Use the partition function for a harmonic oscillator, q = 1 / (1- e^-hv/kBT) , to determine the ratio of populations in the lowest and first excited energy levels, P0/P1 , for the indicated systems at two temperatures, 50 K and 298 K :
(a) The 1700 cm−1 stretching vibrational mode for the carbonyl (C=O) group in the peptide bond of a polypeptide.
(b) the 150 cm−1 torsional vibrational mode for the C=C bond in trans-2-butadiene.

Answer:

1.08g

Explanation:  

Like all other hydrocarbons, methane burns in oxygen to form carbon iv oxide and water. The chemical equation of this equation is shown below;

  CH4 (g)  +  2O2  (g) ----------->  2H2O (l)   +   CO2 (g)

From the reaction equation, we can see that one mole of methane gave 2  moles of water. This is the theoretical yield. We need to note the actual yield.

To get the actual yield, we get the number of moles of methane reacted. To get this, we divide the mass of methane by  the molar mass of methane. The molar mass of methane is 16g/mol. The number of moles is thus 0.481/16 = 0.03 moles.

Since 1 mole methane gave 2 moles of water, this shows that 0.06 moles of water were produced. The mass of water thus produced is 0.06 multiplied by the molar mass of water. The molar mass of water is 18g/mol. The mass produced is 0.06 * 18 = 1.08g

Now, we do same for the mass of oxygen. From the reaction equation, 2 moles of oxygen produced two moles of water. Hence, we can see from here that the number of moles here are equal. We then proceed to calculate the actual number of moles of oxygen produced. This is the mass of the oxygen divided by the molar mass of molecular oxygen. The molar mass of molecular oxygen is 32g/mol. The number of moles thus produced is 0.54/32 = 0.016875 mole. The number of moles are equal, this means that the number of moles of oxygen produced is also 0.016875

Now, to get the mass of water produced, we multiply the number of moles by the molar mass of water. The molar mass of water is 18g/mol.

The mass is thus, 0.016875 * 18 = 0.30375g

1.08g is higher and thus is the maximum mass

Answer:

a. -206,4kJ

b. Surroundings will gain heat.

c. -115kJ are given off.

Explanation:

It is possible to obtain ΔH of a reaction using Hess's law that consist in sum the different ΔH's of other reactions until obtain the reaction you need.

Using:

(1) C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ

(2) H₂(g) + O₂(g) → H₂O₂ (l) ΔH: -187.8 kJ

(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH: -285.8 kJ

It is possible to obtain:

C₆H₄(OH)₂(l) + H₂O₂(l) → C₆H₄O₂(l) + 2H₂O(l)

From (1)-(2)+2×(3). That is:

(1) C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ

-(2) H₂O₂(l) → H₂(g) + O₂(g) ΔH: +187.8 kJ

2x(3) 2H₂(g) + O₂(g) → 2H₂O(l) ΔH: 2×-285.8 kJ

The ΔH you obtain is:

+177,4kJ + 187,8kJ - 2×285.8 kJ = -206,4kJ

b. When ΔH of a reaction is <0, the reaction is exothermic, that means that the reaction produce heat and the surroundings will gain this heat.

c. 20,0g of H₂O are:

20,0g×[tex]\frac{1mol}{18,01g}[/tex] = 1,11 mol H₂O

As 2 moles of H₂O are produced when -206,4kJ are given off, when 1,11mol of H₂O are produced, there are given off:

1,11mol H₂O×[tex]\frac{-206,4kJ}{2mol}[/tex] = -115kJ

I hope it helps!

Final answer:

To calculate the enthalpy change (ΔH°) for the given reaction, use Hess's Law and the enthalpy changes provided. Heat would be gained by the surroundings as the reaction occurs, based on the positive enthalpy change. When 20.0 g of H2O is produced, approximately -317 kJ of heat would be given off.

Explanation:

To calculate the enthalpy change (ΔH°) for the given reaction, we can use the Hess's Law and the enthalpy changes provided. We need to find a combination of reactions whose sum gives us the desired reaction. Here's an approach:

1. Reverse the first equation and multiply it by 2 to get 2CO(g) → 2C(s) + O2(g) with ΔH° = -442.0 kJ.

2. Use the second equation as it is: C(s) + O2(g) → CO2(g) with ΔH° = -393.5 kJ.

3. Add the two equations together: 2CO(g) + C(s) + O2(g) → 2CO2(g) with ΔH° = -835.5 kJ.

This combined equation is equivalent to the given reaction, but we need to multiply it by -2 to match the coefficients. Therefore, the enthalpy change for the given reaction is ΔH° = -2*(-835.5 kJ) = 1671 kJ.

For part (b), based on the positive enthalpy change, heat would be gained by the surroundings as the reaction occurs. In part (c), we can use the enthalpy change value from part (a) to calculate the heat produced. Since the reaction produces 2 moles of H2O, we can use the provided enthalpy change for H2O(l) to find the heat produced when 20.0 g of H2O is produced.

Using the given enthalpy change values:

ΔH° C6H4(OH)2(l) → C6H4O2(l) + H2(g) = +177.4 kJ

ΔH° H2(g) + O2(g) → H2O(l) = -285.8 kJ

We can set up the following equation to find the heat produced:

(20.0 g H2O) x (1 mol H2O / 18.015 g H2O) x (-285.8 kJ / 1 mol H2O) = -317 kJ.

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.

Answer: option d

Explanation:

Project teams must be proactive. Each member of the team should empowered and each member can share ideas and experience in order to make the WBS more precise.

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:

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

Answer: The mass of hydrogen sulfide that can be dissolved is 2.86 grams.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the gas.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{H_2S}=K_H\times p_{liquid}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]0.087M/atm[/tex]

[tex]p_{H_2S}[/tex] = partial pressure of hydrogen sulfide gas = 2.42 atm

Putting values in above equation, we get:

[tex]C_{H_2S}=0.087M/atm\times 2.42atm\\\\C_{H_2S}=0.2105M[/tex]

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

We are given:

Molarity of solution = 0.2105 M

Molar mass of hydrogen sulfide = 34 g/mol

Volume of solution = 400.0 mL

Putting values in above equation, we get:

[tex]0.2105M=\frac{\text{Mass of hydrogen sulfide}\times 1000}{34g/mol\times 400.0mL}\\\\\text{Mass of }H_2S=\frac{0.2105\times 34\times 400}{1000}=2.86g[/tex]

Hence, the mass of hydrogen sulfide that can be dissolved is 2.86 grams.

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:

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: 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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AgI precipitates first due to its smaller Ksp. The I⁻ concentration for the second compound (PbI₂) to begin precipitating is 9.3 x 10⁻⁵ M.

The compound that will precipitate first will be the one with the smallest solubility product constant, Ksp. The given Ksp values for AgI and PbI₂ are 1.5 x 10⁻¹⁶ and 8.7 x 10⁻⁹ respectively; thus, AgI has a smaller Ksp, indicating it is less soluble and precipitates first.

To find the I⁻ concentration for the second compound to precipitate, we set up the Ksp expression for PbI₂ (assuming AgI has already precipitated out, AgNO₃ is gone and the remaining AgI does not affect this calculation): [Pb²⁺][I⁻]² = Ksp. Pb(NO₃)₂ fully dissociates, so [Pb²⁺] is assumed to be 0.10 M, and we can substitute this and the Ksp into our expression, then solve for [I-]: (0.10 M)([I⁻])² = 8.7 x 10⁻⁹ M, from which we find [I-] = sqrt((8.7 x 10⁻⁹ M) / 0.10 M) = 9.3 x 10⁻⁵ M.

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Answer:

Steps

i) In thee mitochondrion, pyruvate carboxylase converts pyruvate to oxloacetate.

ii) Malate dehydrogenase in the mitochondrion reduces oxaloacete to to malate.

iii) Malate dehydrogenase in the cytoplasm oxidizes malate to oxaloacetate.

iv) Phosphoenolpyruvate carboxykinase decarboxylates and phosphorylates oxaloacetate.forming phosphoenolpyruvate.  

Explanation:

Gluconeogenesis is the synthesis of glucose from non - carbohydrate compounds. The substrates for gluconeogenesis are lactate, pyruvate, amino acids, propionate and glycerol.

Gluconeogenesis occurs only in cytosol but the precursor is produced in mitochondria. In the conversion of pyruvte to phosphoenolpyruvate occur in mitochondria and cytosol.

Step -1:

Pyruvate carboxylase is a biotin dependent enzyme located in mitochondria. It converts pyruvate to oxlaoacetate and carbondioxide in the presence of ATP.Oxlaocetate synthesized in mitochondrial matrix has to be transported to cytosol for gluconeogenesis. Oxaloacetate is impermeble, cannot be sent out of mitochondria. So it has to be converted to malate.

Step -2:

Malate dehydrogenase in mitochondria converts oxaloacetate synthesized in mitochondrial matrix to malate. And then  it is transported to cytosol.

Step 3:

Malate dehydrogenase responsible for reversible reaction in cytosol converts malate to oxaloacetate.

Step -4

Phosphoenolpyruvate carboxy-kinase in cytosol converts oxaloacetate to PEP. The enzyme transfer high energy phosphate bond from GTP to oxaloacetate to from PEP and liberated carbondioxide.

Therefore, the steps of glycolysis converts phosphoenolpyruvate to pyruvate are as follows.

i) In thee mitochondrion, pyruvate carboxylase converts pyruvate to oxloacetate.

ii) Malate dehydrogenase in the mitochondrion reduces oxaloacete to to malate.

iii) Malate dehydrogenase in the cytoplasm oxidizes malate to oxaloacetate.

iv) Phosphoenolpyruvate carboxykinase decarboxylates and phosphorylates oxaloacetate.forming phosphoenolpyruvate.  

Gluconeogenesis uses pyruvate as a starting point to generate glucose, going through several steps that largely mirror glycolysis in reverse order. However, the first step in gluconeogenesis varies, converting pyruvate into oxaloacetate before continuing with the conversion into phosphoenolpyruvate. It's a complex process that allows the cell to regulate the production and use of glucose efficiently.

The process of converting pyruvate to phosphoenolpyruvate via gluconeogenesis is not a simple reversal of the glycolysis steps. It starts with the conversion of pyruvate to oxaloacetate. Oxaloacetate serves as a substrate for the enzyme phosphoenolpyruvate carboxykinase (PEPCK), which transforms oxaloacetate into phosphoenolpyruvate (PEP). From this step, gluconeogenesis nearly mimics glycolysis, but in reverse. PEP is converted back into 2-phosphoglycerate, then into 3-phosphoglycerate. Then, 3-phosphoglycerate is converted into 1,3 bisphosphoglycerate and then into glyceraldehyde-3-phosphate. Next, two molecules of glyceraldehyde-3-phosphate combine to form fructose-1-6-bisphosphate, which is then converted into fructose 6-phosphate and then into glucose-6-phosphate. Finally, a series of reactions generates glucose. This sequence enables glycolysis and gluconeogenesis to be regulated independently of each other.

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Final answer:

To calculate the mass of dry ice needed to completely sublime in water, we can use the concept of stoichiometry and the ideal gas law equation. The moles of CO2 can be determined using the ideal gas law equation, and then converted to moles of water using the stoichiometry of the balanced equation. Finally, the mass of dry ice can be calculated using the moles of CO2 and the molar mass of CO2.

Explanation:

In order to calculate the mass of dry ice that should be added to the water, we need to use the concept of stoichiometry. The equation for the sublimation of carbon dioxide is CO2(s) -> CO2(g). We can use the molar ratio between CO2(s) and CO2(g) to convert the volume of water to moles of CO2. From there, we can use the molar mass of CO2 to calculate the mass of dry ice needed.

First, we need to calculate the moles of CO2 that can be produced using the ideal gas law equation:

n = PV / RT

where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature. Since temperature and pressure are not given, we cannot directly solve for moles of CO2. However, we can assume that the pressure and temperature remain constant throughout the process, so the ideal gas equation can still be used to find the moles of CO2.

The next step is to use stoichiometry to convert moles of CO2 to moles of water. The balanced equation is:

CO2(s) + H2O(l) -> H2CO3(aq)

From the balanced equation, we can see that the mole ratio between CO2 and water is 1:1. Therefore, if we know the moles of CO2, we also know the moles of water. The molar mass of CO2 is 44.01 g/mol, so we can calculate the mass of dry ice needed using the equation:

mass = moles * molar mass

Since we know the volume of water (15.0 L) and the final temperature (28°C), we can plug those values into the ideal gas equation and calculate the moles of CO2. Then, we can use the moles of CO2 to calculate the mass of dry ice needed. The final mass will depend on the pressure and temperature at which the dry ice is added, as well as the final temperature of the water.

Final answer:

The student's question involves calculating the mass of dry ice needed to produce fog in water until the temperature drops to 28°C. To perform this calculation, details such as the heat capacity of water, the heat of sublimation of CO2, and the complete temperature value are critical, but these are not provided in the question.

Explanation:

Dry ice, known chemically as solid carbon dioxide (CO2), sublimes at room temperature and pressure, meaning it converts directly from solid to gas. In the context of the question, when dry ice is added to warm water, the heat transfer from the water to the dry ice accelerates this sublimation process. The interaction cools the water down while providing a fog effect until either all the dry ice sublimes or the water temperature drops too much to sustain rapid sublimation.

To calculate the mass of dry ice required to create these special effects until the water cools to 28°C, one would need to determine the amount of heat the water can provide before reaching this temperature, and then use the heat of sublimation for CO2 to find the corresponding mass of dry ice. However, since the question doesn't provide specific heat capacity, heat of sublimation values, or the completion temperature value (presumably 28°C), an exact calculation cannot be performed without this information.

Handling dry ice requires caution due to its extremely cold temperature, and it is essential to be in a well-ventilated area since the resulting CO2 gas is heavier than air and can displace oxygen.

Answer:

The major product of the reaction is (2S,3R)-3-methoxy - 3-methylpentan-2-ol.

Explanation:

Epoxide reacts with methanol in the presence of sulfuric acid.

In this chemicl reaction methanol act as nucleophile and sulfuric acid act as catalyst.

The chemical reaction is as follows.

An epoxide is a ring compound in which the oxygen atom closes the ring.

An epoxide is a ring compound in which the oxygen atom closes the ring. The reaction of the epoxide with methanol in the presence of acid is an example of a ring opening reaction.

The product of this reaction is shown in the image attached to this answer. The acid attacks the epoxide and opens the ring then te methanol attacks subsequently to yield the product.

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Answer: 2948

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]

[tex]k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{5730}=1.21\times 10^{-4}years^{-1}[/tex]

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  = [tex]1.21\times 10^{-4}years^{-1}[/tex]

t = age of sample  = ?

a = let initial amount of the reactant  = 100

a - x = amount left after decay process = [tex]\frac{70}{100}\times 100=70[/tex]

[tex]t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{70}[/tex]

[tex]t=2948years[/tex]

Thus the fossil is 2948 years old.

The age of the fossil with 70% of the Carbon-14 level of living organism is 2948 years old

To determine the age of the fossil, we need to consider the decay of Carbon-14.

Given that the Carbon-14 level in the fossil is 70% that of living organisms, we can calculate the age using the concept of half-life.

Expression for rate law for first order kinetics is given by:

where,

Thus the fossil is 2948 years old.

Answer:Answer: The statements that describe amphibolic characteristics of the citric acid cycle is that both catabolic and anabolic processes occur.

Explanation: the metabolic pathways includes; catabolic, anabolic and amphibolic processes.

A catabolic pathway is an exergonic system that produces chemical energy in the form of Adenosine Triphosphate (ATP) usually from energy containing sources such as carbohydrates, fats, and proteins.

An anabolic pathway is a biosynthetic pathway, whereby smaller molecules combine to form larger and more complex ones.[

An amphibolic pathway is one that can be either catabolic or anabolic based on the need for energy. The citric acid pathway contains both energy producing and utilizing pathway.

The citric acid cycle is amphibolic, participating in both catabolic and anabolic processes, producing intermediates for various biosynthetic pathways and involving oxidation and reduction reactions.

The citric acid cycle, also known as the Krebs cycle, is considered amphibolic because it has both catabolic and anabolic functions. It is a central metabolic pathway in the cell, involved in the oxidation of acetyl-CoA into CO2, and is crucial for energy production in the form of ATP. Furthermore, the cycle provides intermediates for various anabolic pathways, making it an integral part of cellular metabolism. For instance, it produces
-ketoglutarate, which is a precursor for amino acid synthesis, and succinyl-CoA, which is required for heme group synthesis. Also, the cycle involves both oxidation and reduction reactions, further asserting its amphibolic nature.

Anaplerotic and cataplerotic pathways are essential for maintaining the function of the citric acid cycle. Anaplerotic pathways replenish cycle intermediates that are removed for biosynthesis, whereas cataplerotic pathways describe the removal of intermediates from the cycle for the synthesis of other compounds. This balance allows the cycle to efficiently meet the fluctuating needs of the cell for both energy production and the synthesis of crucial molecules.

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:

The coefficients found are: 2,3,10,4,3,2,5

The sum of the coefficients is: 2+3+10 + 4 + 3 + 2 + 5 = 29

Explanation:

K2CrO4+Na2SO3+HCl→KCl+Na2SO4+CrCl3+H2O

On the left side we 2 times K , on the right side we have 1 time K

So we have to multiply KCl by 2

K2CrO4+Na2SO3+HCl→2KCl+Na2SO4+CrCl3+H2O

On the left side wehave 1 Cl, on the right side we have 5 Cl

So the left side we have to multiply by 5

K2CrO4+Na2SO3+5 HCl→2KCl+Na2SO4+CrCl3+H2O

On the left side we have 5 times H, on the right side, we have 2 times H

So on the left side the HCl we should multiply by 10 (instead of 5)

On the right side, we should multiply H2O by 5

K2CrO4+Na2SO3+ 10HCl→2KCl+Na2SO4+CrCl3+ 5H2O

So on the left side we have 10 times Cl, on the right side KCl and CrCL3 should be doubled

K2CrO4+Na2SO3+ 10HCl→4 KCl+Na2SO4+ 2CrCl3+ 5H2O

On the right side we have 4 times K, and 2 times Cr. This mean on the left side K2CrO4 should be multiplied by 2

2K2CrO4+Na2SO3+ 10HCl→4 KCl+Na2SO4+ 2CrCl3+ 5H2O

On the left side we have 8 times O plus 3x times O

On the right side we have 5 times O plus 4x times O

To equal this we should multiply on the left side Na2SO3 by 3

and Na2SO4 on the right side also by 3

Now the entire equation is balanced

2 K2CrO4 + 3 Na2SO3 + 10 HCl → 4 KCl + 3 Na2SO4 + 2 CrCl3 + 5 H2O

The coefficients found are: 2,3,10,4,3,2,5

The sum of the coefficients is: 2+3+10 + 4 + 3 + 2 + 5 = 29

Final answer:

To balance the chemical equation, coefficients are placed in front of the compounds to ensure the same number and types of atoms on both sides. The balanced equation is: 3K2CrO4 + 3Na2SO3 + 14HCl → 3KCl + 3Na2SO4 + 2CrCl3 + 7H2O and the coefficients are 3,3,14,3,3,2,7.

Explanation:

To balance the given chemical equation, we must ensure that the number of atoms of each element is the same on both sides of the equation. Let's add coefficients to balance:

Start with Chrome atoms: There is 1 Cr on the left side and 3 on the right side, so place a coefficient 3 in front of K2CrO4.

Next, look at sulfur atoms: There are 3 S on the left and 3 S on the right, so we are now balanced for sulfur.

For potassium atoms, put a coefficient 3 in front of KCl on the product side.

Balance sodium by having a coefficient 3 in front of Na2SO4 since there are 6 Na on the reactant side.

Hydrogen atoms are balanced by placing a coefficient 6 in front of HCl to get 6 H atoms on both sides.

Now, balance the chloride ions: We have 6 Cl from HCl, but we need 9 Cl on the products side (3 from KCl and 6 from CrCl3). So the final coefficient for HCl is 14.

Finally, place a coefficient 7 in front of H2O to balance out the oxygen atoms.

The balanced equation is therefore: 3K2CrO4 + 3Na2SO3 + 14HCl → 3KCl + 3Na2SO4 + 2CrCl3 + 7H2O. The coefficients in the order that they appear in the equation are: 3,3,14,3,3,2,7.

Answer:

a. 2.041 V

b. 2.116 V

c. 2.043 V

Explanation:

a.

In an electrochemical cell;

reduction occurs at the cathode & oxidation at the anode.

Thus, from the question; reduction at cathode=

[tex]PbO_{2(s)} + 3H_{(aq)}^{+} +HSO_{4}_{(aq)} +2e^{-}[/tex] → [tex]PbSO_{4(s)} + 2H_{2} O_{(l)} E^{0}_{(cathode)} = 1.685V[/tex]

Oxidation at anode;

[tex]Pb_{s} + HSO^{-}_{4}[/tex]  →  [tex]PbSO_{4(s)}+H^{+}+2e^{-} E^{0}_{anode}  =0.356V[/tex]

The net process in the Cell is obtained by the algebraic sum of the two half-cell reactions:

Overall reaction:

[tex]PbO_{2(s)}+Pb_{s}+2HSO^{-}_{4(aq)}+ 2 H^{+}_{(aq)}[/tex]  →  [tex]PbSO_{4(s)} + 2H_{2} O_{(l)}[/tex]

[tex]E^{0} _{cell} = E^{0} _{cathode}- E^{0} _{anode}[/tex]

= 1.685V  -   (- 0.356V)

= 2.041V

b.      

To determine the initial value of E-cell :

In the electrochemical cell; both the anode and the cathode were submerged into 4.30M of Sulfuric acid; therefore:

[tex]H_{2} SO_{4}[/tex] →     [tex]H^{+} + HSO_{4}^{-}[/tex]

Assuming:

[tex]H^{+} = HSO_{4}^{-}[/tex]   such that (;)  [tex]H_{2} SO_{4}[/tex]   = 4.30M

let  : a=  [tex]H^{+}[/tex] & b=  [tex]HSO_{4}^{-}[/tex]

[tex]E^{i} _{cell}[/tex] =  

[tex]E^{0} _{cell}[/tex]  -  [tex]\frac{0.0591}{2}[/tex] ×  log   [tex]\frac{1}{(a)^{2} (b)^{2} }[/tex]

=2.041V -  [tex]\frac{0.0591}{2}[/tex]  ×  log  [tex]\frac{1}{(4.30)^{2} (4.30)^{2} }[/tex]

= 2.041V  -  0.02955  ×  log  [tex]\frac{1}{341.8801}[/tex]

= 2.041V  -  0.02955  ×  log  [tex]{0.0029250026}[/tex]

= 2.041V  -  0.02955  ×  (-2.5339)

= 2.041V   +  0.075V

=2.116V

c.

If concentration of [tex]H^{+}[/tex] is dropped by 76.00%

[tex]H^{+} = HSO_{4}^{-}[/tex]  =  4.30M

= 4.30M  -  4.30M   [tex]\frac{76}{100}[/tex]

= 1.032M

[tex]E_{cell}[/tex]  =  2.041V -  [tex]\frac{0.0591}{2}[/tex]  ×  log  [tex]\frac{1}{(1.032)^{2} (1.032)^{2} }[/tex]

=2.041V  -  0.02955  ×  log  [tex]\frac{1}{1.13427612058}[/tex]

=2.041V  -  0.02955  ×  log  (0.88161954735)

=2.041V  -  0.02955  (- 0.0547)

=2.041V  +   (0.0020)

= 2.043V

The standard electrode potential (E°) of the cell in the lead-acid battery is 2.041 V. The initial value of E_cell in the same condition is also 2.041 V. As [H+] falls by 76%, E_cell can be calculated with the Nernst Equation.

The electrochemical cell in a lead-acid battery consists of an anode made of lead and a cathode made of lead(IV) oxide, both submerged in sulfuric acid. The half reactions are as follows:

(a) You can calculate the cell's standard electrode potential (E°) by subtracting the anode potential from the cathode potential, as per the formula E°cell = E°cathode - E°anode. Therefore, E° = 1.685 V - (-0.356 V) = 2.041 V.

(b) The initial value of E_cell is the same as the standard electrode potential, since initially [H+] = [HSO_4-] = 1M as per the standard conditions, which means E_cell = E° = 2.041 V.

(c) If the H+ concentration drops by 76%, it becomes 0.24M. For calculating the new E_cell we have to utilize the Nernst Equation, which is E=E°-(0.059/n)log(Q), where n is the number of electrons transferred (in this case 2), and Q is the reaction quotient ([Pb][HSO4-])/([PbSO4][H+]^2). Plugging in the values, we can determine the new E_cell.

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Answer:

B) 0.32 %

Explanation:

Given that:

[tex]K_{a}=1.8\times 10^{-5}[/tex]

Concentration = 1.8 M

Considering the ICE table for the dissociation of acid as:-

[tex]\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}[/tex]

The expression for dissociation constant of acid is:

[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}[/tex]

[tex]1.8\times 10^{-5}=\frac{x^2}{1.8-x}[/tex]

[tex]1.8\left(1.8-x\right)=100000x^2[/tex]

Solving for x, we get:

x = 0.00568  M

Percentage ionization = [tex]\frac{0.00568}{1.8}\times 100=0.32 \%[/tex]

Option B is correct.

Final answer:

After setting up the equilibrium expression for acetic acid dissociation and solving the equation, the percent ionization of a 1.8 M HC2H3O2 solution is determined to be approximately 0.1%. The closest answer option given is D) 0.18%.

Explanation:

To calculate the percent ionization for a 1.8 M HC2H3O2 solution, we need to set up an equilibrium expression based on the acid dissociation constant (Ka) and initial concentration of acetic acid. We start with the dissociation reaction:

HC2H3O2 \<=> H+ + C2H3O2-

Let's assume that the concentration of H+ (x) is equal to the concentration of C2H3O2- (x) at equilibrium due to stoichiometry:

Ka = [H+][C2H3O2-]/[HC2H3O2]

If we let x represent the molarity of H+ and C2H3O2- ions at equilibrium and assuming that x is small enough that (1.8 M - x) ≈ 1.8 M, we can simplify the expression to:

Ka = x^2/1.8 M

The equation becomes:

1.8 \ 10^-5 = x^2/1.8

Solving for x gives us:

x ≈ \ 10^-5 \ 1.8)}\> ≈ 1.8 \ 10^-3 M

The percent ionization is then calculated by:

Percent ionization = (x / [HC2H3O2] initial) \ 100%

Percent ionization = (1.8 \ 10^-3 M / 1.8 M) \ 100% = 0.1%

Since none of the options given matches the calculated value strictly, and provided that we've made an approximation, we'll have to choose the closest answer, which is:

D) 0.18%

Answer:

one-quarter as strong

Explanation:

Coulomb's law gives the mathematical expression to calculate the electrical force (F) between the charge of the nucleus (q₊) and the charge of an electron (q₋) separated by a distance (r).

[tex]F=\frac{k.q_{+}.q_{-}}{r^{2} }[/tex]

where,

k is the Coulomb's constant

The force between the nucleus and an electron in the level 1 is:

[tex]F_{1}=\frac{k.q_{+}.q_{-}}{r_{1}^{2} }[/tex]

Considering the distance to an electron from n = 2 is twice as great as the distance to an electron from n = 1, the force between the nucleus and an electron in the level 2 is:

[tex]F_{2}=\frac{k.q_{+}.q_{-}}{r_{2}^{2} }=\frac{k.q_{+}.q_{-}}{(2r_{1})^{2} }=\frac{1}{4} \frac{k.q_{+}.q_{-}}{(r_{1})^{2} }=\frac{1}{4}F_{1}[/tex]

Final answer:

The reaction where 1 mole of H2CO(g) reacts with O2(g) to form CO2(g) and H2O(l) is exothermic, evolving -563 kJ of energy. This is represented in the balanced thermochemical equation with a negative enthalpy change (ΔH = -563 kJ).

Explanation:

When 1 mole of H2CO(g) reacts with O2(g) to form CO2(g) and H2O(l), the reaction is exothermic because energy is released. The value of q in this context represents the amount of energy evolved during the reaction, which is -563 kJ (negative sign indicates energy is released by the system). Therefore, the balanced thermochemical equation incorporating the heat of reaction would be as follows:

H2CO(g) + O2(g) → CO2(g) + H2O(l) ΔH = -563 kJ

An exothermic reaction is characterized by the release of energy to the surroundings, with a negative enthalpy change (ΔH), indicating that the products are at a lower energy level than the reactants. Conversely, an endothermic reaction absorbs energy, with a positive enthalpy change, indicating that the products are at a higher energy level than the reactants.

Final answer:

The reaction in question is exothermic, with the value of q is -563 kJ of energy being evolved. The balanced thermochemical equation reflects this by including a negative ΔH value.

Explanation:

The reaction of [tex]H_2CO[/tex] (g) with [tex]O_2[/tex] (g) to form [tex]CO_2[/tex] (g) and [tex]H_2O[/tex] (l) is an exothermic reaction because energy is released in the process.

The value of q for this reaction is -563 kJ, indicating that 563 kJ of energy are evolved, where the negative sign signifies the direction of energy flow out of the system. The balanced thermochemical equation for the reaction would be:

[tex]H_2CO[/tex] (g) + [tex]O_2[/tex] (g) → [tex]CO_2[/tex] (g) + [tex]H_2O[/tex] (l) ΔH = -563 kJ/mol

Here, the ΔH value represents the enthalpy change for the reaction, and since it is negative, the reaction is considered exothermic.

To calculate the mass of dry ice, we need to consider heat transfer and use the formula q = m × c × ΔT. The mass of dry ice can be found using the equation mdry ice = (q / ΔHsublimation). By plugging in the given values, we can find the mass of dry ice that should be added to the water.

In order to calculate the mass of dry ice that should be added to the water, we need to consider the heat transfer that occurs during the process. The heat gained by the water is equal to the heat lost by the dry ice. We can use the formula:

q = m × c × ΔT

where q is the heat gained by the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

By rearranging the equation, we can solve for the mass of dry ice:

mdry ice = (q / ΔHsublimation)

where mdry ice is the mass of dry ice, q is the heat gained by the water, and ΔHsublimation is the heat of sublimation of CO2.

In this case, the temperature change is the difference between the initial temperature of the water (90°C) and the final temperature (10°C). The specific heat capacity of water is approximately 4.18 J/g°C, and the heat of sublimation of CO2 is 571 J/g. By plugging in the values and solving the equation, we can find the mass of dry ice that should be added to the water.

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Answer: 656.6 nm.

Explanation:

Using Rydberg's Equation for hydrogen atom:

[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]

Where,

[tex]\lambda[/tex] = Wavelength of radiation

[tex]R_H[/tex] = Rydberg's Constant

[tex]n_f[/tex] = Higher energy level  = 3  (least energetic for visible series)

[tex]n_i[/tex]= Lower energy level  = 2

We have:

[tex]n_f=3, n_i=2[/tex]

[tex]R_H=1.097\times 10^7 m^{-1}[/tex]

[tex]\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \left(\frac{1}{2^2}-\frac{1}{3^2} \right )[/tex]

[tex]\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \frac{5}{36}[/tex]

[tex]\frac{1}{\lambda}=0.1523\times 10^{7} m[/tex]

[tex]\lambda=6.566\times 10^{-7}m=656.6nm[/tex]

([tex]1 m= 10^9nm[/tex])

The wavelength of the photon emitted when the hydrogen atom undergoes a transition from n = 2 to n = 3 is 656.6 nm.

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.

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

Answer:

The correct answer is D) 7.081

Explanation:

The water equilibrium is the following:

H₂O ⇄  H⁺ + OH⁻      Kw

Where Kw= [H⁺] x [OH⁻]

In pure water: [H⁺] = [OH⁻]= C

If we introduce C in the expression for Kw:

Kw= [H⁺] x [OH⁻]= C x C= C²

⇒ C= [tex]\sqrt{Kw}[/tex]= [tex]\sqrt{6.88 x 10^{-15} }[/tex]= 8.29 x 10⁻⁸

pH= -log [H⁺] = -log C = -log (8.29 x 10⁻⁸) = 7.081

Final answer:

At 20°C, the pH of pure water is 7.000.

Explanation:

The ion-product constant of water (Kw) is the product of the concentrations of hydrogen ions (H3O+) and hydroxide ions (OH-) in water. At 20°C, the value of Kw is 6.88 x 10^-15. Since pure water is neutral, the concentration of hydrogen ions and hydroxide ions in pure water are equal. Therefore, at 20°C, the concentration of hydrogen ions and hydroxide ions in pure water is 6.88 x 10^-15 M.

The pH of a solution is a measure of the concentration of hydrogen ions. It is calculated as the negative logarithm (base 10) of the hydrogen ion concentration. Therefore, at 20°C, the pH of pure water is 7.000 (approximately) since the hydrogen ion concentration and the hydroxide ion concentration in pure water are both 6.88 x 10^-15 M.

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:

Cr₂O₇⁻²(aq) and ClO₃⁻(aq)

Explanation:

At a redox reaction, one substance must be reduced (gain electrons) and others must be oxidized (lose electrons). To evaluate the potential of the substance to be reduced, it's placed a reaction, in standards conditions, with H₂.

The potential reduction is quantified by E°, and as higher is the value of E°, as easy is to the compound to be reduced. So, at a redox reaction, the compound with the greatest E° will be reduced, and the other will be oxidized, in a spontaneous reaction. The values of E° are:

RuO₄⁻(aq) to RuO₄²⁻(aq) E° = + 0.59 V (the reduction reaction is the opposite of the oxidation reaction).

Ni⁺²(aq) E° = -0.257 V

I₂(s) E° = +0.535 V

Cr₂O₇⁻²(aq) E° = +1.33 V

ClO₃⁻(aq) E° = +0.890 V

Pb²⁺(aq) E° = -0.125 V

So, the substances that have E° higher than the E° of the RuO₄⁻²(aq) are Cr₂O₇⁻²(aq) and ClO₃⁻(aq), which are the substances that can oxidize RuO₄⁻(aq) to RuO₄²⁻(aq).

The correct substances that can oxidize RuO42(aq) to RuO4(aq) under standard conditions are Cr2O72(aq) and ClO3(aq).

To determine which substances can act as oxidizing agents for RuO42−(aq), we need to look at the standard reduction potentials (E°) for the given half-reactions. An oxidizing agent is a substance that accepts electrons, and for a redox reaction to occur spontaneously, the substance with the higher reduction potential will oxidize the substance with the lower reduction potential.

 The half-reaction for the reduction of RuO4’(aq) to RuO4(aq) is as follows:

[tex]\[ \text{RuO}_4(aq) + e^- \rightarrow \text{RuO}_4^{2-}(aq) \][/tex]

 We do not have the standard reduction potential for this half-reaction, but we can compare the potentials of the given substances to determine if they can act as oxidizing agents.

 From Appendix E, we can find the standard reduction potentials for the given substances:

1.[tex]\( \text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \) with \( E^\circ = -0.25 \) V[/tex]

2.[tex]\( \text{I}_2(s) + 2e^- \rightarrow 2\text{I}^-(aq) \) with \( E^\circ = +0.54 \) V[/tex]

3.[tex]\( \text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \) with \( E^\circ = +1.33 \) V[/tex]

4. [tex]\( \text{ClO}_3^-(aq) + 6\text{H}^+(aq) + 6e^- \rightarrow \text{Cl}^-(aq) + 3\text{H}_2\text{O}(l) \) with \( E^\circ = +1.45 \) V[/tex]

5.[tex]\( \text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) \) with \( E^\circ = -0.13 \) V[/tex]

 Comparing these potentials with the half-reaction for RuO42(aq), we can see that Ni2+(aq) and Pb2+(aq) have negative reduction potentials, which means they cannot oxidize RuO42(aq) because they are stronger reducing agents than RuO42(aq).

I2(s) has a positive reduction potential, but it is lower than the reduction potentials of Cr2O72(aq) and ClO3(aq). Therefore, I2(s) can oxidize RuO42’(aq), but there are stronger oxidizing agents available.

Cr2O72(aq) and ClO3(aq) have the highest reduction potentials in the list, which means they are the strongest oxidizing agents and can oxidize RuO4(aq) to RuO4(aq) under standard conditions.

 Thus, the correct substances that can oxidize RuO42(aq) to RuO4(aq) under standard conditions are Cr2O72(aq) and ClO3(aq)

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]