed by the Ostwald process to oxidize ammonia into nitric acid. The net chemical equation represents a combination of these two processes.

The Balanced Equations for each step:

Combining and cancelling out the common elements, we get the net balanced equation for the entire process:

N₂(g) + O₂(g) + 3H₂(g) → 2HNO₃(g) + 2H₂O(l)

Answer:

The option (e) None of the choice are correct.

Explanation:

To find [tex]\frac{m}{z}[/tex] ratio in mass spectrometer,

We start calculating from the first option

(a)

 [tex]CH_{3}- CH_{2}- OH[/tex]

We know mass of carbon [tex]m= 12[/tex] and atomic number of carbon [tex]Z = 6[/tex]

After calculating we come up with,

[tex]\frac{m}{z} = 46[/tex]

Therefore option (a) is incorrect.

(b)

  [tex]CH_{3} -CH_{2}- NH_{2}[/tex]

After calculating we come up with,

[tex]\frac{m}{z} = 45[/tex]

Therefore option (b) is incorrect.

(c)

  [tex]CH_{3}- CH_{2}-CH_{3}[/tex]

[tex]\frac{m}{z} = 44[/tex]

Therefore option (c) is incorrect.

(d)

  [tex]CH_{3}- CH=CH_{2}[/tex]

[tex]\frac{m}{z} = 42[/tex]

Therefore option (d) is incorrect.

Therefore, the option (e) is correct for this problem

Final answer:

The reducing agent in a redox reaction donates electrons, causing the oxidation of another compound and itself being oxidized. Therefore, the correct answer is option B. causes the oxidation of another compound.

Explanation:

In a redox reaction, the reducing agent is a substance that causes reduction by donating electrons. The key function of a reducing agent is option B, as it causes the oxidation of another compound by losing electrons itself. As a result, the reducing agent is itself oxidized in the chemical reaction. To give you an example, in the reaction Zn (s) + S (s) → ZnS (s), zinc is the reducing agent because it gives up electrons to sulfur.

Moreover, as the reducing agent donates electrons, it indirectly causes the oxidation number of the atoms in the other compound to increase. It is also true that the reducing agent itself will undergo an increase in oxidation number, which is synonymous with the loss of electrons.

Answer:

Concentration of phosphate =  635.4 mM

Explanation:

Given Data;

Volume of soft drink = 10.00 ml

Diluted volume = 100.00 ml

Sample absorbance = 0.3317

Slope of the calibration line = 5.22

In calculating the concentration of phosphate in the drink, we use the formula;

Absorbance = slope * concentration

Making concentration subject formula and substituting, we have;

Concentration = Absorbance/slope

                        = 0.3317/5.22

                       = 0.06354 M ( concentration of the sample)

When is has been treated with acid and diluted to 100.00ml, the concentration of phosphate becomes;

concentration of phosphate = 0.0635 * 100/10    

                                                  = 0.0635 * 10

                                                    = 0.635M  

                                                     = 635.4 mM

Answer:

The equilibrium will change in the direction of the reactants

Explanation:

Final answer:

To find out how many grams of Iron (III) oxide (Fe2O3) can be produced from 25.0 g of iron, we calculate the number of moles of iron, use the stoichiometric ratio from the balanced equation, and then convert the moles of Fe2O3 to grams to get the final yield of 35.68 g of Fe2O3.

Explanation:

The question is asking how many grams of Iron (III) oxide (Fe2O3) can be formed from a given mass of iron when reacted with an excess of oxygen. To solve this, we need to use stoichiometry.

First, calculate the number of moles of iron using its molar mass (55.85 g/mol):

25.0 g Fe × (1 mol Fe / 55.85 g Fe) = 0.447 mol Fe

The balanced equation for the reaction is:

4 Fe + 3 O2 → 2 Fe2O3

This indicates that 4 moles of Fe produce 2 moles of Fe2O3. Using this stoichiometric ratio, we calculate the moles of Fe2O3 produced:

0.447 mol Fe × (1 mol Fe2O3 / 2 mol Fe) = 0.2235 mol Fe2O3

Now convert moles of Fe2O3 to grams using its molar mass (159.70 g/mol):

0.2235 mol Fe2O3 × (159.70 g Fe2O3 / 1 mol Fe2O3) = 35.68 g Fe2O3

So, 25.0 g of iron can produce 35.68 g of Iron (III) oxide.

Final answer:

25.0 grams of iron can produce 71.40 grams of Iron (III) oxide (Fe2O3) when reacted with excess oxygen.

Explanation:

The balanced chemical equation for the reaction between iron (Fe) and oxygen (O2) to produce iron (III) oxide (Fe2O3) is:

4 Fe + 3 O2 → 2 Fe2O3

From the balanced equation, we can see that 4 moles of iron reacts with 3 moles of oxygen to produce 2 moles of iron (III) oxide. Therefore, the molar ratio between iron and iron (III) oxide is 4:2 or 2:1.

If we have 25.0 grams of iron, we can convert it to moles using the molar mass of iron (55.85 g/mol).

25.0 g Fe ÷ 55.85 g/mol = 0.447 mol Fe

Since there is an excess of oxygen in the reaction, all the moles of iron will react to form iron (III) oxide. Using the molar ratio of 2:1, we can calculate the moles of iron (III) oxide:

0.447 mol Fe x 2 mol Fe2O3 ÷ 2 mol Fe = 0.447 mol Fe2O3

Now, we can convert moles to grams using the molar mass of iron (III) oxide (159.70 g/mol):

0.447 mol Fe2O3 x 159.70 g/mol = 71.40 grams of Fe2O3 can be produced.

Partial molar properties of components in a mixture can vary and be different from their pure-component properties. In the case of an ethanol-water mixture, the partial molar volumes and enthalpies can be calculated using data from the International Critical Tables.

Partial molar properties of components in a mixture may differ from their corresponding pure-component properties and can vary with composition. In the case of ethanol-water mixture, the partial molar volumes and enthalpies of ethanol and water can be calculated using data from the International Critical Tables. The partial molar volumes can show how the volume changes when the mole fraction of a component is changed, while the partial molar enthalpies can show the change in enthalpy with respect to mole fraction.

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Final answer:

Partial molar volume and partial molar enthalpy are calculated for components in a mixture and represent the change in volume and enthalpy when one mole of a substance is added to a mixture. They differ from pure substances' molar volumes and enthalpies and may vary with composition.

Explanation:

Partial molar properties help us understand how the addition of a component affects a mixture's overall properties, such as volume and enthalpy. The partial molar volume (VM) is the volume change upon adding one mole of a substance to a mixture, at constant temperature and pressure. It differs from molar volume which is the volume occupied by one mole of a pure substance. The notation ∆V/(∆nB)T,p,nA, represents the partial molar volume of methanol (substance B) with respect to water (substance A). An example is given with values of VA = 17.74 cm³ mol⁻¹ for water and VB = 38.76 cm³ mol⁻¹ for methanol at a certain composition.

Similarly, we can define partial molar enthalpy (HM), which is the enthalpy change upon adding one mole of a substance. This property can be complex, as interactions between molecules can cause it to differ significantly from the pure component's molar enthalpy. Without specific data on ethanol-water mixtures, we can only conclude that like partial molar volume, the partial molar enthalpy would have to be measured or calculated for different compositions and cannot be directly determined from this example.

Answer:

Match each exploration mission.

1. explored the Challenger Deep  

5

Apollo 11  

2. took pictures and video of Mars surface  

3

Sputnik 1  

3. first man-made object sent into space  

2

Sojourner  

4. first manned orbit around the earth  

6

Trieste  

5. first men to walk on the moon  

4

Vostok  

6. explored the dark side of Venus  

1

Mariner 5

Explanation:

*

Answer:

Apollo 11 - First man to walk on the moon

Trieste - Explored the challenger deep

Sojourner - First man-made object sent into space

Vostok - First manned orbit around the earth

Mariner 5 - Explored the dark side of venus

Sputnik 1 - Took pictures and videos of Mars surface

Explanation:

The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihydrogen gas. Synthesis gas is one of the most widely used industrial chemicals, and is the major industrial source of hydrogen.

Suppose a chemical engineer studying a new catalyst for the reform reaction finds that 924 liters per second of methane are consumed when the reaction is run at 261°C and 0.96atm. Calculate the rate at which dihydrogen is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits.

Answer: The rate at which dihydrogen is being produced is 0.12 kg/sec

Explanation:

The balanced chemical equation is ;

[tex]CH_4+H_2O\rightarrow 3H_2+CO[/tex]

According to ideal gas equation:

[tex]PV=nRT[/tex]

P = pressure of gas = 0.96 atm

V = Volume of gas = 924 L

n = number of moles

R = gas constant =[tex]0.0821Latm/Kmol[/tex]

T =temperature =[tex]261^0C=(261+273)K=534K[/tex]

[tex]n=\frac{PV}{RT}[/tex]

[tex]n=\frac{0.96atm\times 924L}{0.0820 L atm/K mol\times 534K}=20.2moles[/tex]

According to stoichiometry:

1 mole of methane produces = 3 moles of hydrogen

Thus 20.2 moles of methane produces = [tex]\frac{3}{1}\times 20.2=60.6[/tex] moles of hydrogen

Mass of hydrogen =[tex]moles\times {\text {Molar mass}}=60.6mol\times 2g/mol=121.2g=0.12kg[/tex]

Thus the rate at which dihydrogen is being produced is 0.12 kg/sec

A. Initial pH: 7.00, Final pH: 12.00

B. Initial pH: 4.03, Final pH: 4.09

 A. For pure water:

The initial pH of pure water is 7.00 because water is neutral. When 0.010 mol of NaOH is added to 250.0 mL of pure water, the concentration of OH⁻ ions can be calculated as follows:

[tex]\[ [\text{OH}^-] = \frac{0.010 \text{ mol}}{0.250 \text{ L}} = 0.040 \text{ M} \][/tex]

Using the relationship [tex]\( [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ M}^2 \)[/tex] at 25°C, we can find the final concentration of H^+ ions:

[tex]\[ [\text{H}^+] = \frac{1.0 \times 10^{-14} \text{ M}^2}{0.040 \text{ M}} = 2.5 \times 10^{-13} \text{ M} \][/tex]

The final pH is then calculated as:

[tex]\[ \text{pH} = -\log[\text{H}^+] = -\log(2.5 \times 10^{-13}) \approx 12.00 \][/tex]

B. For the buffer solution (0.195 M in HCHO₂ and 0.275 M in KCHOC):

The initial pH of the buffer solution can be calculated using the Henderson-Hasselbalch equation:

[tex]\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \][/tex]

Given that HCHO₂ (formic acid) and KCHOC (potassium formate) are a conjugate acid-base pair, we can use the provided concentrations:

[tex]\[ \text{pH} = 3.75 + \log\left(\frac{0.275}{0.195}\right) \approx 4.03 \][/tex]

After the addition of 0.010 mol of NaOH, we can assume that most of the NaOH will react with HCHO₂ to form more KCHOC, since the buffer is designed to resist pH changes. The change in concentration of KCHOC can be calculated as:

[tex]\[ \Delta[\text{KCHOC}] = \frac{0.010 \text{ mol}}{0.250 \text{ L}} = 0.040 \text{ M} \][/tex]

The new concentrations will be:

[tex]\[ [\text{HCHO2}] = 0.195 \text{ M} - 0.040 \text{ M} = 0.155 \text{ M} \] \[ [\text{KCHOC}] = 0.275 \text{ M} + 0.040 \text{ M} = 0.315 \text{ M} \][/tex]

The final pH is then calculated using the Henderson-Hasselbalch equation again:

[tex]\[ \text{pH} = 3.75 + \log\left(\frac{0.315}{0.155}\right) \approx 4.09 \][/tex]

 The final pH of the buffer solution changes only slightly after the addition of NaOH, demonstrating the buffer's ability to resist pH changes.

- Initial pH = 10.785

- Final pH = 10.927

To calculate the initial and final pH after the addition of 0.010 mol of NaOH to each solution, we need to consider the different types of solutions involved: pure water, an acidic buffer, and a basic buffer.

### A. 250.0 mL of Pure Water

**Initial pH:**

Pure water has a pH of 7.0 at 25°C because it is neutral.

**Final pH after adding NaOH:**

1. Calculate the concentration of OH\(^-\) ions after adding 0.010 mol of NaOH:

[tex]\[ \text{Concentration of OH}^- = \frac{0.010 \, \text{mol}}{0.250 \, \text{L}} = 0.040 \, \text{M} \][/tex]

2. Use the concentration of OH[tex]\(^-\)[/tex] to find the pOH:

[tex]\[ \text{pOH} = -\log [\text{OH}^-] = -\log(0.040) \approx 1.40 \][/tex]

3. Convert pOH to pH:

[tex]\[ \text{pH} = 14 - \text{pOH} = 14 - 1.40 = 12.60 \][/tex]

**Summary:**

- Initial pH = 7.0

- Final pH = 12.60

### B. 250.0 mL of a Buffer Solution (0.195 M HCHO2 and 0.275 M KCHO2)

HCHO2 is formic acid, and KCHO2 is potassium formate, the salt of formic acid.

**Initial pH:**

Use the Henderson-Hasselbalch equation:

[tex]\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]

Given[tex]\( [\text{HA}] = 0.195 \, \text{M} \) and \( [\text{A}^-] = 0.275 \, \text{M} \). The \( K_a \) of formic acid is \( 1.8 \times 10^{-4} \):[/tex]

[tex]\[ \text{p}K_a = -\log (1.8 \times 10^{-4}) \approx 3.74 \]\[ \text{pH} = 3.74 + \log \left( \frac{0.275}{0.195} \right) \approx 3.74 + 0.155 = 3.895 \][/tex]

**Final pH after adding NaOH:**

1. Determine the moles of HCHO2 and KCHO2 before adding NaOH:

[tex]\[ \text{moles of HCHO2} = 0.195 \, \text{M} \times 0.250 \, \text{L} = 0.04875 \, \text{mol} \]\[ \text{moles of KCHO2} = 0.275 \, \text{M} \times 0.250 \, \text{L} = 0.06875 \, \text{mol} \][/tex]

2. Adding 0.010 mol of NaOH will neutralize 0.010 mol of HCHO2:

[tex]\[ \text{new moles of HCHO2} = 0.04875 - 0.010 = 0.03875 \, \text{mol} \]\[ \text{new moles of KCHO2} = 0.06875 + 0.010 = 0.07875 \, \text{mol} \][/tex]

3. Calculate the new concentrations:

[tex]\[ [\text{HCHO2}] = \frac{0.03875 \, \text{mol}}{0.250 \, \text{L}} = 0.155 \, \text{M} \]\[ [\text{KCHO2}] = \frac{0.07875 \, \text{mol}}{0.250 \, \text{L}} = 0.315 \, \text{M} \][/tex]

4. Use the Henderson-Hasselbalch equation to find the final pH:

[tex]\[ \text{pH} = 3.74 + \log \left( \frac{0.315}{0.155} \right) \approx 3.74 + 0.31 = 4.05 \][/tex]

**Summary:**

- Initial pH = 3.895

- Final pH = 4.05

### C. 250.0 mL of a Buffer Solution (0.255 M CH3CH2NH2 and 0.235 M CH3CH2NH3Cl)

CH3CH2NH2 is ethylamine, and CH3CH2NH3Cl is ethylammonium chloride, the salt of ethylamine.

**Initial pH:**

Use the Henderson-Hasselbalch equation for a base:

[tex]\[ \text{pH} = \text{p}K_b + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \][/tex]

Given [tex]\( [\text{base}] = 0.255 \, \text{M} \) and \( [\text{acid}] = 0.235 \, \text{M} \). The \( K_b \) of ethylamine is \( 5.6 \times 10^{-4} \)[/tex]:

[tex]\[ \text{p}K_b = -\log (5.6 \times 10^{-4}) \approx 3.25 \][/tex]

The [tex]\( \text{p}K_a \)[/tex] of the conjugate acid (CH3CH2NH3\(^+\)) is:

[tex]\[ \text{p}K_a = 14 - \text{p}K_b = 14 - 3.25 = 10.75 \]\[ \text{pH} = 10.75 + \log \left( \frac{0.255}{0.235} \right) \approx 10.75 + 0.035 = 10.785 \][/tex]

**Final pH after adding NaOH:**

1. Determine the moles of CH3CH2NH2 and CH3CH2NH3Cl before adding NaOH:

[tex]\[ \text{moles of CH3CH2NH2} = 0.255 \, \text{M} \times 0.250 \, \text{L} = 0.06375 \, \text{mol} \]\[ \text{moles of CH3CH2NH3Cl} = 0.235 \, \text{M} \times 0.250 \, \text{L} = 0.05875 \, \text{mol} \][/tex]

2. Adding 0.010 mol of NaOH will react with 0.010 mol of CH3CH2NH3\(^+\):

[tex]\[ \text{new moles of CH3CH2NH2} = 0.06375 + 0.010 = 0.07375 \, \text{mol} \]\[ \text{new moles of CH3CH2NH3Cl} = 0.05875 - 0.010 = 0.04875 \, \text{mol} \][/tex]

3. Calculate the new concentrations:

[tex]\[ [\text{CH3CH2NH2}] = \frac{0.07375 \, \text{mol}}{0.250 \, \text{L}} = 0.295 \, \text{M} \]\[ [\text{CH3CH2NH3Cl}] = \frac{0.04875 \, \text{mol}}{0.250 \, \text{L}} = 0.195 \, \text{M} \][/tex]

4. Use the Henderson-Hasselbalch equation to find the final pH:

[tex]\[ \text{pH} = 10.75 + \log \left( \frac{0.295}{0.195} \right) \approx 10.75 + 0.177 = 10.927 \][/tex]

**Summary:**

- Initial pH = 10.785

- Final pH = 10.927

Answer:

ΔH°rxn = - 162.5 kJ

Explanation:

Hello,

In this case, we use the Hess law to compute the required enthalpy of the reaction for the chlorine with ozone.

1. At first, we invert the original (1) equation in order to place the chlorine at the reactants, so the enthalpy sign is inverted to positive:

Cl ( g ) + 2O2 ( g ) ⟶ClO ( g ) + O3 ( g ) ΔH°rxn = 122.8 kJ

2. Then we do not modify the second reaction:

2O3 ( g ) ⟶ 3O2 ( g ) ΔH°rxn = −285.3 kJ

Next, the add the aforementioned reactions:

Cl ( g ) + 2O2 ( g ) + 2O3 ( g )  ⟶ ClO ( g ) + O3 ( g ) + 3O2 ( g )

In order to obtain (3):

O3 ( g ) + Cl ( g ) ⟶ ClO ( g ) + O2 ( g )

And the enthalpy of reaction results:

ΔH°rxn = 122.8 kJ − 285.3 kJ

ΔH°rxn = - 162.5 kJ

Best regards.

To determine the enthalpy of reaction for the reaction of chlorine with ozone, we can use the enthalpy of reaction values for the given reactions. By adding the given reactions, we can obtain the desired reaction and calculate its enthalpy of reaction. The enthalpy of reaction for the desired reaction is -408.1 kJ.

To determine the enthalpy of reaction for the reaction of chlorine with ozone (ClO(g) + O3(g) ⟶ Cl(g) + 2 O2(g)), we can use the enthalpy of reaction values for the given reactions. From the given reactions, we can see that the reaction (1) ClO (g) + O3 (g) ⟶ Cl (g) + 2 O2 (g) has an enthalpy of reaction of -122.8 kJ and the reaction (2) 2 O3 (g) ⟶ 3 O2 (g) has an enthalpy of reaction of -285.3 kJ. By adding these two reactions, we can obtain the desired reaction and calculate its enthalpy of reaction.

To add the reactions, we need to cancel out the common compounds. The Cl (g) in reaction (1) and the ClO (g) in reaction (3) are the common compounds. By cancelling them out, we obtain:

O3 (g) ⟶ O2 (g) + O2 (g)

The enthalpy of reaction for the desired reaction is the sum of the enthalpies of reaction of the added reactions (-122.8 kJ + -285.3 kJ = -408.1 kJ).

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Answer:

111.5 grams of lead (II) oxide would be produced.

Explanation:

[tex]2PbS+3O_2\rightarrow 2PbO+2SO_2[/tex]

Moles of oxygen = 0.750 mol

According to reaction, 3 moles of oxygen gas gives 2 moles of lead(II) oxide ,then 0.750 moles of oxygen gas will give:

[tex]\frac{2}{3}\times 0.750 mol=0.50 mol[/tex] of lead (II) oxide

Mass of 0.50 moles lead(II) oxide:

0.50 mol × 223 g/mol = 111.5 g

111.5 grams of lead (II) oxide would be produced.

Answer:

0.471 mol/L

Explanation:

First, we'll begin by by calculating the number of mole of KMnO4 in 26g of KMnO4.

This is illustrated below:

Molar Mass of KMnO4 = 39 + 55 + (16x4) = 39 + 55 + 64 = 158g/mol

Mass of KMnO4 from the question = 26g

Mole of KMnO4 =?

Number of mole = Mass/Molar Mass

Mole of KMnO4 = 26/158 = 0.165mole

Now we can obtain the concentration of KMnO4 in mol/L as follow:

Volume of the solution = 350mL = 350/1000 = 0.35L

Mole of KMnO4 = 0.165mole

Conc. In mol/L = mole of solute(KMnO4)/volume of solution

Conc. In mol/L = 0.165mol/0.35

conc. in mol/L = 0.471mol/L

Answer:

The concentration of this potassium permanganate solution is 0.470 mol/L or 0.470 M

Explanation:

Step 1: data given

Mass of  potassium permanganate = 26.0 grams

Molar mass of potassium permanganate = 158.034 g/mol

Volume = 350 mL = 0.350 L

Step 2: Calculate moles potassium permanganate

Moles KMnO4 = mass KMnO4 / molar mass KMnO4

Moles KMnO4 = 26.0 grams / 158.034 g/mol

Moles KMnO4 = 0.1645 moles

Step 3: Calculate concentration of KMnO4 solution

Concentration KMnO4 = moles KMnO4 / volume

Concentration KMnO4 = 0.1645 moles / 0.350 L

Concentration KMnO4 = 0.470 M

The concentration of this potassium permanganate solution is 0.470 mol/L or 0.470 M