demonstration of strong electrolytes, weak electrolytes, and nonelectrolytes, Professor Popsnorkle used a lightbulb apparatus that showed how much a solution conducted electricity by the brightness of the lightbulb. When pure water was tested, the bulb did not light. Then Professor Popsnorkle tested the following aqueous solutions. Which one caused the bulb to burn the brightest?

In a galvanic cell, oxidation occurs at the anode, the cathode is the positive electrode, cations flow toward the cathode, and electrons flow from the anode to the cathode.

In a galvanic cell, the following processes occur:

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Answer:

The pH will be 4.5

Explanation:

The mixture of benzoic acid (weak acid) and its salt will make a buffer.

The pH of buffer solution can be calculated using Henderson Hassalbalch's equation, which is

[tex]pH=pKa+log\frac{[salt]}{[acid]}[/tex]

pKa = -logKa

pKa = -log([tex]6.3X10^{-5}[/tex])

pKa = 4.2

[tex]pH=4.2 + log\frac{0.41}{0.22}=4.2+0.27=4.47=4.5[/tex]

Explanation:

The given data is as follows.

Molecular weight of azulene = 128 g/mol

Hence, calculate the number of moles as follows.

      No. of moles = [tex]\frac{mass}{\text{molecular weight}}[/tex]

                            = [tex]\frac{0.392 g}{128 g/mol}[/tex]

                            = 0.0030625 mol of azulene

Also,    [tex]-Q_{rxn} = Q_{solution} + Q_{cal}[/tex]

       [tex]Q_{rxn} = n \times dE[/tex]

         [tex]Q_{solution} = m \times C \times (T_{f} - T_{i})[/tex]

              [tex]Q_{cal} = C_{cal} \times (T_{f} - T_{i})[/tex]

Now, putting the given values as follows.    

     [tex]Q_{solution} = 1.17 \times 10^{3} g \times 10^{3} \times 4.184 J/g^{o}C \times (27.60 - 25.20)^{o}C[/tex]

                   = 11748.67  J

So,  [tex]Q_{cal} = 786 J/^{o}C \times (27.60 - 25.20)^{o}C[/tex]

                    = 1886.4 J

Therefore, heat of reaction will be calculated as follows.

        [tex]-Q_{rxn}[/tex] = (11748.67 + 1886.4) J

                      = 13635.07 J

As,  [tex]Q_{rxn} = n \times dE[/tex]

          13635.07 J = [tex]-n \times dE[/tex]

                dE = [tex]\frac{13635.07 J}{0.0030625 mol}[/tex]

                     = 4452267.75 J/mol

or,                 = 4452.26 kJ/mol       (as 1 kJ = 1000 J)

Thus, we can conclude that [tex]\Delta E[/tex] for the given combustion reaction per mole of azulene burned is 4452.26 kJ/mol.

Answer:

The answer is: 11759 Hz

Explanation:

Given: Chemical shift: δ = 211.5 ppm, Spectrometer frequency = 556 MHz = 556 × 10⁶ Hz

In NMR spectroscopy, the chemical shift (δ), expressed in ppm, of a given nucleus is given by the equation:

[tex]\delta (ppm) = \frac{Observed\,frequency (Hz)}{Frequency\,\, of\,\,the\,Spectrometer (MHz)} \times 10^{6}[/tex]

[tex]\therefore Observed\,frequency (Hz)= \frac{\delta (ppm)\times Frequency\,\, of\,\,the\,Spectrometer (MHz)}{10^{6}}[/tex]

[tex]Observed\,frequency= \frac{211.5 ppm \times 556 \times 10^{6} Hz}{10^{6}} = 11759 Hz[/tex]

Therefore, the signal is at 11759 Hz from the TMS.

Answer:

The change in temperature = 7.65 °C

Explanation:

Step 1: Data given

10.0 grams of ammonium nitrate dissolves in 100.0 grams of water

Hsoln = 25.7 kJ/mol

Molar mass = 80.04 g/mol

Heat capacity of the solution = 4.2 J

Step 2: Calculate moles  ammonium nitrate

Moles = mass / molar mass

Moles = 10.0 grams / 80.04 g/mol

Moles =0.125 moles

Step 3: Calculate q

q = 25.7 kJ/mol * 0.125 moles

q = 3.2125 kJ = 3212.5 J

Step 4: Calculate change in temperature

q = m*c*ΔT

3212.5 J = 100g *4.2 J * ΔT

ΔT= 7.65

The change in temperature = 7.65 °C

Explanation:

The given data is as follows.

Molar mass of ammonium nitrate = 80.0 g/mol

So, we will calculate the number of moles of ammonium nitrate as follows.

     No. of moles = [tex]\frac{\text{given mass}}{\text{molar mass}}[/tex]

                            = [tex]\frac{10.0 g}{80.0 g/mol}[/tex]

                            = 0.125 mol

Heat released due to solution of ammonium nitrate = [tex]\Delta H \times \text{no. of moles}[/tex]

                    = [tex]25.7 kJ/mol \times 0.125  mol[/tex]

                    = 3.2125 KJ

                    = 3212.5 J              (as 1 kJ = 1000 J)

Therefore, calculate the total mass of solution as follows.

   mass of solution(m) = (10.0 + 100.0 ) g

                                    = 110.0 g

Hence, heat released will be calculated as follows.

                 Q = [tex]m \times C \times \Delta T[/tex]

          3212.5 J = [tex]110.0 \times 4.2 J \times \Delta T[/tex]

           [tex]\Delta T = 6.95^{o}C[/tex]

Thus, we can conclude that the change in temperature of the solution is [tex]6.95^{o}C[/tex].

Answer:

Chlorine gas leaked = 20 g        

Explanation:

Given: Molar mass of chlorine gas: m = 70.9 g/mol

Initial Mass of chlorine gas: w₁ = 65.5 g, Initial Absolute pressure: P₁ = 6 × 10⁵ Pa, Initial temperature: T₁ = 73 °C = 73 + 273 = 346 K      (∵ 0°C = 273 K)

Final Absolute pressure: P₂ = 3.70 × 10⁵ Pa, Final temperature: T₂ = 34 °C = 34 + 273 = 307 K

Volume is constant

Final mass of chlorine gas: w₂ = ? g

Chlorine gas have leaked = w₁ - w₂ = ? g

Initial number of moles of chlorine gas: [tex]n_{1}= \frac{w_{1}}{m_{1}} =\frac{65.5 g}{70.9 g/mol} = 0.924 mole[/tex]

According to the Ideal gas law: P.V = n.R.T

∴ at constant volume,

[tex]\frac{n_{1}\times T_{1}}{P_{1}} = \frac{n_{2}\times T_{2}}{P_{2}}[/tex]

[tex]n_{2} = \frac{n_{1}\times T_{1} \times P_{2}}{P_{1} \times T_{2}}[/tex]

[tex]n_{2} = \frac{(0.924)\times (346 K) \times (3.70 \times 10^{5} Pa)}{(6 \times 10^{5} Pa) \times (307 K)} [/tex]

Final number of moles of chlorine gas: [tex]n_{2} = 0.642 = \frac{w_{2}}{m} [/tex]

⇒ Final mass of chlorine: [tex]w_{2} = 0.642 mol \times m = (0.642 mol) \times (70.9 g/mol) = 45.5 g[/tex]

Therefore, Chlorine gas leaked = w₁ - w₂ = 65.5 g - 45.5 g = 20 g

No chlorine gas has leaked out of the tank.

To calculate the amount of chlorine gas leaked out of the tank, we need to use the Ideal Gas Law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, we calculate the initial number of moles using the given mass of chlorine gas (65.5 g) and its molar mass (70.9 g/mol):

65.5 g / 70.9 g/mol = 0.9237 mol

Next, we use the initial conditions (temperature = 73°C = 346 K, pressure = 6.00 × 10^5 Pa) to calculate the initial volume:

V = nRT / P = (0.9237 mol) * (8.314 J/mol K) * (346 K) / (6.00 × 10^5 Pa) ≈ 0.175 m^3

Similarly, we use the final temperature (34°C = 307 K) and pressure (3.70 × 10^5 Pa) to calculate the final volume:

V = nRT / P = (0.9237 mol) * (8.314 J/mol K) * (307 K) / (3.70 × 10^5 Pa) ≈ 0.227 m^3

Finally, we calculate the difference in the volumes to determine the amount of leaked chlorine gas:

Volume leaked = Initial volume - Final volume = 0.175 m^3 - 0.227 m^3 ≈ -0.052 m^3

Since volume cannot be negative, we conclude that no chlorine gas has leaked out of the tank.

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Answer:

See the explanation

Explanation:

In this case, in order to get an elimination reaction we need to have a strong base. In this case, the base is the phenoxide ion produced the phenol (see figure 1).  

Due to the resonance, we will have a more stable anion therefore we will have a less strong base because the negative charge is moving around the molecule (see figure 2).

Finally, the phenoxide will attack the primary carbon attached to the Cl. The C-Cl bond would be broken and the C-O would be produced at the same time to get a substitution (see figure 1).

Final answer:

The phenoxide anion prefers acting as a nucleophile in an SN2 mechanism on 3-chloro-1,2-propanediol because the electronegative chloride creates an electrophilic carbon that is more reactive towards nucleophilic attack than the deprotonation of an alcohol.

Explanation:

The phenoxide anion does not act as a base to deprotonate one of the alcohols on 3-chloro-1,2-propanediol because the presence of the chloride makes that carbon a more electrophilic center, which is highly susceptible to nucleophilic attack.

In an SN2 mechanism, the nucleophile favors attacking an electrophilic carbon, here significantly activated by the chloride leaving group, rather than deprotonating an alcohol which is a less electrophilic and less favorable process.

This is especially true when considering that alcohols are not particularly acidic, and thus their protons are not as easily abstracted by a base as compared to more acidic hydrogens (e.g., hydrogens adjacent to carbonyl groups).

The SN2 reaction is characterized by the simultaneous bond formation by the nucleophile and bond breaking by the leaving group, typically observed in primary alkyl halides. The phenoxide anion is a good nucleophile due to its negative charge, making it highly reactive towards electrophilic carbons, particularly against an atom that bears a good leaving group like chloride.

Answer:

The correct answer is -521.6 KJ/mol

Explanation:

In order to solve the problem, we have to rearrange and to add the reactions with the aim to obtain the requested reaction.

We have:

1) H₂(g) + F₂ (g) → 2HF(g) => ∆H₁ = -546.6 kJ/mol

2) 2H₂(g) + O₂(g) → 2H₂0(l) => ∆H₂ = -571.6 kJ/mol

If we multiply reaction 1 by 2 and add the inverse reaction of 2, we obtain:

2H₂(g) + 2F₂ (g) → 4HF(g) => 4 x ∆H₁ = 4 x (-546.6 kJ/mol)

2H₂0(l) → 2H₂(g) + O₂(g) => (-1) x ∆H₂ = (-1) x (571.6 kJ/mol)

-------------------------------------

2H₂(g) + 2F₂ (g)+  2H₂0(l) → 4HF(g) + 2H₂(g) + O₂(g)

We eliminate 2H₂(g) is repeated in both members of the reaction, and we obtain: 2F₂(g) + 2H₂0(l) -> 4HF(g) + O₂(g)

Finally, the ΔH of the reaction is calculated as follows:

ΔHtotal= (2 x ΔH₁) + ((-1)ΔH₂

ΔHtotal= (2 x (-546.6 KJ/mol)) + ((-1) x (-571.6 KJ/mol)

ΔHtotal= -521.6 KJ/mol

a) Approximately 3.125% of 254/102 No remains 5.0 minutes after it forms.

b) About 0.391% of 254/102 No is left 1.0 hour after it forms.

In nuclear decay the percentage of a substance remaining can be calculated using the formula

[tex]\[ \text{Final amount} = (\frac{1}{2})^{\frac{\text{time elapsed}}{\text{half-life}}} \times 100 \][/tex]

a) For 5.0 minutes after formation we convert the time to seconds (5.0 minutes = 300 seconds) and apply the formula

[tex]\[ \text{Final amount} = (\frac{1}{2})^{\frac{300 \, \text{seconds}}{55 \, \text{seconds}}} \times 100 \approx 3.125\% \][/tex]

b) For 1.0 hour after formation we convert the time to seconds (1.0 hour = 3600 seconds) and use the formula

[tex]\[ \text{Final amount} = (\frac{1}{2})^{\frac{3600 \, \text{seconds}}{55 \, \text{seconds}}} \times 100 \approx 0.391\% \][/tex]

Nuclear decay and half-life involve complex mathematical models that govern the decay of radioactive substances. Understanding these processes is crucial in various scientific fields including nuclear physics medicine and environmental science.

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

               2 ICl(g)   ⇄   I₂(g)    +    Cl₂(g)

Initially      0.20              -               -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React          x              x/2               x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

So in equilibrium I will have

           (0.20 - x)          x/2             x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)

Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).

As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

Kc = (0.07/2 . 0.07/2) / (0.060/2)²

Kc = 1.225x10⁻³ / 9x10⁻⁴

Kc = 1.36

Final answer:

The value of Kc for the decomposition of ICl to I2 and Cl2 is calculated using the equilibrium concentrations and the equilibrium expression. By plugging in the given values and solving the expression, we find Kc to be 0.11.

Explanation:

To calculate the equilibrium constant (Kc) for the decomposition of ICl to I2 and Cl2 at a high temperature using the reaction 2 ICl(g) → I2(g) + Cl2(g), we need to apply the concept of the equilibrium constant expression which is based on the molar concentrations of the products raised to the power of their coefficients divided by the molar concentrations of the reactants raised to the power of their coefficients.

At the start, we have 0.20 mol ICl in a 2.0 L flask, which gives us an initial concentration of 0.10 M (0.20 mol / 2.0 L). At equilibrium, the [ICl] is given as 0.060 M. The change in concentration for ICl (reactant) is the initial concentration minus the equilibrium concentration, which is 0.10 M - 0.060 M = 0.040 M. Since the reaction consumes 2 moles of ICl for every 1 mole of I2 and Cl2 produced, each product will have a concentration change of 0.020 M (0.040 M / 2). Therefore, at equilibrium, both I2 and Cl2 have concentrations of 0.020 M.

The equilibrium constant expression is:

Kc = [I2][Cl2] / [ICl]2

Substituting the equilibrium concentrations we get:

Kc = (0.020 M)(0.020 M) / (0.060 M)2 = 4/9 × 10-3

Calculating the value yields:

Kc = 0.11

Answer: The value of equilibrium constant for the above equation is 36

Explanation:

We are given:

Initial moles of X = 4.00 moles

Initial moles of Y = 4.00 moles

Equilibrium moles of Z = 6.00 moles

Volume of vessel = 1.00 L

Initial concentration of X = [tex]\frac{4}{1}=4[/tex]

Initial concentration of Y = [tex]\frac{4}{1}=4[/tex]

Equilibrium concentration of Z = [tex]\frac{6}{1}=6[/tex]

The given chemical equation follows:

                  [tex]X(g)+Y(g)\rightarrow 2Z(g)[/tex]

Initial:         4         4

At eqllm:     4-x      4-x        2x

Calculating for 'x', we get:

[tex]\Rightarrow 2x=6\\\\\Rightarrow x=3[/tex]

The expression of [tex]K_c[/tex] for above equation follows:

[tex]K_c=\frac{[Z]^2}{[X]\times [Y]}[/tex]

Putting values in above equation, we get:

[tex]K_c=\frac{6^2}{1\times 1}\\\\K_c=36[/tex]

Hence, the value of equilibrium constant for the above equation is 36

The equilibrium constant, Kc, for the given reaction is 36. This is calculated using the ratio of product concentrations to reactant concentrations at equilibrium, each concentration being raised to the power of its stoichiometric coefficient.

The equilibrium constant, Kc, describes the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to the power of its stoichiometric coefficient. Here, the equation is:

X(g) + Y(g) ---> 2 Z(g)

In terms of moles to start, we have 4.00 mol of X and Y, and 0 mol of Z. At equilibrium, we end up with 6.00 mol of Z, which, since 2 mol of Z are generated for every 1 mol of X and Y, means we have lost 3.00 mol of X and Y.

As the volume is 1L, the molar concentrations are the same as the number of moles. So, the concentrations at equilibrium are: [X] = [Y] = 1.00 M and [Z] = 6.00 M.

Applying this to the Kc expression gives: Kc = ([Z]^2) / ([X] * [Y]) = (6.00^2) / (1.00 * 1.00) = 36. Therefore, the equilibrium constant, Kc, is 36.

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Final answer:

The enthalpy change (ΔHreaction) for the given reaction is calculated using Hess's law and is found to be 122 kJ/mol.

Explanation:

To calculate the enthalpy change (ΔHreaction) for the reaction 3NO₂(g) + H₂O(l) → 2HNO3(aq) + NO(g), we will apply Hess's law and use the standard enthalpies of formation (ΔHf) for each compound.

The formula to calculate ΔHreaction is:

ΔHreaction = ∑(ΔHf products) - ∑(ΔHf reactants)

Using the provided ΔHf values:

We can calculate ΔHreaction as follows:

ΔHreaction = [2 × ΔHf(HNO3) + ΔHf(NO)] - [3 × ΔHf(NO2) + ΔHf(H2O)]

ΔHreaction = [(2 × 207 kJ/mol) + (90 kJ/mol)] - [(3 × 32 kJ/mol) + (286 kJ/mol)]

ΔHreaction = (414 kJ/mol + 90 kJ/mol) - (96 kJ/mol + 286 kJ/mol)

ΔHreaction = 504 kJ/mol - 382 kJ/mol

ΔHreaction = 122 kJ/mol

Final answer:

The chemical equation representing the decomposition of hydrogen peroxide into water and oxygen gas is 2 H2O2 → 2 H2O + O2. This represents a stoichiometric relationship with a 2:1 ratio of hydrogen peroxide to oxygen and a 1:1 ratio of hydrogen peroxide to water.

Explanation:

The stoichiometry of the chemical reaction where hydrogen peroxide (H2O2) decomposes into water (H2O) and oxygen gas (O2) can be expressed by the following balanced chemical equation:

2 H2O2 → 2 H2O + O2

According to the equation given by the student, 0.0800 mol of H2O2 decomposes to produce 0.0800 mol of H2O and 0.0400 mol of O2. However, based on stoichiometric coefficients, we should expect a 1:1 ratio between H2O2 and H2O, and a 2:1 ratio between H2O2 and O2. Therefore, the decomposition of 0.0800 mol H2O2 should yield 0.0400 mol O2 according to the equation, which aligns with what was provided by the chemist.

Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

When solid is converted to liquid or gas entropy increases,

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to liquid or gas, the force of attraction between the molecule decreases and hence entropy increases.

So,

The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

And

If the particles are loosely held , the entropy will increase , i.e. , when gas is converted to liquid or solid .

A. boiling water to form steam ,

change of state from liquid to gas ,

and , hence entropy increases .

B. dissolution of solid KCl in water  ,

the number of particles increases ,

and hence , entropy increases .

C. mixing of two gases into one container  ,

the number of particles increases ,

and hence , entropy increases .

D. freezing water to form ice  ,

change of state from liquid to solid ,

and hence , entropy decreases .

E. melting ice to form water .

change of solid to liquid ,

and hence , entropy increases .

Final answer:

The process that results in a decrease in entropy is freezing water to form ice, as it involves water transitioning from a disordered liquid state to an ordered solid state.

Explanation:

The process that produces a decrease in the entropy of the system is freezing water to form ice (Option D). During freezing, water molecules move from a higher state of disorder in the liquid phase to a more ordered solid phase, resulting in reduced entropy. Freezing involves the transition of water from the liquid phase, where the molecules are more disordered, to the solid phase, where they are arranged in a structured pattern. This structured pattern, characterized by a fixed position of the molecules in the crystal lattice of ice, embodies a lower state of entropy.

Even though freezing results in a decrease in the system's entropy, this does not violate the second law of thermodynamics because the surrounding environment's entropy increases when heat is released during the freezing process, ensuring that the total entropy of the system plus its surroundings does not decrease.

Answer: The value of equilibrium constant for reverse reaction is [tex](\frac{1}{K_1})^{1/2}[/tex]

Explanation:

The given chemical equation follows:

[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)[/tex]

The equilibrium constant for the above equation is [tex]K_1[/tex]

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

[tex]SO_3(g)\rightarrow SO_2(g)+\frac{1}{2}O_2(g)[/tex]

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '[tex]\frac{1}{2}[/tex]', the equilibrium constant of the reverse reaction will be the 1/2 power of the equilibrium constant  of initial reaction.

The value of equilibrium constant for reverse reaction is:

[tex]K_{eq}'=(\frac{1}{K_1})^{1/2}[/tex]

Hence, the value of equilibrium constant for reverse reaction is [tex](\frac{1}{K_1})^{1/2}[/tex]

Answer:

Approximately [tex]\rm -249.4\; kJ \cdot mol^{-1}[/tex].

Explanation:

[tex]\rm CO\; (g) + 3\; H_2\; (g) \to CH_4\; (g) + H_2 O\; (g)[/tex].

Note that hydrogen gas [tex]\rm H_2\; (g)[/tex] is the most stable allotrope of hydrogen. Since [tex]\rm H_2[/tex] is naturally a gas under standard conditions, the standard enthalpy of formation of [tex]\rm H_2\; (g)[/tex] would be equal to zero. That is:

Look up the standard enthalpy of formation for the other species:

(Source: CRC Handbook of Chemistry and Physics, 84th Edition (2004).)

[tex]\displaystyle \Delta H^{\circ}_\text{reaction} = \sum \Delta H^{\circ}_f(\text{products}) - \sum \Delta H^{\circ}_f(\text{reactants})[/tex].

In other words, the standard enthalpy change of a reaction is equal to:

In this case,

[tex]\begin{aligned} & \sum \Delta H^{\circ}_f(\text{products}) \\ =& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + \Delta H^{\circ}_f(\mathrm{H_2O\;(g)})\end{aligned}[/tex].

[tex]\begin{aligned} & \sum \Delta H^{\circ}_f(\text{reactants}) \\ =& \Delta H^{\circ}_f(\mathrm{CO\;(g)}) + 3\times \Delta H^{\circ}_f(\mathrm{H_2\;(g)})\end{aligned}[/tex].

Note that the number [tex]3[/tex] in front of [tex]\Delta H^{\circ}_f(\mathrm{H_2\;(g)})[/tex] corresponds to the coefficient of [tex]\rm H_2[/tex] in the chemical equation.

[tex]\begin{aligned}&\Delta H^{\circ}_\text{reaction} \\ =& \sum \Delta H^{\circ}_f(\text{products}) - \sum \Delta H^{\circ}_f(\text{reactants})\\ =& \left(\Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + \Delta H^{\circ}_f(\mathrm{H_2O\;(g)})\right) \\ &- \left(\Delta H^{\circ}_f(\mathrm{CO\;(g)}) + 3\times \Delta H^{\circ}_f(\mathrm{H_2\;(g)})\right) \\ =& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + (-74.6) - (3 \times 0 -110.5)\end{aligned}[/tex].

In other words,

[tex]\begin{aligned} & \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + (-74.6) - (3 \times 0 -110.5) \\=& \Delta H^{\circ}_\text{reaction} = -213.5\; \rm kJ\cdot mol^{-1} \end{aligned}[/tex].

Therefore,

[tex]\begin{aligned}& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) \\ =& -213.5 - ((-74.6) - (3 \times 0 -110.5)) \\=& -249.4\; \rm kJ\cdot mol^{-1} \end{aligned}[/tex].

Answer:

1a) 1.67 atm

1b) 68,330 g/mol

Explanation:

1a) For Boyle's Law, when a state change occurs without a change in the temperature, the product of the pressure by the volume remains constant. For Dalton's Law, the total pressure of a gas mixture is the sum of the partial pressure of the components. Then:

P1V1 + P2V2 = PV

Where P1 is the initial pressure of CO₂, V1 is the initial volume of it, P2 is the initial pressure of O₂, V2 is the initial pressure of it, P is the pressure of the mixture and V is the final volume of the mixture (V1 + V2).

1*250 + 2*500 = P*750

750P = 1250

P = 1.67 atm

1b) Let's call hemoglobin by Hem. The stoichiometry reaction is:

Hem + 4O₂ → HemO₂

So, let's calculate the number of moles of oxygen in the reaction, by the ideal gas law, PV = nRT, where P is the pressure, V is the volume (0.0023 L), n is the number of moles, R is the ideal gas constant (62.3637 L.torr/mol.K), and T is the temperature (37°C = 310 K).

743*0.0023 = n*62.3637*310

19,332.747n = 1.7089

n = 8.84x10⁻⁵ mol

For the reaction, the stoichiometry is:

1 mol of Hem -------------------- 4 mol of O₂

x ------------------- 8.84x10⁻⁵ mol of O₂

By a simple direct three rule

4x = 8.84x10⁻⁵

x = 2.21x10⁻⁵ mol of hemoglobin

The molar mass is the mass divided by the number of moles:

M = 1.51/2.21x10⁻⁵

M = 68,330 g/mol

The total pressure of the mixture of CO2 and O2 is 3.00 atm. The molar mass of hemoglobin is approximately 64.7 g/mol.

To determine the total pressure of the mixture of CO2 and O2, we can use Dalton's Law of Partial Pressures. According to this law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. In this case, the partial pressures of CO2 and O2 are 1.00 atm and 2.00 atm, respectively. Therefore, the total pressure of the mixture is 1.00 atm + 2.00 atm = 3.00 atm.

To find the molar mass of hemoglobin, we can use the ideal gas law. The equation for the ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. We can rearrange this equation to solve for the molar mass (M).

M = (mRT)/(PV), where m is the mass of the substance in grams. Plugging in the given values, we have M = (1.51 g)(0.0821 L·atm/mol·K)(310 K)/(743 torr)(0.0821 L·atm/mol·K)(2.30 mL/1000 L). Simplifying the calculation gives us a molar mass of hemoglobin approximately equal to 64.7 g/mol.

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Answer:

[tex][I^-]=4.6*10^{-8}M[/tex]

Explanation:

The expression of the Ksp:

[tex]Ksp_{AgCl}=[Ag^{+}][Cl^-][/tex]

[tex]Ksp_{AgI}=[Ag^{+}][I^-][/tex]

When the product of the concentrations of both ions equals the Ksp, the salt starts to precipitate.

For the AgCl:

[tex]1.8*10^{-10}M^{2}=[Ag^{+}]*0.1M[/tex]

[tex][Ag^{+}]=1.8*10^{-9}M[/tex]

Initially the concentration of I- was 0.1 M, due to the lower Ksp than the AgCl's, the AgI will precipite before. So, when AgCl starts to precipitate the concentration of I- will be in equilibrium, following the Ksp equation.

[tex]8.3*10^{-17}M^{2}=1.8*10^{-9}M*[I^-][/tex]

[tex][I^-]=4.6*10^{-8}M[/tex]

The standard enthalpy change for the combustion of C3H6(g) + H2(g) → C3H8(g) at 25°C is -219.4 kJ/mol.

The standard enthalpy change for the reaction is obtained by calculating the difference between the enthalpy of the products and the enthalpy of the reactants. In this case, the enthalpy change can be determined using the enthalpy of formation values for the compounds involved. The balanced equation for the combustion of propane is:

C3H6(g) + H2(g) → C3H8(g)

The enthalpy change can be calculated as follows:

∆H° = ∑∆H°f(products) - ∑∆H°f(reactants)

∆H° = [2*(∆H°f(CO2(g))) + ∆H°f(H2O(l))] - [∆H°f(C3H6(g)) + ∆H°f(H2(g))]

Substitute the given values for the enthalpies of formation:

∆H° = [2*(-393.5 kJ/mol) + (-285.8 kJ/mol)] - [(-2058.3 kJ/mol) + 0 kJ/mol]

Simplify the equation:

∆H° = -219.4 kJ/mol

Therefore, the standard enthalpy change for the combustion of C3H6(g) + H2(g) → C3H8(g) at 25°C is -219.4 kJ/mol.

1. The standard free-energy change for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 25 ∘C is -358,000 J (to three significant figures).

2. The standard cell potential for the reaction X(s)+2Y+(aq)→X2+(aq)+2Y(s) at 25 ∘C is 2.90 V (to three significant figures).

1. To calculate the standard free-energy change at 25 ∘C for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s), we can use the following equation:

ΔG∘=−nFE∘

where:

* ΔG∘ is the standard free-energy change (in J)

* n is the number of moles of electrons transferred (2 in this case)

* F is the Faraday constant (96,500 C/(mol e⁻))

* E∘ is the standard cell potential (in V)

The standard cell potential for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) is 1.83 V. Therefore, the standard free-energy change is:

ΔG∘=−(2 mol e⁻)(96,500 C/(mol e⁻))(1.83 V)

ΔG∘=−358,000 J

Therefore, the standard free-energy change for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 25 ∘C is -358,000 J (to three significant figures).

2. To calculate the standard cell potential at 25 ∘C for the reaction X(s)+2Y+(aq)→X2+(aq)+2Y(s), we can use the following equation:

E∘=−ΔG∘nF

where:

* E∘ is the standard cell potential (in V)

* ΔG∘ is the standard free-energy change (in J)

* n is the number of moles of electrons transferred (2 in this case)

* F is the Faraday constant (96,500 C/(mol e⁻))

We are given that ΔH∘ = -675 kJ and ΔS∘ = -357 J/K. We can use these values to calculate ΔG∘ using the following equation:

ΔG∘=ΔH∘−TΔS∘

where T is the temperature in Kelvin (298 K in this case).

ΔG∘=−675 kJ−(298 K)(−357 J/K)

ΔG∘=−559,890 J

Now we can calculate the standard cell potential using the equation above:

E∘=−ΔG∘nF

E∘=−(−559,890 J)/(2 mol e⁻)(96,500 C/(mol e⁻))

E∘=2.90 V

Therefore, the standard cell potential for the reaction X(s)+2Y+(aq)→X2+(aq)+2Y(s) at 25 ∘C is 2.90 V (to three significant figures).

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The standard free-energy change for the reaction Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s) is -372 kJ at 25 °C. The standard cell potential for the reaction X(s) + 2Y⁺(aq) → X²⁺(aq) + 2Y(s) is 2.95 V at 25 °C. These values were determined using proper thermodynamic equations and constants.

1. Calculate the Standard Free-Energy Change:

The given reaction is: Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s).

To find the standard free-energy change (ΔG°), we use the given equation ΔG° = -nFE°.

2. Calculate the Standard Cell Potential:

Given the reaction: X(s) + 2Y⁺(aq) →  X²⁺(aq) + 2Y(s)

And ΔH° = -675 kJ and ΔS° = -357 J/K.

The standard cell potential for this reaction at 25 °C is 2.95 V.

Answer:

The correct answer is A) 1.6 x 10-6

Explanation:

A weak monoprotic acid has the following dissociation equilibrium. At the beggining (t=0), the concentration of the monoprotic acid (HA) is equal to 0.10 M and the concentration of the ions H⁺ and A⁻ is zero (no dissociation). At a time t, dissociation occur and there is x concentration of H⁺ and A⁻ which is given by the dissociation constant Ka.

           HA(aq)    ⇄     H⁺(aq)    +       A⁻(aq)

t=0        0.10 M              0                     0

t               -x                    +x                  +x

eq          0.10 M-x             x                   x

Ka= [tex]\frac{x^{2} }{0.10 - x}[/tex]

As the pH is 3.40, we can calculate the concentration of both H⁺ and A⁻, as follows:

pH= - log (conc H⁺)= -log x

⇒ x = [tex]10^{-3.40}[/tex]= 3.98 x 10⁻⁴

Now we introduce x in the previous equation to calculate Ka:

Ka= [tex]\frac{(3.98 x 10^{-4} )^{2} }{(0.10 - (3.98 x 10^{-4}) }[/tex]

Ka= 1.59 x 10⁻⁶ ≅ 1.60 x 10⁻⁶

Final answer:

To find the acid-ionization constant Ka, the pH is used to calculate the concentration of H+ ions, and this value is squared and divided by the initial concentration of the acid to obtain Ka, which is 1.6 x 10^-6.

Explanation:

The question seeks to determine the acid-ionization constant (Ka) for a weak monoprotic acid with a given molarity and pH. The pH of the solution is 3.40, which means the concentration of hydrogen ions [H+] is 10-3.40 M. Since the acid is weak and monoprotic, its dissociation in water can be represented by HA → H+ + A-. The given concentration of acid (0.10 M) will slightly ionize into H+ and A- ions.

To solve for Ka, the equilibrium expression is Ka = [H+][A-] / [HA]. Given that [H+] = 10-3.40, we can assume that the concentration of A- at equilibrium is also 10-3.40 since the acid donates one proton per molecule (in a 1:1 ratio). The concentration of un-ionized HA will then be approximately ([initial HA] - x) = (0.10 M - 10-3.40 M). However, because the ionization of a weak acid is very small compared to the initial concentration, we can approximate [HA] at equilibrium to be 0.10 M. From this, the Ka can be approximated to be (10-3.40)2 / 0.10 M = 1.6 x 10-6.

Answer:

The half-life of strontium-90 is 28.81 years

Explanation:

Step 1: Data given

Mass of strontium 90 = 1.000 grams

After 2 years there remain 0.953 grams

Step 2: Calculate half-life time

k = (-1/t) * ln (Nt/N0)

⇒ with k = rate constant for the decay

⇒ with N0 = the mass of strontium at the start (t=0) = 1.000 grams

⇒ with Nt =  the mass of strontium after time t (2 years) = 0.953 grams

k = (-1/2) * ln(0.953/1)

k = (-1/2) * (-0.0481) = 0.02405 /yr

t1/2 = 0.693/k= 0.693/ 0.02405 = 28.81  years

The half-life of strontium-90 is 28.81 years

The given data does not provide a clear half-life for strontium-90 since the mass of the substance did not halve in 2 years. Thus, none of the options provided fits the scenario.

The half-life of a radioactive substance is the time taken for half of the atoms in a sample to decay. Here, we have the sample of strontium-90 that has not halved in 2 years since we still have 0.953g out of 1.000g. Given the available options, none of them matches with the provided data because if a substance's mass did not reduce to half in 2 years then its half-life should be greater than 2 years. Possibly, there might be a calculation or data collection error in the problem scenario provided.

The half-life of strontium-90 can be calculated using the given information. If we start with 1.000 g of strontium-90 and after 2.00 years, 0.953 g remains, we can calculate the percentage of strontium-90 remaining: (0.953 g / 1.000 g) x 100% = 95.3%. Since half-life is the time required for half the atoms in a sample to decay, we can conclude that it took 2.00 years for 50% of the strontium-90 to decay. Therefore, the half-life of strontium-90 is 2.00 years.

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Answer:

2.64 × 10⁶ g

Explanation:

We can find the mass of air using the ideal gas equation.

[tex]P.V=n.R.T=\frac{m}{M} .R.T[/tex]

where,

P is the pressure (P = 1.00 atm)

V is the volume (V = 2.95 × 10⁶ L)

n is the number of moles

R is the ideal gas constant (0.08206atm.L/mol.K)

T is the absolute temperature (121°C + 273 = 394 K)

m is the mass

M is the molar mass (28.09 g/mol)

[tex]m=\frac{P.V.M}{R.T} =\frac{1.00atm \times 2.95 \times 10^{6} L \times 28.97g/mol}{(0.08206atm.L/mol.K) \times 394 K } =2.64 \times 10^{6} g[/tex]

Answer:

d) Decreasing the temperature at constant volume

Explanation:

if we assume ideal gas behaviour, since the concentration of the gas is proportional to its pressure at constant temperature

(from the ideal gas law PV =nRT or P= CRT), the equilibrium constant in terms of pressure will be

Kp = (p² HCl *p² I2)/(p² HI *p² CL2)

from Dalton's law : pi = P*xi

Kp = (P²*x² HCl *P* x I2)/(P*x HI *P²x² CL2) =  (x² HCl * x I2)/(x HI *x² CL2)

since Kp does not change because the T is constant and it does not depend on pressure → the equilibrium will not change due to changes in pressure caused by reductions or increases in volume at constant pressure and composition

also since the reaction is exothermic → an increase in temperature will displace the equilibrium towards the reactants , and thus decreasing the moles of I2 at equilibrium

this can be seen from van't hoff equation :

d ln (K) /dT= ΔH/RT² , since ΔH>0 → K diminishes with increase in temperature

on the other hand, a decrease in temperature will displace the equilibrium towards the products , and thus increasing the moles of I2 at equilibrium

Explanation:

There are two type of reaction possible , i.e. , exothermic and endothermic reaction ,

Exothermic reaction -

The type of reaction in which energy in the form of heat or light is released , is known as exothermic reaction ,

The reaction mixture usually get heated after the reaction .( temperature increases ) .

The sign of ΔH of an exothermic reaction is negative .

Similarly ,

Endothermic reaction -

The type of reaction in which energy is absorbed in the form of heat is known as endothermic reaction .

The reaction mixture usually get cooled after the reaction .( temperature decreases ) .

The sign of ΔH of an Endothermic  reaction is positive .

a. dry ice evaporating ,

Endothermic reaction , ΔH = positive .

b. a sparkler burning ,

Exothermic reaction ΔH = negative .

c. the reaction that occurs in a chemical cold pack often used to ice athletic injuries

Endothermic reaction , ΔH = positive .

Dry ice evaporates in an endothermic process with a positive ΔH. A sparkler burning is an exothermic process with a negative ΔH. The reaction in a chemical cold pack is endothermic with a positive ΔH.

In chemistry, understanding whether a process is exothermic or endothermic is crucial. Here’s the breakdown for each scenario:

Answer:

A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M

The precipitate formed is CuI

Explanation:

Step 1: Data given

The solution contains 0.021 M Cl- and 0.017 M I-.

Ksp(CuCl) = 1.0 × 10-6

Ksp(CuI) = 5.1 × 10-12.

Step 2:  Calculate [Cu+]

Ksp(CuCl) = [Cu+] [Cl-]  

1.0 * 10^-6  = [Cu+] [Cl-]  

1.0 * 10^-6  = [Cu+] [0.021]  

[Cu+] = 1.0 * 10^-6 / 0.021

[Cu+] = 4.76 *10^-5 M

Ksp(CuI) = [Cu] [I]  

5.1 * 10^-12  = [Cu+] [I-]  

5.1 * 10^-12 =[Cu+] [0.017]  

[Cu+] = 5.1 * 10^-12 / 0.017

[Cu+] = 3.0 *10^-10 M

[Cu+]from CuI hast the lowest concentration

A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M

The precipitate formed is CuI

To calculate ΔG for the reaction under given conditions, use the equation ΔG = ΔG° + RT ln(Q), substituting given values and solving. Positive ΔG suggests a non-spontaneous reaction, while negative ΔG indicates a spontaneous reaction.

The student's question pertains to the change in Gibbs Free Energy (ΔG) for the reaction CO2(g) + CCl4(g) ⇌ 2 COCl2(g) under specified conditions. The ΔG of a reaction can be calculated using the equation ΔG = ΔG° + RT ln(Q), where ΔG° is the standard free energy change, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (here, 25°C + 273.15 = 298.15 K), and Q is the reaction quotient. The equation thus becomes ΔG = 46.9 kJ + (8.314 J/(mol·K) * 298.15 K * ln((0.653^2)/(0.459 * 0.984))).

Converting R from J to kJ gives us ΔG = 46.9 kJ + (0.008314 kJ/(mol·K) * 298.15 K * ln((0.653^2)/(0.459 * 0.984))). Solving this equation, we obtain a value for ΔG, which represents the Gibbs free energy under the given conditions. A positive ΔG suggests the reaction is non-spontaneous, while a negative ΔG indicates a spontaneous reaction. An equilibrium will lean towards the side with the lower Gibbs free energy.

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Answer:

B, because that was the answer on quizlet.

Explanation:

Answer:

The correct option is: A) N₂

Explanation:

The Bond length of a chemical bond is the length of a chemical bond formed between two given atoms.

Bond length is inversely proportional to the bond order of the chemical bond, which is the total number of bonds between two atoms. Thus as the bond order increases, the bond length decreases.

A) N₂: The nitrogen-nitrogen bond in dinitrogen is a triple bond (N≡N).

Thus the bond order = 3.

B) O₂: The oxygen-oxygen bond in dioxygen is a triple bond (O=O).

Thus the bond order = 2.

C) SO₂: Sulfur dioxide is a resonance stabilized molecule and its resonance hybrid shows that the sulfur-oxygen bond in sulfur dioxide is a partial double bond.

Thus the bond order = 1.5

D) SO₃: Sulfur trioxide is a resonance stabilized molecule and its resonance hybrid shows that the sulfur-oxygen bond in sulfur trioxide is a partial double bond.

Thus the bond order = 1.33

Since the bond order of N₂ is the largest, therefore, the N-N bond length is the shortest.

Answer:

1) Exothermic.

2) It goes up.

3) It moves out.

4) It does work.

5) 94 kJ.

Explanation:

1) An endothermic reaction is a reaction that absorbs heat from the surroundings, and an exothermic reaction is a reaction that releases heat to the surroundings. So the reaction placed is exothermic.

2) Because the heat is flowing to the water, its temperature will go up.

3) By the first law of the thermodynamics:

ΔU = Q - W

Where ΔU is the total energy, Q is the heat, and W is the work. Because the energy and the heat are being released, they are both negative:

-244 = -150 - W

W = 94 kJ

The work is positive, so it's being doing by the system, it means that the system is expanding, and the piston moves out.

4) As explained above, the gas mixture (the system) does work.

5) As shown above, W = 94 kJ