20 Which of the following will complete the reaction below? 1) 2 CH3MgBr/ether 2)NaOH 1) PC13 2) 2 CH3OH 1) H30+ 2) LIAIH4 1) PC13 2) 2 CH3MgBr/ether 3)H30+ OH o of OH CH3 CH3

1. 1200 atoms

2. 1/4 or 25% of the original amount

1) Undecayed atoms = Initial atoms * (1/2)^(Number of half-lives)

Given:

Initial atoms = 2400

Number of half-lives = 1

Undecayed atoms = 2400 * (1/2)^(1) = 2400 * (1/2) = 1200 atoms

2) Remaining amount = Initial amount * (1/2)^(Number of half-lives)

Given:

Number of half-lives = 2

Remaining amount = Initial amount * (1/2)^(2) = Initial amount * (1/2)^2 = Initial amount * 1/4 = 1/4 of the Initial amount

Since one half-life represents 36 years, two half-lives would represent 2 * 36 = 72 years. After 72 years, the remaining amount would be 1/4 or 25% of the initial amount.

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To determine the specific rotation [a] of optically active 2-butanol is -14°·dm³·g⁻¹.

The specific rotation [a] is a measure of the optical activity of a compound and is defined as the observed optical rotation (in degrees) divided by the concentration of the solution (in g/mL) and the length of the sample container (in dm or cm).

To calculate the specific rotation [a], we use the formula:

[a] = observed rotation / (concentration * path length)

Given that the observed optical rotation is -0.56°, the concentration of the solution is 0.4 g in water, and the path length is 10 cm (converted to 0.1 dm), we can substitute these values into the formula:

[a] = (-0.56°) / (0.4 g * 0.1 dm)

[a] = -0.56° / 0.04 g·dm⁻³

Simplifying the expression, we find:

[a] = -14°·dm³·g⁻¹

Therefore, the specific rotation [a] of the given solution of optically active 2-butanol is -14°·dm³·g⁻¹.

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The pH of a 0.29 M solution of carbonic acid (H₂CO3) is approximately 4.

Carbonic acid is a weak acid, and its ionization reactions contribute to the presence of H⁺ ions in solution, resulting in an acidic pH.

Carbonic acid is a diprotic acid, meaning it can donate two protons (H⁺ ions) in separate steps. The equilibrium expressions for the ionization reactions of carbonic acid are as follows:

Ka1 = [HCO₃⁻][H⁺]/[H₂CO₃]

Ka2 = [CO₃²⁻][H⁺]/[HCO₃⁻]

Given the values of Ka1 and Ka2, we can set up an equilibrium table to determine the concentrations of the species involved:

Species Initial Concentration Change Equilibrium Concentration

H₂CO₃ 0.29 M -x 0.29 - x M

HCO₃⁻ 0 M +x x M

CO₃²⁻ 0 M +x x M

H⁺ 0 M +x x M

We can assume that x is small compared to 0.29, so we can neglect x when subtracting it from 0.29 to get the equilibrium concentration of H₂CO₃.

Since the pH is defined as -log[H⁺], we can calculate the pH using the concentration of H⁺ at equilibrium. From the equilibrium table, we see that [H⁺] = x.

Taking the negative logarithm of x, we find that the pH is approximately 4.

The pH of a 0.29 M solution of carbonic acid is approximately 4. Carbonic acid is a weak acid, and its ionization reactions contribute to the presence of H⁺ ions in solution, resulting in an acidic pH.

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None of the provided options (NaBr, Br2, light, HOBr, HBr, PBr3) are suitable for the given transformation.

Based on the provided options, NaBr is a compound (sodium bromide), Br2 represents molecular bromine, light typically indicates the use of light as a reagent or condition, HOBr is hypobromous acid, HBr is hydrobromic acid, and PBr3 is phosphorus tribromide. None of these options directly relate to the specific transformation described in the question.

Without additional information about the desired reaction or outcome, it is not possible to determine the correct method for the transformation.

Please provide more details about the specific reaction or desired outcome to determine the appropriate method.

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The pH of the solution with an H+ concentration of 2.90×10-12 is approximately 11.54, indicating that the solution is basic.

The pH scale is a measure of the acidity or basicity of a solution. It ranges from 0 to 14, where values below 7 indicate acidity, values above 7 indicate basicity, and a pH of 7 represents a neutral solution. To calculate the pH of a solution, we can use the formula:

pH = -log[H+]

In this case, the given H+ concentration is 2.90×10-12. Taking the negative logarithm of this concentration gives us:

pH = -log(2.90×10-12)

Using the logarithm properties, we can rewrite this equation as:

pH = -log(2.90) - log(10-12)

Since log(10-12) is equal to -12, we can simplify further:

pH = -log(2.90) - (-12)

  = -log(2.90) + 12

Using a calculator or logarithmic tables, we can evaluate -log(2.90) to be approximately 11.54. Adding 12 to this value gives us:

pH ≈ 11.54 + 12

     = 23.54

Therefore, the pH of the solution is approximately 11.54, indicating that it is basic.

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Upon complete reaction of the ammonium chloride (NH4Cl) solution with the NaOH solution, ammonia, water, and NaCl remain. If the container is left open for a long time, the ammonia will evaporate.

When ammonium chloride (NH4Cl) reacts with sodium hydroxide (NaOH), the following reaction occurs:

NH4Cl + NaOH → NH3 + H2O + NaCl

This means that ammonium chloride reacts with sodium hydroxide to produce ammonia (NH3), water (H2O), and sodium chloride (NaCl). The reaction is a double displacement reaction where the ammonium ion (NH4+) is replaced by the sodium ion (Na+), resulting in the formation of ammonia gas, water, and salt.

If the container is left open for a long time, the ammonia gas will gradually evaporate into the air. Ammonia is a highly volatile compound with a strong smell, and it easily turns into a gas at room temperature. As a result, over time, the ammonia gas will escape from the open container, leaving behind water and sodium chloride.

It's important to note that ammonia gas can be harmful if inhaled in large quantities, as it is an irritant to the respiratory system. Therefore, proper ventilation or containment measures should be taken when working with or storing ammonia solutions.

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approximately 20.0 mL of thiosulfate solution will be consumed during the titration.

To determine the volume of thiosulfate solution consumed, we need to use the given information and perform calculations based on the stoichiometry of the reaction.

From the information provided, we know that the potassium dichromate is being titrated with a 0.2 M thiosulfate solution, and the factor for the titration is 1.02. The molar mass of potassium dichromate (K2Cr2O7) is 294.18 g/mol.

To begin the calculation, we convert the given mass of potassium dichromate to moles by dividing it by its molar mass:

Number of moles of potassium dichromate = 200 mg / 294.18 g/mol = 0.68 mmol

Since the stoichiometry of the reaction is not provided, we assume a 1:6 ratio between potassium dichromate and thiosulfate based on the common redox reaction between them:

K2Cr2O7 + 6 Na2S2O3 + 8 H2SO4 → 2 Cr2(SO4)3 + 6 Na2SO4 + K2SO4 + 8 H2O

Therefore, 0.68 mmol of potassium dichromate would react with 4.08 mmol of thiosulfate solution (0.68 mmol × 6).

Volume of thiosulfate solution consumed = (4.08 mmol) / (0.2 mol/L) = 20.4 mL

However, we need to consider the titration factor of 1.02. Therefore, the final volume of thiosulfate solution consumed would be:

Volume of thiosulfate solution consumed = 20.4 mL / 1.02 = 20.0 mL

Hence, approximately 20.0 mL of thiosulfate solution will be consumed during the titration.


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The gas is Krypton gas. Answer: Krypton gas

The given time of effusion for the unknown gas is 155 s and for Neon, it is 76 s. Thus, the rate of effusion for the unknown gas is 76/155 times the rate of effusion of neon gas, which is equal to 0.4903. Mathematically, we can write this as: Rate of effusion of unknown gas/rate of effusion of Neon gas = t(Neon gas)/t(unknown gas)

Therefore, Rate of effusion of unknown gas/0.4903 = Rate of effusion of Neon gas/1Rate of effusion of unknown gas = 0.4903 × Rate of effusion of Neon gas

Now, since both the gases belong to the noble gases, their molecular weights will differ only by the atomic mass of their atoms. Atomic mass of Neon = 20.2 g/mol Atomic mass of Krypton = 83.8 g/mol

Now, since the molecular weights of the two noble gases are in the ratio of their atomic masses, we can write the following relation :Molecular weight of Krypton/Molecular weight of Neon = Atomic mass of Krypton/Atomic mass of Neon Or, Molecular weight of Krypton/83.8 = Molecular weight of Neon/20.2Or, Molecular weight of Krypton = (83.8/20.2) × Molecular weight of Neon Or, Molecular weight of Krypton = 4.152 × Molecular weight of Neon Since, the two gases contain equal number of atoms, so the molecular weight is directly proportional to the molar mass of the gas.

Therefore, Molar mass of Krypton = 4.152 × Molar mass of Neon = 4.152 × 20.18 = 84.09 g/mol

Now, we know that the rate of effusion of Krypton gas is given by: Rate of effusion of Krypton gas = (Rate of effusion of Neon gas) × sqrt(Molar mass of Neon/Molar mass of Krypton)= 4.85 g/mol. Thus, the gas is Krypton gas. Answer: Krypton gas

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The half-life (in min) of a radioactive isotope if the activity of a sample drops from 3,184 cpm to 199 cpm in 11.0 min is 2.34 min.

Given that the activity of a sample drops from 3,184 cpm to 199 cpm in 11.0 min.We are to determine the half-life of the radioactive isotope. We can use the following formula:

A = A0 (1/2)^(t/T)

A0 = initial activity

A = activity after time t

T = half-life of the radioactive isotope

t = time taken

(3,184) = A0(1/2)^(11.0/T)199 = A0(1/2)^(T/T)

Let us divide the second equation by the first equation:(199)/(3,184) = (1/2)^(11.0/T)×(1/2)^(-T/T)(199)/(3,184)

= (1/2)^(11.0/T-T/T)(199)/(3,184)

= (1/2)^(11.0/T-1)(199)/(3,184)

= 2^(-11/T+1)

Taking natural logarithms on both sides of the equation:

ln(199/3,184) = ln(2^(-11/T+1))ln(199/3,184)

= (-11/T+1)ln(2)ln(199/3,184) / ln(2) - 1 = -11/T1/T

= [ln(2) - ln(199/3,184)] / ln(2)T = 2.34 min

Therefore, the half-life (in min) of a radioactive isotope if the activity of a sample drops from 3,184 cpm to 199 cpm in 11.0 min is 2.34 min.

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To calculate the pH of a buffer solution before and after the addition of a base, we need to consider the equilibrium between the weak acid (acetic acid, CH3COOH) and its conjugate base (acetate ion, CH3COO-).

Given:

Volume (V) = 0.342 L

Initial concentration of acetic acid (CH3COOH) = 0.25 M

Initial concentration of sodium acetate (CH3COONa) = 0.26 M

Amount of KOH added = 0.0057 mol

Step 1: Calculate the initial moles of acetic acid and acetate ion:

moles of CH3COOH = initial concentration * volume = 0.25 M * 0.342 L

moles of CH3COO- = initial concentration * volume = 0.26 M * 0.342 L

Step 2: Calculate the change in moles of CH3COOH and CH3COO- after the addition of KOH:

moles of CH3COOH remaining = initial moles of CH3COOH - moles of KOH added

moles of CH3COO- formed = initial moles of CH3COOH - moles of CH3COOH remaining

Step 3: Calculate the new concentrations of CH3COOH and CH3COO- after the addition of KOH:

new concentration of CH3COOH = moles of CH3COOH remaining / volume

new concentration of CH3COO- = moles of CH3COO- formed / volume

Step 4: Calculate the pH before and after the addition of KOH using the Henderson-Hasselbalch equation:

pH1 = pKa + log([CH3COO-] / [CH3COOH])

pH2 = pKa + log([CH3COO-] / [CH3COOH])

Note: The pKa value of acetic acid (CH3COOH) is typically around 4.75.

Substitute the values into the equations to calculate pH1 and pH2.

Please provide the pKa value of acetic acid for a more accurate calculation.

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The number of moles of lead nitrate required to react with 2.5 moles of lithium iodide is 1.25 moles of lead nitrate.

The balanced chemical equation for the given chemical reaction is:

Pb(NO3)2(aq) + 2 LiI(aq) → PbI2(s) + 2 LiNO3(aq)

The balanced chemical equation shows that 1 mole of Pb(NO3)2 reacts with 2 moles of LiI.

So, 2.5 moles of LiI will react with (2.5/2) moles of Pb(NO3)2.

Number of moles of Pb(NO3)2 required = (2.5/2) moles

= 1.25 moles.

Moles of Pb(NO3)2 required to react with 2.5 moles of LiI = 1.25 moles of Pb(NO3)2.

howing the calculation work;

2 LiI(aq) = Pb(NO3)2(aq)

==> PbI2(s) + 2 LiNO3(aq)Moles of LiI

= 2.5Moles of Pb(NO3)2

Using the balanced equation, we know that the mole ratio of LiI to Pb(NO3)2 is 2:

1.2 LiI = 1 Pb(NO3)2

Therefore:1 LiI = 1/2 Pb(NO3)22.5 mol LiI

= (1/2)2.5 mol Pb(NO3)22.5 mol LiI

= 1.25 mol Pb(NO3)2

So, the number of moles of lead nitrate required to react with 2.5 moles of lithium iodide is 1.25 moles of lead nitrate.

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The decrease in the water content of the air is 0.5161 kg/ s.

Given values Initial state: Dry-bulb temperature (T1) = 30°C Relative humidity (φ1) = 80%Pressure (P) = 101.325 k Pa Mass flow rate of air (ma) = 39.7 kg/ s

Final state: Dry-bulb temperature (T2) = 20°CRelative humidity (φ2) = 60%First, we will determine the moisture content at the initial and final states of the moist air using the psychrometric chart:

From the chart, we have the following: Moisture content at initial state = 0.0228 kg moisture / kg dry air Moisture content at final state = 0.0098 kg moisture / kg dry air

Now, we can determine the mass flow rate of moisture at the initial and final states using the mass flow rate of air:

Mass flow rate of moisture at initial state = ma × (w1) = 39.7 × 0.0228 = 0.90516 kg/ s Mass flow rate of moisture at final state = ma × (w2) = 39.7 × 0.0098 = 0.38906 kg/ s

Thus, the decrease in the water content of the air = Mass flow rate of moisture at the initial state - Mass flow rate of moisture at the final state= 0.90516 - 0.38906= 0.5161 kg/ s

Therefore, the decrease in the water content of the air is 0.5161 kg/ s.

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The number of tube passes for the shell-and-tube cross-flow heat exchanger is 2, with 39 tubes per pass, and the length of the tubes is 1.89 meters, satisfying the space constraints.

To calculate the number of tube passes, we need to consider the flow rates and temperature differences on both the tube side and the shell side. In this case, the flow rate of water on the tube side is given as 18000 kg/h, and the temperature change is from 27°C to 43°C. On the shell side, water flows at a rate of 14000 kg/h, and its temperature is constant at 80°C.

The overall heat transfer coefficient of the heat exchanger is provided as 1250 W/(m².K). By using the formula for heat transfer rate, Q = U × A × ΔT, where Q is the heat transfer rate, U is the overall heat transfer coefficient, A is the heat transfer area, and ΔT is the temperature difference, we can calculate the heat transfer area.

By rearranging the formula to solve for A, we have A = Q / (U × ΔT). Plugging in the given values, we can find the heat transfer area. Using the average velocity of water flowing through the pipe, we can calculate the cross-sectional area of flow, which is then used to determine the number of tubes required for the desired flow rate.

In this case, the space constraint limits the tube length below 2.5 m. Therefore, we need to find a length that satisfies this constraint. By dividing the total required heat transfer area by the product of the number of tubes per pass and the tube length, we can calculate the required number of tube passes.

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The correct answer is c. There is an increase in the number of molecules in solution.

In hydrolysis reactions, such as the hydrolysis of ATP, a molecule is broken down by the addition of water. In the case of ATP hydrolysis, ATP (adenosine triphosphate) is converted to ADP (adenosine diphosphate) and inorganic phosphate (Pi) by the addition of water. This reaction results in an increase in the number of molecules in solution because ATP is a single molecule while ADP and Pi are two separate molecules.

Entropy is a measure of the disorder or randomness of a system. An increase in the number of molecules in solution leads to a greater degree of disorder, resulting in an increase in entropy. Therefore, the hydrolysis of ATP above pH 7 is entropically favored due to an increase in the number of molecules in solution.

The completed question is given as,

The hydrolysis of ATP above pH 7 is entropically favored because

a. The electronic strain between the negative charges is reduced.

b. The released phosphate group can exist in multiple resonance forms

c. There is an increase in the number of molecules in solution

d. There is a large change in the enthalpy.

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a. The electronic strain between the negative charges is reduced.

The hydrolysis of ATP above pH 7 is entropically favored because of the reduction in the electronic strain between the negative charges. The electronic strain between the negative charges is reduced because the hydrolysis of ATP results in the breaking of the bonds between the phosphate groups, leading to the release of energy. This energy causes the phosphate groups to move further apart from each other, thus reducing the electronic strain between the negative charges.

The hydrolysis of ATP above pH 7 is also favored due to the release of a highly reactive phosphate group that can exist in multiple resonance forms. This allows for the formation of many different chemical reactions that can be utilized by the cell to carry out its various metabolic functions. The hydrolysis of ATP is important in many cellular processes, including muscle contraction, nerve impulse transmission, and protein synthesis. In addition, the energy released from ATP hydrolysis is used to power many other cellular processes, such as active transport of molecules across membranes and cell division.

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The correct answer for the given question is (D) H2, Pd. H2 and Pd are the reagents for the following reaction.

What is the hydrogenation reaction?The addition of hydrogen to a molecule is referred to as hydrogenation.

An unsaturated hydrocarbon is converted to a saturated hydrocarbon during this chemical reaction.

A chemical reaction occurs when atoms of one element or compound are rearranged and combined with atoms of another element or compound.

This reaction is usually represented by the equation;C=C + H2 → C-C Hydrogenation is a crucial reaction in the food industry.

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The temperature change of the sodium hydroxide solution is given as

ΔT = [tex]8.319^{0} C[/tex].

To calculate the temperature change of the sodium hydroxide solution, we can use the formula:

Q = mcΔT

Where, Q is the heat energy absorbed (1.876 kJ), m is the mass of the solution (calculated as density × volume), c is the specific heat capacity of the solution, and ΔT is the change in temperature.

First, we need to calculate the mass of the solution:

mass = density × volume = 1.10 g/mL × 50.0 mL = 55.0 g

Next, we rearrange the formula to solve for ΔT:

ΔT = Q / (mc)

Plugging in the given values:

ΔT = (1.876 kJ) / (55.0 g × 4.10 J/gºC)

Converting the heat energy to J:

ΔT = (1.876 × 10^3 J) / (55.0 g × 4.10 J/gºC)= [tex]8.319^{0}[/tex] C

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To calculate the theoretical yield and percent yield, we need to use the given information and perform the necessary calculations. From this, the theoretical yield of C₅H₅Cl is 6.945 g And the percent yield of C₂H₅Cl is approximately 155.64%.

(a) Calculate the theoretical yield of C₅H₅Cl:

Calculate the molar mass of C₅H₅Cl:

C: 5 × 12.01 g/mol = 60.05 g/mol

H: 5 × 1.01 g/mol = 5.05 g/mol

Cl: 1 × 35.45 g/mol = 35.45 g/mol

Total: 60.05 g/mol + 5.05 g/mol + 35.45 g/mol = 100.55 g/mol

Determine the number of moles of C₅H₅Cl produced:

Given mass of C₅H₅Cl = 10.8 g

Moles of C₅H₅Cl = 10.8 g / 100.55 g/mol ≈ 0.1074 mol

Use stoichiometry to relate C₅H₅Cl to C₂H₅Cl:

From the balanced equation, the mole ratio is 1:1. So, the moles of C₂H₅Cl produced would also be approximately 0.1074 mol.

Calculate the theoretical yield of C₂H₅Cl:

The molar mass of C₂H₅Cl is 64.52 g/mol.

Theoretical yield = 0.1074 mol × 64.52 g/mol = 6.945 g

(b) Calculate the percent yield of C₂H₅Cl:

Given actual yield = 10.8 g

Percent yield = (actual yield / theoretical yield) × 100%

Percent yield = (10.8 g / 6.945 g) × 100% ≈ 155.64%

Hence, the answers are:

(a) The theoretical yield of C₅H₅Cl is 6.945 g.

(b) The percent yield of C₂H₅Cl is approximately 155.64%.

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The predicted products P1, P2, and P3 can be determined by considering the reagent lists A-F. Among the predicted products, P3 is identified as an enamine.

To predict the products P1-P3, we need to analyze the reagent lists A-F and their compatibility with the given reaction conditions. Without specific information on the reagents and reaction conditions, it is challenging to provide precise predictions. However, we can discuss a general approach.

Reagent lists A-F may contain a variety of compounds that can participate in different reactions. Depending on the reaction conditions and reactants involved, different products can be formed. In the absence of specific details, it is difficult to determine the exact products.

Regarding enamine formation, an enamine is typically generated by the reaction of a secondary amine with a carbonyl compound, such as an aldehyde or ketone, under appropriate reaction conditions. If one of the reagents in the given lists A-F corresponds to a secondary amine and another reagent corresponds to a carbonyl compound, the resulting product involving these two reagents could potentially be an enamine.

In summary, without more specific information about the reagents and reaction conditions in lists A-F, it is not possible to provide precise predictions for the products P1-P3. However, based on the general knowledge of reactions, an enamine product, identified as P3, could potentially be formed if the reagents corresponding to a secondary amine and a carbonyl compound are present.

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#Note, The complete question is :

Predict the products P1-P3 from Reagent List A-F, also identify which product you predicted is enamine P3 Reagent. List Predict the products P1-P4 with the Reagent list A-H.

If the concentration of [tex]Sn(NO_3)_2[/tex] is changed to 0.11 M and that of [tex]FeCl_2[/tex]to 0.011 M, there will be a change in the the Ecell value. the resulting change in [tex]E_{cell}[/tex] is approximately -0.55 V.

We can calculate the change in [tex]E_{cell}[/tex] when the concentration of [tex]Sn(NO_3)_2[/tex] is changed to 0.11 M and that of [tex]FeCl_2[/tex]is changed to 0.011 M. The change in [tex]E_{cell}[/tex] can be calculated using the Nernst equation.

The Nernst equation relates the cell potential ([tex]E_{cell}[/tex]) to the standard reduction potentials (E°), concentration of reactants ([R]), and the number of electrons involved in the half-reaction (n):

Ecell = E° - (0.0592/n) * log([R])

For the oxidation half-reaction of [tex]Sn^2^+[/tex]:

E°ox = 0.14 V

[R]ox = 0.11 M

n = 2 (since it involves the transfer of 2 electrons)

For the reduction half-reaction of [tex]Fe^2^+[/tex]:

E°red = -0.44 V

[R]red = 0.011 M

n = 2 (since it involves the transfer of 2 electrons)

Using the Nernst equation, we can calculate the new [tex]E_{cell}[/tex]:

Ecell = E°red - E°ox - (0.0592/n) * log([R]red/[R]ox)

Substituting the values:

[tex]E_{cell}[/tex] = (-0.44 V) - (0.14 V) - (0.0592/2) * log(0.011 M/0.11 M)

Calculating the expression inside the log:

log(0.011 M/0.11 M) = log(0.1) = -1

[tex]E_{cell}[/tex] = (-0.44 V) - (0.14 V) - (0.0592/2) * (-1)

[tex]E_{cell}[/tex] = -0.44 V - 0.14 V + 0.0296 V

[tex]E_{cell}[/tex] = -0.55 V

Therefore, the change in [tex]E_{cell}[/tex] when the concentration of [tex]Sn(NO_3)_2[/tex] is changed to 0.11 M and that of [tex]FeCl_2[/tex]is changed to 0.011 M is approximately -0.55 V.

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c) 1s22s22p63s23p63d44s2 is the proper electron configuration out of the given configuration.

In this electron configuration, the numbers represent the principal energy levels (n), and the letters and superscripts represent the sublevels (s, p, d) and the number of electrons in each sublevel.

The electron configuration follows the Aufbau principle, which states that electrons fill the orbitals in order of increasing energy. The "1s2" represents the filling of the 1s orbital with two electrons. The "2s2" represents the filling of the 2s orbital with two electrons. The "2p6" represents the filling of the 2p orbitals with six electrons. The "3s2" represents the filling of the 3s orbital with two electrons. The "3p6" represents the filling of the 3p orbitals with six electrons. The "3d4" represents the filling of the 3d orbitals with four electrons. Finally, the "4s2" represents the filling of the 4s orbital with two electrons.

This electron configuration is in accordance with the rules and principles of electron filling order and the maximum number of electrons allowed in each sublevel.

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The missing reagents used in the synthesis of the pharmaceutical intermediate are 1: NaH and 2: Br2, HBr. These reagents are used in the two steps of the synthesis process.

Based on the multiple-choice options provided, the missing reagents in the synthesis of the pharmaceutical intermediate are 1: NaH and 2: Br2, HBr. In the first step, NaH (sodium hydride) is used as the reagent. Sodium hydride is commonly used as a strong base in organic synthesis to deprotonate acidic hydrogen atoms.

In the second step, Br2 (bromine) and HBr (hydrogen bromide) are used as reagents. Bromine is an oxidizing agent that can introduce bromine atoms into the molecule, while hydrogen bromide serves as a source of bromine and can also act as an acid catalyst.

The combination of NaH and Br2, HBr suggests that the synthesis involves a deprotonation reaction followed by bromination.

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The complete question is:

What are the missing reagents used in the synthesis of this pharmaceutical intermediate? Multiple Choice 1: NaH and 2: NaBr HBr in both steps 1: H

2

​

O and 2: Br

2

​

,HBr 1: NaH and 2: Br

2

​

,HBr 1: H

2

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O and 2: NaBr

The correct answer is option A: 37 moles, 3629 g. The number of moles in 2.00 L of 18.5-M H₂SO4 is 37 moles and the mass of the solute in the solution is 3629 g.

Molarity (M) is a unit of concentration used to express the amount of solute dissolved in a given volume of solution. It is defined as the number of moles of solute per liter of solution (mol/L or M).

Number of moles = Molarity (mol/L) × Volume (L)

First, let's calculate the number of moles:

Number of moles = 18.5 mol/L × 2.00 L = 37.0 moles

Therefore, the number of moles of sulfuric acid (H₂SO₄) in the solution is 37.0 moles.

To find the mass of the solute, we need to consider the molar mass of sulfuric acid (H₂SO₄), which is:

H₂SO₄ molar mass = (2 × atomic mass of hydrogen) + atomic mass of sulfur + (4 × atomic mass of oxygen)

= (2 × 1.008 g/mol) + 32.06 g/mol + (4 × 16.00 g/mol)

= 2.016 g/mol + 32.06 g/mol + 64.00 g/mol

= 98.076 g/mol

Mass of solute = Number of moles × Molar mass

= 37.0 moles × 98.076 g/mol

= 3622.41 g

Therefore, the mass of the solute (concentrated sulfuric acid) in the given solution is approximately 3622.41 grams.

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The humidity ratio of the moist air can be calculated using the given information: temperature, pressure, and mole fraction of water vapor. The humidity ratio is approximately 0.0155 kg v/kg da.

The humidity ratio, also known as the specific humidity, is the ratio of the mass of water vapor to the mass of dry air in a mixture. To calculate the humidity ratio, we need to determine the mass of water vapor and the mass of dry air.

Given:

- Temperature of the moist air (T) = 45°C = 45 + 273.15 K = 318.15 K

- Pressure of the moist air (P) = 1.38 bar

- Mole fraction of water vapor (x) = 4.7% = 0.047

First, we need to determine the mole fraction of dry air (xd) in the mixture. Since the mole fractions of all components in a mixture must sum up to 1, we have:

xd + x = 1

Solving for xd, we find:

xd = 1 - x = 1 - 0.047 = 0.953

Next, we need to determine the partial pressure of water vapor (Pv) and the partial pressure of dry air (Pd). The partial pressure of each component is given by:

Pv = x * P = 0.047 * 1.38 bar = 0.06486 bar

Pd = xd * P = 0.953 * 1.38 bar = 1.31514 bar

Now, we can use the ideal gas law to calculate the mass of water vapor (mv) and the mass of dry air (md) in the mixture. The ideal gas law states:

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.

Rearranging the equation, we have:

n = PV / RT

For water vapor, using the given values of Pv and T, we can calculate the number of moles (nv) of water vapor:

nv = Pv / (R * T)

Similarly, for dry air, using the given values of Pd and T, we can calculate the number of moles (nd) of dry air:

nd = Pd / (R * T)

The mass of water vapor (mv) and the mass of dry air (md) can be calculated using the molecular weight of water vapor (Mv) and the molecular weight of dry air (Md), respectively:

mv = nv * Mv

md = nd * Md

Finally, the humidity ratio (W) is given by the ratio of the mass of water vapor to the mass of dry air:

W = mv / md

By substituting the calculated values, we can find the humidity ratio. The approximate value is 0.0155 kg v/kg da.

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[Co(en)2Cl2]Cl has a tetrahedral geometry, with two chlorides occupying trans positions and two en molecules occupying cis positions. [Co(en)2Cl2]Cl is a coordination compound that contains a chelate ligand.

En has a bidentate character and thus, forms a chelate complex with Co(III) ion, stabilizing it.

Thus, [Co(en)2Cl2]Cl has cis-trans isomerism, but it does not have geometric isomerism, also known as coordination isomerism.

Coordination isomerism, also known as geometric isomerism, is the kind of stereoisomerism seen in coordination compounds.

A coordination compound that exhibits coordination isomerism contains two or more coordination isomers, each with a different number or types of ligands associated with the central metal atom or ion.

The coordinated groups may be the same or different, and they may be arranged in different ways around the central atom.

However, the arrangement of the coordinated groups is the only thing that varies between the isomers. The number of coordinated groups and the identity of the central atom remain constant.

[Co(en)2Cl2]Cl is a coordination compound that contains a chelate ligand.

A chelate ligand is a ligand that binds to a central metal ion through two or more atoms.

The bidentate ethylenediamine (en) ligand binds to the cobalt ion in the [Co(en)2Cl2]Cl complex via two nitrogen atoms.

The en ligand is capable of forming a chelate complex with cobalt because it has two donor atoms separated by a distance equal to the metal's coordination number.

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The equilibrium constant, Kc, obtained for this reaction at 532 K is 2.90×10^(-2).

The balanced equation for the reaction is: COCl₂(g) ⇌ CO(g) + Cl₂(g)

Initial concentration of COCl₂(g): 1.05 moles

Equilibrium concentration of Cl₂(g): 3.04×10^(-2) M

Volume of the container: 1.00 liter

To calculate the equilibrium constant, Kc, we need to use the equilibrium concentrations of the species involved in the reaction. Since the reaction is in the gas phase, we can use the concentration of Cl₂(g) to determine the equilibrium constant.

Kc = [CO(g)][Cl₂(g)] / [COCl₂(g)]

Substituting the given equilibrium concentrations into the equation:

Kc = (3.04×10^(-2) M) / (1.05 moles / 1.00 L)

Note that we divide the moles of COCl₂(g) by the volume of the container to convert it into concentration.

Kc = 2.90×10^(-2)

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The complete question is:

Use the References to access important values if needed for this question. A student ran the following reaction in the laboratory at 532 K: cocl₂(g) co(g) + Cl₂(g) When she introduced 1.05 moles of COCl₂(g) into a 1.00 liter container, she found the equilibrium concentration of Cl₂(g) to be 3.04×10-2 M. Calculate the equilibrium constant, K, she obtained for this reaction. Kc =

The amount of theoretical air required for the combustion of 10 kg of ethane C2H6 is 26 m3. Combustion is the process of burning a fuel substance with air or oxygen to produce heat. When complete combustion occurs, fuel burns entirely, which means that all the carbon in the fuel becomes CO2 while all the hydrogen turns into H2O.

Hence, air is required to support combustion in the right ratio with the fuel for complete combustion to occur. Therefore, it is necessary to know the amount of air required for a given quantity of fuel to burn completely. One method to calculate the amount of theoretical air required for the combustion of 10 kg of ethane C2H6 is as follows: Ethane C2H6 is made up of carbon (C) and hydrogen (H).Therefore, the molar mass of ethane is calculated by adding the molar masses of carbon and hydrogen:

2 x (1.008 g/mol) + 6 x (12.01 g/mol) = 30.07 g/mol

The balanced chemical equation for the combustion of ethane is:

C2H6 + 3.5 O2 → 2 CO2 + 3 H2O

From the balanced equation, we can determine that 3.5 moles of oxygen are required for every 1 mole of ethane burned completely. Therefore, the number of moles of ethane in 10 kg is calculated by dividing the mass by the molar mass:

n = m/M = 10,000 g/30.07 g/mol = 332.6 mol

Therefore, the number of moles of oxygen required for the combustion of 10 kg of ethane is:

332.6 mol x 3.5 mol O2/1 mol

ethane = 1164.1 mol O2 Finally,

the amount of theoretical air required is calculated by multiplying the moles of oxygen by the molar volume of air (22.4 L/mol):

1164.1 mol O2 x 22.4 L/mol = 26,044.6 L or approximately 26 m3 of air.

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Scenario B would produce the largest moment (torque) about the lower back.  The moment (torque) about a point is calculated by multiplying the force applied by the perpendicular distance from the point to the line of action of the force.

In this case, the point of interest is the lower back, and the force is the weight of the 10 kg mass. In scenario A, the mass is held 0.5 meters from the lower back. The perpendicular distance from the lower back to the line of action of the force is 0.5 meters. Therefore, the moment is calculated as the force (weight) multiplied by the distance, resulting in a certain value.

In scenario B, the mass is held 1 meter from the lower back. The perpendicular distance from the lower back to the line of action of the force is 1 meter. Since the distance is greater in scenario B, the moment will be larger when calculated using the same force (weight).

Hence, holding a 10 kg mass 1 meter from the lower back (scenario B) would produce the largest moment (torque) about the lower back compared to holding the same mass 0.5 meters from the lower back (scenario A).

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The vapor pressure of the solution made from biphenyl and benzene is 100.84 Torr, which is the same as the vapor pressure of pure benzene.

To calculate the vapor pressure of a solution made from biphenyl (C₁₂H₁) and benzene (C₆H₆), we need to apply Raoult's law, which states that the vapor pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent in the solution.

Let's assume we have a solution where biphenyl is dissolved in benzene. Biphenyl is considered a nonvolatile solute, meaning it does not easily evaporate and contribute to the vapor pressure. Therefore, we can assume that the vapor pressure of the solution is primarily determined by the benzene component.

The vapor pressure of pure benzene is given as 100.84 Torr at 25 °C. This value represents the vapor pressure of pure benzene.

Now, let's consider the solution of biphenyl and benzene. Since biphenyl is nonvolatile, it does not contribute significantly to the vapor pressure. Therefore, the mole fraction of benzene in the solution is effectively 1.

According to Raoult's law, the vapor pressure of the solution is equal to the vapor pressure of the pure solvent (benzene) multiplied by its mole fraction:

Vapor pressure of solution = Vapor pressure of pure benzene × Mole fraction of benzene

Vapor pressure of solution = 100.84 Torr × 1

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By following serial dilution method, you can achieve the desired concentrations using small volumes while ensuring accurate dilution ratios. It is essential to handle the small volumes carefully and accurately to maintain the desired concentrations throughout the dilution process.

To perform a serial dilution with small volumes, such as in this case where measuring less than 1µL is not possible, we can use a stepwise dilution approach.

Start with the initial concentration of 14.2mM in 1mL of stock media.

To prepare the first dilution of 10µM, transfer 1µL from the stock solution and add it to 99µL of a diluent (such as water or buffer). This results in a 100µL solution with a concentration of 10µM.

For subsequent dilutions, repeat the same process. Take 1µL from the previous dilution and add it to 99µL of diluent.

Repeat step 3 for each desired concentration. For example, to obtain a concentration of 5µM, take 1µL from the 10µM solution and add it to 99µL of diluent.

Continue this stepwise dilution process until you reach the final desired concentrations: 2.5µM, 1µM, 750nM, 500nM, 250nM, 100nM, 50nM, and 10nM.

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10.18 grams of sodium chloride will be formed upon the complete reaction of 26.2 grams of sodium iodide with excess chlorine gas.

The balanced equation for the reaction of chlorine gas and sodium iodide is given as:

Cl2 (g) + 2NaI (s) → 2NaCl (s) + I2 (s)

According to the balanced equation:

1 mole of chlorine gas reacts with 2 moles of sodium iodide to give 2 moles of sodium chloride.

The molar mass of sodium iodide is 149.89 g/mol.

Thus, 26.2 g of sodium iodide will be equal to:

26.2g NaI x (1mol NaI/149.89g NaI) = 0.1745 moles NaI

According to the balanced equation, 2 moles of NaI are needed to produce 2 moles of NaCl.

Therefore, the number of moles of NaCl produced is:

0.1745 moles NaI x (2 moles NaCl/2 moles NaI)

= 0.1745 moles NaCl

The molar mass of NaCl is 58.44 g/mol.

Thus, 0.1745 moles of NaCl will be equal to:

0.1745 moles NaCl x (58.44 g NaCl/1 mol NaCl)

= 10.18 grams NaCl

Therefore, 10.18 grams of sodium chloride will be formed upon the complete reaction of 26.2 grams of sodium iodide with excess chlorine gas.

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