Use the electron-transfer method to balance this equation:


solid copper and dilute nitric acid react to produce copper(ii) nitrate, water, and nitrogen monoxide gas (no)

The enthalpy change when 8.00 g of nitrogen oxide react is -15.162 kJ for the given chemical reaction.

The molar mass of NO =  30.01 g/mol

8.00 g of NO  = 8.00 g / 30.01 g/mol

8.00 g of NO = 0.266 mol of NO

Heat rejection = 57. 0 kJ

Here, 1 mole of NO reacts with 1/2 mole of Oxygen to produce 1 mole of [tex]NO_{2}[/tex]

The amount of Oxygen required for 0.266 mol of NO is calculated as:

The amount of Oxygen = 0.266 mol NO x (1/2) mol [tex]O_{2}[/tex]    / 1 mol NO

The amount of Oxygen required = 0.133 mol [tex]O_{2}[/tex]

The heat reaction will be:

-57.0 kJ/mol x 0.266 mol NO = -15.162 kJ

Therefore, we can conclude that the enthalpy change is -15.162 kJ.

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In the given chemical equation:

Fe(NO3)2 + Al → Fe + Al(NO3)3

Iron (Fe) is being reduced because it is gaining electrons and its oxidation state is decreasing from +2 to 0 (elemental state).

Aluminum (Al) is being oxidized because it is losing electrons and its oxidation state is increasing from 0 (elemental state) to +3.

Therefore, Fe(NO3)2 is the oxidizing agent, and Al is the reducing agent in this reaction.

First, we need to use stoichiometry to find out how many moles of hydrogen gas are produced. From the balanced chemical equation, we can see that for every 1 mole of zinc (Zn), 1 mole of hydrogen gas (H2) is produced. Therefore, if we have 20.0 mol of Zn, we will also produce 20.0 mol of H2.

Next, we can use the formula for the mass of a gas:

mass = molar mass x number of moles

The molar mass of hydrogen gas is approximately 2.02 g/mol. Therefore, the mass of 20.0 mol of hydrogen gas would be:

mass = 2.02 g/mol x 20.0 mol
mass = 40.4 g

So, 40.4 grams of hydrogen gas are produced when 20.0 mol of Zn are added to excess hydrochloric acid.

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The normal boiling point of the 3.45 mol solution of KBr is 104.7384°C.

The normal boiling point of a 3.45 mol solution of KBr with a density of 1.10 g/mL can be calculated using the formula:

ΔT = Kb * molality

where ΔT is the boiling point elevation, Kb is the molal boiling point elevation constant for water (0.512°C kg/mol), and molality is the number of moles of solute per kilogram of solvent.

First, we need to calculate the mass of the solvent (water) required to dissolve 3.45 mol of KBr. The molar mass of KBr is 119 g/mol, so 3.45 mol of KBr would weigh 409.55 g.

Since the density of the solution is given as 1.10 g/mL, the volume of the solution is:

V = m / ρ = 409.55 g / 1.10 g/mL = 372.32 mL

So, the mass of the water is:

mH2O = V * ρH2O = 372.32 mL * 1 g/mL = 372.32 g

The molality of the solution can be calculated as follows:

molality = moles of solute / mass of solvent (in kg) = 3.45 mol / 0.37232 kg = 9.27 mol/kg

Substituting the values in the formula for boiling point elevation:

ΔT = 0.512°C kg/mol * 9.27 mol/kg = 4.7384°C

The normal boiling point of pure water is 100°C, so the boiling point of the KBr solution would be:

Boiling point = 100°C + ΔT = 100°C + 4.7384°C = 104.7384°C

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We can use the ideal gas law to calculate the pressure of the bromine gas in the 1.5 L container. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

Since we know the temperature and the volume, we can rearrange the ideal gas law to solve for P, the pressure. We can use the pressure and volume from the first container to calculate the number of moles. Plugging in all of the known values, we get:

P1V1 = nRT

n = P1V1/RT

P2 = (P1V1/RT) * (V2/V1)

Using the values from the question, we get:

P2 = (1.20 atm * 2.65 L)/(0.08206 L·atm·mol-1·K-1 * 298 K) * (1.50 L/2.65 L)

This gives us a pressure of 1.04 atm in the 1.5 L container, which is equal to 1040 kPa.

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We need to dissolve 6.85 grams of sucrose in 2000 grams of water to make a 0.1 M solution.

To calculate the mass of sucrose needed to make a 0.1 molar solution in 2000 grams of water, we need to use the formula:

[tex]m = n *M * MW[/tex]

Step 1: Calculate the number of moles of sucrose needed

Molarity (M) = 0.1 mol/L

volume of solution = 2000 grams of water ÷ density of water = 2000 mL

We need to calculate the number of moles of sucrose that would be present in 2000 mL of a 0.1 M solution:

moles of solute (n) = [tex]M * V = 0.1 mol/L *2.0 L = 0.2 moles[/tex]

Step 2: Calculate the mass of sucrose needed

Molecular weight of sucrose is 342.3 g/mol.

We can use the formula:

[tex]m = n * M * MW \\m = 0.2 moles *0.1 mol/L * 342.3 g/mol = 6.85 g[/tex]

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The final pressure of the gas when the volume changes from 720 mL to 820 mL is approximately 1.22 atm.


To solve this problem, we can use Boyle's Law, which states that the product of the initial pressure and volume (P1V1) is equal to the product of the final pressure and volume (P2V2) for a gas at a constant temperature:

P1V1 = P2V2

Given the initial pressure (P1) is 1.4 atm and the initial volume (V1) is 720 mL, we need to find the final pressure (P2) when the volume (V2) changes to 820 mL. Rearrange the formula to solve for P2:

P2 = P1V1 / V2

Substitute the given values:

P2 = (1.4 atm × 720 mL) / 820 mL
P2 ≈ 1.22 atm

Therefore, the final pressure of the gas is approximately 1.22 atm.

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If 44.0 grams of sodium reacts with 10.0 grams of chlorine gas, 54.0 grams of sodium chloride could potentially be formed in the reaction: 2 Na(s) + Cl₂(g) ⟶ 2 NaCl(s).


1. Calculate the moles of sodium and chlorine:
  - moles of Na = mass (g) / molar mass = 44.0 g / 22.99 g/mol = 1.91 mol
  - moles of Cl₂ = mass (g) / molar mass = 10.0 g / 70.90 g/mol = 0.141 mol

2. Determine the limiting reactant by dividing the moles of each reactant by their stoichiometric coefficients:
  - Na: 1.91 mol / 2 = 0.955
  - Cl₂: 0.141 mol / 1 = 0.141

3. Since the value for Cl₂ is lower, chlorine gas is the limiting reactant.

4. Calculate the moles of NaCl produced using the stoichiometry of the reaction:
  - moles of NaCl = moles of Cl₂ × (2 moles of NaCl / 1 mole of Cl₂) = 0.141 × 2 = 0.282 mol

5. Calculate the mass of NaCl produced:
  - mass of NaCl = moles × molar mass = 0.282 mol × 58.44 g/mol = 54.0 g

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The compounds A, B and C are ammonium chloride, ammonia gas and silver chloride respectively.

Ammonium chloride,  is a white crystalline solid which is soluble in water. On heating with sodium hydroxide it will produce ammonia gas, which is a colorless gas and has a odor or pungent smell.

So,  Ammonia gas will turn red litmus into blue as it's pH is 11.6.

When Ammonium chloride reacts with silver nitrate  in  presence of dilute Nitric acid ,  it produces Silver chloride and Ammonium nitrate

AgCl is soluble in Ammonia because it can form complexes which

makes it behave like an ion, making it soluble.

Therefore, Compound A is Ammonium chloride,

B is ammonia gas

and C is Silver chloride.

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The complete question is

A white solid A when heated with sodium hydroxide solution gives a pungent gas B which turns red litmus paper blue. The solid when dissolved in dilute nitric acid is treated with Silver nitrate solution to give white precipitate C, which is soluble in ammonia.

(A) What are the substance A, B, and C?

The E° for cell reaction is - 2.37 V and  2.23 V and E for cell reaction  = 2.22V and ΔG = - 428.39kJ/mol.

The formula for solving the equation for given cell is as follows :

                      E°cell , Ecell and Δ[tex]G_{rnx}[/tex]

The standard cell potential is the potential of cell at standard condition of 1MConcentration and pressure 1 atm E°cell

calculation :

                E°cell = E° cathode - E° anode          it is calculated using the Nernst equation which is discussed below :

                  Ecell = E°cell -- [tex]\frac{RT}{nF}[/tex] 1n K  = E°cell -- [tex]\frac{0.0591}{n}[/tex]log [tex]\frac{Products}{Reactants}[/tex]

Here, F is the Faraday's constant, R is the gas constant, T is the temperature, and n is the number of transferred electrons. K is the equilibrium constant.

The Gibbs free energy is the greatest work that is finished by a framework . The standard cell potential is without like energy by the recipe as follows;and F is Faraday's steady.

A system's maximum amount of work is referred to as its Gibbs free energy. The standard cell potential is connected with the free energy by the recipe as follows:    Δ G = -n F Ecell

Here, E cell is cell potential

Δ G is the free energy n is the quantity of electrons moved and F is Faraday's steady.

 Oxidation :                                        

Mg ⇒ Mg ²⁺ + 2e⁻ E⁰anode = - 2.37 V

Reduction:Sn²⁺ + 2e⁻⇒ Sn E⁰

So,  cathode = - 0.14V

The standard cell potential is calculated as follows:E⁰ cell = - 0.14 V- (- 2.37 V ) =  2.23 V

The half reaction potentials for the oxidation and reduction are determined. They are subbed in the equation and the standard cell potential is determined.

Number of electrons transferred ,      n = 2   ,[Mg²⁺]   = 0.055M   ,  [ Sn²⁺ ]  = 0.030 M               The Nernst equation for reaction :

Ecell = E °cell = [tex]\frac{0.0591}{n}[/tex]log Mg ²⁺ / Sn²⁺

The cell potential for reaction is :

                       Ecell = 2.23V - [tex]\frac{0.0591}{2}[/tex]log[tex]\frac{0.055M}{0.030M}[/tex]= 2.22V

 The values are substituted for the reaction calculated here in the Nernst equation and cell potential.     

Calculation for the free energy for reaction ,

                                             ] Δ[tex]G_{rxn}[/tex] = -nFE cell

       = - 2 × 96485 C/ mol ×2.22 V

                          = --428393J/mol × [tex]\frac{1KJ}{1000J}[/tex]  = - 428.39kJ/mol

The cell potential for the response is subbed in the recipe and free energy for the response is determined

Nernst equation :

The standard electrode potential, absolute temperature, the number of electrons involved in the redox reaction, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation, respectively, can all be used to calculate the reduction potential of a half-cell or full cell reaction using the Nernst equation, a chemical thermodynamic relationship.

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Translating the given balanced chemical equation into words :A.)Phosphorus pentachloride and water yield phosphoric acid and hydrochloric acid.

Phosphorus pentachloride and water react to yield phosphoric acid and hydrochloric acid. Balanced chemical equation shows that for every one mole of PCl₅ and four moles of H₂O that react, one mole of H₃PO₄ and five moles of HCl are produced.

Phosphorus pentachloride (PCl₅) is a chemical compound composed of one phosphorus atom and five chlorine atoms. It is yellowish-white crystalline solid that is highly reactive and can decompose violently when exposed to water or moist air.

PCl₅ is primarily used as a chlorinating agent in organic chemistry, where it is used to convert alcohols, carboxylic acids, and other functional groups into the corresponding chlorides.

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Answer:

Welcome to the project on communicating design details for the properties and uses of unsaturated hydrocarbons. This project aims to enhance your understanding of the characteristics and applications of unsaturated hydrocarbons.

Here are the steps to complete this project:

Step 1: Research

Research the different types of unsaturated hydrocarbons, including alkenes and alkynes. Find out their general properties, such as their reactivity, flammability, and solubility. Also, identify their uses in various industries, such as plastics, rubber, and fuel.

Step 2: Create a Design

Using your research findings, create a design to visually communicate the properties and uses of unsaturated hydrocarbons. You can use tools like Canva, PowerPoint, or other design software to create infographics, posters, or slideshows.

Step 3: Incorporate Key Information

Incorporate the key information you gathered in step 1 into your design. Make sure to include the following details:

Definitions of unsaturated hydrocarbons, alkenes, and alkynes

Properties of unsaturated hydrocarbons, including reactivity, flammability, and solubility

Applications of unsaturated hydrocarbons in various industries, such as plastics, rubber, and fuel

Examples of unsaturated hydrocarbons, such as ethene and propene for alkenes, and ethyne for alkynes

Step 4: Review and Refine

Review your design and refine it to make sure it effectively communicates the properties and uses of unsaturated hydrocarbons. Check for spelling and grammar errors, and ensure that the information is accurate and easy to understand.

Step 5: Present Your Design

Present your design to your class or teacher, and explain the properties and uses of unsaturated hydrocarbons. You can also invite feedback and questions to enhance your understanding of the topic.

In conclusion, the properties and uses of unsaturated hydrocarbons are essential for many industries. Through this project, you will gain a better understanding of unsaturated hydrocarbons and develop your communication skills to effectively present your findings. Good luck!

Explanation:

Answer:

Explanation:

The three types of unsaturated hydrocarbons is alkynes, alkenes, and aromatic hydrocarbons. Which is composed of alkynes? acetylene. brainlist

The weight/volume percentage of sodium acetate in the solution is 4.6%.

The weight/volume percentage of sodium acetate in the solution can be calculated using the formula:
Weight/volume percentage = (Weight of solute ÷ Volume of solution) x 100%

In this case, the weight of sodium acetate is 13.8 grams and the volume of solution is 300 ml.

Therefore,
Weight/volume percentage = (13.8 g ÷ 300 ml) x 100%
Weight/volume percentage = 0.046 x 100%
Weight/volume percentage = 4.6%

Therefore, the weight/volume percentage of sodium acetate in the solution is 4.6%.

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The angle of a light beam affects the intensity and the amount of light reflected or transmitted through a process known as the "angle of incidence." When a light beam strikes a surface, the angle between the incoming light beam and the surface is called the angle of incidence. This angle plays a crucial role in determining the amount of light reflected or transmitted.

When the angle of incidence is small (light beam nearly perpendicular to the surface), more light is transmitted through the surface, and less is reflected. As the angle of incidence increases (light beam more parallel to the surface), the amount of light reflected also increases, while the intensity of the transmitted light decreases.

This phenomenon occurs due to the interaction of light with the surface material, which can either absorb, transmit, or reflect the incoming light, depending on the angle of incidence and the material's properties. The angle at which the light beam is incident on the surface also affects the intensity of the reflected light.

At a specific angle, called the "critical angle," the light beam is no longer transmitted but is entirely reflected, a phenomenon called "total internal reflection." The critical angle depends on the refractive indices of the two materials at the interface. When the angle of incidence is greater than the critical angle, all the light is reflected, and none is transmitted.

In summary, the angle of a light beam significantly influences the intensity and the amount of light reflected or transmitted by a surface. The angle of incidence determines the amount of light reflection, with a smaller angle leading to more transmission and a larger angle leading to increased reflection. The critical angle, in particular, plays a crucial role in determining the behavior of the light beam at the surface.

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The pressure at the bottom of a lake is given as 2.35 atm, and we are asked to find the pressure of the water. Since water is the fluid in question, we can assume that it is incompressible and that its density is constant. To find the pressure of the water, we can use the following formula:

Pressure = Density x Acceleration due to gravity x Height

Here, the height refers to the depth of the lake, which we can assume to be the same as the height of the water column. The acceleration due to gravity is a constant, and the density of water is given as 0.21 g/L.

Substituting these values in the formula, we get:

Pressure = 0.21 g/L x 9.8 m/s^2 x Depth

Since the pressure at the bottom of the lake is given as 2.35 atm, we can convert this to SI units using the conversion factor:

1 atm = 101325 Pa

Therefore, 2.35 atm = 2.35 x 101325 Pa = 2.38 x 10^5 Pa

Substituting this value in the formula, we can solve for the depth:

2.38 x 10^5 Pa = 0.21 g/L x 9.8 m/s^2 x Depth

Depth = 114.7 m

Therefore, the pressure of the water at this depth is:

Pressure = 0.21 g/L x 9.8 m/s^2 x 114.7 m = 240.3 kPa

In conclusion, the pressure of the water at the bottom of the lake is 240.3 kPa. This is the pressure exerted by the water column due to its weight, and it is in addition to the atmospheric pressure. Understanding the pressure of fluids is important in many fields, such as hydrology, engineering, and physics.

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The new volume of the gas, assuming constant temperature and a change in pressure from 760 torr to 730 torr, is 2.09 L.

Using the Boyle's Law equation,

P₁V₁ = P₂V₂,

where P is pressure and V is volume, we can solve for V₂ by plugging in the given values in the equation:

(760 torr)(2.00 L) = (730 torr)(V₂)

Solving for V₂, we get:

V₂ = (760 torr)(2.00 L) / (730 torr) = 2.09 L

Therefore, the new volume of the gas is 2.09 L.

This result makes sense because according to Boyle's Law, as pressure decreases, volume increases proportionally, assuming a constant temperature.

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The heat energy transferred to the copper can be calculated using the formula:

Q = m × c × ΔT

where Q is the heat energy transferred, m is the mass of the copper, c is the specific heat capacity of copper, and ΔT is the change in temperature.

Substituting the given values:

m = 2.3 g

c = 0.385 J/g·°C

ΔT = 174.0°C - 23.0°C = 151.0°C

Q = 2.3 g × 0.385 J/g·°C × 151.0°C = 131.38 J

Therefore, the heat energy transferred to 2.3 g of copper is 131.38 J.

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While balancing equation write the physical state of reactants and products as well as any reaction conditions.

Reactants are the substances that are present at the start of a chemical reaction. They are typically the substances that are used up during the reaction and are converted into different products. Reactants are usually written on the left side of a chemical equation, while the products are written on the right side. Reactants are essential components of any chemical reaction and are essential in order for the reaction to take place. Reactants are also known as substrates or starting materials.

Balancing the Following Equations:

1) CuSO4(s) + 2KI(aq) → Cu2I2(s) + K2SO4(aq) + I2(g)

2) 2NH3(g) + O2(g) → 2NO(g) + 2H2O(g)

3) 3Fe2O3(s) + 4CO(g) → 6Fe(s) + 3CO2(g)

4) Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)

5) 2Pb(NO3)2(aq) + H2SO4(aq) → PbSO4(s) + 2HNO3(aq)

6) CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

7) 2MnO2(s) + 4HCl(aq) → 2MnCl2(aq) + 2H2O(l) + Cl2(g)

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Magnesium metal will conduct electricity via mobile electrons whether it is in the solid or liquid state.

Magnesium oxide will not conduct electricity in the solid state as they are no mobile charge carriers.

Molten (liquid) magnesium oxide has mobile ions and these can transfer electrons via mobile ions. This is electrolysis and the compound is turned back into its elements (magnesium and oxygen).

H2O⋅⋅⋅⋅⋅⋅H−CH3 shows intermolecular attraction between water molecules and methane molecules, but not a hydrogen bond specifically.

A hydrogen bond is a type of intermolecular attraction that occurs when a hydrogen atom is bonded to an electronegative atom such as nitrogen, oxygen, or fluorine, and it is attracted to another electronegative atom in a nearby molecule. This attraction is represented by a dotted line.

Looking at the choices provided, the only option that shows a hydrogen bond is H3N⋅⋅⋅⋅⋅⋅H−O−H. In this molecule, the hydrogen atom in the H−O−H group is bonded to the highly electronegative oxygen atom, and it forms a hydrogen bond with the lone pair of electrons on the nitrogen atom in the H3N group.

The dotted line between the H and N represents the hydrogen bond.

In contrast, the other options do not show a hydrogen bond. H−H represents a simple covalent bond between two hydrogen atoms, while H4C⋅⋅⋅⋅⋅⋅H−F represents a covalent bond between a carbon atom and a fluorine atom, with no electronegative atoms capable of forming a hydrogen bond. H2O⋅⋅⋅⋅⋅⋅H−CH3 shows intermolecular attraction between water molecules and methane molecules, but not a hydrogen bond specifically.

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The molality of the naphthalene solution in chloroform is approximately 0.551 mol/kg.

To calculate the molality of a solution of naphthalene dissolved in chloroform, we need to use the boiling point elevation formula: ΔTb = Kb * molality, where ΔTb is the change in boiling point, Kb is the boiling point elevation constant, and molality is the moles of solute per kilogram of solvent.

First, we need to find the change in boiling point (ΔTb) by subtracting the normal boiling point of chloroform from the given boiling point. The normal boiling point of chloroform is 61.2°C. Therefore, ΔTb = 63.2°C - 61.2°C = 2.0°C.

Next, we need to find the boiling point elevation constant (Kb) for chloroform. The Kb value for chloroform is 3.63 °C/kg/mol.

Now, we can use the boiling point elevation formula to solve for molality:
2.0°C = 3.63 °C/kg/mol * molality

Rearranging the equation and solving for molality, we get:
molality = 2.0°C / 3.63 °C/kg/mol = 0.551 mol/kg

Therefore, the molality of the naphthalene solution in chloroform is approximately 0.551 mol/kg.

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The volume of dichloromethane [tex](CH_2Cl_2)[/tex] produced when 149 liters of methane [tex](CH_4)[/tex] react according to the given reaction is approximately 6.224 x [tex]10^5 J/K*m^3[/tex].  

The volume of dichloromethane [tex](CH_2Cl_2)[/tex] produced when 149 liters of methane [tex](CH_4)[/tex] react according to the given reaction is not immediately apparent from the reaction stoichiometry.

The balanced equation for the reaction between methane  [tex](CH_4)[/tex] and carbon tetrachloride (CCl4) to form dichloromethane  [tex](CH_2Cl_2)[/tex] and carbon dioxide (CO2) is:

[tex](CH_4)[/tex] +  [tex]CO_2[/tex] →  [tex](CH_2Cl_2)[/tex] +  [tex]CO_2[/tex]

The balanced equation shows that 1 mole reacts with 1 mole of CCl4 to produce 1 mole of  [tex](CH_2Cl_2)[/tex] and 1 mole of  [tex]CO_2[/tex].

The volume of the gas can be calculated using the ideal gas law:

PV = nRT

To find the number of moles of gas, we can use the molecular masses of the reactants and products:

Molar mass of  [tex](CH_4)[/tex] = 16.04 g/mol

Molar mass of  [tex]CCl_4[/tex] = 89.9 g/mol

Molar mass of   [tex](CH_2Cl_2)[/tex] = 70.1 g/mol

Molar mass of  [tex]CO_2[/tex] = 44.01 g/mol

The number of moles of  [tex](CH_4)[/tex] can be calculated from the initial amount of gas:

149 L of CH4 = 149 x 16.04 g/mol = 2432 g

The number of moles of CCl4 can be calculated from the given volume:

149 L of  [tex](CH_4)[/tex] +  [tex]CCl_4[/tex] →   [tex](CH_2Cl_2)[/tex] +  [tex]CO_2[/tex]

The volume of the gas is given as 149 L, so the number of moles of  [tex]CCl_4[/tex] can be calculated as:

149 L = 149 x 89.9 g/mol = 13,277 g

The number of moles  can be calculated from the given volume and the desired amount of product

149 L of  [tex](CH_4)[/tex] + [tex]CCl_4[/tex] →   [tex](CH_2Cl_2)[/tex] + [tex]CO_2[/tex]

149 L of  [tex](CH_4)[/tex] + [tex]CCl_4[/tex] → 149 x 70.1 g/mol + 13,277 g x 1 mol/13.277 g = 43,691 g

V = nRT

V = 43,691 g x 8.314 J/mol·K = 364,617.5 J/K

1 J/K = 1/1000 L·K

Therefore, the volume of the gas is:

V = 364,617.5 J/K x (1/1000 L·K) = 3.646 x 10^4 L

substitute this value for V in the equation for the volume of  [tex](CH_2Cl_2)[/tex] :

PV = nRT

PV = 149 x 8.314 J/mol·K x (3.646 x [tex]10^4[/tex] L)

PV = 6.224 x   [tex]10^5 J/K*m^3[/tex].  

Therefore, The volume of dichloromethane [tex](CH_2Cl_2)[/tex] produced when 149 liters of methane [tex](CH_4)[/tex] react according to the given reaction is approximately 6.224 x [tex]10^5 J/K*m^3[/tex].  

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Answer:

4 valence electrons.

Explanation:

Carbon has 4 valence electrons because it is in the 14th group on the Periodic Table.

Approximately 0.35175 moles of C₁₂H₂₆ were reacted to produce 4.2105 moles of CO₂.

To find the moles of C₁₂H₂₆ that reacted to produce 4.2105 moles of CO₂, you can use the stoichiometry of the balanced chemical equation: 2 C₁₂H₂₆ + 37 O₂ → 24 CO₂ + 26 H₂O.

Step 1: Identify the mole-to-mole ratio between C₁₂H₂₆ and CO₂ in the balanced equation.
In this case, the ratio is 2 moles of C₁₂H₂₆ to 24 moles of CO₂.

Step 2: Set up a proportion to find the moles of C₁₂H₂₆.
(2 moles C₁₂H₂₆) / (24 moles CO₂) = (x moles C₁₂H₂₆) / (4.2105 moles CO₂)

Step 3: Solve for x, which represents the moles of C₁₂H₂₆.
x moles C₁₂H₂₆  = (2 moles C₁₂H₂₆) * (4.2105 moles CO₂) / (24 moles CO₂)

Step 4: Calculate the value of x.
x = (2 * 4.2105) / 24
x ≈ 0.35175

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175.16 grams of[tex]CaCO3[/tex]will be produced when 98.2 grams of [tex]CaO[/tex] are reacted with an excess of [tex]CO2[/tex].

The balanced chemical equation for the reaction between[tex]CaO and CO2[/tex]is:

[tex]CaO + CO2 → CaCO3[/tex]

According to the equation, one mole of[tex]CaO[/tex] reacts with one mole of [tex]CO2[/tex]to produce one mole of [tex]CaCO3[/tex].

The molar mass of [tex]CaO[/tex]is 56.08 g/mol, and the molar mass of [tex]CO2[/tex] is 44.01 g/mol. Therefore, the number of moles of [tex]CaO[/tex] present in 98.2 g can be calculated as:

moles of [tex]CaO[/tex] = mass / molar mass = 98.2 g / 56.08 g/mol = 1.75 mol

Since the reaction is with an excess of [tex]CO2[/tex], all the [tex]CaO[/tex]will react. Therefore, the number of moles of CaCO3 produced will be the same as the number of moles of [tex]CaO[/tex] used, which is 1.75 mol.

The molar mass of [tex]CaCO3[/tex]is 100.09 g/mol. Therefore, the mass of [tex]CaCO3[/tex] produced can be calculated as:

mass of [tex]CaCO3[/tex] = moles of [tex]CaCO3[/tex] × molar mass = 1.75 mol × 100.09 g/mol = 175.16 g

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The correct interpretation is  89.6 L of [tex]NH_{3}[/tex](g) react with 156.8 L of [tex]O_{2}[/tex](g) to produce 89.6 L of [tex]NO_{2}[/tex](g) and 134.4 L of [tex]H_{2} O[/tex](g).

We know that one mole of a gas does occupy 22.4 L. We can now use this to obtain the number of volumes of the gas based on the stoichiometric coefficient that has been given in the problem.

Molar volume of a gas refers to the volume occupied by one mole of a gas at a specific temperature and pressure. This value is useful in many applications of chemistry, such as in stoichiometric calculations and the determination of gas densities.

Using the stoichiometric coefficients we can see that the volume of the gases are;

Ammonia - 89.6 L

Oxygen -  156.8 L

Nitrogen dioxide - 89.6 L

Water -  134.4 L

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Answer:

because they are poorly made

Explanation: