Calcium Carbonate reacts with HCl to produce Calcium chloride, carbon dioxide and water.(I) calculate the number of moles of CO2 produced from 36.5g of HCl (ii) Calculate the amount of Calcium chloride produced (in g) when 3 moles of calcium Carbonate reacts with HCl

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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On a rock in coastal Maine, the fungus, the algae, and the bacteria make up part of an ecosystem.

An ecosystem consists of all the organisms and the physical environment with which they interact.

In the yellow scale lichen, the fungus provides protection, moisture, and nutrients for the algae. The algae carry out photosynthesis to produce food that is used by the fungus.

Ecosystem function is generally described as the capacity of natural processes and components to provide goods and services that satisfy human needs, either directly or indirectly.

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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:

4 valence electrons.

Explanation:

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

The concentration of the solution containing 25.0 g NaOH in 500 cm³ of water is approximately 1.25 M (moles per liter).

To find the concentration of a solution containing 25.0 g NaOH in 500 cm³ of water, follow these steps:

1. Convert grams of NaOH to moles. The molar mass of NaOH is approximately 40 g/mol (Na = 23 g/mol, O = 16 g/mol, H = 1 g/mol).

25.0 g NaOH × (1 mol NaOH / 40 g NaOH) ≈ 0.625 mol NaOH

2. Convert the volume of water from cm³ to liters (L).

500 cm³ × (1 L / 1000 cm³) = 0.5 L

3. Calculate the concentration of the solution in moles per liter (M).

Concentration = moles of solute/volume of solvent (in liters)
Concentration = 0.625 mol NaOH / 0.5 L ≈ 1.25 M

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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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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 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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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 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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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.

At equilibrium, the concentrations of [tex]H_{3} O+, C_{6} H_{5}COO[/tex]-, and [tex]C_{6} H_{5}COO[/tex] in the solution will be 0.038 M, 0.038 M, and 0.012 M, respectively.

First, we can write the chemical equation for the dissociation of benzoic acid in water as follows:  [tex]C_{6}H5COOH + H_{2}O[/tex] ⇌[tex]C_{6}H_{5}COO- + H_{3}O[/tex]

The acid dissociation constant, Ka, is given as 6.3 × 10^-5.

[tex]Ka = [C_{6}H_{5}COO-][H_{3}O+] / [C_{6}H_{5}COOH][/tex]

We can assume that the initial concentration of [tex]C_{6}H_{5}COOH[/tex] is equal to its concentration at equilibrium, x. Thus, at equilibrium:

[tex][C_{6}H_{5}COOH] = x M \\[/tex]

[tex][C_{6}H_{5}COO-] = y M \\[/tex]

[tex][H_{3}O+] = y M\\[/tex]

Using the equilibrium expression and the given value of Ka, we can solve for the values of x and y:

[tex]Ka = [C_{6}H_{5}COO-][H_{3}O+] / [C_{6}H_{5}COOH]\\6.3 * 10^-5 = y^2 / x[/tex]

Since we know that the initial concentration of benzoic acid is 0.050 M, we can write: [tex]x + y = 0.050 M[/tex]

Now we have two equations and two unknowns. Solving for x and y:

[tex]x = 0.012 M\\y = 0.038 M[/tex]

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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 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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The temperature when the thermometer reads a pressure of 800 mmHg is approximately 282.2 K (or 9.1 °C).

To solve this problem, we can use the ideal gas law:

PV = nRT

We can use this equation to calculate the temperature of the gas when the pressure is 800 mmHg.

First, we need to convert the pressures from mmHg to atm, since R is in units of L·atm/K·mol.

1 atm = 760 mmHg

780 mmHg = 1.026 atm

800 mmHg = 1.053 atm

Next, we can set up a ratio of the two pressures and temperatures:

P1/T1 = P2/T2

[tex](1.026 atm) / (273.15 K) = (1.053 atm) / T2[/tex]

Solving for T2, we get:

[tex]T2 = (1.053 atm) / (1.026 atm/273.15 K) \\T2 = 282.2 K[/tex]

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The moles of hydrogen added to the balloon to give it a volume of 17.1 L were 8.8 mol.

Determine the initial ratio of moles to volume:
Initial moles of hydrogen = 2.6 mol
Initial volume of the balloon (V₁) = 3.9 L
Ratio of moles to volume: 2.6 mol / 3.9 L = 0.6667 mol/L

Final volume of the balloon (V₂) = 17.1 L

Calculate the final moles of hydrogen in the balloon using the initial ratio of moles to volume.
Final moles of hydrogen (H₂) = Ratio of moles to volume * V₂
Final moles of hydrogen (H₂) = 0.6667 mol/L * 17.1 L = 11.4 mol

The number of moles of hydrogen added thus are:
Moles of hydrogen added = Final moles of hydrogen - Initial moles of hydrogen
Moles of hydrogen added = 11.4 mol - 2.6 mol = 8.8 mol

So, 8.8 moles of hydrogen were added to the balloon to give it a volume of 17.1 L.

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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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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).

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 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 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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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?

Explanation:

3120 j / (2000 g * (212-200 C) )  = .13 j /( g C)  

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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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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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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