consider the following reaction: 2 c2h6(g) 7 o2(g) → 4 co2(g) 6 h2o(g) at stp, what is the total volume of co2 formed when 6.0 liters of c2h6 are combusted?

Phosphorus can be prepared by electrolysis of an aqueous phosphate solution. The method of extraction of phosphorus from the phosphate solution involves the process of electrolysis.

The electrolysis process is defined as a process in which a chemical reaction is caused by an electric current passing through an electrolyte. The electrolysis of the aqueous solution of phosphate is the process in which phosphorus is extracted from the solution. The procedure of electrolysis of an aqueous phosphate solution is as follows: Two electrodes are placed into the phosphate solution, one is the anode (positive electrode) and the other is the cathode (negative electrode). The anode is made up of graphite and the cathode is made up of mercury. A direct current is then passed through the solution. When the current is passed, it causes the oxygen to be released at the anode, which combines with graphite to produce carbon dioxide.

In addition, at the cathode, hydrogen gas is released along with mercury, and the mercury is mixed with the phosphorus to produce an alloy. The mixture is then filtered, and the impurities are removed with the help of hydrochloric acid. The obtained phosphorus is solid and can be melted and moulded into sticks. The process of electrolysis is a good method for the extraction of phosphorus from aqueous solutions and the process is economically viable. This process is also good because the phosphate solution can be obtained from rock phosphates, and this process is a good alternative to the use of non-renewable phosphorus sources. The main advantages of this process are that it is cost-effective, does not require high temperatures, and is environmentally friendly.

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a. has a sour taste - this statement is for acid.

b. neutralizes bases - this statement is for acid.

c. produces ions in water - this statement is for both acid and base.

d. is named barium hydroxide -  this statement is for base

e. is an electrolyte -  this statement is for both acid and base.

An acid, any substance that in water solution tastes sour, changes the colour of certain indicators.

Acids are ionic compounds that, when dissolved in water, produce positive hydrogen ions ( H⁺ ).

We will indicate whether each of the following statements is characteristic of an acid, a base, or both as follows;

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Answer:

Some human activities that cause damage (either directly or indirectly) to the environment on a global scale include population growth, overconsumption, overexploitation, pollution, and deforestation, to name but a few.

Explanation:

Hope this helped!!!

To prepare a [tex]HC_{2} H_{3}O_{2}[/tex]  buffer with a pH of 4.24 and [tex]pK_{a}[/tex] of 4.74 ,the ratio  [tex]\frac{{C_{2} H_{3} O_{2}} ^{-}}{HC_{2} H_{3} O_{2} } }[/tex] is 0.3162.

A buffer is a solution containing an acid and its conjugate base or a base and its conjugate acid .

Eg: Acidic buffer : Acetic acid and sodium acetate

     Basic buffer : Ammonium hydroxide and ammonium chloride

Buffer is used to resist changes in the pH of the solution to which it is added.

given, pH of [tex]HC_{2} H_{3}O_{2}[/tex]  = 4.24

            [tex]pK_{a}[/tex] of [tex]HC_{2} H_{3} O_{2}[/tex]  = 4.74

According to Henderson -Hasselbalch equation.

pH                 =  [tex]pK_{a}[/tex]  + log [tex]\frac{conjugate base}{acid}[/tex]

4.24               =  4.74  + log [tex]\frac{{C_{2} H_{3} O_{2}} ^{-}}{HC_{2} H_{3} O_{2} } }[/tex]

4.24 - 4.74     =  log [tex]\frac{{C_{2} H_{3} O_{2}} ^{-}}{HC_{2} H_{3} O_{2} } }[/tex]

-0.50              =   log[tex]\frac{{C_{2} H_{3} O_{2}} ^{-}}{HC_{2} H_{3} O_{2} } }[/tex]

[tex]log^{-1}[/tex](- 0.50)   =   [tex]\frac{{C_{2} H_{3} O_{2}} ^{-}}{HC_{2} H_{3} O_{2} } }[/tex]

0.3162            =  [tex]\frac{{C_{2} H_{3} O_{2}} ^{-}}{HC_{2} H_{3} O_{2} } }[/tex]

Therefore, the ratio  [tex]\frac{{C_{2} H_{3} O_{2}} ^{-}}{HC_{2} H_{3} O_{2} } }[/tex]  is 0.3162 .

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This ratio can be determined using the Henderson-Hasselbalch equation.

To prepare a HC2H3O2 buffer with a pH of 4.24 and a pKa of 4.74, calculate the ratio of [tex]C2H3O2^-[/tex](conjugate base) to HC2H3O2 (acid) that will result in the desired pH.

The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the acid and the ratio of the conjugate base to the acid. It is given by:

pH = pKa + log([[tex]C2H3O2^-[/tex]]/[HC2H3O2])

In this case, the desired pH is 4.24, and the pKa of HC2H3O2 is 4.74. We can rearrange the Henderson-Hasselbalch equation to solve for the ratio [[tex]C2H3O2^-[/tex]]/[HC2H3O2]:

[C2H3O2^-]/[HC2H3O2] = 10^(pH - pKa)

Substituting the values, we have:

[[tex]C2H3O2^-[/tex]]/[HC2H3O2] = 10^(4.24 - 4.74)

Simplifying the equation and calculating the ratio, you can determine the appropriate ratio of [tex]C2H3O2^-[/tex] to HC2H3O2 needed to prepare the buffer with the desired pH of 4.24.

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The increasing order of reactivity of the given compounds in nucleophilic addition reactions is: Butanone < Propanone < Propanal < Ethanal

The reactivity of a carbonyl compound in nucleophilic addition reactions depends on the electron density at the carbonyl carbon. The more electron density at the carbonyl carbon, the less reactive it is towards nucleophilic attack.

In the given compounds, the electron density at the carbonyl carbon decreases with increasing number of alkyl groups. This is because alkyl groups are electron-releasing groups and they donate electrons to the carbonyl carbon.

The more alkyl groups there are, the more electrons are donated to the carbonyl carbon, and the less reactive it is towards nucleophilic attack.

Therefore, butanone, which has the fewest alkyl groups, is the most reactive towards nucleophilic attack. Propanone, which has one alkyl group, is less reactive than butanone.

Propanal, which has two alkyl groups, is less reactive than propanone. And ethanal, which has three alkyl groups, is the least reactive towards nucleophilic attack.

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For the sequential deprotonations of a polyprotic acid, the equilibrium constants typically decrease with successive ionization. Therefore, the correct answer is option c: decrease with successive ionization.

For the sequential deprotonations of a polyprotic acid, the equilibrium constants typically decrease with each successive ionization. This is due to several factors. As protons are successively removed, the remaining ions become more negatively charged, leading to increased electrostatic repulsion and reduced stability. Additionally, the loss of each proton becomes energetically less favorable as the acid becomes more negatively charged. Furthermore, the availability of electrons for stabilization through resonance or induction decreases as protons are lost. These factors contribute to a decrease in the strength of the subsequent acids formed, leading to lower equilibrium constants for each successive ionization.

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The volume of which the stock calcium solution would you need to deliver the 8 mEg dose in the 250-mL of D5W is 2 mL

Therefore, the correct answer is A.

.In this case, CaCl₂ produces Ca₂⁺ and 2Cl⁻.So the equivalent weight of CaCl₂ is:

1Ca₂⁺ + 2Cl⁻ → 1 mol of CaCl2 has 1Ca₂⁺ and 2Cl⁻ions

Equivalent weight of CaCl₂ = Molecular weight / n

Equivalent weight of CaCl₂ = 111/ 3 = 37

The molecular weight of CaCl₂ is 111 gm/mole and n = 3 because it gives 3 ions when it dissociates in a solution.

Number of milliequivalents in 8 mmol of calcium = 8 x 2 = 16 mEq

The number of milliequivalents in the stock solution is 4 x 2 = 8 mEq/ml.

Volume of the stock calcium solution required = (16/8) x 1= 2 mL

So, the Answer: A. 2 mL.

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Answers:

Aluminum Frame - The aluminum frame is used to protect all the parts of the solar panel. It basically works as a shield against damaging items (hail, ice.)

Tempered Glass - The tempered glass keeps pressure inside the solar panel and keeps compressure outside of the solar panel to protect it from breaking.

EVA - EVA is a great alternative for electric radiational heat. It has a small amount of degradability to sunlight which keeps the panel from burning.

Solar Cells - Solar Cells are the part of the solar panel which work to transform the solar energy into electricity and not just use raw sunlight.

Back Sheet - Back sheets keep solar cells and the top and the bottom of the panel together. They make sure nothing inside the panel comes off.

Junction Box - The junction box makes sure all the solar cells inside of the solar panel are kept together so that the machine works as a whole and the back sheet's function does not get destroyed.

Answer: C. 0.57 M

Explanation:

Molarity= Moles/Vol

Answer: 0.391

Explanation:

Ideal gas law is valid only for ideal gas not for vanderwaal gas. Therefore, 0.38moles of gas are in the sample​. Ideal gas is a hypothetical gas.

Ideal gas equation is the mathematical expression that relates pressure volume and temperature. There is no force of attraction between the particles.

Mathematically the relation between Pressure, volume and temperature can be given as

PV=nRT

where,

P = pressure of gas sample = 660.0mmHg=0.86atm

V= volume of gas sample =11.0 L

n =number of moles of gas sample=?

T =temperature of gas =298K

R = Gas constant = 0.0821 L.atm/K.mol

Substituting all the given values, we get

0.86atm×11.0 L =n× 0.0821×298

On calculation we get

9.46=n×24.4

number of moles of sample=0.38moles

Therefore, 0.38moles  of gas are in the sample​.

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In a multistep mechanism, a complex reaction is broken down into a series of elementary steps, each involving the collision and transformation of reactant molecules. The correct answer is (b) the rate of the overall reaction is slower than the slowest step.

The overall rate of the reaction is determined by the rate of the slowest (rate-determining) step. This step limits the overall reaction rate because it takes the longest time to occur or requires the highest activation energy.

The rate of the overall reaction cannot be faster than the slowest step because the slowest step sets the maximum rate at which the reaction can proceed.

Similarly, the rate of the overall reaction is not equal to the fastest step or the average of all elementary steps. It is solely determined by the slowest step in the mechanism. The correct answer is (b) the rate of the overall reaction is slower than the slowest step.

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Answer:

experiments

Explanation:

I am not sure but heres my take

Answer:

It was only after the invention of electron microscope that the idea developed

Answer:

If this is an idea gas then 1mol takes up 22.4L.

So, knowing how many L you have you can figure out how many  mole syou have by doing a simple equation:

[tex]\frac{1mol}{y} =\frac{22.4L}{925.4L}[/tex]

Solve for y.

Then, since you know how many moles you have use the ptable https://ptable.com/#Properties to figure out the mass in grams.

NOTE: The ptable tells you that 1mol of H = 1g.....so this should be an easy calculation :) enjoy

In any organic redox reaction, you can recognize the reduced and oxidized organic molecules by the net change in electrons between products and reactants. Reduction corresponds to the gain of electrons, and oxidation corresponds to the loss of electrons.

The redox reactions refer to chemical reactions in which the oxidation state of atoms or molecules is altered. In organic redox reactions, one organic compound is reduced (gains electrons) while another is oxidized (loses electrons).The term 'oxidized' means that the molecule is losing electrons or increasing in oxidation state. In contrast, the term 'reduced' means that the molecule is gaining electrons or decreasing in oxidation state. In any organic redox reaction, the reduced and oxidized organic molecules can be recognized by the net change in electrons between products and reactants.The products have lower potential energy than the reactants in a spontaneous redox reaction. When electrons are transferred, energy is either released or absorbed.

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Answer:

The correct answer is Choice D.

(Headaches or backaches)​

Explanation:

Hope this helps!

Please mark me as Brainlinieast.

The lime and soda ash dosages necessary for precipitation softening are 373/1000 kg/m^3 and 374/1000 kg/m^3 respectively.

Given information:

Municipality treats 15 x 10^6 gal/day of groundwater containing the following:

CO2 = 17.6mg/L,

Ca^2+ = 80 mg/L,

Mg^2+ = 48.8 mg/L,

Na^+ = 23 mg/L,

Alk(HCO3^-) = 270 mg/L

as CaCO3,

SO4^2- = 125 mg/L,

and Cl^- = 35 mg/L.

The soda ash is 90% sodium carbonate, and the lime is 85% weight CaO.

Softening by excess lime treatment needs to be determined.

Concept used:

Soda ash dosage = 1.4 (Alkalinity as CaCO3 mg/L) - 1.2 (CO2 as CaCO3 mg/L)

Lime dosage = 2.2 (Alkalinity as CaCO3 mg/L) - 1.2 (Calcium hardness as CaCO3 mg/L) - 1.7 (Magnesium hardness as CaCO3 mg/L) + 0.7 (Iron and manganese hardness as CaCO3 mg/L)

Soda ash dosage = 1.4 (Alkalinity as CaCO3 mg/L) - 1.2 (CO2 as CaCO3 mg/L)CO2 as CaCO3 mg/L

= 17.6 × (50/44) = 20 mg/L

Alkalinity as CaCO3 mg/L = 270 mg/L

Soda ash dosage

= 1.4 × 270 - 1.2 × 20

= 374 mg/L (or) 374/1000 kg/m^3

Lime dosage = 2.2 (Alkalinity as CaCO3 mg/L) - 1.2 (Calcium hardness as CaCO3 mg/L) - 1.7 (Magnesium hardness as CaCO3 mg/L) + 0.7 (Iron and manganese hardness as CaCO3 mg/L)

Calcium hardness as CaCO3 mg/L = 80 mg/L

Magnesium hardness as CaCO3 mg/L

= 48.8 × 2.5

= 122 mg/L (or) 0.122 kg/m^3

Iron and manganese hardness as CaCO3 mg/L = 0 mg/L

Lime dosage

= 2.2 × 270 - 1.2 × 80 - 1.7 × 0.122 + 0.7 × 0

= 373 mg/L (or) 373/1000 kg/m^3

Soda ash dosage required for precipitation softening = 374/1000 kg/m^3

Lime dosage required for precipitation softening = 373/1000 kg/m^3

Therefore, the lime and soda ash dosages necessary for precipitation softening are 373/1000 kg/m^3 and 374/1000 kg/m^3 respectively.

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The time until nuclear waste cesium 137 is deemed safe can be determined by calculating the number of half-lives required for the amount of cesium 137 to decrease to a safe level.

The half-life of cesium 137 is approximately 30 years. This means that every 30 years, the amount of cesium 137 will be reduced by half. To determine the time until the waste is deemed safe, we need to calculate the number of half-lives required for the amount of cesium 137 to decrease to a safe level. Given that there are 88.44 g of cesium 137 initially, we can calculate the number of half-lives using the formula:

Number of Half-Lives = (ln(initial amount) - ln(final amount)) / ln(0.5)

Using the given values, we can substitute the initial amount as 88.44 g and the final amount as the safe level of cesium 137 (determined by regulatory standards). By solving the equation, we can find the number of half-lives required. Once we have the number of half-lives, we can multiply it by the half-life of cesium 137 (30 years) to determine the total time until the waste is deemed safe. It is important to note that the specific safe level of cesium 137 is not provided in the question. Regulatory bodies and authorities establish safe levels based on various factors, such as environmental and health considerations.

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False. Viscosity of a polymer is not typically measured hourly; measurements depend on specific requirements.

False. The viscosity of a polymer is not typically measured on an hourly basis. Viscosity refers to the resistance of a fluid to flow, and it is generally measured under specific conditions, such as at a particular temperature or shear rate.

The viscosity of a polymer can be influenced by various factors, including molecular weight, temperature, and shear stress.

It is more common to measure the viscosity of a polymer at specific time intervals relevant to the experiment or process being studied. The frequency of measurements depends on the specific objectives and requirements of the study.

These measurements are often recorded at regular intervals, such as daily, weekly, or at specific milestones during a polymerization process.

Therefore, the statement that the viscosity of a polymer is measured hourly is false.

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Answer:

a women standing in high heels

Explanation:

Answer:

A woman standing in high heels

A gas expands in volume from 29.3 mL to 80.1 mL at constant temperature.

(a)  The work done (in joules) if the gas expands against a vacuum is 0.

(b) The work done (in joules) against a constant pressure of 3.5 atm is -12.7 J.

(c) The work done (in joules) against a constant pressure of 10.1 atm is -36.6 J.

To calculate the work done during the expansion of a gas, we can use the formula:

Work (W) = -PΔV

Where P is the pressure and ΔV is the change in volume.

(a) If the gas expands against a vacuum, it means there is no external pressure opposing the expansion. In this case, the work done is zero because there is no pressure acting against the gas.

W = 0 (no work done against a vacuum)

(b) If the gas expands against a constant pressure of 3.5 atm, we need to convert the pressure to SI units (Pascals) before calculating the work.

Given:

Initial volume (V1) = 29.3 mL = 29.3 × 10⁻⁶L

Final volume (V2) = 80.1 mL = 80.1 × 10⁻⁶ L

Pressure (P) = 3.5 atm = 3.5 × 101325 Pa

ΔV = V2 - V1 = (80.1 × 10⁻⁶ L) - (29.3 × 10⁻⁶ L)

W = -PΔV = -(3.5 × 101325 Pa) × [(80.1 - 29.3) × 10⁻⁶) L]

W ≈ -12.7 J.

Therefore, the work done against a constant pressure of 3.5 atm is approximately -12.7 J.

(c) Similarly, for a constant pressure of 10.1 atm:

Pressure (P) = 10.1 atm = 10.1 × 101325 Pa

ΔV = V2 - V1 = (80.1 × 10⁻⁶ L) - (29.3 × 10⁻⁶) L)

W = -PΔV = -(10.1 × 101325 Pa) × [(80.1 - 29.3) × 10⁻⁶L]

W ≈ -36.6 J.

Therefore, the work done against a constant pressure of 10.1 atm is approximately -36.6 J.

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Experimental errors or variations can significantly impact the results of the experiment. In this case, inadequate stirring, incomplete mixing of layers, incorrect extraction solution, and improper pH measurement can lead to inaccurate component percentages in the final product.

(a) Inadequate stirring or shaking of the mixture after adding DCM to Pan acetin can result in incomplete dissolution or extraction of certain components. This would lead to lower percentages of the components that require proper mixing for their extraction.

(b) Failure to thoroughly mix the aqueous and organic layers during the NaOH extraction can cause incomplete transfer of target components from one layer to another. As a result, the percentages of the desired components may be lower than expected, indicating incomplete extraction.

(c) Mistakenly using 5% HCL instead of 5% NaOH for the DCM extraction can affect the selectivity of the extraction process. Different solvents have varying affinities for specific components, so using the wrong extraction solution can lead to incorrect percentages of the components in the final product.

(d) Instead of using pH paper, if litmus paper is used to neutralize the [tex]NaHCO_{3}[/tex] solution to pH 7, the accuracy of pH measurement may be compromised. Litmus paper provides a visual color change indication but lacks the precision of pH paper. As a result, the pH adjustment may not be accurate, potentially leading to deviations in the final component percentages.

In summary, these experimental errors or variations can introduce inaccuracies in the component percentages of the final product due to inadequate mixing, incorrect extraction solution, and imprecise pH measurement. It is essential to carefully follow the experimental procedure to minimize such errors and ensure reliable results.

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The rate of consumption of I- in the first 15.0 seconds of the reaction is 0.227 M/s.

The given balanced chemical equation is:

H2O2(aq) + 3I-(aq) + 2H+(aq) → I3-(aq) + 2H2O(l)

From the stoichiometry of the balanced equation, we can see that the ratio between I- and H2O2 is 3:1. This means that for every 1 mole of H2O2 reacted, 3 moles of I- are consumed.

In the given time interval of 15.0 seconds, the concentration of I- decreases from 1.000 M to 0.773 M. The change in concentration is:

Δ[I-] = [I-]final - [I-]initial = 0.773 M - 1.000 M = -0.227 M

To find the rate of consumption of I-, we divide the change in concentration by the time interval:

Rate = Δ[I-] / Δt = -0.227 M / 15.0 s = -0.0151 M/s

The negative sign indicates the decrease in concentration of I-. However, since rate values are usually reported as positive values, we take the absolute value:

Rate = 0.0151 M/s

Therefore, the rate of consumption of I- in the first 15.0 seconds of the reaction is 0.0151 M/s or approximately 0.227 M/s.

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The number of water molecules formed from the given starting materials is 3.99 × 10-21 molecules.

Balanced equation is:2H2 + O2 ---> 2H2OIf you start with 24 molecules of hydrogen gas, how many molecules of oxygen gas do you need?From the above-balanced chemical equation, we can see that 2 molecules of hydrogen react with 1 molecule of oxygen gas to form 2 molecules of water.Thus, to form 24 molecules of hydrogen gas, we need 12 molecules of oxygen gas. How many molecules of water are made from the amount of starting materials described above?To determine the number of molecules of water formed from the starting materials, we have to determine the limiting reagent. The reactant that gives the least amount of product is called the limiting reagent.To determine the limiting reagent, we can use the concept of mole. The number of moles of a substance is determined by dividing the given mass of the substance by its molar mass.Therefore, the given information of the number of molecules is required to be converted to the number of moles.Number of molecules of hydrogen gas = 24The number of moles of hydrogen gas = (24/6.023 × 1023) = 3.989 × 10-21The number of moles of oxygen gas = (3.989 × 10-21)/2 = 1.995 × 10-21Since there are only 1.995 × 10-21 moles of oxygen gas, it will get consumed in the reaction before all the hydrogen gas is consumed. Therefore, oxygen gas is the limiting reagent.The number of moles of water formed is given by the stoichiometric coefficient of water in the balanced equation, which is 2.Number of molecules of water formed from 24 molecules of hydrogen and 12 molecules of oxygen = 2 × (1.995 × 10-21) = 3.99 × 10-21Therefore, the number of water molecules formed from the given starting materials is 3.99 × 10-21 molecules.

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Answer:

0.205M

Explanation:

Molarity is an unit of concentration used in chemistry defined as the ratio between moles of solute (In this case, LiF) and liters of solution.

To solve this question, we must find the moles of LiF and the liters of the solution:

Moles LiF -Molar mass: 25.939g/mol-:

2.39g * (1mol / 25.939g) = 0.0921moles

Liters solution:

450mL * (1L / 1000mL) = 0.450L

Molarity is:

0.0921 moles / 0.450L =

When, a 20.0 ml of 0.20 m HNO₃ was titrated with 10.0 ml of 0.20 m NaOH. Then, the pH of the resulting solution is approximately 1.18.

To determine the pH of the solution resulting from the titration of 20.0 mL of 0.20 M HNO₃ with 10.0 mL of 0.20 M NaOH, we need to calculate the concentration of the resulting solution and then find the pH using the appropriate equations.

Let's start by calculating the moles of HNO₃ and NaOH used in the titration:

Moles of HNO₃ = concentration of HNO₃ × volume of HNO₃ used

= 0.20 mol/L × 0.0200 L

= 0.0040 mol

Moles of NaOH = concentration of NaOH × volume of NaOH used

= 0.20 mol/L × 0.0100 L

= 0.0020 mol

Since HNO₃ and NaOH have a 1:1 stoichiometric ratio, the moles of HNO₃ remaining after the reaction are:

Moles of HNO₃ remaining = Moles of HNO₃ initial - Moles  of NaOH used

= 0.0040 mol - 0.0020 mol

= 0.0020 mol

Now, let's calculate the new concentration of HNO₃ in the resulting solution:

Concentration of HNO₃ = Moles of HNO₃ remaining / Total volume of resulting solution

= 0.0020 mol / (20.0 mL + 10.0 mL) = 0.0020 mol / 0.0300 L

= 0.067 M

To find the pH of the resulting solution, we can use the fact that HNO₃ is a strong acid and completely dissociates in water. Therefore, the concentration of H⁺ ions in the solution is equal to the concentration of HNO₃;

[H⁺] = 0.067 M

Now, we can calculate the pH;

pH = -log10([H⁺])

= -log10(0.067)

≈ 1.18

Therefore, the pH of the resulting solution is approximately 1.18 (rounded to the appropriate number of significant figures).

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The final molarity of the zinc cation from the calculation is 0.144 M

Stoichiometry is an essential tool for chemists to understand and predict the quantitative aspects of chemical reactions. It allows for precise calculations of reactant quantities, product yields, and the optimization of reaction conditions for industrial processes.

Number of moles of the zinc chloride = 4.84 g/136 g/mol

= 0.036 moles

Number of moles of potassium carbonate = 250/1000 * 0.10

= 0.025 moles

The equation of the reaction is;

ZnCl2(s) + K2CO3(aq) ----> ZnCO3(aq) + 2KCl(aq)

Final molarity of the Zinc cations = 0.036 moles * 1000/250 L

= 0.144 M

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Answer:

1.06L = V₂

Explanation:

Charle's Law states that the volume of a gas is directly proportional to the absolute temperature when pressure remains constant.

The equation is:

V₁T₂ = V₂T₁

Where V is volume and T absolute temperature of 1, initial state and 2, final state of the gas.

V₁ = 1.83L

T₂ = -100°C + 273.15 = 173.15K

V₂ = ?

T₁ = 27°C + 273.15 = 300.15K

V₁T₂ = V₂T₁

1.83L*173.15K = V₂*300.15K

Explanation:

The balanced half-reaction for the reduction of nitrate, NO, (aq), to nitrogen gas, N.(g), in a basic solution is given below:Balanced half-reaction equationNO₃¯ → N2(g)

Step 1: Write the half-reaction equation and balance it without adding water or H⁺ or OH⁻.NO₃¯ → N2(g)

Step 2: Add water to the right-hand side of the equation to balance the oxygen.NO₃¯ → N2(g) + H2O

Step 3: Add sufficient H⁺ ions to balance the hydrogen.NO₃¯ + 10H⁺ → N2(g) + 5H2O

Step 4: Add electrons (e⁻) to balance the charges.NO₃¯ + 10H⁺ + 8e⁻ → N2(g) + 5H2OThe half-reaction for the reduction of nitrate, NO, (aq), to nitrogen gas, N.(g), in a basic solution is given by NO₃¯ + 10H⁺ + 8e⁻ → N2(g) + 5H2O.

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Answer:

300

Explanation:

at least 300 molecules

When considering the structure for 4POC, Parallel β sheets are often found in the protein interior due to the arrangement of nonpolar amino acids on both sides of the sheet.  Thus, correct option is (A).

Option (A) is the proper conclusion. Because they are stabilized by interactions between nonpolar amino acids on both sides of the sheet, parallel sheets are frequently found inside proteins. These nonpolar residues can interact well with one another outside of an aqueous environment due to their hydrophobic nature. The protein structure is stabilized by this configuration.

Contrarily, parallel sheets with nonpolar amino acids on either one side of the sheet (option B) or on both sides of the sheet in the protein exterior (option C) are less frequent because doing so would expose hydrophobic residues unfavorably to the solvent and cause the protein to become instabilized.

To know more about β sheets here https://brainly.com/question/14728237

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