A sample of nitrogen gas is inside a sealed container. The container is slowly compressed, while the temperature is kept constant. This is a(n) ________ process. A sample of nitrogen gas is inside a sealed container. The container is slowly compressed, while the temperature is kept constant. This is a(n) ________ process.

a. adiabatic
b. constant-volume
c. isobaric
d. isothermal

The solubility of CuX in a 0.200 M NaCN solution cannot be calculated solely based on the Ksp of CuX, because NaCN forms a complex with Cu2+, significantly affecting its solubility. Additional information on the stability constant of the copper-cyanide complex is needed

To calculate the solubility of CuX in a solution that is 0.200 M in NaCN, we must consider the common ion effect due to the presence of the cyanide ion, CN-. The equation for the solubility product (Ksp) is given by Ksp = [Cu2+][X2−]. Since cyanide ions form a complex with copper ions, the direct precipitation of CuX is suppressed, and we cannot simply take the square root of the Ksp to find its molar solubility.

Instead, we would write the reaction of copper with cyanide: CuX(s) + 4CN−
ightleftharpoons [Cu(CN)4]3− + X2−. This complexation reaction would vastly reduce the concentration of free Cu2+ ions in solution, thereby affecting the solubility of CuX. To find the actual solubility, we would need to know the stability constant (Kf) for the copper-cyanide complex, and then set up an equilibrium calculation that includes both the Ksp of CuX and the Kf of [Cu(CN)4]3−. Since the problem doesn't provide Kf, we would not be able to calculate the solubility without additional information

The presence of NaCN significantly decreases the solubility of CuX due to the common ion effect. Therefore, the solubility of CuX in a solution that is 0.200 M in NaCN remains extremely low, approximately 1.27×10 ⁻¹⁸ M.

To calculate the solubility of CuX in a solution that is 0.200 M in NaCN, we need to consider the common ion effect. When NaCN dissolves, it produces CN⁻ ions, which can interact with Cu²⁺ ions, reducing the solubility of CuX.

Given that the solution is 0.200 M in NaCN, we can assume that the concentration of CN⁻ ions ([CN⁻]) is 0.200 M.

Now, let's denote the solubility of CuX as x M. Since CuX dissociates into Cu²⁺ and X²⁻ ions, the concentration of Cu²⁺ ions ([Cu²⁺]) and X²⁻ ions ([X²⁻]) will both be equal to x M at equilibrium.

The solubility product constant (K sp​ ) expression for CuX is:

K sp​ =[Cu²⁺ ][X²⁻ ]

Given that K sp​ =1.27×10⁻³⁶ , we can substitute the concentrations into the expression:

1.27×10⁻³⁶ =(x)(x)

1.27×10⁻³⁶ =x²

Now, we'll solve this equation for x to find the solubility of CuX. Let's proceed with the calculations.

To solve for x, we take the square root of both sides of the equation:

x= 1.27×10⁻³⁶

x=1.127×10⁻¹⁸

So, the solubility of CuX in a solution that is 0.200 M in NaCN is

1.127×10⁻¹⁸  M.

In chemical reactions, a change in volume affects the position of equilibrium based on stoichiometry. Increasing volume results in a shift towards the side with more gaseous molecules for reactions with Δn not equal to zero, while there is no shift if Δn equals zero. The direction of the shift is to decrease the reaction quotient (Q) in order to re-establish equilibrium with the equilibrium constant (K).

Explanation:

The effect of volume change on the position of equilibrium in chemical reactions is dependent on the stoichiometry and the reaction in question. For a reaction where Δn = -1, increasing the volume would result in a decrease in pressure and a shift towards the side with more moles of gas to re-establish equilibrium, typically the side with more molecules. However, if Δn = 0, an increase in volume has no effect on the equilibrium as there is no change in moles of gaseous substances on either side of the reaction. When Δn = +1, increasing the volume leads the equilibrium to shift towards the products, as this increases the total number of gaseous molecules which tends to lower the pressure.

When volume is increased and the reaction quotient Q becomes greater than the equilibrium constant K (Q > K), the reaction tends to shift towards the reactants to re-establish equilibrium. Conversely, when volume is decreased, and the pressure is increased, the reaction tends to shift towards the side of the reaction with fewer moles of gas.

Answer:

Electrons

Explanation:

In an atom there would be three subatomic particles: Neutrons, electrons, protons. The smallest and lightest in terms of mass is electrons. This is because the nucleus is comprised of the protons and the neutrons, these have a greater mass than electrons as electrons has very little mass that can considered to be 0.

Answer:

we are given the 3-methyl2 butanone and upon the reduction with LiAIH4 there is formed alcohol, there are two possible side attack,

therefore, whenever the front side attacks then the -CH3 moves back  the plain and the H will be above the plane. More so, when  attack from the back side the H moves below the plane and -CH3 moves above the plane. -OH is evident in the plane. see the attachment below to view the structure.

Explanation:

Answer:

hydrogen ions

Explanation:

Acids are substances that when dissolved in water release hydrogen ions, H+(aq). Bases are substances that react with and neutralise acids, producing water. When dissolved, bases release hydroxide ions, OH-(aq) into solution. Water is the product of an acid and base reacting.

When acids dissolve in water, they produce hydrogen ions (H+). This is known as the process of ionization.

The hydrogen ions are responsible for the acidic properties of the solution.When HCl is added to water, it dissociates into hydrogen ions (H+) and chloride ions (Cl-). The presence of hydrogen ions gives the solution acidic properties.

Similarly, other acids such as sulfuric acid, nitric acid, and acetic acid also produce hydrogen ions when dissolved in water.

It's important to note that not all substances that dissolve in water are acids. Acids have a pH value less than 7, and their properties can vary depending on their strength. Strong acids, like hydrochloric acid, completely dissociate in water, producing a large number of hydrogen ions. Weak acids, like acetic acid, only partially ionize, producing fewer hydrogen ions.

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Answer:

13 grams Na₂SO₄ (2 sig.figs.)

Explanation:

1st convert mass values to moles and solve yield using limiting reactant principles and reaction ratio of balance equation.

Determine limiting reactant and

Complete problem by converting yield into grams.

H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O

moles H₂SO₄ = 8.83g/98g·mol⁻¹ = 0.0901 mole H₂SO₄

moles NaOH = 9.1g/40g·mol⁻¹ = 0.2275 mole NaOH

Determine Limiting Reactant by dividing each mole value by the respective coefficient in the balanced equation. The smaller value is always the limiting reactant.

H₂SO₄ => (0.0901/1) = 0.0901 <= Limiting Reactant (smaller value)

NaOH =>  (0.2275/2) = 0.1138

NOTE: when working the problem use the calculated moles of reactant NOT the LR test number. In this problem, use 0.0901 mole H₂SO₄. (yeah, it is the same but this does not occur for the LR in many other problems). Anyways...

                    H₂SO₄        +        2NaOH       =>       Na₂SO₄      +      2H₂O

moles      0.0901 mole          0.2275 mol            0.0901 mol      2(0.0901 mol)

mass (g) Na₂SO₄ = 0.0901 mole x 142.04 g/mol = 12.798 grams ≅ 13 grams                                    Na₂SO₄ (2 sig.figs.)

Answer:

13 grams Na₂SO₄ (2 sig.figs.)

Explanation:

Firstly

We convert mass values to moles and solve yield using limiting reactant principles and reaction ratio of balance equation.

To determine limiting reactant and

Complete problem by converting yield into grams. We write the equations for the reaction

H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O

To give

moles H₂SO₄ = 8.83g/98g·mol⁻¹ = 0.0901 mole H₂SO₄

Then,

moles NaOH = 9.1g/40g·mol⁻¹ = 0.2275 mole NaOH

Determine Limiting Reactant by dividing each mole value by the respective coefficient in the balanced equation. The smaller value is always the limiting reactant.

H₂SO₄ => (0.0901/1) = 0.0901 <= Limiting Reactant (smaller value)

NaOH =>  (0.2275/2) = 0.1138

N/B: when working the problem use the calculated moles of reactant NOT the LR test number. In this problem, use 0.0901 mole H₂SO₄. (yeah, it is the same but this does not occur for the LR in many other problems). Anyways...

H₂SO₄        +        2NaOH       =>       Na₂SO₄      +      2H₂O

moles      0.0901 mole          0.2275 mol            0.0901 mol      2(0.0901 mol)mass (g)

Na₂SO₄ = 0.0901 mole x 142.04 g/mol = 1

2.798 grams ≅ 13 grams                                    Na₂SO₄ (2 sig.figs.)

Answer:

Zinc is the reducing agent.

Explanation:

Zn is the reducing agent because its oxidation state increases from 0 to 2 and hydrochloric acid, HCl is the oxidizing agent because it loses hydrogen ions, H+ based on the following balanced equation:

Zn (s) + 2HCl (aq)  ----->  ZnCl2 (aq) + H2 (g)

In oxidation, there is the loss of electrons. In reduction, there is a gain of electrons. A nemonic device that might help is LEO (the lion said) GER.

Losses

Electrons

Oxidation

Gains

Electrons

Reduction

Answer:

Zinc

The reducing agent in these reaction is Zinc (Zn)

Explanation:

Zn is the reducing agent because its oxidation state increases from 0 to 2 and hydrochloric acid, HCl is the oxidizing agent because it loses hydrogen ions, H+ based on the following balanced equation:

Zn (s) + 2HCl (aq) -----> ZnCl2 (aq) + H2 (g)

In oxidation, there is the loss of electrons.

In reduction, there is a gain of electrons.

Answer:

Yes

Explanation:

It causes excess bromine to be given off as a gas. -It is used to absorb excess heat generated by the exothermic reaction.

Answer

The desired product should contain acidic group like -COOH group.

Explanation:

Saturated sodium bicarbonate solution will react with a acid compound which contain in reaction mixture and produce CO2 gas.

We have the reaction has;  

NaHCO3 (aq) + R-COOH (aq) ..............> R-COONa (aq) + H2O (l) + CO2 (g)

Therefore.

The desired product will contain acidic group like -COOH (carboxylic acid group).

Answer:

The mole fraction of [tex]KCl[/tex] is [tex]N_{KCl}=0.099[/tex]

Explanation:

Generally number of mole is mathematically represented as

                   [tex]n = \frac{mass}{Molar mass }[/tex]

 The number of mole of  [tex]KCl[/tex] is

               [tex]n_{KCl} = \frac{mass \ of \ KCl}{Molar\ mass \ of \ KCl }[/tex]

                   [tex]n_{KCl} = \frac{125}{74.6}[/tex]

                            [tex]=1.676 \ moles[/tex]

 The number of mole of  [tex]H_2O[/tex] is

         [tex]n_{H_2O} = \frac{mass \ of \ H_2O}{Molar\ mass \ of \ H_2O }[/tex]

         [tex]n_{H_2O} = \frac{275}{18}[/tex]

                   [tex]= 15.28 \ moles[/tex]

Mole fraction of     [tex]N_{KCl}= \frac{n_{KCl}}{n_{KCl} + n_{H_2O}}[/tex]

                                        [tex]= \frac{1.676}{15.28 +1.676}[/tex]

                                        [tex]N_{KCl}=0.099[/tex]

Answer:

The mole fraction of KCl is 0.13

Explanation:

no of moles of KCl (n KCl)    = W/G.F.Wt

                                                = 135/74.6  

                                                = 1.81moles

no of moles of H2O (nH2O) = W/G.F.Wt

                                              = 225/18   = 12.5 moles

mole fraction of KCl ( XKCl)   = nKCl/nKCl + nH2O

                                                = 1.81/(1.81+12.5)

                                                 = 1.81/14.31  

      The mole fraction of KCl is   = 0.13

Answer:

The time required for the coating is 105 s

Explanation:

Zinc undergoes reduction reaction and absorbs two (2) electron ions.

The expression for the mass change at electrode [tex](m_{ch})[/tex] is given as :

[tex]\frac{m_{ch}}{M} ZF = It[/tex]

where;

M = molar mass

Z = ions charge at electrodes

F = Faraday's constant

I = current

A = area

t = time

also; [tex](m_{ch})[/tex] = [tex](Ad) \rho[/tex] ; replacing that into above equation; we have:

[tex]\frac{(Ad) \rho}{M} ZF = It[/tex]  ---- equation (1)

where;

A = area

d = thickness

[tex]\rho[/tex] = density

From the above equation (1); The time required for coating can be calculated as;

[tex][ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons } ] = (20 A) t[/tex]

[tex]t = \frac{2100}{20}[/tex]

= 105 s

Answer:

A. 11.26

B. 12.37

Explanation:

A. Step 1:

Dissociation of Mg(OH)2. This is illustrated below below:

Mg(OH)2 <==> Mg2+ + 2OH-

A. Step 2:

Determination of the concentration of the OH-

From the above equation,

1 mole of Mg(OH)2 produce 2 moles of OH-

Therefore, 9.01x10^-4 M Mg(OH)2 will produce = 9.01x10^-4 x 2 = 1.802x10^-3 M of OH-

A. Step 3:

Determination of the pOH. This is illustrated below:

pOH = - Log [OH-]

[OH-] = 1.802x10^-3 M

pOH = - Log [OH-]

pOH = - Log 1.802x10^-3

pOH = 2.74

A. Step 4:

Determination of the pH.

pH + pOH = 14

pOH = 2.74

pH + 2.74 = 14

Collect like terms

pH = 14 - 2.74

pH = 11.26

B. Step 1:

Dissociation of NH4OH. This is illustrated below below:

NH4OH <==> NH4+ + OH-

B. Step 2:

Determination of the concentration of the OH-

From the above equation,

1 mole of NH4OH produce 1 moles of OH-

Therefore, 2.33x10^-2M of NH4OH will also produce 2.33x10^-2M of OH-

B. Step 3:

Determination of the pOH. This is illustrated below:

pOH = - Log [OH-]

[OH-] = 2.33x10^-2M

pOH = - Log [OH-]

pOH = - Log 2.33x10^-2M

pOH = 1.63

B. Step 4:

Determination of the pH.

pH + pOH = 14

pOH = 1.63

pH + 1.63 = 14

Collect like terms

pH = 14 - 1.63

pH = 12.37

Answer : The molarity of KF solution is, 1 M

Explanation : Given,

Mass of [tex]KF[/tex] = 116 g

Volume of solution = 2 L

Molar mass of [tex]KF[/tex] = 58 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

[tex]\text{Molarity}=\frac{\text{Mass of }KF}{\text{Molar mass of }KF\times \text{Volume of solution (in L)}}[/tex]

Now put all the given values in this formula, we get:

[tex]\text{Molarity}=\frac{116g}{58g/mole\times 2L}=1mole/L=1M[/tex]

Therefore, the molarity of KF solution is, 1 M

Answer:

3,5-dimethyl-2-octene

Explanation:

Please note that there is no H at carbon 3 less carbon becomes penta hydra.

The compound is:

CH3

I

C = CH2 - CH3

I

CH2

I

CH - CH3

I

CH2

I

CH2

I

CH3

To name the above compound, do the following:

1. Locate the longest continuous chain i.e octene

2. Start counting from the side that gives the double bond the lowest low count since the double bond is the functional group. In doing this, the double bond is at carbon 2.

3. Locate the substituent groups attached and their position in the parent chain. In doing so, you will see that there are two CH3 group attached and they are at carbon 3 and carbon 5. Since the substituents attached are the same, we'll name them as 'dimethyl' indicate that they are two methyl groups

Now, we'll combine the above findings in order to obtain the name. Therefore, the name of the compound is:

3,5-dimethyl-2-octene

Answer:

First, Second, Second-to-last, and last choice

Explanation:

Chemistry student. The other options are incorrect. If you would like a more thorough explanation, please reply to this comment.

pH according to Arrhenius definition is the measure of hydrogen ion concentration in solution:

[tex]pH=-log[H^{+}][/tex] Acids release H+ ions into solution, while bases release OH- ions into solution. This explains why second-to-last choice is correct.

Litmus paper are thin strips of paper that have been manufactured with indicators, examples being red cabbage or phenolphthalein. Indicators react to changes in pH with an according color change. This explains why the last choice is also correct.  

Acid strength is better the lower the number it is, hence why pH 2 is stronger acid than 5, but in turn is a weaker base. This explains why second choice is correct.

The first option can be explained by the pH diagram. HCl is acidic, and will turn blue (basic) litmus paper more acidic, and move towards an acidic pH, hence more "red."

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thanks,

Answer:

I dont know if im correct number 1 is 2

Explanation:

ps im not good at this subject

Answer:

Mole Fraction of O2 --> 0.42

Mole Fraction of Ar --> 0.037

Explanation:

Answer:

Mole Fraction of O2 --> 0.42

Mole Fraction of Ar --> 0.037

Explanation:

Answer:

Explanation:

find the solution below

The organic compounds carboxylic acid, amine, and alcohol contain different numbers of sigma and pi bonds resulting from the overlapping of atomic orbitals. In these provided examples, the carboxylic acid contains nine sigma and two pi bonds, the amine has twelve sigma and two pi bonds, and the alcohol has nine sigma and 0 pi bonds.

Sigma

and

pi bonds

occur in various types of organic molecules, including carboxylic acids, amines, and alcohols. A sigma (σ) bond is formed by the head-on overlapping of atomic orbitals, whereas a pi (π) bond is formed by the lateral overlap of two p orbitals. In the carboxylic acid H2CCHCH2COOH, there are nine sigma and two pi bonds. As for the amine HCCCH2CHCHCH2NH2, there are twelve sigma and two pi bonds. Lastly, an alcohol molecule, such as CH3OH, consists of 9 sigma bonds and 0 pi bonds. It's essential to note that a double bond contains one sigma and one pi bond, while a triple bond contains one sigma and two pi bonds.

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Answer:The Zinc Reacts With The Hydrochloric Acid Producing Zinc Chloride And Hydrogen Gas, And Leaving The Copper Behind. A. If 25.0 G Of Zinc ... Zn+ 2 HCI --> ZnCl2 + H2 (answer .771 G H2) B. If The Reaction Yields . ... If 25.0 g of zinc are in a sample of bronze, determine the theoretical yield of hydrogen gas. Zn+ 2 HCI

Answer:

Lemon

HCI

Blood

Saliva

Bleach

NaOH

Explanation:

Blood 7.35-7.45

Bleach 12.6

Saliva 6.2-7.6

Lemon 2-3

HCI 3.01

NaOH 13

Answer:

31.5mL

Explanation:

The following were obtained from the question:

C1 (concentration of stock solution) = 2M

V1 (volume of stock solution) =.?

C2 (concentration of diluted solution) = 0.630M

V2 (volume of diluted solution) = 100mL

Using the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:

C1V1 = C2V2

2 x V1 = 0.630 x 100

Divide both side by 2

V1 = (0.630 x 100) /2

V1 = 31.5mL

Therefore, 31.5mL of 2M solution of FeCl2 required

Answer:

We need 31.5 mL of the 2.0 M FeCl2 solution

Explanation:

Step 1: Data given

Molarity of a FeCl2 solution = 2.0 M

Initial volume of FeCl2 = 100 mL

Initial molarity of FeCl2 = 0.630 M

Step 2: Calculate volume of the stock solution

C1V1 = C2V2

⇒with C1 = the initial molarity FeCl2 = 0.630 M

⇒with V1 = the initial volume = 100 mL = 0.100 L

⇒with C2 = the new molarity FeCl2 = 2.0 M

⇒with V2 = the new volume = TO BE DETERMINED

0.630M * 0.100 L = 2.0 M * V2

V2 = (0.630 * 0.100) / 2.0

V2 = 0.0315 L = 31.5 mL

We need 31.5 mL of the 2.0 M FeCl2 solution

Answer:

i have an answer but i can only show you because my teacher helped my on it and wrote it down for me to remember! hope this helps!!!

Explanation:

A 1.00 g sample ofn-hexane (C6H14) under-goes complete combustion with excess O2ina bomb calorimeter. The temperature of the1502 g of water surrounding the bomb risesfrom 22.64◦C to 29.30◦C. The heat capacityof the hardware component of the calorimeter(everything that is not water) is 4042 J/◦C.What is ΔUfor the combustion ofn-C6H14?One mole ofn-C6H14is 86.1 g.The specificheat of water is 4.184 J/g·◦C.1.-9.96×103kJ/mol2.-7.40×104kJ/mol3.-1.15×104kJ/mol4.-4.52×103kJ/mol5.-5.92×103kJ/molcorrectExplanation:mC6H8= 1.00 gmwater= 1502 gSH = 4.184 J/g·◦CHC = 4042 J/◦CΔT= 29.30◦C-22.64◦C = 6.66◦CThe increase in the water temperature is29.30◦C-22.64◦C = 6.66◦C. The amount ofheat responsible for this increase in tempera-ture for 1502 g of water isq= (6.66◦C)parenleftbigg4.184Jg·◦Cparenrightbigg(1502 g)= 41854 J = 41.85 kJThe amount of heat responsible for the warm-ing of the calorimeter isq= (6.66◦C)(4042 J/◦C)= 26920 J = 26.92 kJ

Answer:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Explanation:

Step 1: Data given

gaseous methane = CH4(g)

Combustion reaction is adding O2. The products will be carbondioxide (CO2) and water vapor (H2O)

Step 2: The unbalanced equation

CH4(g) + O2(g) → CO2(g) + H2O(g)

Step 3: Balancing the equation

CH4(g) + O2(g) → CO2(g) + H2O(g)

On the left side we have 4x H (in CH4), on the right side we have 2x H (in H2O). To balance the amount H on both sides, we have to multiply H2O by 2.

CH4(g) + O2(g) → CO2(g) + 2H2O(g)

On the left side we have 2x O (in O2), on the right side we have 4x O (2x in CO2 and 2x in 2H2O). To balance the amount of O on both sides, we have to multiply O2 on the left side, by 2. Now the equation is balanced.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

The balanced chemical equation for the combustion of gaseous methane (CH₄) with gaseous oxygen (O₂) to form gaseous carbon dioxide (CO₂) and gaseous water (H₂O) is:

CH₄ + 2O₂ → CO₂ + 2H₂O

The combustion of gaseous methane (CH₄) with gaseous oxygen (O₂) to form gaseous carbon dioxide (CO₂) and gaseous water (H₂O) is a fundamental chemical reaction that occurs in natural gas combustion and many other combustion processes. To write a balanced chemical equation for this reaction, we must ensure that the number of atoms of each element on both sides of the equation is the same.

The unbalanced equation for the combustion of methane is:

CH₄ + O₂ → CO₂ + H₂O

Now, let's balance the equation:

Balance the carbon (C) atoms:

There is one carbon atom on the left and one on the right, so carbon is already balanced.

Balance the hydrogen (H) atoms:

There are four hydrogen atoms on the left (in CH₄) and two on the right (in H₂O). To balance hydrogen, we need to place a coefficient of 2 in front of H₂O on the right side.

CH₄ + O₂ → CO₂ + 2H₂O

Balance the oxygen (O) atoms:

On the left side, there are two oxygen atoms in CH₄ and two in O₂, making a total of four oxygen atoms. On the right side, there are two oxygen atoms in CO₂ and four in 2H₂O, making a total of six oxygen atoms. To balance the oxygen atoms, we need to adjust the coefficient of O₂ on the left side.

CH₄ + 2O₂ → CO₂ + 2H₂O

Now, the equation is balanced with an equal number of atoms of each element on both sides:

CH₄ + 2O₂ → CO₂ + 2H₂O

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Answer:

pH → 1.13

Explanation:

Our solution is pure HCl

HCl(aq) + H₂O(l)  →  H₃O⁺(aq) + Cl⁻(aq)

As a strong acid, it is completely dissociated.

1 mol of HCl, can give 1 mol of H⁺ to the medium. Water does not participate. Let's find out M for the acid.

1st step: We convert the mass from mg to g → 327 mg . 1g /1000mg = 0.327 g

2nd step: We convert the mass(g) to moles: 0.327 g / 36.45 g/mol = 8.97×10⁻³ moles

3rd step: We convert the volume from mL to L → 120mL . 1L /1000 mL = 0.120L

Molarity (mol/L) = 8.97×10⁻³ mol / 0.120L = 0.075M

We propose: HCl(aq) + H₂O(l)  →  H₃O⁺(aq) + Cl⁻(aq)

                     0.075M                    0.075M

pH = - log [H₃O⁺] → - log 0.075 = 1.13 → pH

The mass of AgI in the precipitate is approximately 38.65 grams (rounded to two significant figures) with the appropriate units.

To find the mass of AgI in the precipitate, we first need to determine the moles of AgI formed. We'll use the information provided and follow these steps:

1. Write the balanced chemical equation for the precipitation reaction:

[tex]\[ \text{Ag}^+ + \text{I}^- \rightarrow \text{AgI} \][/tex]

[tex]\[ \text{Hg}_2^{2+} + 2\text{I}^- \rightarrow \text{HgI}_2 \][/tex]

2. Determine the limiting reactant:

  - For AgI: [tex]\( \text{Ag}^+ + \text{I}^- \)[/tex] (1 mole of Ag per mole of I)

  - For HgI2: [tex]\( \frac{1}{2}\text{Hg}_2^{2+} + \text{I}^- \)[/tex] (1 mole of Hg per 2 moles of I)

  The limiting reactant is the one that produces the fewer moles of I^-, as it determines the amount of AgI formed.

  [tex]\[ \text{moles of I}^- = 0.100 \, \text{L} \times 1.71 \, \text{mol/L} = 0.171 \, \text{mol} \][/tex]

  The limiting reactant is AgI, as it requires 0.171 moles of I^-, while HgI2 would require 0.342 moles of I^-.

3. Calculate the moles of AgI formed:

  [tex]\[ \text{moles of AgI} = \text{moles of I}^- \times \frac{1 \, \text{mol AgI}}{1 \, \text{mol I}^-} = 0.171 \, \text{mol} \][/tex]

4. Calculate the mass of AgI formed:

  [tex]\[ \text{mass of AgI} = \text{moles of AgI} \times \text{molar mass of AgI} \][/tex]

  The molar mass of AgI is the sum of the atomic masses of Ag (107.87 g/mol) and I (126.904 g/mol).

  [tex]\[ \text{mass of AgI} = 0.171 \, \text{mol} \times (107.87 + 126.904) \, \text{g/mol} \][/tex]

  [tex]\[ \text{mass of AgI} \approx 38.65 \, \text{g} \][/tex]

Therefore, the mass of AgI in the precipitate is approximately 38.65 grams (rounded to two significant figures) with the appropriate units.

Answer:

992.302 K

Explanation:

V(rms) = 750 m/s

V(rms) = √(3RT / M)

V = velocity of the gas

R = ideal gas constant = 8.314 J/mol.K

T = temperature of the gas

M = molar mass of the gas

Molar mass of CO₂ = [12 + (16*2)] = 12+32 = 44g/mol

Molar mass = 0.044kg/mol

From

½ M*V² = 3 / 2 RT

MV² = 3RT

K = constant

V² = 3RT / M

V = √(3RT / M)

So, from V = √(3RT / M)

V² = 3RT / M

V² * M = 3RT

T = (V² * M) / 3R

T = (750² * 0.044) / 3 * 8.314

T = 24750000 / 24.942

T = 992.302K

The temperature of the gas is 992.302K

Note : molar mass of the gas was converted from g/mol to kg/mol so the value can change depending on whichever one you use.

The temperature of a gas can be calculated from the root-mean-square speed of its molecules using the kinetic theory of gases. For CO2 gas with the rms speed of 750 m/s, the derived temperature is approximately 485.4 K.

The temperature of a gas can be calculated from the root-mean-square (rms) speed of its molecules using the kinetic theory of gases. This theory develops a relationship between the average kinetic energy of the gas molecules and the temperature of the gas through the equation K = 3/2kBT = mv²/2, where kB is Boltzmann’s constant, T is the temperature, m is the mass of a gas molecule, and v is the rms speed. Thus, temperature can be derived as T = mv² / (3kB).

For CO2, the molar mass is 0.044 kg/mol. Knowing that the number of molecules (n) is the number of moles times Avogadro's number (6.022 × 10²³), we derive the molecular mass m = 0.044 kg/mol / (6.022 × 10²³) = 7.3 × 10^-26 kg. Plugging in the values for v (750 m/s), m, and kB (1.38 × 10^-23 J/K), we get an approximate temperature of 485.4 K.

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Answer:

5.0 %

Explanation:

Given data

Step 1: Calculate the volume of solution

If we assume that the volumes are additive, the volume of the solution is equal to the sum of the volume of the solute and the solvent.

V = 50 mL + 950 mL = 1000 mL

Step 2: Calculate the percent by volume

We will use the following expression.

[tex]\% v/v = \frac{volume\ of\ solute}{volume\ of\ solution} \times 100 \% = \frac{50mL}{1000mL} \times 100 \% = 5.0\%[/tex]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

A

The percentage of water of hydration is   [tex]P_h= 11.01[/tex]%

Mass of [tex]Fe^{3+}[/tex] in 100mg is 10.60mg

Moles of  [tex]Fe^{3+}[/tex] in 100mg is [tex]n_i= 0.19[/tex]

mol / mol Fe (3 sig figs) is [tex]= 1.00[/tex]

mol / mol Fe (whole number) is = 1

B

Mass of [tex]K^{+}[/tex] in 100mg is 27.70mg

Moles of  [tex]K^{+}[/tex] in 100mg is [tex]n_i= 0.581 moles[/tex]

mol  of K / mol of Fe (3 sig figs) is [tex]= 3.05[/tex]

mol  of K / mol of Fe (whole number) is [tex]=3[/tex]

C

Mass of [tex]C_2O_4^{-2}[/tex] in 100mg is 55.69 mg

Moles of [tex]C_2O_4^{-2}[/tex]  in 100mg is [tex]n_i= 0.633 moles[/tex]

mol  of  [tex]C_2O_4^{-2}[/tex] / mol of Fe (3 sig figs) is [tex]= 3.33[/tex]

mol  of [tex]C_2O_4^{-2}[/tex] / mol of Fe (whole number) is [tex]=3[/tex]

D

Mass of water in 100mg is 11.01 mg

Moles of water  in 100mg is [tex]n_i= 0.611 moles[/tex]

mol  of  water / mol of Fe (3 sig figs) is [tex]= 3.21[/tex]

mol  of water / mol of Fe (whole number) is [tex]=3[/tex]

Explanation:

The percentage of water of hydration is mathematically represented as

        [tex]P_h = 100 - (Pi + P_p + P_o)[/tex]

Now substituting 10.60% for [tex]P_i[/tex] (percentage of iron ) , 22.70% for [tex]P_p[/tex](Percentage of potassium) , 55.69% for [tex]P_o[/tex] (percentage of Oxlate)

        [tex]P_h =100 - (10.60 + 22.70+55.69)[/tex]

             [tex]P_h= 11.01[/tex]%

For IRON

Since the percentage of [tex]Fe^{3+}[/tex] is 10.60% then in a 100 mg of the sample the amount of [tex]Fe^{3+}[/tex] would be 10.60 mg

  Now the no of moles is mathematically denoted as

            [tex]n = \frac{mass}{molar \ mass }[/tex]

The molar mass of [tex]Fe[/tex] is  55.485 g/mol

     So the number of moles of [tex]Fe^{3+}[/tex] in 100mg of he sample is

                  [tex]n_i = \frac{10.60}{55.485}[/tex]

                       [tex]n_i= 0.19[/tex]

mol / mol Fe (3 sig figs) is [tex]= \frac{0.19}{0.19} = 1.00[/tex]

FOR POTASSIUM

Since the percentage of [tex]K^{+}[/tex] is 22.70% then in a 100mg of the sample the amount of [tex]K^{+}[/tex] would be 22.70mg

The molar mass of [tex]K[/tex] is  39.1 g/mol

   So the number of moles of [tex]K^{+}[/tex] in 100mg of he sample is

                  [tex]n_i = \frac{22.70}{39.1}[/tex]

                      [tex]=0.581 moles[/tex]

mol  of K / mol of Fe (3 sig figs) is [tex]= \frac{0.581}{0.19} = 3.05[/tex]

FOR OXILATE [tex]C_2O_4^{-2}[/tex]

Since the percentage of [tex]C_2O_4^{-2}[/tex]  is 55.69% then in a 100mg of the sample the amount of [tex]C_2O_4^{-2}[/tex] would be 55.69 mg

The molar mass of [tex]C_2O_4^{-2}[/tex] is  88.02 g/mol

  So the number of moles of [tex]C_2O_4^{-2}[/tex]   in 100mg of he sample is

                  [tex]n_i = \frac{55.69}{88.02}[/tex]

                         [tex]=0.633 moles[/tex]

mol  of [tex]C_2O_4^{-2}[/tex] / mol of Fe (3 sig figs) is  [tex]= \frac{0.633}{0.19} = 3.33[/tex]

FOR WATER OF HYDRATION

        Since the percentage of water  is 11.01% then in a 100mg of the sample the amount of water would be 11.0 mg

   The molar mass of water  is  18.0 g/mol

  So the number of moles of water   in 100mg of he sample is

                  [tex]n_i = \frac{11.01}{18.0}[/tex]

                      [tex]=0.611 moles[/tex]

mol  of  water / mol of Fe (3 sig figs) is  [tex]= \frac{0.611}{0.19} = 3.21[/tex]

The equilibrium constant, Kc, for the given reaction can be calculated using the solubility product constant, Ksp, and the overall formation constant, Kf. Substituting the given values, we find that Kc is equal to 3.186 × 10⁻⁵.

The equilibrium constant, Kc, for the reaction AgCl (s) + Cl⁻ (aq) ⇌ AgCl⁻₂ (aq) can be calculated using the solubility product constant, Ksp, for AgCl and the overall formation constant, Kf, for AgCl⁻₂.

The equilibrium constant is given by the product of Ksp and Kf:

Kc = Ksp * Kf

Substituting the values given, Ksp = 1.77 × 10⁻¹⁰ and Kf = 1.8 × 10⁵, we can calculate the value of Kc:

Kc = (1.77 × 10⁻¹⁰) * (1.8 × 10⁵) = 3.186 × 10⁻⁵

Answer:

Water of crystallization, X = 4.

Explanation:

Molar mass of [tex]CuSO_{4}.XH_{2} O[/tex]

64 + 32 + (4x18) + x ( 1 × 2 + 16)

= 160 + 18x

Given: % water of crystallization (decrease in mass after heating) = 30%

⇒ [tex]\frac{18x}{160 + 18x} =\frac{31}{100}[/tex]

1800x = 31 (160 + 18x)

58.0645x = 160 + 18x

(58.0645 - 18)x = 160

x = [tex]\frac{160}{40.0645}[/tex] = 3.99 ≅ 4.

Water of crystallization, X = 4.

Answer: The volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L

Explanation:

According to ideal gas equation:

[tex]PV=nRT[/tex]

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = ?

n = number of moles = 0.684

R = gas constant = [tex]0.0821Latm/Kmol[/tex]

T =temperature =[tex]273K[/tex]   (at STP)

[tex]V=\frac{nRT}{P}[/tex]

[tex]V=\frac{0.684\times 0.0821L atm/K mol\times 273K}{1atm}[/tex]

[tex]V=15.3L[/tex]

Thus the volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L

Answer:

The stopper is not fitted on the flask quickly.

Explanation:

The reaction has to do with a gas. The predicted value of the slope is 1. The value of slope obtained from the experiment varies slightly from the predicted value of 1. This discrepancy may be caused by not fitting the stopper on the flask quickly enough. This means that some gas may still be left in the flask leading a discrepancy in the slope obtained.