A 5.36–g sample of NH4Cl was added to 25.0 mL of 1.00 M NaOH and the resulting solution diluted to 0.100 L. (a) What is the pH of this buffer solution? (b) Is the solution acidic or basic? (c) What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to the solution?

Answer: The correct answer is Option c.

Explanation:

Weak acid is defined as the acid which does not get completely dissociated into its ions when dissolved in water. For Example: [tex]CH_3COOH,H_2CO_3[/tex] etc..

Strong acid is defined as the acid which gets completely dissociated into its ions when dissolved in water. For Example: [tex]HCl,HNO_3[/tex] etc..

Weak base is defined as the base which does not get completely dissociated into its ions when dissolved in water. For Example: [tex]NH_3,NH_4OH[/tex] etc..

Strong base is defined as the base which gets completely dissociated into its ions when dissolved in water. For Example: [tex]NaOH,KOH[/tex] etc..

From the above information, it is clearly visible that the correct answer is Option c.

The correct statement is that C. NH3 is a weak base, and HCl is a strong acid.

When a weak base, such as ammonia (NH3), reacts with a strong acid, like hydrochloric acid (HCl), a chemical reaction occurs. The acid donates protons (H+) to the base, forming water and the conjugate acid of the weak base. For example, in the reaction between NH3 and HCl, [tex]NH3 + HCl -- NH4+ + Cl-,[/tex]  ammonium chloride is formed. The resulting solution is acidic due to the presence of excess H+ ions.

Thus, the correct statement among the options is c. NH3 is a weak base, and HCl is a strong acid. Ammonia (NH3) is a weak base because it donates only a partial amount of its lone pair of electrons. Hydrochloric acid (HCl) is a strong acid because it completely ionizes in water to produce a high concentration of hydrogen ions.

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Answer: The correct answer is Option B.

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant volume.

Mathematically,

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]        (at constant volume)

where,

[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.

[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.

We are given:

Conversion factor:  [tex]T(K)=T(^oC)+273[/tex]

[tex]P_1=1.85atm\\T_1=25^oC=(25+273)K=298K\\P_2=?atm\\T_2=7.5^oC=(7.5+273)K=280.5[/tex]

Putting values in above equation, we get:

[tex]\frac{1.85atm}{298K}=\frac{P_2}{280.5K}\\\\P_2=1.74atm[/tex]

Hence, the correct answer is Option B.

Final answer:

Using Gay-Lussac's Law, after converting the temperatures to Kelvin and applying the given initial pressure and temperatures, the new pressure of the sports ball when the temperature drops to 7.5 °C is calculated to be B) 1.74 atm.

Explanation:

The problem you've presented involves the concept of gas laws, specifically Gay-Lussac's Law, which states that the pressure of a gas varies directly with its absolute temperature, provided the volume does not change. This law can be mathematically represented as P1/T1 = P2/T2, where P1 and P2 are the initial and final pressures, and T1 and T2 are the initial and final temperatures in Kelvin.

To solve for the new pressure (P2), we first convert the temperatures from °C to Kelvin: T1 = 25 °C + 273.15 = 298.15 K, and T2 = 7.5 °C + 273.15 = 280.65 K. Then we rearrange the formula to solve for P2: P2 = P1 * (T2/T1). Substituting the given values, P2 = 1.85 atm * (280.65 K / 298.15 K) = 1.74 atm.

Answer: The molarity of hydroiodic acid in the titration is 0.485 M.

Explanation:

To calculate the molarity of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HI[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ca(OH)_2[/tex]

We are given:

[tex]n_1=1\\M_1=?M\\V_1=20.3mL\\n_2=2\\M_2=0.186M\\V_2=26.5mL[/tex]

Putting values in above equation, we get:

[tex]1\times M_1\times 20.3=2\times 0.186\times 26.5\\\\M_1=0.485M[/tex]

Hence, the molarity of hydroiodic acid is 0.485M.

Final answer:

The molarity of the hydroiodic acid solution is calculated using the stoichiometry of its reaction with calcium hydroxide and the volumes and molarity of the titrant, resulting in a molarity of 0.485 M for the hydroiodic acid.

Explanation:

To find the molarity of the hydroiodic acid solution, we need to first understand the reaction occurring between hydroiodic acid (HI) and calcium hydroxide (Ca(OH)₂). The balanced chemical equation for this reaction is: 2HI(aq) + Ca(OH)₂(aq) → CaI₂(aq) + 2H₂O(l). This equation indicates that two moles of HI react with one mole of Ca(OH)₂.

Using the titration information provided: 26.5 mL of 0.186 M Ca(OH)₂ were required to neutralize 20.3 mL of HI. From this, we can calculate the moles of Ca(OH)₂ used (Moles = Molarity × Volume in L), which is (0.186 M) × (0.0265 L) = 0.004929 moles of Ca(OH)₂. Given the stoichiometry of the reaction, there are twice as many moles of HI, so 0.009858 moles of HI were neutralized.

To find the molarity of the hydroiodic acid solution, we use the formula Molarity = Moles/Volume in L. Here, it is 0.009858 moles / 0.0203 L = 0.485 M HI.

Answer:

Kc = 9.52.

Explanation:

A + 2B ⇌ C,

Kc = [C]/[A][B]²,

Concentration:     [A]                [B]              [C]

At start:               0.3 M         1.05 M        0.55 M

At equilibrium:   0.3 - x        1.05 - 2x     0.55 + x

                            0.14 M        1.05 - 2x      0.71 M

∵ 0.3 M - x = 0.14 M.

∴ x = 0.3 M - 0.14 M = 0.16 M.

∴ [B] at equilibrium = 1.05 - 2x = 1.05 M -2(0.16) = 0.73 M.

∵ Kc = [C]/[A][B]²

∴ Kc = (0.71)/(0.14)(0.73)² = 9.5166 ≅ 9.52.

To determine the number of moles of HNO3 formed from 8.44 moles of NO2, we need to use the mole ratio from the balanced chemical equation.

To determine the number of moles of HNO3 formed from 8.44 moles of NO2, we need to use the mole ratio from the balanced chemical equation. The ratio of HNO3 to NO2 is 2:3, which means for every 3 moles of NO2, 2 moles of HNO3 are formed. Therefore, we can calculate:

(8.44 mol NO2) / (3 mol NO2) * (2 mol HNO3) = 5.63 mol HNO3

So, 5.63 moles of HNO3 are formed from 8.44 moles of NO2.

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To find the moles of HNO3 formed, we can use the stoichiometric coefficients and set up a proportion. The answer is 5.63 moles HNO3.

To determine the number of moles of HNO3 formed from 8.44 moles of NO2, we can use the stoichiometric coefficients in the balanced equation. From the equation, we see that 3 moles of NO2 produce 2 moles of HNO3. Therefore, if we have 8.44 moles of NO2, we can set up the following proportion:

(3 moles NO2) / (2 moles HNO3) = (8.44 moles NO2) / (x moles HNO3)

Solving for x, we find that x = (2 moles HNO3) * (8.44 moles NO2) / (3 moles NO2) = 5.63 moles HNO3.

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Final answer:

The theoretical yield of MnO2, given 2 moles of Mn and 2 moles of O2, is calculated using stoichiometry and is found to be approximately 1.33 moles based on O2 being the limiting reactant.

Explanation:

To calculate the theoretical yield of MnO₂ from the reaction 2 Mn(s) + 3 O₂ (g) → MnO₂(s), with the initial quantities of 2 mol Mn and 2 mol O₂, we will use stoichiometry. First, we write the balanced chemical equation:

2 Mn(s) + 3 O₂(g) → 2 MnO₂(s)

This equation tells us that 2 moles of Mn react with 3 moles of O₂ to produce 2 moles of MnO₂. Therefore, Mn and O₂ react in a 2:3 mole ratio to produce MnO₂.

The stoichiometry shows the molar ratio of Mn to MnO₂ is 1:1. Since we have 2 moles of Mn, we can produce 2 moles of MnO₂, assuming Mn is the limiting reactant. However, we must also consider O₂. With 2 moles of O₂, the stoichiometry suggests every 3 moles of O₂ produce 2 moles of MnO₂. The limiting reactant is O₂ since it will limit the formation of MnO₂ to 4/3 moles (which is the amount of MnO₂ formed from 2 moles of O₂ based on the 3:2 ratio of O₂ to MnO₂).

Therefore, the theoretical yield of MnO₂ in moles is 4/3 moles or approximately 1.33 moles. This is based on the stoichiometric calculations from the balanced equation taking into account the initial quantities of both reactants and identifying the limiting reactant.

Answer: The [tex]\Delta H_f[/tex] for [tex]C_3H_5N_3O_9[/tex] in the reaction is -365.5 kJ/mol.

Explanation:

Enthalpy change of a reaction is defined as the difference in enthalpy of all the products and the reactants each multiplied with their respective number of moles. The equation that is used to calculate the enthalpy change of a reaction is:

[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}][/tex]

For the given chemical reaction:

[tex]4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+10H_2O(g)+6N_2(g)+O_2(g)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H_{rxn}=[(12\times \Delta H_f_{(CO_2)})+(10\times \Delta H_f_{(H_2O)})+(6\times \Delta H_f_{(N_2)})+(1\times \Delta H_f_{(O_2)})]-[(4\times \Delta H_f_{(C_3H_5N_3O_9)})][/tex]

We are given:

[tex]\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(N_2)}=-0kJ/mol\\\Delta H_f_{(CO_2)}=-393.5kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-5678kJ[/tex]

Putting values in above equation, we get:

[tex]-5678=[(12\times (-393.5))+(10\times (-241.8))+(6\times (0))+(1\times (0))]-[(4\times \Delta H_f_{(C_3H_5N_3O_9)})]\\\\\Delta H_f_{(C_3H_5N_3O_9)}=-365.5kJ/mol[/tex]

Hence, the [tex]\Delta H_f[/tex] for [tex]C_3H_5N_3O_9[/tex] in the reaction is -365.5 kJ/mol.

The standard enthalpy of formation (ΔfHo) for nitroglycerin which decomposes by the given equation can be calculated using the formula for enthalpy change and the provided enthalpies of formation for CO2(g) and H2O(g). Nitrogen and Oxygen are in their standard states so their enthalpies of formation are zero. The calculated ΔfHo for Nitroglycerin is about -364.6 kJ/mol.

To calculate the standard enthalpy of formation (ΔfHo) for nitroglycerin, we first need to know that the formation reactions are those that produce 1 mol of a substance from its elements in their standard states. Here, the combustion reaction for nitroglycerin given is: 4 C3H5N3O9 (l) → 12 CO2 (g) + 10 H2O (g) + 6 N2 (g) + O2 (g). And the provided ΔrxnHo = −5678 kJ.

Also, the standard enthalpy of formations for CO2(g) and H2O(g) are given as -393.5 kJ/mol and -241.8 kJ/mol, respectively. But, the enthalpy of formation for elements in their standard state (here N2 and O2) is zero.

Using the equation: ΔrxnHo = Σ ΔfHo(products) - Σ ΔfHo(reactants), and the knowledge of stoichiometry we get: -5678 kJ = [12(-393.5 kJ/mol) + 10(-241.8 kJ/mol) + 6 * 0 + 1 * 0] - [4 * ΔfHo (C3H5N3O9)]

By calculating we get the ΔfHo for Nitroglycerin = -364.6 kJ/mol (approx)

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Answer: [tex][Al^{3+}][/tex] = 1.834 M

[tex][F^-][/tex] =  0.004 M

[tex][AlF_6^{3-}][/tex] = 0.166 M

Explanation:

[tex]Al^{3+}+6F^-\rightleftharpoons AlF_6^{3-}[/tex]

Initial concentration of [tex]Al^{3+}[/tex] = 0.15 M

Initial concentration of [tex]F^-[/tex] = 2.0 M

The given balanced equilibrium reaction is,

                           [tex]Al^{3+}+6F^-\rightleftharpoons AlF_6^{3-}[/tex]

Initial conc.           2 M           0.15 M                         0

At eqm. conc.    (2-x) M     (1-6x) M                     (x) M

The expression for equilibrium constant for this reaction will be,

[tex]K_f=\frac{[AlF_6^{3-}]}{[Al^{3+}][F^-]^6}[/tex]

Now put all the given values in this expression, we get :

[tex]4.0\times 10^{19}=\frac{(x)}{(2-x)\times (1-6x)^6}[/tex]

By solving the term 'x', we get :

[tex]x=0.166[/tex]

[tex][Al^{3+}][/tex] = (2-x) = 2-0.166 = 1.834 M

[tex][F^-][/tex] = (1-6x) = 1-6(0.1660)=  0.004 M

[tex][AlF_6^{3-}][/tex] = x = 0.166 M

Try this suggested solution (the figures are not provided).

The answer is marked with red colour.

Answer: The specific heat of substance A is 1.1 J/g°C

Explanation:

When substance A is mixed with substance B, the amount of heat released by substance B (initially present at high temperature) will be equal to the amount of heat absorbed by substance A (initially present at low temperature)

[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]

The equation used to calculate heat released or absorbed follows:

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]       ......(1)

where,

q = heat absorbed or released

[tex]m_1[/tex] = mass of substance A = 6.07 g

[tex]m_2[/tex] = mass of substance B = 26.1 g

tex]T_{final}[/tex] = final temperature = 47.0°C

[tex]T_1[/tex] = initial temperature of substance A = 20.7°C

[tex]T_2[/tex] = initial temperature of substance B = 52.8°C

[tex]c_1[/tex] = specific heat of substance A = ?

[tex]c_2[/tex] = specific heat of substance B = 1.17 J/g°C

Putting values in equation 1, we get:

[tex]6.07\times c_1\times (47-20.7)=-[26.1\times 1.17\times (47-52.8)][/tex]

[tex]c_1=1.1J/g^oC[/tex]

Hence, the specific heat of substance A is 1.1 J/g°C

Answer:

E = 1.602v

Explanation:

Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …

             Zn⁰(s) => Zn⁺²(aq) + 2 eˉ

2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)          

_____________________________

Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)

Given E⁰ = 1.562v

Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044

E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v

Answer:

E = 1.602 V

Explanation:

Let's consider the following galvanic cell.

Zn(s) | Zn²⁺(aq, 1.00 × 10⁻³ M) || Ag⁺(aq, 0.150 M) | Ag(s)

The corresponding half-reactions are:

Zn(s) → Zn²⁺(aq, 1.00 × 10⁻³ M) + 2 e⁻

2 Ag⁺(aq, 0.150 M) + 2 e⁻ → 2 Ag(s)

The overall reaction is:

Zn(s) + 2 Ag⁺(aq, 0.150 M) → Zn²⁺(aq, 1.00 × 10⁻³ M) + 2 Ag(s)

We can find the cell potential (E) using the Nernst equation.

E = E° - (0.05916/n) . log Q

where,

E°: standard cell potential

n: moles of electrons transferred

Q: reaction quotient

E = E° - (0.05916/n) . log [Zn²⁺]/[Ag⁺]²

E = 1.562 V - (0.05916/2) . log (1.00 × 10⁻³)/(0.150)²

E = 1.602 V

To synthesize the given ether, you would need ethyl bromide as the alkyl halide and ethoxide as the nucleophile.

To synthesize the ether CH3CH2OCH2CH2CHCH3CH3, you would need an alkyl halide and a nucleophile. One possible set of reactants that could be used is ethyl bromide (CH3CH2Br) as the alkyl halide and ethoxide (CH3CH2O-) as the nucleophile. The reaction can be represented as:

CH3CH2Br + CH3CH2O- → CH3CH2OCH2CH2CHCH3CH3 + Br-

Explanation:

Moles of nitrogen gas = [tex]n_1=\frac{1.20 g}{28 g/mol}=0.0428 mol[/tex]

Moles of oxygen gas = [tex]n_2=\frac{0.77 g}{32 g/mol}=0.0240 mol[/tex]

Mole fraction of nitrogen gas=[tex]\chi_1=\frac{n_1}{n_1+n_2}[/tex]

[tex]\chi_1=\frac{0.0428 mol}{0.0428 mol+0.0240 mol}=0.6407\approx 0.64[/tex]

Mole fraction of oxygen gas=[tex]\chi_2=1-\chi_1=1-0.6407=0.3593\approx 0.36[/tex]

Total  umber of moles in container :

n =[tex]n_1+n_2[/tex]= 0.0428 mol + 0.0240 mol = 0.0668 mol

Volume of the container = V = 1.65 L

Temperature of the container = T = 15°C = 288.15 K

Total pressure in the container = P

Using an ideal gas equation:

[tex]PV=nRT[/tex]

[tex]P=\frac{0.0668 mol\times 0.0821 atm L/mol k\times 288.15 K}{1.65 L}[/tex]

P = 0.9577 atm

Partial pressure of nitrogen gas = [tex]p^{o}_1[/tex]

Partial pressure of nitrogen gas = [tex]P^{o}_2[/tex]

Partial pressure of nitrogen gas and oxygen gas can be calculated by using Dalton's law of partial pressure:

[tex]p^{o}_i=p_{total}\times \chi_i[/tex]

[tex]p^{o}_1=P\times \chi_1=0.6135 atm\approx 0.61 atm[/tex]

[tex]p^{o}_2=P\times \chi_2=0.3441 atm\approx 0.34 atm[/tex]

No net shift in the reactants and products is an equilibrium constant. In the first reaction, there is no effect, in second shifts to the left and third shifts to right.

Equilibrium shift can be explained by the rules of Le and the change in the volume affects the system as the number of the particles gets varied.

For the first reaction [tex]\rm CO(g) + H_{2}O(g) \Leftrightarrow CO_{2}(g) + H_{2}(g)[/tex] when the volume is decreased then the pressure will increase then the equilibrium will shift towards the fewer and then more number of the gas moles and hence, will have no net effect.

In the second reaction [tex]\rm PCl_{3}(g) + Cl_{2}(g) \Leftrightarrow PCl_{5}(g)[/tex] when the volume is increased then the pressure will decrease and the reaction will shift towards more moles of the gas present that is the left side.

In the third reaction, [tex]\rm CaCO_{3}(s) \Leftrightarrow CaO(s) + CO_{2}(g)[/tex] when the volume is increased then the pressure will decrease and the equilibrium will shift towards the side where more moles are present. Hence the increase in the volume will shift the equilibrium towards the right.

Therefore, a decrease in volume will have no effect, while the increase in volume will shift the reaction towards the left and right.

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Answer:

VP(solution) = 186.3 Torr

Explanation:

Given 24g Lactose (C₁₂H₂₂O₁₁) IN 200g H₂O @338K (65°C) & 187.5Torr

VP(soln) = VP(solvent) - [mole fraction of solute(X)·VP(solvent)]  => Raoult's Law

VP(H₂O)@65°C&187.5Torr = 187.5Torr

moles Lactose = (24g/342.3g/mol) = 0.0701mole Lactose

moles Water = (200g/18g/mol) = 11.11mole Water

Total moles = (11.11 + 0.0701)mole = 11.181mole

mole fraction Lac = n(lac)/[n(lac) + n(H₂O)] = (0.0701/11.181) = 6.27 x 10⁻³

VP(solution) = 187.5Torr - (6.27 x 10⁻³)187.5Torr = 186.3Torr

Final answer:

To find the vapor pressure of water above a lactose solution, calculate the mole fraction of water in the solution using moles of lactose and water, then apply Raoult’s Law by multiplying the mole fraction by the vapor pressure of pure water at 338 K, resulting in a vapor pressure of 186.3 torr.

Explanation:

To calculate the vapor pressure of water above a solution prepared by adding 24 g of lactose (C12H22O11) to 200 g of water at 338 K, we use Raoult’s Law. Raoult's Law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution. First, calculate the mole fraction of water in the solution:

Moles of lactose = mass (g) / molar mass (g/mol). For lactose, molar mass = 342.3 g/mol, so moles of lactose = 24 g / 342.3 g/mol = 0.0701 mol.

Moles of water = mass (g) / molar mass (g/mol). For water, molar mass = 18.015 g/mol, so moles of water = 200 g / 18.015 g/mol = 11.10 mol.

Total moles = moles of lactose + moles of water = 0.0701 mol + 11.10 mol = 11.1701 mol.

Mole fraction of water = moles of water / total moles = 11.10 / 11.1701 = 0.9937.

Finally, calculate the vapor pressure of water: Vapor pressure = mole fraction of water × vapor pressure of pure water = 0.9937 × 187.5 torr = 186.3 torr.

Answer:

Cohesion

Explanation:

Think of it like this. The water molecules STICK TOGETHER, so they COoperate.

COhesion     COoperate

Cohesion is the tendency of water molecules to stick together, due to hydrogen bonding. It is a unique property of water that plays a vital role in living organisms and the environment.

The tendency of water molecules to stick together is referred to as cohesion. This process occurs due to hydrogen bonding between the water molecules, where the slightly positive hydrogen of one water molecule is attracted to the slightly negative oxygen of a neighboring water molecule. Thus, creating a force that holds these molecules together. It is one of the unique properties of water that contributes to its important role in living organisms and the environment. Some other terms related to water are adhesion, which is the tendency of water to stick to other substances; polarity, which is the property of having oppositely charged ends; transpiration, which is the process by which water evaporates from the leaves of plants; and evaporation, which is the change of state from liquid to gas.

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Answer:

[H₃O⁺] = 1.82 x 10⁻¹⁰M

Explanation:

[H₃O⁺][OH⁻] = Kw = 1.0 x 10⁻¹⁴ = [H₃O⁺](5.5 x 10⁻⁵) => [H₃O⁺] = (1.0 x 10⁻¹⁴/5.5 x 10⁻⁵)M = 1.82 x 10⁻¹⁰M

Answer:

The rate of the reaction is [tex]6.24\times 10^{-6} M/s[/tex].

Explanation:

[tex]NH4^+(aq) + NO_2^-(aq)\rightarrow N_2(g) + 2H_2O(l)[/tex]

Concentration of [tex][NH_4^{+}]=0.26 M[/tex]

Concentration of [tex][NO_2^{-}]=0.080 M[/tex]

Rate constant of the reaction = k= [tex]3.0\times 10^{-4} M^{-1} s^{-1}[/tex]

[tex]R= k[NH_{4}^+][NO_{2}^-][/tex]

[tex]R=3.0\times 10^{-4} M^{-1} s^{-1}\times 0.26 M\times 0.080 M[/tex]

[tex]R=6.24\times 10^{-6} M/s[/tex]

The rate of the reaction is [tex]6.24\times 10^{-6} M/s[/tex].

Answer:

Rate of Reaction = 6.24 x 10–6 M/s

Explanation:

Rate of reaction = k[NH4+][NO2–]

Concentration of [NH4+] = 0.26 M

Concentration of [NO2–] = 0.080 M

k= 3.0 × 10–4/ M · s

Rate of Reaction = (3.0 × 10–4/ M · s)( 0.26 M)(0.080 M)

The total heat energy required to convert 25 grams of ice at -4.00°C to steam at 110.0 °C can be calculated in several steps involving warming of ice to 0°C, melting of ice, heating water to 100°C, vaporization of water, and heating of steam to the final temperature. Each step requires the use of specific heat values, molar heats of fusion and vaporization, and the mass of the starting ice chunk. The total energy calculated would be the sum of these steps, yields 75579 Joules.

The question is about calculating the total energy, or enthalpy change, needed to convert ice at -4.00°C to steam at 110.0°C. We need to take into account the warming of ice, the melting of ice, the heating of water, the vaporization of water, and the heating of steam. This process involves several steps and requires using the principle of conservation of energy and heat transfer equations.

First, the ice is warmed to 0°C : ΔH1 = mcΔT = (25.0 grams) (2.09 J/g-K)(4 K) = 209 J

Then, the ice is melted: ΔH2 = n(ΔHfus) = (25.0 grams/18.015 g/mol)(6.01 kJ/mol)= 8.33 kJ = 8330 J

The water is heated to 100°C: ΔH3 = mcΔT = (25.0 grams) (4.18 J/g-K)(100 K) = 10450 J

The water is vaporized: ΔH4 = n(ΔHvap) = (25.0 grams/18.015 g/mol)(40.67 kJ/mol) = 56.13 kJ = 56130 J

Finally, the steam is heated to 110°C: ΔH5 = mcΔT = (25.0 grams) (1.84 J/g-K)(10 K) = 460 J

The total energy change is the sum of all these changes: ΔHtotal = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5 = 209J + 8330J + 10450J + 56130J + 460J = 75579 J

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Answer: The [tex]\Delta G[/tex] for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

[tex]CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)[/tex]

We are given:

[tex]\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol[/tex]

To calculate [tex]\Delta G^o_{rxn}[/tex] for the reaction, we use the equation:

[tex]\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)][/tex]

For the given equation:

[tex]\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})][/tex]

Putting values in above equation, we get:

[tex]\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J[/tex]

Conversion factor used = 1 kJ = 1000 J

The expression of [tex]K_p[/tex] for the given reaction:

[tex]K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}[/tex]

We are given:

[tex]p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm[/tex]

Putting values in above equation, we get:

[tex]K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85[/tex]

To calculate the gibbs free energy of the reaction, we use the equation:

[tex]\Delta G=\Delta G^o+RT\ln K_p[/tex]

where,

[tex]\Delta G[/tex] = Gibbs' free energy of the reaction = ?

[tex]\Delta G^o[/tex] = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = [tex]8.314J/K mol[/tex]

T = Temperature = [tex]25^oC=[25+273]K=298K[/tex]

[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

[tex]\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol[/tex]

Hence, the [tex]\Delta G[/tex] for the reaction is 54.425 kJ/mol

The change in free energy for the reaction is 54.4  kJ/mol.

The change in free energy under nonstandard conditions can be obtained from the chage in free energy under standard conditions using the formula;

ΔG = ΔG° + RTlnK

Now ΔG° is obtained from;

ΔG°reaction = 2[−204.9kJ/mol] - [(−394.4kJ/mol) + (−62.3kJ/mol)]

ΔG°reaction =(-409.8) + 456.7

ΔG°reaction = 46.9 kJ/mol

Q =PCOCl2^2/ PCO2 * PCCl4

Q = (0.735)^2/ 0.140 * 0.185

Q = 20.8

ΔG = 46.9 * 10^3 + [8.314 * 298 * ln(20.8)]

ΔG = 54.4  kJ/mol

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Answer: The mass of nitrogen gas produced will be 45.64 grams.

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 [tex]N_2O_4=50.0g[/tex]

Molar mass of [tex]N_2O_4=92.02g/mol[/tex]

Putting values in equation 1, we get:

[tex]\text{Moles of }N_2O_4=\frac{50g}{92.02g/mol}=0.543mol[/tex]

Given mass of [tex]N_2H_4=45.0g[/tex]

Molar mass of [tex]N_2O_4=32.05g/mol[/tex]

Putting values in  equation 1, we get:

[tex]\text{Moles of }N_2H_4=\frac{45g}{32.05g/mol}=1.40mol[/tex]

For the given chemical reaction:

[tex]N_2O_4(l)+2N_2H_4(l)\rightarrow 3N_2(g)+4H_2O(g)[/tex]

By stoichiometry of the reaction:

1 mole of [tex]N_2O_4[/tex] reacts with 2 moles of [tex]N_2H_4[/tex]

So, 0.543 moles of [tex]N_2O_4[/tex] will react with = [tex]\frac{2}{1}\times 0.543=1.086moles[/tex] of [tex]N_2H_4[/tex]

As, the given amount of [tex]N_2H_4[/tex] is more than the required amount. Thus, it is considered as an excess reagent.

Hence, [tex]N_2O_4[/tex] is the limiting reagent.

By Stoichiometry of the reaction:

1 mole of [tex]N_2O_4[/tex] produces 3 moles of nitrogen gas

So, 0.543 moles of [tex]N_2O_4[/tex] will produce = [tex]\frac{3}{1}\times 0.543=1.629moles[/tex] of nitrogen gas.

Now, calculating the mass of nitrogen gas from equation 1, we get:

Molar mass of nitrogen gas = 28.02 g/mol

Moles of nitrogen gas = 1.629 moles

Putting values in equation 1, we get:

[tex]1.629mol=\frac{\text{Mass of nitrogen gas}}{28.02g/mol}\\\\\text{Mass of nitrogen gas}=45.64g[/tex]

Hence, the mass of nitrogen gas produced will be 45.64 grams.

Answer : The mass of copper produced will be, 11.796 grams

Explanation : Given,

Mass of [tex]Al[/tex] = 5 g

Molar mass of [tex]Al[/tex] = 26.98 g/mole

Molar mass of [tex]Cu[/tex] = 63.66 g/mole

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

[tex]\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=\frac{5g}{26.98g/mole}=0.185moles[/tex]

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

The balanced chemical reaction is,

[tex]2Al+3CuSO_4\rightarrow Al_2(SO_4)_3+3Cu[/tex]

From the balanced reaction we conclude that

As, 2 moles of [tex]Al[/tex] react to give 3 moles of [tex]Cu[/tex]

So, 0.185 moles of [tex]Al[/tex] react to give [tex]\frac{3}{2}\times 0.185=0.2775[/tex] moles of [tex]Cu[/tex]

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

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

[tex]\text{Mass of }Cu=(0.2775mole)\times (63.66g/mole)=17.66g[/tex]

The theoretical yield of Cu = 17.66 grams

Now we have to calculate the actual yield of Cu.

[tex]\%\text{ yield of }Cu=\frac{\text{Actual yield of }Cu}{\text{Theoretical yield of }Cu}\times 100[/tex]

Now put all the given values in this formula, we get the actual yield of Cu.

[tex]63.4=\frac{\text{Actual yield of }Cu}{17.66}\times 100[/tex]

[tex]\text{Actual yield of }Cu=11.796g[/tex]

Therefore, the mass of copper produced will be, 11.796 grams

The first part of the answer is in the photo attached. I hope i can attach a second photo...

To find the change in pH, use the Henderson-Hasselbalch equation with the initial concentrations of NH3 and NH4+ to find the initial pH. Adjust concentrations after addition of HCl to find the final pH. Subtract initial pH from final to find change.

The question asks us to calculate the change in pH when 5.00 ml of 0.100 M HCl is added to 75.0 ml of a buffer solution that is 0.100 M in NH3 and 0.100 M in NH4Cl. This is a problem related to acid-base chemistry. We have a strong acid (HCl) being added to a buffer solution composed of NH3 (a weak base) and NH4Cl (the conjugate acid of the base).

First, it's important to use the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), where pKa is the -log of the Ka, [A-] is the concentration of base (NH3 in this case) and [HA] is the concentration of conjugate acid (NH4+ in this case). Plug the initial concentrations into the equation to find initial pH. Then, calculate the moles of HCl being added and adjust the concentrations of NH4+ and NH3 accordingly. Using the new concentrations in Henderson-Hasselbalch equation will give the final pH. The difference between the initial and final pH is the change in pH.

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Answer:

[tex]\boxed{\text{4.62 kJ}}[/tex]

Explanation:

There are two heat transfers to consider:

[tex]\begin{array}{ccccc}\text{Heat released by reaction} & + &\text{heat absorbed by water} & =& 0\\q_{1}& + & q_{2} & = & 0\\q_{1}& + & mC_{p}\Delta T & = & 0\\\end{array}[/tex]

Calculate q₂

 m = 120.0 g

 C = 4.184 J·°C⁻¹g⁻¹

 T₂ = 29.20 °C

 T₁ = 20.00 °C

ΔT = T₂ - T₁ =(29.20 – 20.00) °C =9.20 °C

 q₂ = 120.0 g × 4.184 J·°C⁻¹g⁻¹ × 9.20 °C = 4620 J = 4.62 kJ

Calculate q₁

q₁ + 4.62 kJ = 0

q₁ = -4.62 kJ

The negative sign shows that heat is given off.

[tex]\text{The reaction gives off }\boxed{\textbf{4.62 kJ}}[/tex]

Answer:

A. 4.62kJ

Explanation:

Answer:

60.00% is the volume percent the ethylene glycol in the antifreeze mixture.

Explanation:

Volume of the ethylene glycol in the mixture = 3.00 gal

Volume of the water in the mixture = 2.00 gal

Total volume of the mixture =3.00 gal + 2.00 gal = 5.00 gal

Volume percent the ethylene glycol :

[tex]\frac{\text{Volume of solute}}{\text{Volume of solution or mixture}}\times 100[/tex]

[tex]\%=\frac{3.00 gal}{5.00 gal}\times 100=60.00\%[/tex]

60.00% is the volume percent the ethylene glycol in the antifreeze mixture.

Answer: The mass of magnesium consumed will be 14.731 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 nitrogen gas = 5.65 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of }N_2=\frac{5.65g}{28g/mol}=0.202mol[/tex]

For the given chemical equation:

[tex]3Mg+N_2\rightarrow Mg_3N_2[/tex]

By Stoichiometry of the reaction:

1 mole of nitrogen gas reacts with 3 moles of magnesium.

So, 0.202 moles of nitrogen gas will react with = [tex]\frac{3}{1}\times 0.202=0.606mol[/tex] of magnesium.

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

Moles of magnesium = 0.606 moles

Molar mass of magnesium = 24.31 g/mol

Putting values in equation 1, we get:

[tex]0.606mol=\frac{\text{Mass of magneisum}}{24.31g/mol}\\\\\text{Mass of magnesium}=14.731g[/tex]

Hence, the mass of magnesium consumed will be 14.731 g.

Answer:

K in KClO2 = +1

Cl in KClO2 = +3

O in KClO2 = -2

K in KCl = +1

Cl in KCl = -1

O in O2 = 0

Chlorine is going from +3 to -1 so it is being reduced

Oxygen is going from -2 to 0 so it is being oxidized

Explanation:

Potassium is a constant +1

Chlorine could be -1, +1, +3, +5, or +7

Oxygen can be either -2 or +2

Every reactant has to equal zero if there is not a given charge.

Each element can only have certain charges, for example in the previous answer K = +5. Potassium can only be the charge of +1

Chlorine was reduced and oxygen was oxidized in the reaction as shown.

The oxidation number of an atom in a compound is the charge that the atom appears to have as determined by a certain set of rules.

On the left hand side, the oxidation numbers of the elements are;

K = +1

Cl = +3

O = -2

On the right hand side;

K = +1

Cl = -1

O = zero

The element that was oxidized is oxygen. Its oxidation number increased from -2 to zero. The element that was reduced is chlorine. Its oxidation number decreased from +3 to -1.

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Answer:

In osmosis, the water has a net movement to the side with lower water concentration.

Explanation:

It moves in the direction of a solution with higher solute ( therefore  lower water) concentration.

Answer:

Exhibit a net movement to the side with lower water concentration.

Explanation:

Hello,

Osmosis is a process by which the equilibrium between the concentration of a solute placed at two sides separated by a permeable membrane is reached by moving the solvent's molecules.

For the considered example, it is seen that the water will exhibit a net movement to the side with lower water concentration as long as the place with more water will have a higher solute's concentration which implies that the water move from such side to other one in order to equal both the solutions concentrations.

Best regards.

Answer:

[tex]\boxed{ \text{0.303 mol}}[/tex]

Explanation:

(a) Balanced equation

2Fe + 3Cl₂ ⟶2FeCl₃

(b) Calculation

You want to convert moles of Cl₂ to moles of Fe.

The molar ratio is 2 mol Fe:3 mol Cl₂

[tex]\text{Moles of Fe} =\text{0.455 mol Cl$_{2}$} \times \dfrac{\text{2 mol Fe}}{\text{3 mol Cl$_{2}$}} = \textbf{0.303 mol Fe}\\\\\boxed{ \textbf{0.303 mol of Fe}}\text{ will react.}[/tex]

Answer: The empirical formula of the ether will be [tex]C_4H_{10}O[/tex]

Explanation:

The chemical equation for the combustion of ether follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=3.565g[/tex]

Mass of [tex]H_2O=1.824g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 3.565 g of carbon dioxide, [tex]\frac{12}{44}\times 3.565=0.972g[/tex] of carbon will be contained.

In 18g of water, 2 g of hydrogen is contained.

So, in 1.824 g of water, [tex]\frac{2}{18}\times 1.824=0.202g[/tex] of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.972g}{12g/mole}=0.081moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.202g}{1g/mole}=0.202moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.327g}{16g/mole}=0.0204moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0204 moles.

For Carbon = [tex]\frac{0.081}{0.0204}=3.97\approx 4[/tex]

For Hydrogen  = [tex]\frac{0.202}{0.0204}=9.9\approx 10[/tex]

For Oxygen  = [tex]\frac{0.0204}{0.0204}=1[/tex]

The ratio of C : H : O = 4 : 10 : 1

Hence, the empirical formula for the given compound is [tex]C_4H_{10}O_1=C_4H_{10}O[/tex]

In the reaction Mg + Cl₂-> MgCl₂, Mg is oxidized, and Cl₂ is reduced. Mg serves as the reducing agent, and Cl₂ is the oxidizing agent.

In the redox reaction Mg(s) + Cl₂(g)
ightarrow MgCl₂(s), we can identify the substances that are oxidized and reduced by looking at the changes in oxidation states. Magnesium (Mg) starts with an oxidation number of 0 in elemental form and increases to +2 when it forms Mg⁺², indicating that Mg is oxidized. Chlorine (Cl₂) begins with an oxidation number of 0 and is reduced to -1 in Cl-, showing that Cl₂ is reduced.

The substance that gets oxidized works as the reducing agent, which in this case is Mg. The substance that gets reduced acts as the oxidizing agent, which is Cl₂ in this reaction. Therefore, Mg is the reducing agent because it provides electrons, and Cl₂ is the oxidizing agent because it accepts electrons, facilitating the oxidation of Mg.