Given the following equation: |4 NH3 (g) + 5 О2 (g) —> 4 NO (g) + 6 H20 () How many grams of H20 is produced if 21.1 grams of NH3 reacts with 73.9 grams of O2?

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:

what I dont get this bro

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:

7.186

Explanation:

The mean is the average of some given data from a sample point.

To calculate the mean, we use the formula below:

             Mean = ∑fx/∑f

Where f = frequency

           x = sample data

From the given pH of the solutions, we can form a table:

      x                                     f                                          fx

    7.15                                  1                                         7.15

    7.16                                  1                                         7.16

    7.18                                  1                                         7.18

    7.19                                  1                                         7.19

    7.20                                 3                                        21.6

    7.21                                  1                                         7.21

Now ∑fx = 7.15 +7.16 +7.18 + 7.19 + 7.20 + 7.21 = 57.49

         ∑f = 1 + 1 + 1 + 1 + 3 + 1 = 8

The mean = [tex]\frac{57.49}{8}[/tex] = 7.18625 = 7.186

Answer : The mass of sodium iodide used to produced must be, 105.22 grams.

Explanation : Given,

Mass of [tex]I_2[/tex] = 89.1 g

Molar mass of [tex]I_2[/tex] = 253.8 g/mole

Molar mass of [tex]NaI[/tex] = 149.89 g/mole

First we have to calculate the moles of [tex]I_2[/tex].

[tex]\text{Moles of }I_2=\frac{\text{Mass of }I_2}{\text{Molar mass of }I_2}=\frac{89.1g}{253.8g/mole}=0.351moles[/tex]

Now we have to calculate the moles of [tex]NaI[/tex].

The balanced chemical reaction is,

[tex]2NaI(aq)+Cl_2(g)\rightarrow I_2(s)+2NaCl(aq)[/tex]

From the balanced chemical reaction, we conclude that

As, 1 mole of [tex]I_2[/tex] obtained from 2 moles of [tex]NaI[/tex]

So, 0.351 moles of [tex]I_2[/tex] obtained from [tex]2\times 0.351=0.702[/tex] moles of [tex]NaI[/tex]

Now we have to calculate the mass of [tex]NaI[/tex].

[tex]\text{Mass of }NaI=\text{Moles of }NaI\times \text{Molar mass of }NaI[/tex]

[tex]\text{Mass of }NaI=(0.702mole)\times (149.89g/mole)=105.22g[/tex]

Therefore, the mass of sodium iodide used to produced must be, 105.22 grams.

To produce 89.1 grams of iodine (I2), 178.2 grams of sodium iodide (NaI) must be used.

To find the number of grams of sodium iodide (NaI) needed to produce 89.1 g of iodine (I2), we need to use the stoichiometric coefficients from the balanced chemical equation:

2NaI(aq) + Cl2(g) ⟶ I2(s) + 2NaCl(aq)

From the equation, we can see that 2 moles of NaI react to produce 1 mole of I2. The molar mass of NaI is 149.89 g/mol. We can set up a proportion:

(2 mol NaI / 1 mol I2) = (x g NaI / 89.1 g I2)

Solving for x:

x = (2 mol NaI / 1 mol I2) * (89.1 g I2 / 1 mol I2) = 178.2 g NaI

Therefore, 178.2 grams of sodium iodide (NaI) must be used to produce 89.1 grams of iodine (I2).

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Answer: The [tex]\Deltas H^o_{formation}[/tex] for the reaction is -1406.8 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of [tex]AlCl_3[/tex] is:

[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)[/tex]    [tex]\Delta H^o_{formation}=?[/tex]

The intermediate balanced chemical reaction are:

(1) [tex]HCl(g)\rightarrow HCl(aq.)[/tex]    [tex]\Delta H_1=-74.8kJ[/tex]    ( ×  6)

(2) [tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]    [tex]\Delta H_2=-185kJ[/tex]     ( ×  3)

(3) [tex]AlCl_3(aq.)\rightarrow AlCl_3(s)[/tex]    [tex]\Delta H_3=+323kJ[/tex]     ( ×  2)

(4) [tex]2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)[/tex]    [tex]\Delta H_4=-1049kJ[/tex]

The expression for enthalpy of formation of [tex]AlCl_3[/tex] is,

[tex]\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4][/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ[/tex]

Hence, the [tex]\Deltas H^o_{formation}[/tex] for the reaction is -1406.8 kJ.

The change in enthalpy for the reaction 2Al(s) + 3Cl2(g) ⟶ 2AlCl3(s) is -832.2 kJ, calculated by manipulating and summing the enthalpy changes of the given reactions using Hess's Law.

The change in enthalpy (∆H°) of the reaction 2Al(s) + 3Cl2(g) ⟶ 2AlCl3(s) can be calculated using Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction. In this case, we want to find the enthalpy change of reaction (i) using reactions (ii) - (v).

To achieve this, let's manipulate the reactions as follows:
1. Reverse reaction (ii) and multiply by 6. This will give it the same number of HCl (moles) as in reaction (v): 6HCl(aq) ⟶ 6HCl(g), ∆H′ = +74.8 kJ * 6 = +448.8 kJ.
2. Multiply reaction (iii) by 3 to match the number of H2 (moles) and HCl in reaction (v): 3H2 (g) + 3Cl2 (g) ⟶ 6HCl(g), ∆H′′ = -185 kJ * 3 = -555 kJ.
Reaction (iv) remains unchanged: AlCl3 (aq) ⟶ AlCl3 (s), ∆H′′′ = +323 kJ.
3. Using reaction (v) in the forward direction: 2Al(s) + 6HCl(aq) ⟶ 2AlCl3 (aq) + 3H2 (g), ∆H′′′′ = -1049 kJ.

Summing these manipulated reactions will give you reaction (i): ∆H° = ∆H′ + ∆H′′ + ∆H′′′ + ∆H′′′′ = 448.8 kJ - 555 kJ + 323 kJ - 1049 kJ = -832.2 kJ

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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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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:

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:

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

Explanation:

Answer: The number of moles of [tex]Na_2SO_4[/tex] is 0.05 moles.

Explanation:

To calculate the molarity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

Molarity of solution = 0.10 mol/L

Volume of solution = 1 L

Putting values in above equation, we get:

[tex]0.10mol/L=\frac{\text{Moles of sodium}}{1.0L}\\\\\text{Moles of sodium}=0.10mol[/tex]

The chemical reaction for the ionization of sodium sulfate follows the equation:

[tex]Na_2SO_4(s)\rightarrow 2Na^+(aq.)+SO_4^{2-}(aq.)[/tex]

By Stoichiometry of the reaction:

2 moles of sodium ions are produced by 1 mole of sodium sulfate

So, 0.10 moles of sodium ions will be produced by = [tex]\frac{1}{2}\times 0.1=0.05moles[/tex] of sodium sulfate.

Hence, the number of moles of [tex]Na_2SO_4[/tex] is 0.05 moles.

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

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.

Answer : The metal used was iron (the specific heat capacity is [tex]0.44J/g^oC[/tex]).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

[tex]q_1=-q_2[/tex]

[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]

where,

[tex]C_1[/tex] = specific heat of metal = ?

[tex]C_1[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

[tex]m_1[/tex] = mass of metal = 47.1 g

[tex]m_2[/tex] = mass of water = 120 g

[tex]T_f[/tex] = final temperature of water = [tex]24.5^oC[/tex]

[tex]T_1[/tex] = initial temperature of metal = [tex]99^oC[/tex]

[tex]T_2[/tex] = initial temperature of water = [tex]21.4^oC[/tex]

Now put all the given values in the above formula, we get

[tex]47.1g\times c_1\times (24.5-99)^oC=-120g\times 4.18J/g^oC\times (24.5-21.4)^oC[/tex]

[tex]c_1=0.44J/g^oC[/tex]

Form the value of specific heat of metal, we conclude that the metal used in this was iron.

Therefore, the metal used was iron (the specific heat capacity is [tex]0.44J/g^oC[/tex]).

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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Answer:

4B + 3O₂ => 2B₂O₃; ΔH° = -3673Kj

Explanation:

Work these type problems in pairs of rxns… That is, add Rxn-1 & Rxn-2 => Rxn-1,2; then add Rxn-3 to Rxn-1,2 => Rxn- 1,2,3. Rxn-4 is not needed to obtain target rxn.  

Target Rxn => 4B + 3O₂ => 2B₂O₃

Given …

(1) B₂O₃ + 2H₂O => 3O₂ + B₂H₆

       => reverse and double

       => 2B₂H₆ + 6O₂ => 2B₂O₃ + 6H₂O

(2) 2B + 3H₂ => B₂H₆ => double and add to Rxn-1 => 4B + 6H₂ => 2B₂H₆

         2B₂H₆ + 6O₂ => 2B₂O₃ + 6H₂O

              4B + 6H₂ => 2B₂H₆

              ________________________

∑(1,2)    4B + 6O₂ + 6H₂ => 2B₂O₃ + 6H₂O; ΔH°₁₂ = -2035Kj + (+72Kj) = -1963Kj

(3) => H₂ + ½O₂ => H₂O

       => reverse and multiply by 6, then add to (1,2) => 6H₂O => 6H₂ + 3O₂

                           6H₂O => 6H₂ + 3O₂

           4B + 6O₂ + 6H₂ => 2B₂O₃ + 6H₂O  

   ________________________

∑[(1,2,3) 4B + 3O₂ => 2B₂O₃; ΔH°₁₂₃ = -1963Kj + 6(-285Kj) = -3673Kj

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]

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.

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.

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.

Answer:

184.46

Explanation:

(84.7°c x 9/5)+32=184.46 °F

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.

Explanation:

Since, it is given that density is 1.16 grams per milliliter and molar mass of copper sulfate pentahydrate is 249.68 g/mol.

Now, as we known that density is the amount of mass present in a unit volume.

Mathematically,         Density = [tex]\frac{\text{molar mass}}{volume}[/tex]

Hence, calculate the volume as follws.

                    Density = [tex]\frac{\text{molar mass}}{volume}[/tex]

                  1.16 g/mL = [tex]\frac{249.68 g/mol}{volume}[/tex]          

                 volume = 215.24 mL

As, 215.24 mL can dissolve in 249.68 g/mol of complex. So, in 100 mL volume amount dissolved will be calculated as follows.

                       [tex]\frac{249.68 g/mol}{215.24 mL} \times 100[/tex]

                    = 116 grams        

Thus, we can conclude that 116 grams of copper sulfate pentahydrate that will dissolve in 100 g of water at 0[tex]^{o}C[/tex].                              

To calculate the grams of copper sulfate pentahydrate that dissolve in 100 g of water at 0°C, solubility data at that temperature is needed, which is not provided in the question.

The student is asking to calculate the grams of copper sulfate pentahydrate that will dissolve in 100 grams of water at 0°C, given that the density of the saturated solution is 1.16 g/mL and the molar mass of copper sulfate pentahydrate is 249.68 g/mol. Since the volume of solution can be derived from the mass of the water and the density of the solution, we can calculate the amount of copper sulfate pentahydrate dissolved in this volume by using dimensional analysis. However, to perform this calculation, we need the solubility data for copper sulfate pentahydrate at 0°C, which is not provided in the question. Without this information, we cannot proceed with the calculation.

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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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:

In this chemistry problem, we first calculate the heat produced by the combustion of C6H6, and then we use the heat equation to find the final temperature of the water which comes out to be 53.14 °C.

The subject of this question pertains to the field of chemistry, specifically physical chemistry dealing with energy changes in chemical reactions, and this seems to be a high school-level problem based on the complexities involved. The principle used here is q=mc∆T, which allows us to calculate how much heat is transferred when the water temperature changes.

First, the amount of heat produced by the combustion of 8.6 g of C6H6 should be calculated based on its enthalpy change (heat produced per mole). Given the heat of combustion, -6542 kJ for 2 moles of C6H6, the heat of combustion for 8.6g of C6H6 can be calculated by (8.6 g / (78.11 g/mol) mol ) * (-6542 kJ / 2 mol) = -227.11 kJ.

The heat absorbed by the water can then be calculated using the heat equation: q=mc∆T. Here, m = mass of water = 5691 g, c = specific heat capacity of water = 4.18 J/g°C. The heat is converted to kilojoules: q = -227.11 kJ * 1000 = -227110 J. Hence, the equation becomes -227110 = 5691*4.18*∆T. After rearranging and solving, the final temperature (T) will be 53.14 °C.

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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.

Final answer:

The net ionic equation for the reaction between barium hydroxide and sulfuric acid is Ba2+(aq) + SO42-(aq) → BaSO4(s), which shows the formation of an insoluble precipitate of barium sulfate.

Explanation:

The correct net ionic equation for the reaction between barium hydroxide and sulfuric acid is as follows:

Ba2+(aq) + SO42−(aq) → BaSO4(s)

In this equation, Ba2+(aq) represents the barium ion in aqueous solution, SO42−(aq) denotes the sulfate ion in aqueous solution, and BaSO4(s) is the precipitate of barium sulfate. It is important to remember that net ionic equations must be balanced by both mass and charge, and in this case, the equation follows these rules with a product that is insoluble in water.