how might you minimize the hydrogen evolution when using NaBH4as a reducing agent in water?

The molar solubility of iron(II) sulfide in a 0.121 M iron(II) nitrate solution is approximately: 4.96 x 10⁻¹⁷ M.

To find the molar solubility of iron(II) sulfide in a 0.121 M iron(II) nitrate solution, you'll need to follow these steps:
1. Write the balanced chemical equation:
FeS(s) <=> Fe²⁺(aq) + S²⁻(aq)

2. Set up the solubility product constant expression (Ksp) for iron(II) sulfide:
Ksp = [Fe²⁺][S²⁻]

3. Find the Ksp value for iron(II) sulfide from a reliable source (e.g., a textbook or database). Let's assume the Ksp = 6 x 10⁻¹⁸.

4. Since the solution already contains 0.121 M iron(II) nitrate, the initial concentration of Fe²⁺ is 0.121 M.

5. Let x represent the molar solubility of iron(II) sulfide. The change in concentration for Fe²⁺ and S²⁻ can be represented as:
Fe²⁺: 0.121 + x
S²⁻: x

6. Substitute the concentrations into the Ksp expression:
Ksp = (0.121 + x)(x)

7. Plug in the Ksp value and solve for x:
6 x 10⁻¹⁸ = (0.121 + x)(x)

8. Since Ksp is very small, you can assume x is much smaller than 0.121, so the equation can be approximated as:
6 x 10⁻¹⁸ ≈ (0.121)(x)

9. Solve for x:
x ≈ 6 x 10⁻¹⁸ / 0.121 ≈ 4.96 x 10⁻¹⁷

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When considering the reaction at equilibrium: CH3COOH (aq) ⇄ H+ (aq) + CH3COO- (aq), and NaCH3COO is added to the solution, the correct answer is D) Ka remains the same and pH increases.

Adding NaCH3COO introduces more CH3COO- ions into the solution, which causes a shift in the equilibrium to the left, according to Le Chatelier's principle. This results in the formation of more CH3COOH and a decrease in H+ ion concentration. The decrease in H+ ions leads to an increase in pH, as pH is inversely proportional to the concentration of H+ ions.

However, the value of Ka, the acid dissociation constant, remains constant as it is only affected by temperature and not by the concentration of reactants or products. Ka is a measure of the strength of an acid, and since the acid (CH3COOH) itself does not change, its Ka value remains the same. When considering the reaction at equilibrium: CH3COOH (aq) ⇄ H+ (aq) + CH3COO- (aq), and NaCH3COO is added to the solution, the correct answer is D) Ka remains the same and pH increases.

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It is essential to use an indicator during a titration to ensure accurate and precise measurements. If you forget to use the indicator, it is best to start the titration again from the beginning.

When performing a titration, an indicator is used to signal the endpoint of the reaction. The endpoint is the point where the amount of reactant added is equal to the amount of reactant required for the reaction to occur completely. If you forget to use the indicator during a titration, it can be challenging to identify the endpoint accurately, and you may end up adding too much or too little of the reactant. This can result in inaccurate measurements and unreliable results. Therefore, it is essential to use an indicator during a titration to ensure accurate and precise measurements. If you forget to use the indicator, it is best to start the titration again from the beginning.

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To titrate and find the molar mass of an unknown acid, we need to first standardize a NaOH solution. This means that we need to determine the exact concentration of the NaOH solution using a known amount of an acid such as potassium hydrogen phthalate (KHP).

If the standardized NaOH solution is exposed to air for a long period of time, it can react with carbon dioxide in the air to form sodium carbonate. This can cause the concentration of the NaOH solution to decrease, resulting in a lower calculated molar mass of the unknown acid. Therefore, the calculated molar mass would be too low.

On the other hand, if the NaOH solution is exposed to air before standardization, it can absorb moisture and carbon dioxide from the air, which can also cause a decrease in concentration. This means that the calculated molar mass would be too high.

In conclusion, it is important to standardize the NaOH solution as soon as possible after preparation and to store it in a sealed container to prevent exposure to air. This will ensure accurate and reliable results when determining the molar mass of an unknown acid through titration.

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The mass of GaN formed will be 4812.75 g or 4.81 Kg.

The limiting reagent in the reaction is Ga, and the balanced chemical equation is:

2Ga + N2 → 2GaN

The moles of Ga present is 4 Kg / 69.723 g/mol = 57.42 mol.

According to the stoichiometry of the reaction, 2 moles of Ga react with 1 mole of N2 to produce 2 moles of GaN. Therefore, the moles of GaN formed will be 57.42 mol / 2 * 2 = 57.42 mol.

The mass of GaN formed will be 57.42 mol * 83.73 g/mol = 4812.75 g or 4.81 Kg.

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The mass of GaN that will be formed if 4 Kg of Gallium (Ga) is reacted with 11 Kg of Nitrogen (N) is 4,802 grams.

To solve this problem, we first need to determine the limiting reactant. We can do this by comparing the moles of each reactant:

Moles of Ga = (4,000 g Ga) / (69.723 g/mol Ga) = 57.35 mol Ga
Moles of N = (11,000 g N) / (14.0067 g/mol N) = 785.32 mol N

The balanced chemical equation for the reaction is:

2 Ga + N₂ → 2 GaN

From the balanced equation, we can see that 2 moles of Ga react with 1 moles of N₂. We can now calculate the amount of GaN that will be produced in each mass of reactant.

57.35 mol Ga (2 mol GaN/2 mol Ga) = 57.35 mol GaN
785.32 mol N(1 mol N₂/ 2 mol N)(2 mol GaN/1 mol N₂) = 785.32 mol GaN

Since the amount of product produced from Ga is smaller than the amount produced from N, then ga is the limiting reactant.

Now we can determine the mass of GaN formed by using the stoichiometry of the balanced equation:

Moles of GaN formed = 1/1 * Moles of limiting reactant (Ga) = 57.35 mol GaN
Mass of GaN formed = (57.35 mol GaN) * (83.73 g/mol GaN) = 4,802 g GaN

So, the mass of GaN formed is 4,802 grams.

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The reaction of acetanilide with bromine is a substitution reaction where bromine replaces a hydrogen atom on the aromatic ring of acetanilide.

This reaction is commonly used in organic chemistry to prepare brominated derivatives of aromatic compounds. The resulting product is known as 4-bromacetanilide, which has various applications in the pharmaceutical and chemical industries. The reaction of acetanilide with bromine is known as bromination, which is an electrophilic aromatic substitution reaction. In this reaction, bromine reacts with acetanilide to form para-bromo acetanilide as the major product, and ortho-bromo acetanilide as a minor product.

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The carbonyl group is transferred from C-2 of ribulose to C-1 of ribose 5-phosphate in the Calvin cycle.

The transformation you're portraying includes moving a carbonyl gathering (- C=O) from the subsequent carbon (C-2) of ribulose to the principal carbon (C-1) of ribose 5-phosphate. This response is a key stage in the Calvin cycle, which is the metabolic pathway that plants use to fix carbon dioxide (CO2) and produce glucose.

The carbonyl gathering is moved through a progression of protein catalyzed responses that include a few intermediates, including an enediol transitional framed by the expansion of ribulose 5-phosphate to the carbonyl gathering.

The enediol then goes through a revision to shape another carbonyl gathering at C-1 of ribose 5-phosphate, while delivering the first carbonyl gathering at C-2 of ribulose.

This change is fundamental for the legitimate working of the Calvin cycle, as it takes into consideration the creation of two atoms of glyceraldehyde 3-phosphate (G3P) from three particles of CO2, which can then be utilized to orchestrate glucose and different sugars.

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The complete question is:

What is the process that involves moving a carbonyl group from the second carbon of ribulose to the first carbon of ribose 5-phosphate, and what is its significance in the Calvin cycle?

The maximum amount of calcium sulfide that will dissolve in a 0.161 M ammonium sulfide solution is approximately 4.97 × 10⁻²⁵ M.

To find the maximum amount of calcium sulfide (CaS) that will dissolve in a 0.161 M ammonium sulfide (NH4)2S solution, we'll use the solubility product constant (Ksp) and the common ion effect.

1. Write the balanced dissolution equation for calcium sulfide:
CaS(s) ⇌ Ca²⁺(aq) + S²⁻(aq)

2. Find the Ksp value for CaS. The Ksp for calcium sulfide is approximately 8.0 × 10⁻²⁶.

3. Determine the initial concentration of S²⁻ ions in the 0.161 M ammonium sulfide solution. In (NH4)2S, there is one S²⁻ ion for everyone (NH4)2S, so the concentration of S²⁻ is 0.161 M.

4. Set up the Ksp expression:
Ksp = [Ca²⁺][S²⁻]

5. Since we know the initial concentration of S²⁻ and the Ksp, we can solve for the concentration of Ca²⁺ ions.
8.0 × 10⁻²⁶ = [Ca²⁺](0.161)

6. Solve for [Ca²⁺]:
[Ca²⁺] = (8.0 × 10⁻²⁶) / (0.161) ≈ 4.97 × 10⁻²⁵ M

The maximum amount of calcium sulfide that will dissolve in a 0.161 M ammonium sulfide solution is approximately 4.97 × 10⁻²⁵ M.

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The value for ΔG° for the given reaction comes out to be -9.32533 KJ /(mol K). Since ΔG° is negative, the reaction is spontaneous at the given temperature of 715 K.

How we calculated it?

To estimate ΔG° for the reaction: 2 NO(g) + O2(g) → 2 NO2(g) at a given temperature of 715 K, we need to use the equation:

ΔG° = ΔH° - TΔS°

where ΔG° is the gibbs free energy change , ΔH° is the standard enthalpy change, ΔS° is the standard entropy change, and T is the temperature in Kelvin.

From the balanced chemical equation, we can see that the reaction involves the formation of two moles of NO2 from two moles of NO and one mole of O2. Thus, we can write the reaction as:

2 NO(g) + O2(g) → 2 NO2(g)

The standard enthalpy change for this reaction, ΔH°, can be found in a thermochemical data table and is -114.14 kJ/mol.

The standard entropy change, ΔS° for NO, O2, NO2 can also be found in the same table and are as follows:

for NO2, ΔS° = 239.9 J/(mol K), for NO, ΔS° = 210.76 J/(mol K), for O2, ΔS° = 205.138 J (mol K)
for the entire reaction , ΔS° = 2*ΔS°(NO2) - [2*ΔS°(NO) + ΔS°(O2)]

ΔS°(reaction) = 2* 239.9J/(mol K) -[2*210.76 J/(mol K) + 205.138 J /(mol K)]

ΔS°(reaction) = -146.538 J/(mol K)

Converting ΔS° into KJ/mol :

ΔS° =  -(146.538 /1000) KJ/mol = -0.146538 KJ/mol
Plugging in these values and the given temperature of 715 K into the equation above, we get:

ΔG° = -114.1 kJ/mol - (715 K)(-0.146538 KJ/(mol K) = -114.1 KJ/mol + 104.77467 KJ /(mol K) = -9.32533 KJ /(mol K)

ΔG° = -9.32533 KJ /(mol K)
Since ΔG° is negative, the reaction is spontaneous at the given temperature of 715 K. In other words, the reaction will proceed in the forward direction from left to right under standard conditions.

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Using Hess's law to calculate the δG°rxn for the reaction CO(g) → C(s) + 1/2 O2(g) is 651.6 kJ.

To use Hess's law to calculate the δG°rxn for the reaction CO(g) → C(s) + 1/2 O2(g), you will need to manipulate the given reactions and their respective δG° rxn values so that the target reaction is the sum of the given reactions.

Here are the given reactions:

1) CO2(g) → C(s) + O2(g), δG°rxn = 394.4 kJ
2) CO(g) + 1/2 O2(g) → CO2(g), δG°rxn = -257.2 kJ

First, reverse reaction 2 to obtain the following reaction:

2') CO2(g) → CO(g) + 1/2 O2(g), δG°rxn = 257.2 kJ

Now, add reactions 1 and 2' together:

1) CO2(g) → C(s) + O2(g), δG°rxn = 394.4 kJ
2') CO2(g) → CO(g) + 1/2 O2(g), δG°rxn = 257.2 kJ
------------------------------------------
Sum) CO(g) → C(s) + 1/2 O2(g), δG°rxn = ?

Add the δG°rxn values for reactions 1 and 2':

δG°rxn (Sum) = δG°rxn (1) + δG°rxn (2')
δG°rxn (Sum) = 394.4 kJ + 257.2 kJ
δG°rxn (Sum) = 651.6 kJ

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The synthesis of silver nanoparticles can involve various methods and conditions, but a general order of steps in a typical synthesis process might be as follows:

Preparation of the silver precursor: Silver nitrate (AgNO3) is commonly used as a silver precursor. It is dissolved in water or other suitable solvents to prepare a silver precursor solution.

Preparation of the reducing agent: A reducing agent, such as sodium borohydride (NaBH4) or sodium citrate, is prepared separately. The reducing agent will react with the silver precursor to form silver nanoparticles.

Mixing the silver precursor and reducing agent: The silver precursor solution and the reducing agent solution are mixed together under suitable conditions, such as controlled temperature and stirring, to allow the reduction reaction to occur.

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The initial concentration of NOCl was approximately 0.207 M (Moles).

To solve this problem, we need to use the integrated rate law for a second-order reaction:

1/[A]t = kt + 1/[A]₀

Where:
[A]t is the concentration of NOCl at time t (0.231 M)
[A]₀ is the initial concentration of NOCl
k is the rate constant (8.00 × 10⁻² 1/M·s)
t is the time (5.00 seconds)

Rearrange the formula to find [A]₀:

[A]₀ = 1/(kt + 1/[A]t)

Now plug in the values:

[A]₀ = 1/((8.00 × 10⁻² 1/M·s)(5.00 s) + 1/0.231 M)

[A]₀ = 1/(0.4 + 4.329)

[A]₀ ≈ 0.207 M

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The mean-square radius of a polyethylene chain of molecular weight 4200 g⋅mol−1, if the C−C bond length is 0.154 nm, is approximately 1.28 nm.

To find the mean-square radius of a polyethylene chain of molecular weight 4200 g⋅mol−1, we can use the Flory's equation:
R^2 = n * b^2 * N
where R is the mean-square radius, n is the number of segments in the chain, b is the bond length, and N is the degree of polymerization (i.e. the number of monomers in the chain).
To solve for R, we need to know the degree of polymerization, which can be calculated from the molecular weight as:
N = M / m
where M is the molecular weight and m is the monomer weight. For polyethylene, the monomer weight is 28 g⋅mol−1.
N = 4200 g⋅mol−1 / 28 g⋅mol−1 = 150
Now we can plug in the values for n, b, and N into Flory's equation:
R^2 = n * b^2 * N
R^2 = 149 * (0.154 nm)^2 * 150
R^2 = 1.64 nm^2
R = sqrt(1.64 nm^2)
R = 1.28 nm
Therefore, the mean-square radius of a polyethylene chain of molecular weight 4200 g⋅mol−1, if the C−C bond length is 0.154 nm, is approximately 1.28 nm.

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a. There are four peaks in the chromatogram due to the presence of four different dyes. b. A gradient elution would result in sharper and more distinct peaks for each dye

a. There are four tops in the chromatogram since there are four distinct colors present in the example. Each pinnacle compares to an alternate color with an alternate maintenance time, which is the time it takes for the color to go through the HPLC section and be distinguished by the identifier.

b. In the event that a rising slope of water (i.e., expanded extremity of the portable stage) were utilized after some time, the chromatogram would show a more isolated and settled top for each color. This is on the grounds that the more polar versatile stage would assist with isolating the colors all the more really as they travel through the section, bringing about a higher goal and more clear partition between the pinnacles. The subsequent chromatogram would show more honed and more unmistakable tops, with each pinnacle addressing a solitary color in the example. The inclination elution would take more time than the isocratic technique, yet would yield more exact and itemized results.

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The number of sigma and pi bonds in propene is a total of 8 and 1 respectively.

A step-by-step explanation is given below:
1. Propene has the chemical formula C3H6.
2. There are three carbon atoms and six hydrogen atoms.
3. Each carbon-hydrogen bond is a sigma bond, totaling 6 sigma bonds.
4. There is a single bond between the first two carbon atoms, which is a sigma bond, adding 1 sigma bond to the total.
5. The double bond between the second and third carbon atoms consists of one sigma bond and one pi bond.

So, It concludes that in propene, there are 8 sigma bonds and 1 pi bond.

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The sum of the coefficients in the balanced equation for the reaction between zinc hydrogen sulfate and manganese (IV) chlorite is 6.

1. Write the formulas for the reactants and products:
  - Zinc hydrogen sulfate: Zn(HSO₄)₂
  - Manganese (IV) chlorite: Mn(ClO₂)₄
  - Assume the products are zinc chlorite and manganese (IV) sulfate: Zn(ClO₂)₂ and Mn(SO₄)₂

2. Write the unbalanced chemical equation:
  Zn(HSO₄)₂ + Mn(ClO₂)₄ → Zn(ClO₂)₂ + Mn(SO₄)₂

3. Balance the chemical equation:
  2Zn(HSO₄)₂ + Mn(ClO₂)₄ → 2Zn(ClO₂)₂ + Mn(SO₄)₂

4. Add the coefficients in the balanced equation:
  Sum = (Coefficient of Zn(HSO₄)₂) + (Coefficient of Mn(ClO₂)₄) + (Coefficient of Zn(ClO₂)₂) + (Coefficient of Mn(SO₄)₂)
  Sum = 2 + 1 + 2 + 1
  Sum = 6

The sum of the coefficients in the balanced equation for the reaction between zinc hydrogen sulfate and manganese (IV) chlorite is 6.

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The substance having the chemical formula N2O5 is known as dinitrogen pentoxide.  The Rate laws provide a mathematical description of how changes in the amount of a substance affect the rate of a chemical reaction.

Use the rate law for the reaction:
Rate = - Δ[N2O5] / Δt = k[N2O5]^2
Here, k is the rate constant.

We can rearrange this equation to solve for [N2O5]: [N2O5] = - Δt / (kΔ[N2O5])

The data are given to calculate the initial rate of the reaction:
Rate = (-0.92 x 10^-2 mol/L - 1.24 x 10^-2 mol/L) / (20 min - 0 min) = 1.6 x 10^-4 mol/L/min

We can also use the stoichiometry of the reaction to find the rate of change of [N2O5] concerning time:
Δ[N2O5] / Δt = - 2 rate

Substituting the values we have: Δ[N2O5] / Δt = - 2 (1.6 x 10^-4 mol/L/min) = -3.2 x 10^-4 mol/L/min

Now we can plug in the given values to find the concentration of N2O5 at 100 minutes:
[N2O5] = - (100 min - 20 min) / (kΔ[N2O5]) = 80 min / (3.2 x 10^-4 mol/L/min x Δ[N2O5])

To find Δ[N2O5], we can subtract the concentration at 20 minutes from the concentration at 0 minutes:
Δ[N2O5] = 1.24 x 10^-2 mol/L - 0.92 x 10^-2 mol/L = 0.32 x 10^-2 mol/L

Substituting this value into the equation for [N2O5], we get:
[N2O5] = 80 min / (3.2 x 10^-4 mol/L/min x 0.32 x 10^-2 mol/L) ≈ 0.10 x 10^-2 mol/L

Therefore, the answer is (C) 0.10 x 10^-2 mol/L.

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There are approximately 65.4 calories in each serving of the soup.

To calculate the number of calories in each serving of the soup, we'll use the given nutrition label information: 2.2 g of fat, 8.0 g of carbohydrates, and 3.4 g of protein.

1. Calculate the calories from fat: 2.2 g of fat × 9 calories/gram = 19.8 calories
2. Calculate the calories from carbohydrates: 8.0 g of carbohydrates × 4 calories/gram = 32 calories
3. Calculate the calories from protein: 3.4 g of protein × 4 calories/gram = 13.6 calories

Now, add up the calories from fat, carbohydrates, and protein:
19.8 calories (from fat) + 32 calories (from carbohydrates) + 13.6 calories (from protein) = 65.4 calories

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No further experiments are needed to determine the number of types of charges that exist, as the evidence we have is sufficient.

Based on the information provided, I cannot determine the specifics of the experiments you are referring to.

From historical experiments conducted by various scientists, it has been concluded that there are two types of charges: positive and negative. This conclusion is supported by sufficient evidence from numerous experiments, such as the ones conducted by Benjamin Franklin, Charles-Augustin de Coulomb, and Michael Faraday.

These experiments observed the interactions between charged objects and established the fundamental principle that like charges repel each other while opposite charges attract each other. Furthermore, the concept of charge quantization, as demonstrated in the Millikan oil-drop experiment, provides strong evidence for the existence of only two types of charges.

Based on this well-established knowledge and the vast experimental data available, it is highly unlikely that there are more types of charges than the two we already know: positive and negative. Therefore, no further experiments are needed to determine the number of types of charges that exist, as the evidence we have is sufficient.

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La(OH)3 precipitates at a pH of approximately 7.21 in this acidic solution when treated with NaOH.

To determine the pH at which La(OH)3 precipitates in an acidic solution containing 0.017 M La3+ treated with NaOH, we can follow these steps:

1. Write the balanced chemical equation for the reaction:
La3+ + 3OH- → La(OH)3 (s)

2. Write the solubility product (Ksp) expression for La(OH)3:
Ksp = [La3+][OH-]^3
Given that Ksp for La(OH)3 is 2×10^-21.

3. Calculate the concentration of OH- ions needed to precipitate La(OH)3:
[OH-] = (Ksp / [La3+])^(1/3)
[OH-] = (2×10^-21 / 0.017)^(1/3)
[OH-] ≈ 1.62×10^-7 M

4. Determine the pOH:
pOH = -log([OH-])
pOH = -log(1.62×10^-7) = -(log 1.62 +(-7*log10)) = -(0.209 +(-7*1)) = -(0.209-7) = -(-6.791) = 6.791
pOH ≈ 6.79

5. Calculate the pH:
pH = 14 - pOH
pH ≈ 14 - 6.79
pH ≈ 7.21

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After analyzing every option given for the required question when Al3+ and SO42− react the ionic formula is Al2(SO4)3.

An ionic formula refers to a formula that is considered an empirical formula that shows the given ratio of ions in an ionic compound. This form formula contains the same and counted number of positive and negative changes so that the compound is neutral overall.

This is a form of the equation that shows only ions that participate in a reaction.

Examples of ionic formula

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Part A: The Arrhenius acid that contains the fluoride anion is hydrofluoric acid (HF).
Part B: There is no Arrhenius acid that contains the ascorbate anion.

Ascorbic acid is a weak organic acid that can act as a reducing agent.
Part C: The Arrhenius acid that contains the bromite anion is hypobromous acid (HBrO).
Part D: The Arrhenius base that contains the lithium cation is lithium hydroxide (LiOH). he name of the Arrhenius acidthat contains the fluoride anion. he name of the Arrhenius acid thatcontains the bromite anion. the name of the Arrhenius acid thatcontains the bromite .

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The mass of magnesium metal that can be produced by the electrolysis of molten MgCl2 for 2.47 hr with an electrical current of 10.1 A is approximately 11.3 g.

To find the mass of magnesium metal produced by the electrolysis of molten MgCl2 for 2.47 hours with an electrical current of 10.1 A, you can follow these steps:

Step 1: Determine the total charge (in coulombs) that has passed through the system:
Q = current × time = 10.1 A × 2.47 hr × (3600 s/hr) = 89667 C

Step 2: Calculate the moles of electrons transferred:
Moles of electrons = Q / (Faraday's constant)
Faraday's constant = 96485 C/mol
Moles of electrons = 89667 C / 96485 C/mol = 0.929 mol

Step 3: Use the stoichiometry of the electrolysis reaction to find the moles of Mg produced:
Mg2+ + 2e- → Mg (s)
Moles of Mg = 0.929 mol e- / 2 = 0.465 mol Mg

Step 4: Convert moles of Mg to grams using the molar mass of Mg (24.31 g/mol):
Mass of Mg = 0.465 mol × 24.31 g/mol = 11.3 g

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A highly conserved protein called actin is essential for many biological functions include cell movement, cell division, and intracellular transport.

Monomeric actin molecules assemble into filamentous actin (F-actin) structures through an essential process known as actin polymerization. Nucleators, elongators, and capping proteins are just a few of the proteins that control the dynamic actin polymerization process.

The barbed end of the F-actin filament is the edge connected to actin polymerization. The plus end, also known as the barbed end, is the developing end of the actin filament.

The pointed end, also known as the non-growing or minus end, is distinguishable from the barbed end by its barbs. Most actin polymerization takes place at the F-actin filament's barb-like end.

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The K_{p} of the reaction, 2 HClO₄ (g) ⇄ H₂ (g) + Cl₂ (g) + 4 O₂ (g), is option (E) 2.4 * 10¹¹.

To find the  of this reaction, we will first use the equation:

K_{p} = K_{c} * (RT)^{Δn}

where R is the universal gas constant (8.314 J/mol*K), T is the temperature in Kelvin (273 + 155 = 428K), and Δn is the change in the number of moles of gas.

1: Calculate Δn :

Δn = (moles of products) - (moles of reactants)

Δn = (1 + 1 + 4) - (2) = 4

2: Calculate (RT)^{Δn}:

(RT)^{Δn} = [(8.314 J/mol*K) * (428K)]⁴ = 1.60 * 10¹⁴

3: Calculate :

[tex]K_{p}[/tex] = [tex]K_{c}[/tex] * [tex](RT)^{Δn}[/tex]

Kp  = (1.5 * 10⁻³) * (1.60 * 10¹⁴) = 2.4 * 10¹¹

Therefore, the correct option is E.

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For the specified reaction, K is roughly equal to 0.0219.

To calculate the value of the equilibrium constant (K) for the given reaction, we'll use the expression for K, which is the ratio of the product of the equilibrium concentrations of the products to the product of the equilibrium concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients.

For the reaction [tex]N_2(g) + 3 H_2(g)[/tex] ⇌ [tex]2 NH_3(g)[/tex], the equilibrium constant expression is:

[tex]K = [NH_3]^2[/tex] / [tex][N_2] * [H_2]^3[/tex]

Given the equilibrium concentrations: [tex][N_2][/tex] = 1.30 [tex]M[/tex], [tex][H_2][/tex] = 1.30[tex]M[/tex], and [tex][NH_3][/tex] = 0.250 M, we can now substitute these values into the expression:

K = [tex](0.250)^2[/tex] / [tex](1.30 * (1.30)^3)[/tex]

Now, we can perform the calculations:

K = 0.0625 / (1.30 * 2.197)

K ≈ 0.0625 / 2.8541

K ≈ 0.0219

Thus, the value of K for the given reaction is approximately 0.0219.

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6N NaOH = 6M NaOH 6M NaOH are 6 moles in 1L MW (NaOH) = 39.88 gr/mole so: m = n x MW = 6 x 39.88 = 239.28 gram NaOH.

[tex] \: [/tex]

The entropy change for the system (ΔSsys) is -688.9 J/K. The total entropy change for the universe (ΔSuniv) is 4026.5 J/K.

a. To calculate ΔSsys, we can use the equation:

ΔSsys = ΣS(products) - ΣS(reactants)

ΔSsys = [2S(AlCl3) - 2S(Al) - 3S(Cl2)]
Note that we need to use the molar entropy values for each substance, which can be found in a table of thermodynamic data.

Using these values and substituting them into the equation, we get:

ΔSsys = [2(186.3 J/K/mol) - 2(28.3 J/K/mol) - 3(223.1 J/K/mol)]
ΔSsys = -688.9 J/K

Therefore, the entropy change for the system (ΔSsys) is -688.9 J/K.

b. To calculate ΔSuniv, we can use the equation:

ΔSuniv = ΔSsys + ΔSsurr

Where ΔSsurr is the entropy change for the surroundings. At constant pressure and temperature, we know that:

ΔSsurr = -ΔH/T

Substituting the given values and solving, we get:

ΔSsurr = -(-1408.4 kJ)/(298 K)
ΔSsurr = 4715.4 J/K

Therefore,

ΔSuniv = ΔSsys + ΔSsurr
ΔSuniv = -688.9 J/K + 4715.4 J/K
ΔSuniv = 4026.5 J/K

Therefore, the total entropy change for the universe (ΔSuniv) is 4026.5 J/K.

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Increasing the polarity of the developing solvent in TLC generally leads to a decrease in the Rf value of a compound.

This is because polar compounds tend to interact more strongly with the polar stationary phase (silica gel) and therefore move more slowly with the developing solvent. In contrast, nonpolar compounds move more easily with the developing solvent and have higher Rf values on the TLC plate.

One way colorless compounds can be visualized on a TLC plate is through the use of a UV lamp. Many organic compounds absorb UV light, and so they appear as dark spots on a fluorescent background when viewed under a UV lamp. Another option is to use a chemical reagent that reacts specifically with the compound of interest, producing a visible color or fluorescent product. TLC is the more appropriate method for separating compounds of a mixture analytically. This is because it allows for the rapid separation and identification of compounds based on their different Rf values. Column chromatography, on the other hand, is a preparative method that is used to isolate larger quantities of a specific compound from a mixture. In column chromatography, the stationary phase is packed into a column and the mixture is passed through it under pressure, allowing for the separation of individual compounds based on their differential interactions with the stationary phase.

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To find the Kc value for the given equation, we need to use the Kc values of the two given equations and apply the law of chemical equilibrium.

The given equation can be written as a combination of the two given equations:

2PCl5(g) + 2H2O(g) ⇌ 2PCl3(g) + Cl2(g) + 2HCl(g)     (1)

4HCl(g) + O2(g) ⇌ 2Cl2(g) + 2H2O(g)                          (2)

Multiplying equation (2) by 2 and adding it to equation (1), we get:

2PCl5(g) + 2H2O(g) ⇌ 2PCl3(g) + 2Cl2(g) + 4HCl(g) + O2(g)

Now, applying the law of chemical equilibrium, we can write:

Kc1 = [PCl3][Cl2]/[PCl5]

Kc2 = [Cl2]^2[H2O]^2/[HCl]^4[O2]
Kc3 = [PCl3]^2[HCl]^4[O2]/[PCl5]^2[Cl2]^2[H2O]^2

Substituting the given Kc values in the above equations, we get:

0.0231 = [PCl3][Cl2]/[PCl5]
12.9 = [Cl2]^2[H2O]^2/[HCl]^4[O2]
Kc3 = [PCl3]^2[HCl]^4[O2]/[PCl5]^2[Cl2]^2[H2O]^2
Multiplying the two given equations, we get:
2PCl5(g) + 4HCl(g) + 2O2(g) ⇌ 4Cl2(g) + 4H2O(g)
Now, applying the law of chemical equilibrium, we can write:

Kc4 = [Cl2]^4[H2O]^4/[HCl]^8[O2]^2

Substituting the given Kc values in the above equation, we get:

Kc4 = (12.9)^2/(0.0231)^4

Kc4 = 3.12 × 10^9

Therefore, the Kc value for the given equation is 3.12 × 10^9.

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