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Answer : The line-angle formula for [tex](CH_3)_2CHCH(CH_3)_2[/tex] is shown below.

Explanation :

Condensed structural formula : It is a formula in which the lines are used between the bonded atoms and the atoms are also shown in the structural formula.

Line-angle formula : In the line-angle formula, the carbon atoms are represented at the corner and ends of the lines with some angle and the hydrogen atoms are hidden.

The line-angle formula for [tex](CH_3)_2CHCH(CH_3)_2[/tex] is shown below.

The line-angle formula for (CH3)2CHCH(CH3)2 is CH3 CH3   CH2  CH3 CH3.

The line-angle formula for (CH3)2CHCH(CH3)2 can be written as:

CH3         CH3

  |

 CH2

 |

CH3         CH3

This formula represents the structural arrangement of atoms in the molecule, with each line representing a bond between carbon atoms. The ends of the lines represent carbon atoms, and any additional atoms or groups attached to those carbon atoms are shown as branches off the main line.

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The removal of acid gases from natural gas is critical for ensuring the safety and longevity of the gas transportation and storage infrastructure, protecting human health and the environment, complying with regulations, and improving the quality and economic value of the natural gas product.

The removal of acid gases from natural gases is essential for several reasons:

1. Pipeline Corrosion Prevention: Acid gases, such as hydrogen sulfide (H2S) and carbon dioxide (CO2), can react with water to form acids. These acids, particularly sulfuric acid from H2S, can cause severe corrosion in pipelines and other transportation and storage facilities. Removing these gases prevents this corrosion, extending the lifespan of the infrastructure and reducing maintenance costs.

2. Safety and Health Concerns: Hydrogen sulfide is highly toxic and can be lethal at relatively low concentrations. It also has a strong, unpleasant odor. Removing H2S from natural gas is crucial to ensure the safety of workers and the public, as well as to comply with environmental and health regulations.

3. Product Quality: Acid gases can interfere with the burning properties of natural gas, leading to inefficient combustion and the potential for equipment malfunction. Removing these impurities improves the quality of the natural gas, ensuring it meets the specifications required for various applications, including its use as a fuel for power generation, industrial processes, and residential heating.

4. Environmental Protection: CO2 is a greenhouse gas, and its release into the atmosphere contributes to global warming. While natural gas combustion produces less CO2 than other fossil fuels like coal and oil, reducing the CO2 content in natural gas can further lessen its environmental impact. Additionally, the removal of sulfur compounds helps prevent the release of sulfur dioxide (SO2), a precursor to acid rain, during combustion.

5. Regulatory Compliance: Many countries have regulations that limit the amount of acid gases in natural gas. These regulations are in place to protect the environment and public health. Gas processing facilities must remove acid gases to comply with these standards.

6. Economic Value: Sulfur, which can be extracted from acid gases, has economic value and can be sold as a byproduct. The Claus process is a common method used to convert hydrogen sulfide into elemental sulfur. By recovering sulfur, the economic viability of natural gas processing is improved.

In summary, the removal of acid gases from natural gas is critical for ensuring the safety and longevity of the gas transportation and storage infrastructure, protecting human health and the environment, complying with regulations, and improving the quality and economic value of the natural gas product.

Answer:I can't art but I envision a comic of a swimming pool with chlorine water in it. The bottom of the pool is black. The chlorine is happy and excited to protect the people going to swim in it. Then the sun comes out, warms the black tile, the water is heated and the chlorine is boiled into gas form. Unable to control its movement through the atmosphere, the large amount of chlorine from the in-ground pool infiltrates the home of its lovely owners, and they die from chlorine gas inhalation, as well as half the neighborhood.

The end.

Here is a comic strip created to explain the transformation of a substance as it changes phases.

Title: The Adventures of Mr. Ice Cube

Introduction: Mr. Ice Cube is a solid block of water. He lives in a freezer, where it is very cold.

Source of Thermal Energy: One day, Mr. Ice Cube is taken out of the freezer and placed in a hot cup of coffee. The hot coffee transfers thermal energy to Mr. Ice Cube, causing him to melt.

Particle Transformation from Solid to Liquid Phase: As Mr. Ice Cube melts, the particles in his solid structure start to move faster. They move so fast that they break free from the solid structure and become liquid particles.

Particle Transformation from Liquid to Gas Phase: As Mr. Ice Cube continues to heat up, the liquid particles start to move even faster. They move so fast that they escape from the liquid state and become gas particles.

Conclusion: Mr. Ice Cube has now transformed from a solid to a liquid to a gas. He is now a cloud of water vapor, floating in the air above the hot cup of coffee.

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Answer:

9.7 x 10⁻⁴

Explanation:

              HA     ⇄           H⁺     +        A⁻

C(eq)   0.0174             10⁻²·³⁹         10⁻²·³⁹

                              =0.0041M     =0.0041M

Ka = [H⁺][A⁻]/[HA] = (0.0014)²/(0.0174) = 9.7 x 10⁻⁴

Answer:

Na₂CO₃.2H₂O

Explanation:

For the hydrated compound, let us denote is by Na₂CO₃.xH₂O

The unknown is the value of x which is the amount of water of crystallisation.

Given values:

Starting mass of hydrate i.e Na₂CO₃.xH₂O = 4.31g

Mass after heating (Na₂CO₃) = 3.22g

Mass of the water of crystallisation = (4.31-3.22)g = 1.09g

To determine the integer x, we find the number of moles of the anhydrous Na₂CO₃ and that of the water of crystallisation:

        Number of moles  = [tex]\frac{mass }{molar mass }[/tex]

Molar mass of Na₂CO₃ =[(23x2) + 12 + (16x3)]  = 106gmol⁻¹

Molar mass of H₂O = [(1x2) + (16)] = 18gmol⁻¹

Number of moles of Na₂CO₃ = [tex]\frac{3.22}{106}[/tex] = 0.03mole

Number of moles of H₂O =  [tex]\frac{1.09}{18}[/tex] = 0.06mole

From the obtained number of moles:

                          Na₂CO₃                               H₂O

                           0.03                                    0.06

Simplest

Ratio                  0.03/0.03                         0.03/0.06

                                 1                                      2

Therefore, x = 2    

After heating, the mass loss corresponds to 1.09 g of water, which, when compared to the moles of anhydrous [tex]\( \text{Na}_2\text{CO}_3 \)[/tex], gives a mole ratio of approximately 2:1, making [tex]\( x = 2 \)[/tex].

To determine the integer x in the hydrate [tex]\( \text{Na}_2\text{CO}_3 \cdot x \text{H}_2\text{O} \)[/tex], we need to follow these steps:

1. Calculate the mass of the water lost during heating:

[tex]\[\text{Mass of water} = \text{Mass of hydrate} - \text{Mass of anhydrous compound}\][/tex]

Given:

[tex]\[\text{Mass of hydrate} = 4.31 \, \text{g}\][/tex]

[tex]\[\text{Mass of anhydrous compound (Na}_2\text{CO}_3\text{)} = 3.22 \, \text{g}\][/tex]

Thus:

[tex]\[\text{Mass of water} = 4.31 \, \text{g} - 3.22 \, \text{g} = 1.09 \, \text{g}\][/tex]

2. Calculate the moles of anhydrous [tex]\( \text{Na}_2\text{CO}_3 \)[/tex]:

The molar mass of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] is calculated as follows:

[tex]\[\text{Molar mass of Na}_2\text{CO}_3 = 2 \times 23.0 + 12.0 + 3 \times 16.0 = 46.0 + 12.0 + 48.0 = 106.0 \, \text{g/mol}\][/tex]

[tex]\[\text{Moles of Na}_2\text{CO}_3 = \frac{\text{Mass of Na}_2\text{CO}_3}{\text{Molar mass of Na}_2\text{CO}_3} = \frac{3.22 \, \text{g}}{106.0 \, \text{g/mol}} = 0.0304 \, \text{mol}\][/tex]

3. Calculate the moles of water lost:

The molar mass of water [tex](\( \text{H}_2\text{O} \))[/tex] is:

[tex]\[\text{Molar mass of H}_2\text{O} = 2 \times 1.0 + 16.0 = 18.0 \, \text{g/mol}\][/tex]

[tex]\[\text{Moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of H}_2\text{O}} = \frac{1.09 \, \text{g}}{18.0 \, \text{g/mol}} = 0.0606 \, \text{mol}\][/tex]

4. Determine the mole ratio of water to [tex]\( \text{Na}_2\text{CO}_3 \)[/tex]:

[tex]\[x = \frac{\text{Moles of water}}{\text{Moles of Na}_2\text{CO}_3} = \frac{0.0606 \, \text{mol}}{0.0304 \, \text{mol}} = 1.993 \approx 2\][/tex]

Thus, the integer x in the hydrate[tex]\( \text{Na}_2\text{CO}_3 \cdot x \text{H}_2\text{O} \)[/tex]is 2.

In the balanced molecular equation between CuCl₂(aq) and K₃PO₄(aq), the cations and anions exchanged, resulting in 6KCl(aq) and Cu₃(PO₄)₂(s). The net ionic equation, focusing only on participating substances is: 3Cu²⁺(aq) + 2PO₄³⁻(aq) → Cu₃(PO₄)₂(s).

The reaction between aqueous copper(II) chloride (CuCl2) and aqueous potassium phosphate (K3PO4) is a double displacement reaction, where the cation of one compound combines with the anion of the other compound. The balanced molecular equation for this reaction is: 2CuCl2(aq) + 3K3PO4(aq) → 6KCl(aq) + Cu3(PO4)2(s).

Now, for the net ionic equation, you would only include the species that participate in the reaction and exclude the spectator ions (ions that do not participate in the reaction). After breaking down the molecular equation to ions (excluding spectator ions) the net ionic equation of the reaction would be: 3Cu^2+(aq) + 2PO4^3-(aq) → Cu3(PO4)2(s).

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The balanced molecular equation is 3CuCl₂(aq) + 2K₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6KCl(aq). The net ionic equation is 3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s). This represents the formation of copper(II) phosphate precipitate.

Complete and Balance the Molecular Equation

Let's complete and balance the molecular equation for the reaction of aqueous copper(II) chloride (CuCl₂) and aqueous potassium phosphate (K₃PO₄).

Molecular Equation:

CuCl₂(aq) + K₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6KCl(aq)

First, we need to balance this equation:

Balance the copper (Cu) atoms: 3 CuCl₂(aq) + K₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6KCl(aq)

Balance the potassium (K) atoms: 3 CuCl₂(aq) + 2 K₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6KCl(aq)

Write the Net Ionic Equation

To write the balanced net ionic equation, first write the complete ionic equation:

3 Cu²⁺(aq) + 6 Cl⁻(aq) + 6 K⁺(aq) + 2 PO₄³⁻(aq) ⟶ 6 K⁺(aq) + 6 Cl⁻(aq) + Cu₃(PO₄)₂(s)

Next, identify and remove the spectator ions (K⁺ and Cl⁻), which are ions that do not participate in the reaction:

Net Ionic Equation:

3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

This net ionic equation represents the formation of copper(II) phosphate precipitate from copper(II) ions and phosphate ions in solution.

Answer:

True => (3, 2, 0, -1/2) => 3d₀² electron

Explanation:

The enthalpy change of the reaction per mole of limiting reactant is -568 kJ/mol.

The enthalpy change of a reaction can be calculated using the amount of reactant used. In this case, 0.0500 mol of HCl was used, and the reaction absorbed 104 J of heat. Since AH is an extensive property, it is proportional to the amount of acid neutralized. Therefore, the enthalpy change for 1 mol of HCl is -2.9 kJ. To find the enthalpy of the reaction per mole of limiting reactant, we can use the molar mass of the metal and the stoichiometric ratio between HCl and the metal in the balanced equation.

To calculate the enthalpy change per mole of limiting reactant, we need to determine the number of moles of the metal. Since the molar mass of the metal is 57.78 g/mol, we can calculate the number of moles by dividing the mass of the metal (0.295 g) by its molar mass: 0.295 g ÷ 57.78 g/mol = 0.00510 mol.

Next, we use the stoichiometric ratio in the balanced equation to relate the moles of HCl reacted to the moles of metal reacted. From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of metal. Therefore, the moles of HCl reacted is also 0.00510 mol. Now we can calculate the enthalpy change per mole of limiting reactant: -2.9 kJ ÷ 0.00510 mol = -568 kJ/mol (rounded to three significant figures).

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Answer:

1) The no. of grams of Na₂CO₃ = 1.96 g.

2) The no. of grams of AgNO₃ = 0.0 g, because it is the limiting reactant that is consumed completely.

3) The no. of grams of Ag₂CO₃ = 3.998 g ≅ 4.0 g.

4) The no. of grams of NaNO₃ = 2.498 g ≅ 2.5 g.

Explanation:

Na₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃(s) + 2NaNO₃,

It is clear that 1 mol of Na₂CO₃ and 2 mol of AgNO₃ to produce 1 mol of Ag₂CO₃ and 2 mol of NaNO₃.

no. of moles of Na₂CO₃ = mass/molar mass = (3.5 g)/(105.9888 g/mol) = 0.033 mol.

no. of moles of AgNO₃ = mass/molar mass = (5.0 g)/(169.87 g/mol) = 0.0294 mol.

1) the no. of grams of sodium carbonate, and silver nitrate:

1 mol of Na₂CO₃ reacts completely with 2 mol of AgNO₃ with (1: 2 molar ratio).

∴ 0.0145 mol of Na₂CO₃ "the excess reactant, and the remaining is in excess (0.033 mol - 0.0145 mol = 0.0185 mol)" reacts completely with (0.0294 mol) of AgNO₃ "limiting reactant".

∴ The no. of grams of Na₂CO₃ = (no. of moles remaining)(molar mass) = (0.0185 mol)((105.9888 g/mol) = 1.96 g.

∴ The no. of grams of AgNO₃ = 0.0 g, because it is the limiting reactant that is consumed completely.

2) the no. of grams of silver carbonate:

Using cross multiplication:

1 mol of Na₂CO₃ produces → 1 mol of Ag₂CO₃, from stichiometry.

∴ 0.0145 mol of Na₂CO₃ produces → 0.0145 mol of Ag₂CO₃.

∴ The no. of grams of Ag₂CO₃ = (no. of moles of Ag₂CO₃)(molar mass of Ag₂CO₃) = (0.0145 mol)(275.7453 g/mol) = 3.998 g ≅ 4.0 g.

3) the no. of grams of sodium nitrate:

Using cross multiplication:

2 mol of AgNO₃ produces → 2 mol of NaNO₃, from stichiometry.

∴ 0.0294 mol of AgNO₃ produces → 0.0294 mol of NaNO₃.

∴ The no. of grams of NaNO₃ = (no. of moles of NaNO₃)(molar mass of NaNO₃) = (0.0294 mol)(84.9947 g/mol) = 2.498 g ≅ 2.5 g.

After the reaction of 3.50 g sodium carbonate and 5.00 g silver nitrate, approximately 0.42 g of sodium carbonate, 0 g of silver nitrate, 7.8 g of silver carbonate, and 2.3 g of sodium nitrate will be present.

The reaction between sodium carbonate (Na2CO3) and silver nitrate (AgNO3) forms solid silver carbonate (Ag2CO3) and a solution of sodium nitrate (NaNO3). The balanced chemical equation for this reaction is 2AgNO3 + Na2CO3 yields 2Ag2CO3 (s) + 2NaNO3 (aq). To determine the amount of each compound present after the reaction, we first need to calculate the moles of sodium carbonate and silver nitrate. We find that 3.50 g Na2CO3 amounts to 0.033 moles and 5.00 g AgNO3 equals 0.029 moles. Since AgNO3 is the limiting reagent, it will be completely consumed in the reaction, producing 0.029 moles of Ag2CO3 and 0.029 moles of NaNO3. This results 0.004 moles of Na2CO3 left unreacted. Converting these moles back into grams gives approximately 0.42 g Na2CO3, 0 g AgNO3, 7.8 g Ag2CO3, and 2.3 g NaNO3 present after the reaction.

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Answer: The mass of aluminium chloride that can be formed are 46.3 g

Explanation:

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

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

Given mass of aluminium = 32 g  

Molar mass of aluminium = 26.98 g/mol

Putting values in above equation, we get:  

[tex]\text{Moles of aluminium}=\frac{32g}{26.98g/mol}=1.186mol[/tex]

Given mass of chlorine = 37 g  

Molar mass of chlorine = 71 g/mol

Putting values in above equation, we get:  

[tex]\text{Moles of chlorine gas}=\frac{37g}{71g/mol}=0.521mol[/tex]

For the given chemical equation:

[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]

By Stoichiometry of the reaction:

3 moles of chlorine gas is reacting with 2 moles of aluminium.

So, 0.521 moles of chlorine gas will react with = [tex]\frac{2}{3}\times 0.521=0.347moles[/tex] of aluminium.

As, given amount of aluminium is more than the required amount. Thus, it is considered as an excess reagent.

So, chlorine gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of chlorine gas is producing 2 moles of aluminium chloride

So,  0.521 moles of chlorine gas will react with = [tex]\frac{2}{3}\times 0.521=0.347moles[/tex] of aluminium chloride.

Now, calculating the mass of aluminium chloride by using equation 1, we get:

Moles of aluminium chloride = 0.347 moles

Molar mass of aluminium chloride = 133.34 g/mol

Putting all the values in equation 1, we get:

[tex]0.347mol=\frac{\text{Mass of aluminium chloride}}{133.34g/mol}\\\\\text{Mass of aluminium chloride}=46.3g[/tex]

Hence, the mass of aluminium chloride that can be formed are 46.3 g

The quantity of the substance is given by the mass. The mass of aluminium chloride formed when reacting aluminium with chlorine will be 46.3 gm.

Mass is a quantitative factor that determines the amount of substance or matter present in the sample.

The chemical reaction can be shown as,

[tex]\rm 2 Al + 3Cl_{2} \rightarrow 2AlCl_{3}[/tex]

Calculate the number of moles of Aluminium:

[tex]\begin{aligned}\rm Moles & = \dfrac {\rm Mass }{\rm Molar\; mass}\\\\& = \dfrac{32}{26.98}\\\\& = 1.186\;\rm moles\end{aligned}[/tex]

Calculate the number of moles of Chlorine:

[tex]\begin{aligned}\rm Moles & = \dfrac {\rm Mass }{\rm Molar\; mass}\\\\& = \dfrac{37}{71}\\\\& = 0.521 \;\rm moles\end{aligned}[/tex]

From the stoichiometry of the reaction above:

2 moles of aluminium reacts with 3 moles of chlorine

So, moles of aluminium will react with 0.521 moles chorine is,

[tex]\dfrac{2}{3} \times 0.521 = 0.347 \;\rm moles[/tex]

From this, it can infer that chlorine gas is a limiting reagent and aluminium is an excess reagent.

From the stoichiometry of the reaction,

3 moles of chlorine =  2 moles of aluminium chloride

So, 0.521 moles chorine will give,

[tex]\dfrac{2}{3} \times 0.521 = 0.347 \;\text{ moles of aluminium chloride.}[/tex]

Calculate the mass of the aluminium chloride as:

[tex]\begin{aligned} \rm mass &= \rm moles \times molar\; mass\\\\&= 0.347 \times 133.34\\\\&= 46.3 \;\rm g\end{aligned}[/tex]

Therefore, the mass of the aluminium chloride is 46.3 gm.

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Answer:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

α =2

β = 19

γ = 12

δ = 14

53.2moles of O₂

Explanation:

Proper equation of the reaction:

                    αC₆H₁₄ + βO₂ → γCO₂ + δH₂O

This is a combustion reaction for a hydrocarbon. For the combustion of a hydrocarbon, the combustion equation is given below:

         CₓHₙ + (x + [tex]\frac{n}{4}[/tex])O₂ → xCO₂ + [tex]\frac{n}{2}[/tex]H₂O

From the given combustion equation, x = 6 and n = 14

Therefore:

β = x + [tex]\frac{n}{4}[/tex] = 6 + [tex]\frac{14}{4}[/tex] = 6 + 3.5 = 9[tex]\frac{1}{2}[/tex]

γ = 6

δ = [tex]\frac{n}{2}[/tex] = [tex]\frac{14}{2}[/tex] = 7

The complete reaction equation is therefore given as:

                   C₆H₁₄ + 9[tex]\frac{1}{2}[/tex]O₂ → 6CO₂ + 7H₂O

To express as whole number integers, we multiply the coefficients through by 2:

                  2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

Problem 2

           From the reaction:

2 moles of hexane are required to completely react with 19 moles of O₂

∴ 5.6 moles of hexane would react with k moles of O₂

This gives:     5.6 x 19 = 2k

                        k = [tex]\frac{5.6 x 19}{2}[/tex]

                        k = 53.2moles of O₂

Answer:

[tex]\boxed{\text{47.4 g}}[/tex]

Explanation:

We are given the mass of two reactants, so this is a limiting reactant problem.

We know that we will need mases, moles, and molar masses, so, let's assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:    17.03   32.00                 18.02  

           4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

m/g:     70.1      70.1

Step 1. Calculate the moles of each reactant

[tex]\text{Moles of CO } = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{17.03 g}} = \text{4.116 mol}\\\\\text{Moles of H$_{2}$O} = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{32.00 g}} = \text{2.191 mol}[/tex]

Step 2. Identify the limiting reactant  

Calculate the moles of H₂O we can obtain from each reactant.

From NH₃:

The molar ratio of H₂O:NH₃ is 6:4.

[tex]\text{Moles of H$_{2}$O} = \text{4.116 mol NH$_{3}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{4 mol NH$_{3}$}} = \text{6.174 mol H$_{2}$O}[/tex]

From O₂:  

The molar ratio of H₂O:O₂ is 6:5.  

[tex]\text{Moles of H$_{2}$O} = \text{2.191 mol O$_{2}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{5 mol O$_{2}$}} = \text{2.629 mol H$_{2}$O}[/tex]

O₂ is the limiting reactant because it gives the smaller amount of H₂O.  

Step 3. Calculate the theoretical yield.

[tex]\text{Theor. yield } = \text{2.629 mol H$_{2}$O}\times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{47.4 g H$_{2}$O}\\\\\text{The maximum yield of H$_{2}$O is }\boxed{\textbf{47.4 g}}[/tex]

Answer: [tex]0.67\times 10^{23}[/tex]  molecules are contained in the flask.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atm respectively.

According to the ideal gas equation:'

[tex]PV=nRT[/tex]

P= Pressure of the gas = 1 atm

V= Volume of the gas = 2.50 L

T= Temperature of the gas = 273 K

R= Value of gas constant = 0.0821 Latm/K mol

[tex]n=\frac{PV}{RT}=\frac{1\times 2.50L}{0.0821 \times 273}=0.11moles[/tex]

1 mole of gas contains=[tex]6.022\times 10^{23}[/tex]  molecules  

0.11 moles of gas contains=[tex]\frac{6.022\times 10^{23}}{1}\times 0.11=0.67\times 10^{23}[/tex]  molecules

[tex]0.67\times 10^{23}[/tex]  molecules are contained in the flask.

Answer: The coefficients for balancing the given chemical equation are 2, 7, 4 and 6

Explanation:

Every balanced chemical equation follows law of conservation of mass.

This law states that mass can neither be created nor be destroyed, but it can only be transformed from one form to another form. This also means that total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.

The given balanced chemical equation follows:

[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]

On reactant side:

Number of carbon atoms = 4

Number of hydrogen atoms = 12

Number of oxygen atoms = 14

On product side:

Number of carbon atoms = 4

Number of hydrogen atoms = 12

Number of oxygen atoms = 14

Hence, the coefficients for balancing the given chemical equation are 2, 7, 4 and 6

D. radioactive isotopes are one of the environmental waste products of nuclear energy.

One of the environmental side effects of nuclear energy is radioactive isotopes.

A chemical element in an unstable state that emits radiation as it decomposes and becomes more stable.

Radioisotopes can be created in a lab or in the natural world. They are utilized in imaging studies and therapy in medicine.

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Answer:

The carbon is sp² hybridized and the unpaired electron occupies a 2p orbital.  

Explanation:

The central C atom has a trigonal planar geometry, so it is sp² hybridized.

The sp² bonds are formed by the hybridization of the 2s orbital with two 2p orbitals.

The unpaired electron goes into the unhybridized 2p orbital.

Answer:

Here's what I find.

Explanation:

a. Structure

Acebutolol is a secondary amine (basic). It forms a substituted ammonium salt when treated with hydrochloric acid.

The structure of the salt is shown below, with a red arrow pointing toward the positive charge on the N atom.

b. Solubility

The formula of acebutolol is C₁₈H₂₈N₂O₄.

The amide, acetyl, and ether groups confer little solubility to the molecule.

The alcohol and secondary amine do confer some solubility, because they can donate and accept hydrogen bonds.

However, they can each overcome the hydrophobic properties of only three to five carbons, and acebutolol has 18 of them.

The free amine would be preferentially soluble in lipid material (fats)

The protonated amine is ionic and therefore much more soluble in aqueous media (e.g., blood).

c. Marketing

The drug must be delivered to the tissues of the heart, where it blocks the effects of adrenalin. The best way to do this is through the blood, so acebutolol is marketed as the hydrochloride salt.

Answer:

what I dont get this bro

By equating the heat gained by the melting ice to the heat lost by the water, and using the enthalpy of fusion for ice and the specific heat of water, the calculation reveals that there were approximately 595 grams of water in the sample.

To find out how many grams of water were in the sample before adding the ice, we need to equate the heat gained by the melting ice to the heat lost by the water when its temperature decreased from 7.4 °C to 0.0 °C. The enthalpy of fusion of ice, which is the amount of heat required to melt the ice without changing its temperature, is a key concept in this problem. The specific heat of water is also an important factor as it influences the amount of heat water can lose or absorb.

The enthalpy of fusion of ice is typically given as 334 J/g. Using this, the heat gained by the ice as it melts can be calculated by multiplying the mass of the ice (48 g) by the enthalpy of fusion. The heat lost by the water can be calculated using the specific heat of water (4.184 J/g°C), the change in temperature (7.4 °C), and the unknown mass of water.

Heat gained by ice = Heat lost by water (Qice = Qwater)

334 J/g × 48 g = 4.184 J/g°C × (mass of water) × 7.4 °C

By rearranging the equation, we find the mass of water:

mass of water = (334 J/g × 48 g) / (4.184 J/g°C × 7.4 °C)

After solving, the mass of the water is approximately 595 g.

To calculate the mass of the solid CaCO3 in the vessel once equilibrium is reached, we need to use the equilibrium constant (K) and the given information. The mass of the solid CaCO3 can be calculated by subtracting the mass of the solid CaO from the initial mass of CaCO3. The molar concentration of CaCO3 can be calculated using the given mass and volume, and the equilibrium constant can be used to determine the concentrations of CaO and CO2.

To calculate the mass of the solid CaCO3 in the vessel once equilibrium is reached, we need to use the equilibrium constant (K) and the given information. The equilibrium constant (K) is 1.16 at 800 °C. We are given a 5.00 L vessel containing 10.0 g of CaCO3(s) and the volume occupied by the solid is ignored. To calculate the mass of the solid in the vessel, we can use the equation n = m/M to calculate the number of moles of CaCO3 and then multiply it by the molar mass of CaCO3 to get the mass.

Given that the volume of the vessel is 5.00 L, we can calculate the initial concentration of CaCO3 (Molar concentration = moles/volume) as [CaCO3] = (10.0g / 100.09 g/mol) / 5.00 L = 0.02 mol/L.

Now we can use the equation Q = [CaO][CO2] / [CaCO3] = 1.16 (as it's at equilibrium) and substitute the concentrations as 0.02 mol/L and calculate the concentration of [CaO] and [CO2]. Since the volume occupied by the solid is ignored, the concentrations of CaO and CO2 will be the same. Now we can multiply the concentration of CaO by the volume to get the moles of CaO. Finally, multiply the moles of CaO by its molar mass to calculate the mass of the solid CaO.

Therefore, the mass of the solid CaCO3 in the vessel once equilibrium is reached is the initial mass of CaCO3 minus the mass of the solid CaO, which is calculated above.

Answer:

K(net) = K₁ x K₂ = 1.51 x 10⁻²⁶

Explanation:

The given equations (a)-(d) represent synthesis, decomposition, neutralization, and precipitation reactions, respectively.

(a) The given equation represents a synthesis reaction, where hydrogen gas (H2) reacts with carbon solid (C) to form carbon monoxide gas (CO) and hydrogen gas (H2).

(b) The given equation represents a decomposition reaction, where potassium chlorate solid (KClO3) decomposes to form potassium chloride solid (KCl) and oxygen gas (O2).

(c) The given equation represents a neutralization reaction, where aluminum hydroxide (Al(OH)3) reacts with hydrochloric acid (HCl) to form aluminum chloride (AlCl3) and water (H2O).

(d) The given equation represents a precipitation reaction, where lead nitrate (Pb(NO3)2) reacts with sulfuric acid (H2SO4) to form lead sulfate solid (PbSO4) and nitric acid (HNO3) in aqueous form.

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The given equations (a)-(d) represent, a) Synthesis reaction, b) Decomposition reaction, c) neutralization reaction and d) precipitation reaction.

Reaction (a): H₂O(g) + C(s) ⟶ CO(g) + H₂(g) is a synthesis reaction, also known as a combination reaction, where two or more reactants combine to form a single product. In this case, hydrogen gas (H₂) and carbon monoxide (CO) are produced from the combination of water vapor (H₂O) and carbon (C).

Reaction (b): 2KClO₃(s) ⟶ 2KCl(s) + 3O₂(g) is a decomposition reaction, where a single compound breaks down into two or more simpler substances.

Reaction (c): Al(OH)₃(aq) + 3HCl(aq) ⟶ AlCl₃(aq) + 3H₂O(l) is an acid-base neutralization reaction, in which an acid (HCl) reacts with a base (Al(OH)₃) to produce a salt (AlCl₃) and water.

Reaction (d): Pb(NO₃)₂(aq) + H₂SO₃(aq) ⟶ PbSO₄(s) + 2HNO₃(aq) is a precipitation reaction, where two solutions react to form an insoluble solid (precipitate) and a soluble compound.

Answer:

(a) 5Ca(OH)₂ + 3H₃PO₄ = Ca₅(PO₄)₃(OH) + 9H₂O.

(b) 135.62 g.

Explanation:

(a) Write a balanced equation for this preparation.

5Ca(OH)₂ + 3H₃PO₄ = Ca₅(PO₄)₃(OH) + 9H₂O,

that 5 mol of Ca(OH)₂ react with 3 mol of H₃PO₄ to produce 1 mol of Ca₅(PO₄)₃(OH) "hydroxyapatite" and 9 mol of H₂O.

(b) What mass of hydroxyapatite could form from 100.0 g of 85% phosphoric acid and 100.0 g of calcium hydroxide?

The no. of moles of 100.0 g of 85% phosphoric acid:

∵ The percent of phosphoric acid = 85%.

∴ The actual mass of phosphoric acid in solution = 85.0 g.

∴ no. of moles of phosphoric acid = mass/molar mass = (85.0 g)/(97.994 g/mol) = 0.87 mol.

The no. of moles of 100.0 g of calcium hydroxide:

∴ no. of moles of calcium hydroxide = mass/molar mass = (100.0 g)/(74.093 g/mol) = 1.35 mol.

From the stichiometry, Ca(OH)₂ react with H₃PO₄ with (5: 3) molar ratio.

∴ 1.35 mol of calcium hydroxide "limiting reactant" react completely with 0.813 mol of phosphoric acid "excess reactant" with (5: 3) molar ratio.

Using cross multiplication:

5 mol of Ca(OH)₂ produce → 1 mol Ca₅(PO₄)₃(OH), from stichiometry.

1.35 mol of Ca(OH)₂ produce → ??? mol Ca₅(PO₄)₃(OH).

∴ The no. of moles of produced Ca₅(PO₄)₃(OH) "hydroxyapatite" = (1.35 mol)(1 mol)/(5 mol) = 0.27 mol.

Finally, we can get the mass of produced Ca₅(PO₄)₃(OH) "hydroxyapatite" = (no. of moles)(molar mass) = (0.27 mol)(502.3 g/mol) = 135.62 g.

The equation of the reaction is:

Based on the data provided, the mass of hydroxyapatite formed is  135.62 g.

Minerals are substances which contain metallic elements combined with other elements found in the earth's crust.

Example of a mineral is Hydroxyapatite, Ca5(PO4)3(OH).

The balanced equation for the  preparation of hydroxyapatite is given below:

that 5 mol of Ca(OH)₂ react with 3 mol of H₃PO₄ to produce 1 mol of Ca₅(PO₄)₃(OH) and 9 mol of H₂O.

The mass of hydroxyapatite prepared from 100.0 g of 85% phosphoric acid and 100.0 g of calcium hydroxide? is calculated as follows:

100.0 g of 85% phosphoric acid contains 100 * 85/100 g of  phosphoric acid = 85.0 g.

moles of phosphoric acid in 85 g = mass/molar mass

molar mass of phosphoric acid = 97.994 g/mol

moles of phosphoric acid in 85 g = 85.0 g/97.994 g/mol

moles of phosphoric acid in 85 g = 0.87 moles

moles of calcium hydroxide in 100.0 g = mass/molar mass

molar mass of calcium hydroxide = 74.093 g/mol

moles of calcium hydroxide in 100.0 g = 100.0 g/74.093 g/mol

moles of calcium hydroxide in 100.0 g = 1.35 moles

From the equation of the reaction, Ca(OH)₂ react with H₃PO₄ in a 5: 3 molar ratio.

5 mol of Ca(OH)₂ produce 1 mole Ca₅(PO₄)₃(OH)

1.35 mol of Ca(OH)₂ will  produce 1.35 * 1/5 moles of Ca₅(PO₄)₃(OH) = 0.27 moles

mass of 0.27 moles of hydroxyapatite = no. of moles * molar mass

molar mass of hydroxyapatite = 502.3 g/mol

mass of hydroxyapatite =  0.27 mol * 502.3 g/mol

mass of hydroxyapatite = 135.62 g

Therefore, the mass of hydroxyapatite produced is 135.62 g

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The solution with a pH of 11 has a hydrogen ion concentration of 1 x 10^-11. Hence, the correct option is B.

The question involves understanding the relationship between pH and hydrogen ion concentration ([H+]) in solutions. The pH of a solution is a measure of its acidity or basicity, with values less than 7 being acidic, around 7 being neutral, and greater than 7 being basic. The [H+] is related to pH through the formula [H+] = 10-pH.

For a solution with a pH of 11, we can calculate its [H+] concentration using the formula mentioned. Converting the pH into the hydrogen ion concentration gives us [H+] = 10-11. Thus, the correct answer is B. 1 x 10-11.

This formula demonstrates the logarithmic relationship between pH and hydrogen ion concentration, highlighting that small changes in pH equate to exponential changes in [H+]. For instance, a pH difference of 1 corresponds to a tenfold difference in hydrogen ion concentration. The question essentially tests the understanding of this critical concept in Chemistry.

Answer:

-112 Kj/mole (diprotic acid)

Explanation:

The enthalpy of reaction is -55 KJ/mol.

We must first write down the equation of the reaction;

2KOH(aq) + H2SO4(aq) ------> K2SO4(aq) + 2H2O(l)

Then we compute the number of moles of H2SO4 = 23.6/1000 × 0.500 M  = 0.012 moles

And the number of moles of KOH = 23.6/1000 × 1.00 M = 0.024 moles

From what we know in the reaction equation;

2 moles of KOH produces 2 moles of water

Therefore, 0.0026 moles of KOH produces 0.024 moles of water.

The total volume of solution is obtained by adding  = 23.6 mL + 23.6 mL = 47.2 mL

Mass of water = density × volume = 1.00 g/mL ×  47.2 mL =47.2 g

Using the formula;

ΔH = mcθ

Mass of solution (m) = 47.2 g

Specific heat capacity of solution (c) =  4.184 J/g·°C

Temperature difference(θ) =  30.17°C - 23.50°C = 6.67°C

Substituting values;

ΔH = -( 47.2 g × 4.184 J/g·°C  × 6.67°C)/ 0.024 moles  

ΔH =  -(1.32 KJ/0.024 moles)

ΔH = -55 KJ/mol

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Answer:

0.2208 is the equilibrium constant,[tex]K_p[/tex].

Explanation:

Equilibrium constant is defined as ratio of concentration of products to the concentration of reactants raised to the power equal to their stoichiometric coefficients in balanced chemical equation. It is expressed as [tex]K_c[/tex]

If the equilibrium is in gaseous phase then instead of concentration take partial pressure of each compound.The It is expressed as [tex]K_p[/tex].

[tex]CO(g)+Cl_2(g)\rightarrow COCl_2(g)[/tex]

Partial pressure of the[tex] p_{[CO]} = 0.820 atm[/tex]

Partial pressure of the[tex] p_{[Cl_2]} = 1.27 atm[/tex]

Partial pressure of the [tex]p_{[COCl_2]} = 0.230 atm[/tex]

The expression of an equilibrium constant is given as:

[tex]K_p=\frac{p_{[COCl_2]}}{p_{[CO]}p_{[Cl_2]}}[/tex]

[tex]K_p=\frac{0.230 atm}{0.820 atm \times 1.27 atm}=0.2208 [/tex]

Answer:

[tex]\boxed{\text{2.17 g/cm}^{3}}[/tex]

Explanation:

1. Ions per unit cell

(a) Chloride

8 corners + 6 faces

[tex]\text{No. of Cl$^{-}$ ions}\\\\= \text{8 corners} \times \dfrac{\frac{1 }{ 8} \text{ ion}} {\text{1 corner}} + \text{6 faces}\times \dfrac{\frac{1}{2} \text{ ion}}{\text{1 face}} = \text{1 ion + 3 ions = 4 ions}}[/tex]

(b) Chloride

12 edges + 1 centre

[tex]\text{No. of Na$^{+}$ ions}\\\\= \text{12 edges} \times \dfrac{\frac{1 }{ 4} \text{ ion}} {\text{1 edge}} + \text{1 centre}\times\dfrac{\text{1 ion}}{\text{1 centre}} = \text{3 ions + 1 ion = 4 ions}[/tex]

There are four formula units of NaCl in a unit cell.

2. Mass of unit cell

m = 4 × NaCl = 4 × 58.44 u = 233.76 u

[tex]m = \text{233.76 g} \times \dfrac{\text{1 g} }{6.022 \times 10^{23} \text{ u} } = 3.882 \times 10^{-22}\text{ g}[/tex]

3. Volume of unit cell

(a) Edge length

[tex]a = 2d_{\text{Na-Cl}} = 2 \times \text{2.819 \AA} = 5.638 \times10^{-10} \text{ m} = 5.638 \times10^{-8} \text{ cm}[/tex]

(b) Volume

[tex]V = a^{3} = \left( 5.638 \times (10^{-8} \text{ cm}\right)^{3} = 1.792 \times10^{-22} \text{ cm}^{3}[/tex]

4. Density

[tex]\rho = \dfrac{\text{mass}}{\text{volume}} = \dfrac{3.882 \times 10^{-22}\text{ g}}{1.792 \times10^{-22} \text{ cm}^{3}}} = \text{2.17 g/cm}^{3}\\\\\text{The density of NaCl is }\boxed{\textbf{2.17 g/cm}^{3}}[/tex]

Final answer:

Using the half-life of Potassium-40, which is 1.25 billion years, and given the 8:1 ratio of K-40 to Ar-40, the rock sample is determined to be approximately 3.75 billion years old.

Explanation:

To date the rock sample to the time it contained only Potassium-40 (K-40), we must analyze the ratio of K-40 to Argon-40 (Ar-40). Given the information that there are eight times as many K-40 atoms as Ar-40 atoms in the sample, we can use the concept of half-lives to date the sample. One half-life (1.25 billion years) would result in an equal number of K-40 and Ar-40 atoms.

Since we have eight times as many K-40 atoms, the sample has gone through three half-lives (since 2³ = 8), which equates to 3.75 billion years (three half-lives of 1.25 billion years each). The rock sample would therefore be approximately 3.75 billion years old, which is the time it would have contained only K-40 atoms.

Answer:

184.46

Explanation:

(84.7°c x 9/5)+32=184.46 °F