Calculate the [H+] in a solution that is 0.803 M in NaX and 0.677 M in HX given that the Ka of HX is 8.64 ⋅ 10 − 7 8.64⋅10-7. Report your answer in scientific notation to 3 sig figs.

Answer:

The correct answer is B. 734 g

Explanation:

The chemical formulae of lithium sulfate is Li₂SO₄. With this, we can calculate the molecular weight (MM) of lithium sulfate as follows:

MM(Li₂SO₄) = (2 x molar mass Li) + molar mass S + (4 x molar mass O)

                   = (2 x 6.9 g/mol) + 32 g/mol + (4 x 16 g/mol)

                   = 109.9 g/mol

We need to prepare a solution with a molarity of 2.67 M. That means that the solution has to have 2.67 moles of Li₂SO₄ per liter of solution. We can convert from mol to grams with the calculated molecular weight and then we have to multiply by the volume of solution (2500 ml= 2.5 L), as follows:

Mass = 2.67 mol/L x 109.9 g/mol x 2.5 L ≅ 734 g

Answer:

2-hexanone

Explanation:

First, we'll begin by:

1. Locating the longest continuous chain i.e haxane

2. Determine the functional group in the compound. The functional group in ketone (C=O). This changes the name from hexane to hexanone i.e replacing the -e in at the end in hexane with -one to make it hexanone.

3. Give the functional group the lowest low count. In doing this, we'll start counting from the left. The functional group is at carbon 2.

Note: no substitute group is attached.

Now, in naming the compound, we must indicate the position of functional group as illustrated below:

The name of the compound is:

2-hexanone

The compound is called 2-hexanone. This is because there is a six carbon chain with a ketone functional group attached to the second carbon.

You are dealing with an organic compound formula. Let's identify the compound based on its structure. The compound you have given is CH3 - C - CH2 - CH2 - CH2 - CH3. This is a 6 carbon chain and there is double bond between Oxygen and the second carbon. Now, when naming compounds in chemistry we firstly count the length of the chain, it is hexane. Then, we look for any functional group and where it is attached in the chain. A ketone functional group is CO, so this would make it a hexanone. As the CO is attached at the second carbon, it is 2-hexanone. So the correct answer for the name of the compound would be 2-hexanone.

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Answer:

Check the explanation

Explanation:

Going by the question above for parta, the feed flow rate is stated to be 9072g/hr.& result is offered for part a considering the same. if the unit is kg/hr, then kindly let me know & i shall make slight modification that might be required in solution which in that case, area & steam consumption answer shall differ.

In the b part of the question, reduced flow rate is higher than the initial flow rate therefore same has been considered in gram /hr.

Kindly check the attached images below to see the step by step solution to the above question.

In this exercise we have to use the knowledge of mechanics to calculate the steamused of the solution, in this way we will have to:

a) [tex]M_S=\frac{H_S}{\lambda_S} = 8.43 KG/h[/tex] , [tex]A= 0.0526 m^2[/tex]

b) We can assume that all water shall evaporate and produd shell contain solid only.

So from the data informed in the exercise we have that:

So apply overall mass balance, will be:

[tex]F=L+V\\9072=1814.4-V\\V=7.2576 kg/h[/tex]

Now, for enthalpy balance we shall make assumption that, all liquid phase enthalphy is that heet capacity of water is constant, so the data will be:

So, find the enthalpy of liquid, we have;

[tex]H_L= L*C_P*(T_P-T_F)\\=(1.8144)(4.18)(60.0586-15.6)\\H_v=V*(P(t_P-T_F)+\lambda)\\=(7.2576)(4.18(60.0586-15.6)+2357.5477))H_s=H_L+H_V\\H_S= (1.8144)(4.18)(60.0586-15.6)+(7.2576)(4.18(60.0586-15.6)+2357.5477))\\=18796.051 KJ/h[/tex]

Now, we can find the mass flow rate, that will be;

[tex]M_S=\frac{H_S}{\lambda_S} = 8.43 KG/h[/tex]

Now we can find the area of evaporator, that will be:

[tex]H_S=UA=(T_S-T_P)\\A=0.0526m^2[/tex]

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Answer:

Explanation:

Attach is the solution

Final answer:

The formation of an imine involves nucleophilic addition, carbinolamine formation, and subsequent deprotonations leading to the imine, which is then reduced to an amine by sodium borohydride through nucleophilic attack.

Explanation:

Arrow-pushing Mechanism for Imine and Amine Formation

The question involves describing the arrow-pushing mechanisms for the formation of an imine from an aldehyde and ammonia, followed by the conversion of the imine to an amine using sodium borohydride. Let's address this step by step.

Step 1: Formation of Imine

The formation of an imine involves several key steps:

Nucleophilic addition of the amine to the carbonyl carbon of the aldehyde.Formation of carbinolamine via proton transfer.Protonation of carbinolamine oxygen turns it into a better leaving group, facilitating its elimination as water and resulting in the formation of an iminium ion.Finally, deprotonation of the nitrogen atom yields the imine.

Step 2: Conversion of Imine to Amine

In the next step, sodium borohydride (NaBH4) is used as a reducing agent to convert the imine into an amine. This involves the nucleophilic attack of hydride (H:-) from NaBH4 on the carbon atom of the iminium ion, leading to the formation of an amine.

Answer:

calories burned = 202 calories

Explanation:

22 minutes = 22/60th of an hour => calories burned in 22 min = (22/60)550 calories = 202 calories.

Answer:

202 calories is the estimate of calories burned

Explanation:

22 minutes = 22/60th of an hour => calories burned in 22 min = (22/60)550 calories = 202 calories.

Explanation:

10mil 76 the experllon is the most numbers

Answer:

Ater adding 10.0 mL the pH = 4.74

After adding 20.0 mL the pH = 8.75

After adding 30.0 mL the pH = 12.36

Explanation:

Step 1: Data given

Volume of a 0.100 M acetic acid solution = 25.0 mL

Molarity of NaOH solution = 0.125 M

A) 25.0mL of HAc(acetic acid) and 10.0mL of NaOH

Calculate moles

Equation:   NaOH    +   HAc    -->   NaAc + H20

Moles HAc =  0.025 L*  0.100 mol /L = 0.00250 moles

Moles NaOH = 0.01 L * 0.125mol/L = 0.00125 moles

Initial moles

NaOH = 0.00125 moles

HAc = 0.00250 moles

NaAC = 0 moles

Moles at the equilibrium

NaOH = 0.00125 - 0.00125 = 0 moles

HAc = 0.00250 - 0.00125 = 0.00125 moles

NaAC = 0.00125 moles

Calculate molarity

[HAc] = [Ac-] =  0.00125 moles / 0.035 L = 0.0357M

Concentration at equilibrium

HAc   <--> Ac-     +  H+         Ka = 1.76 * 10^-5

[HAc] = 0.0357 - x M

[Ac-] = 0.0357 + X M

[H+] = XM

Ka=1.76*10^-5 = [Ac-][H+]/[HAc] = [0.00357+x][x]/[0.00357-x]

Ka=1.76*10^-5 = [0.00357][x]/[0.0357] usually good approx.

Ka = 1.76* 10-5 = x = [H+]

pH = -log [H+] = -log [1.76*10^-5] = 4.74

B) 25.0mL HAc and 20.0mL NaOH

Calculate moles

Equation:   NaOH    +   HAc    -->   NaAc + H20

Moles HAc =  0.025 L*  0.100 mol /L = 0.00250 moles

Moles NaOH = 0.02 L * 0.125mol/L = 0.00250 moles

Initial moles

NaOH = 0.00250 moles

HAc = 0.00250 moles

NaAC = 0 moles

Moles at the equilibrium

NaOH = 0.00250 - 0.00250 = 0 moles

HAc = 0.00250 - 0.00250 = 0 moles

NaAC = 0.00250 moles

Calculate molarity

[NaAC] =  0.00250 moles / 0.045 L = 0.0556M

Concentration at equilibrium

Ac- + H20   <--> HAc     +  OH-        Ka = 1.76 * 10^-5

[Ac] = 0.0556 - x M

[OH-] = XM

[HAc] = XM

**Kw = 1*10^-14 at STP I'll assume this for this problem

**Kw = Ka*Kb ⇒ Kb = Kw/Ka

**Kb = (1*10^-14)/(1.76*10^-5)= 5.68*10^-10  

Kb = ([HAc][OH-])/[Ac-] =  ([x][x])/[0.0556]

Kb =  x² /0.0556

5.68*10^-10 = x² /0.0556

so x =5.62*10^-6

pOH = -log [OH-] = 5.250

pH = 14 - pOH

pH 14 - 5.25 = 8.75

C)25.0mL HAc and 30.0mL NaOH

NaOH    +   HAc    -->   NaAc + H20

Initial numbers of moles

Moles NaOH = 0.125M * 0.03 L = 0.00375 moles

Moles HAc =  0.00250 moles

moles NaAC = 0 moles

Moles at the equilibrium

Moles NaOH = 0.00375 moles - 0.00250 = 0.00125 moles

Moles HAc =  0.00250 moles - 0.00250 = 0 moles

moles NaAC = 0.00250 moles

After the reaction there is some NaOH left over so it is the only thing that matters for the pH as it is a stong base.

Calculate NaOH molarity

(0.00125moles/0.055 L= 0.0227M NaOH

Strong Bases dissociate 100% so [OH-] = 0.0227M

pOH= -log[0.0227] = 1.644

pH = 14-pOH = 12.36

Answer:

C

Explanation: you have to observe to have an educated guess about something you test it

Answer:

44.63g

Explanation:

First, let us calculate the number of mole of KBr in 1.50M KBr solution.

This is illustrated below:

Data obtained from the question include:

Volume of solution = 250mL = 250/1000 = 0.25L

Molarity of solution = 1.50M

Mole of solute (KBr) =.?

Molarity is simply mole of solute per unit litre of solution

Molarity = mole /Volume

Mole = Molarity x Volume

Mole of solute (KBr) = 1.50 x 0.25

Mole of solute (KBr) = 0.375 mole

Now, we calculate the mass of KBr needed to make the solution as follow:

Molar Mass of KBr = 39 + 80 = 119g/mol

Mole of KBr = 0.375 mole

Mass of KBr =?

Mass = number of mole x molar Mass

Mass of KBr = 0.375 x 119

Mass of KBr = 44.63g

Therefore, 44.63g of KBr is needed to make 250.0mL of 1.50 M potassium bromide (KBr) solution

Final answer:

To make a 1.50 M potassium bromide solution with a volume of 250.0 mL, you would need approximately 44.625 grams of potassium bromide.

Explanation:

To calculate the mass of potassium bromide required, we can use the formula:

mass (g) = concentration (M) x volume (L) x molar mass (g/mol)

The concentration is given as 1.50 M, the volume is 250.0 mL (which can be converted to 0.250 L), and the molar mass of potassium bromide (KBr) is 119.0 g/mol. Plugging in these values, we get:

mass (g) = 1.50 M x 0.250 L x 119.0 g/mol = 44.625 g

Therefore, you would need approximately 44.625 grams of potassium bromide to make 250.0 mL of a 1.50 M potassium bromide solution.

Answer:

1. C9H20(l) + 14O2(g) —> 9CO2(g) + 10H2O(g)

2. 213.70L of CO2

Explanation:

1. The equation for the combustion of liquid nonane into gaseous carbon dioxide and gaseous water is given below:

C9H20(l) + O2(g) —> CO2(g) + H2O(g)

The equation can be balanced as follow:

There are 9 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 9 in front of CO2 as shown below:

C9H20(l) + O2(g) —> 9CO2(g) + H2O(g)

There are 20 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 10 in front of H2O as shown below:

C9H20(l) + O2(g) —> 9CO2(g) + 10H2O(g)

Now, there are a total of 28 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 14 in front of O2 as shown below:

C9H20(l) + 14O2(g) —> 9CO2(g) + 10H2O(g)

Now the equation is balanced.

2. Let us convert 0.130 kg of nonane (C9H20) to mole. This is illustrated below:

Molar Mass of C9H20 = (12x9) + (20x1) = 108 + 20 = 128g/mol

Mass of C9H20 from the question =

0.130kg = 0.130 x 1000 = 130g

Mole of C9H20 =?

Number of mole = Mass/Molar Mass

Mole of C9H20 = 130/128 = 1.016 mole

The equation for the reaction is:

C9H20(l) + 14O2(g) —> 9CO2(g) + 10H2O(g)

From the balanced equation above,

1 mole of C9H20 produced 9 moles of CO2.

Therefore, 1.016 mole of C9H20 will produce = 1.016 x 9 = 9.144 moles

Now, let us calculate the volume of CO2 formed. This is illustrated below:

Data obtained from the question include:

P (pressure) = 1atm

T (temperature) = 12°C = 285K

n (number of mole of CO2) = 9.144 moles

R (gas constant) = 0.082atm.L/Kmol

V (volume of CO2) =?

Using the ideal gas equation PV = nRT, the volume of CO2 can be obtained as follow:

PV = nRT

1 x V = 9.144 x 0.082 x 285

V = 213.70L

Therefore, 213.70L of CO2 is produced

The combustion of nonane, C9H20(l) + 14O2(g) → 9CO2(g) + 10H2O(g), produces carbon dioxide and water. Using stoichiometry and the ideal gas law, the volume of carbon dioxide produced from 0.130 kg of nonane is calculated to be approximately 216.5 L.

The first step is to write a balanced chemical equation for the combustion of nonane. The complete combustion of hydrocarbons, like nonane, in the presence of oxygen (O2) yields carbon dioxide (CO2) and water (H2O). The correct balanced equation is:

C9H20(l) + 14O2(g) → 9CO2(g) + 10H2O(g).

To calculate the volume of carbon dioxide produced from the amount of nonane combusted, one must apply the concepts of stoichiometry and gas laws. According to the equation, 9 moles of CO2 is produced from 1 mole of nonane. As the molar mass of nonane is 128.26 g/mol, 0.130 kg or 130 g of nonane is approximately 1.013 moles. Therefore, we will have around 9.117 moles of carbon dioxide produced.

Applying the ideal gas law, PV = nRT, at standard temperature (12°C = 285.15 K) and pressure (1 atm), and R = 0.0821 L.atm / K.mol, we can calculate the volume as V = nRT/P. The volume of CO2 produced would be approximately 216.5 L, with correct significant digits.

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Answer:

The Ksp at 25°C is 1.6 * 10^-5

Explanation:

Step 1: Data given

Temperature = 25°C

Solubility of lead(II) chloride = 1.59 * 10^-2 mol/L

Step 2: The balanced equation

PbCl2(s) <===> Pb2+(aq) + 2Cl-(aq)

Step 3: Calculate the Ksp

Ksp = [Pb2+][Cl-]²

Ksp = [Pb2+][Cl-]²

[Pb2+] = 1.59 *10-2  = 0.0159 M

[Br-] = 2 x 1.59*10-2 = 3.18 *10-2 M

Ksp = (1.59*10-2)(3.18*10-2)²

Ksp =1.6 * 10^-5

The Ksp at 25°C is 1.6 * 10^-5

The value of solubility constant (Ksp) at 25°C in scientific notation is 1.6 × 10-⁵.

According to this question, the following parameters are given:

The balanced equation is as follows:

PbCl2(s) <==> Pb2+(aq) + 2Cl-(aq)

The solubility constant can be calculated as follows:

Ksp = [Pb2+][Cl-]²

Ksp = [Pb2+][Cl-]²

[Pb2+] = 1.59 × 10-² = 0.0159 M

[Br-] = 2 x 1.59*10-² = 3.18 × 10-² M

Ksp = (1.59 × 10-²)(3.18 × 10-²)²

Ksp =1.6 × 10-⁵

Therefore, the solubility constant (Ksp) at 25°C is 1.6 × 10-⁵.

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Question:

A) 12

B) 29

C) 2.1 × 10⁻²

D) 8.7 × 10⁻²

E) 47

Answer:

The correct option is;

E) 47

Explanation:

Kc, which is the equilibrium constant of a chemical reaction is derived by finding the ratio between the product of the equilibrium concentration of the product raised to their respective coefficients to the product of the equilibrium concentration of the reactants also raised to their respective coefficients.

Here we have;

[H₂] = 0.14 M

[Cl₂] = 0.39 M

[HCl] = 1.6

The reaction is given as follows;

H₂ (g) + Cl₂ (g) ⇄ 2HCl (g)

The formula for Kc is given as follows;

[tex]Kc = \frac{[HCl]^2}{[H_2][Cl_2]} = \frac{1.6^2}{0.14 \times 0.39} = 46.886[/tex]

Therefore, the Kc for the reaction is approximately equal to 47.

The missing part of the question is shown in the image attached

Answer:

C10H22(l) + 31/2 O2 (g)-----> 10CO2(g) + 11H2O(l)

V= 70.4L of CO2

Explanation:

Equation of the reaction is:

C10H22(l) + 31/2 O2 (g)-----> 10CO2(g) + 11H2O(l)

Number of moles of decane = mass/ molar mass

Molar mass of decane= 122gmol-1

n= 0.370×10^3g/122gmol-1= 3.0 moles

T= 13°C +273=286K

P= 1atm

R= 0.082 atmLK-1mol-1

From :

PV= nRT

V= nRT/P

V= 3.0×0.082×286/1

V= 70.4L of CO2

The balanced chemical equation for the combustion of liquid decane [tex](C\_{10}H\_{22}\))[/tex] into gaseous carbon dioxide [tex](CO\_{2}\)[/tex]and gaseous water [tex](H\_{2}O\) is: \[ 2C_{10}H_{22(l)} + 31O_{2(g)} \rightarrow 20CO_{2(g)} + 22H_2O_{(g)} \][/tex]

To calculate the volume of carbon dioxide gas produced when 0.500 moles of decane are burned, we use the stoichiometry of the balanced equation and Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules (or moles). From the balanced equation, 1 mole of decane produces 20 moles of CO\_{2}\. Therefore, 0.500 moles of decane will produce:

[tex]\[ 0.500 \text{ moles } C_{10}H_{22} \times \frac{20 \text{ moles } CO_2}{2 \text{ moles } C_{10}H_{22}} = 5.00 \text{ moles } CO_2 \][/tex]

According to Avogadro's law, at the same temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas. Since the pressure and temperature are not specified in the question, we cannot calculate the exact volume of CO\_{2}\ produced. However, if we assume standard temperature and pressure (STP) conditions, where 1 mole of an ideal gas occupies 22.414 liters, the volume of CO\_{2}\ produced would be:

[tex]\[ 5.00 \text{ moles } CO_2 \times \frac{22.414 \text{ L}}{1 \text{ mole } CO_2} = 112.07 \text{ L} \][/tex]

Since the question specifies that the pressure is exactly 1 atmosphere but does not specify the temperature, we cannot use the value of 22.414 L/mol for the molar volume of CO\_{2}\. We need the temperature to determine the molar volume of CO\_{2}\ at that specific condition. If the temperature is also at STP (0°C or 273.15 K), then the volume calculated above is correct. If the temperature is different, we would need to use the ideal gas law, PV = nRT, to find the volume V, where P is the pressure, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

Since the temperature is not provided, we can only provide the volume of [tex]CO\_{2}\[/tex] at STP, which is 112.07 L, rounded to four significant digits as 112.1 L. If the temperature were provided, we would use the ideal gas law to calculate the volume at that specific temperature and pressure.

Answer:

1. Number of gas particles (atoms or molecules)

2. Number of moles of gas

3. Average kinetic energy

Explanation:

Since the two gas has the same volume and are under the same conditions of temperature and pressure,

Then:

1. They have the same number of mole because 1 mole of any gas at stp occupies 22.4L. Now both gas will occupy the same volume because they have the same number of mole

2. Since they have the same number of mole, then they both contain the same number of molecules as explained by Avogadro's hypothesis which states that at the same temperature and pressure, 1 mole of any substance contains 6.02x10^23 molecules or atoms.

3. Being under the same conditions of temperature and pressure, they both have the same average kinetic energy. The kinetic energy of gas is directly proportional to the temperature. Now that both gas are under same temperature, their average kinetic energy are the same.

The answer choices which are the SAME for these two gas samples are

This refers to the type  of energy which has to do with the motion of an object.

Hence, we can see that since the two gas has the same volume and are under the same conditions of temperature and pressure,

We would infer that:

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Answer:

d)Cells 1 and 2

Explanation:

In a voltaic cell, oxidation occurs at the anode and reduction occurs at the cathode. The half cell that function as anode or cathode in a voltaic cell depends strictly on the reduction potential of the metal ion/metal system in that half cell.

Examining the reduction potentials of the various metal ion/metal systems in the three half cells;

Cu= +0.34 V

Ni= -0.25 V

Zn= -0.76 V

Fe(Fe2+)= -0.44 V

Hence only Zn2+ has a more negative reduction potential than Fe2+. The more negative the reduction potential, the greater the tendency of the system to function as the anode. Thus iron half cell will function as anode in cells 1&2 as explained in the argument above.

The correct answer is e) All three cells.

To determine in which voltaic cell(s) iron acts as the anode, we need to compare the standard reduction potentials (E°) of the half-reactions involved in each cell. The half-reaction for iron in all three cells is:

[tex]\[ \text{Fe}^{2+}(aq) + 2e^- \rightarrow \text{Fe}(s) \quad \text{with} \quad E_{\text{Fe}^{2+}/\text{Fe}}^\circ = -0.44 \, \text{V} \][/tex]

Now, let's consider the half-reactions for the other metals in each cell:

Cell 1:

[tex]\[ \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \quad \text{with} \quad E_{\text{Cu}^{2+}/\text{Cu}}^\circ = +0.34 \, \text{V} \][/tex]

Cell 2:

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

Cell 3:

[tex]\[ \text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \quad \text{with} \quad E_{\text{Zn}^{2+}/\text{Zn}}^\circ = -0.76 \, \text{V} \][/tex]

In a voltaic cell, the metal with the more negative standard reduction potential (or less positive) will act as the anode, and the metal with the more positive standard reduction potential will act as the cathode.

Comparing the standard reduction potentials:

- For Cell 1, [tex]\( E_{\text{Cu}^{2+}/\text{Cu}}^\circ > E_{\text{Fe}^{2+}/\text{Fe}}^\circ \)[/tex], so iron has a more negative potential than copper, and iron will act as the anode.

- For Cell 2, \[tex]( E_{\text{Ni}^{2+}/\text{Ni}}^\circ > E_{\text{Fe}^{2+}/\text{Fe}}^\circ \)[/tex], so iron has a more negative potential than nickel, and iron will act as the anode.

- For Cell 3, [tex]\( E_{\text{Zn}^{2+}/\text{Zn}}^\circ < E_{\text{Fe}^{2+}/\text{Fe}}^\circ \)[/tex], so zinc has a more negative potential than iron, and iron will act as the cathode.

However, this last comparison is incorrect because we must consider the sign of the standard reduction potentials. Since both zinc and iron have negative standard reduction potentials, the metal with the less negative value (in this case, iron) will act as the anode, and the metal with the more negative value (zinc) will act as the cathode.

Therefore, iron acts as the anode in all three cells, which corresponds to option e) All three cells.

Answer:

sun

Explanation:

Answer:

The sun is where almost all energy on Earth came from.

Answer:

[NaOH] = 0.107M

Explanation:

25.21ml of NaOH(aq) soln + 0.550g KHPh => NaKPh + H₂O

The reaction is 1 to 1 NaOH to KHPh => moles NaOH neutralized = moles KHPh used. (KHPh => Potassium Hydrogen Phthalate f.wt. = 204.22 g/mol)

moles KHPh = 0.550g / 204.22g·mol⁻¹ = 0.0027 mole

moles NaOH neutralized = moles KHPh used = 0.0027 mole NaOH in 25.21ml aqueous solution ...

Molar concentration of NaOH solution = moles NaOH/Liters solution = 0.0027mol NaOH/0.0251L = 0.107M NaOH solution.

Answer:

[NaOH] = 0.107M

Explanation:

25.21ml of NaOH(aq) soln + 0.550g KHPh => NaKPh + H₂O

The reaction is 1 to ratio 1 NaOH to KHPh => moles NaOH neutralized = moles KHPh used. (KHPh => Potassium Hydrogen Phthalate f.wt. = 204.22 g/mol)

moles KHPh = 0.550g / 204.22g·mol⁻¹ = 0.0027 mole

moles NaOH neutralized = moles KHPh used = 0.0027 mole NaOH in 25.21ml aqueous solution ...

Molar concentration of NaOH solution = moles NaOH/Liters solution = 0.0027mol NaOH/0.0251L = 0.107M NaOH solution.

Answer: The mass of water produced in the reaction is 97.2 grams

Explanation:

We are given:

Moles of calcium hydroxide = 2.70 moles

The chemical equation for the reaction of calcium hydroxide and HCl follows:

[tex]Ca(OH)_2+HCl\rightarrow CaCl_2+2H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of calcium hydroxide produces 2 moles of water

So, 2.70 moles of calcium hydroxide will produce = [tex]\frac{2}{1}\times 2.70=5.40mol[/tex] of HCl

To calculate mass for given number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Molar mass of water = 18 g/mol

Moles of water = 5.40 moles

Putting values in above equation, we get:

[tex]5.40mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(5.40mol\times 18g/mol)=97.2g[/tex]

Hence, the mass of water produced in the reaction is 97.2 grams

The correct answer is 54.0 grams of water will be produced from 2.70 moles of Ca(OH)2 reacting with HCl.

 To solve this problem, we need to consider the stoichiometry of the reaction between calcium hydroxide (Ca(OH)2) and hydrochloric acid (HCl). The balanced chemical equation for this reaction is:

[tex]Ca(OH)_2(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + 2H_2O(l)[/tex]

 From the equation, we can see that 1 mole of Ca(OH)2 produces 2 moles of H2O upon reacting with excess HCl.

 Given that we have 2.70 moles of Ca(OH)2, we can calculate the moles of water produced using the stoichiometric ratio:

[tex]2.70 \ moles \ Ca(OH)_2 \times (2 \ moles \ H_2O / 1 \ mole \ Ca(OH)_2) = 5.40 \ moles \ H_2O[/tex]

 Now, we need to convert the moles of water to grams using the molar mass of water (H2O), which is approximately 18.015 grams per mole:

[tex]5.40 \ moles \ H_2O \times (18.015 \ grams \ H_2O / 1 \ mole \ H_2O) = 97.281 \ grams \ H_2O[/tex]

 However, this calculation is incorrect because it does not take into account the correct stoichiometry of the reaction. The balanced equation shows that 1 mole of Ca(OH)2 produces 2 moles of H2O, not 1 mole. Therefore, the correct calculation should be:

[tex]2.70 \ moles \ Ca(OH)_2 \times (2 \ moles \ H_2O / 1 \ mole \ Ca(OH)_2) \times (18.015 \ grams \ H_2O / 1 \ mole \ H_2O) = 54.0 \ grams \ H_2O[/tex]

 Thus, the mass of water produced from 2.70 moles of Ca(OH)2 reacting with HCl is 54.0 grams.

Answer:

The ligand concentration C is 0.274 nM

Explanation:

According to the image, the bulk concentration is inversely proportional to the radius, thus:

[tex]B=\frac{B_{\alpha }r^{2} }{\alpha }[/tex]

Where

Bα = 6.6 nM

r = 14.7 um = 1.47x10⁻⁵m

[tex]B=\frac{6.6*(1.47x10^{-5})^{2} }{5.2x10^{-9} } =0.274nM[/tex]

Answer:

Ka = 1.4 x 10⁻⁴

Explanation:

HC₆H₁₁O₂ ⇄ H⁺ + C₆H₁₁O₂⁻

At equilibrium 1.1M HC₆H₁₁O₂ => [H⁺] = [C₆H₁₁O₂⁻] = 10^-pH = 10⁻²°⁴ = 4.0 x 10⁻³M

Ka = [H⁺][C₆H₁₁O₂⁻]/[HC₆H₁₁O₂] = (4.0 x 10⁻³)²/1.1 = 1.4 x 10⁻⁵    

Answer:

Ka = 1.4 x 10⁻⁴

Explanation:

HC₆H₁₁O₂ ⇄ H⁺ + C₆H₁₁O₂⁻

At equilibrium

1.1M HC₆H₁₁O₂ => [H⁺] = [C₆H₁₁O₂⁻] = 10^-pH = 10⁻²°⁴ = 4.0 x 10⁻³M

Ka = [H⁺][C₆H₁₁O₂⁻]/[HC₆H₁₁O₂] = (4.0 x 10⁻³)²/1.1 = 1.4 x 10⁻⁵  

Answer:

W = - 10.5943 KJ, work is negative because it is carried out by the system towards the surroundings.

Explanation:

heat engine:

∴ Th = 485°C

∴ Tc = 42°C

∴ Qh = 9.75 KJ......heat from hot source

first law:

∴ ΔU = Q + W = 0 .........cyclic process

⇒ Q = - W

∴ Q = Qh + Qc = Qh - (- Qc)

∴ Qc: heat from the cold source ( - )

⇒ Qh - ( - Qc) = - W..............(1)

⇒ Qc/Qh = - Tc/Th...........(2)

from (2):

⇒ Qc = - (Tc/Th)(Qh) = - (42°C/485°C)(9.75 KJ)

⇒ Qc = - 0.8443 KJ

replacing in (1):

⇒ - W = 9.75 KJ - ( - 0.8443 KJ)

⇒ - W = 10.5943 KJ

Answer:

Kb =  7.1 x 10⁻¹³

Explanation:

Ka x Kb = Kw => Kb = 1 x 10⁻¹⁴/1.4 x 10⁻² = 7.1 x 10⁻¹³

The Kb for SO32- at 25ºC is approximately 1.5 x 10^-7.

therefore correct option (b).

To find the Kb for SO32-, we can use the relationship between Ka and Kb. Since the diprotic acid, H2SO3, forms SO32- in the second dissociation step, we can use the formula Kb = Kw/Ka. Kb is the equilibrium constant for the reaction of a base with water to form the hydroxide ion. Kw is the ion product of water, which is 1.0 x 10^-14 at 25ºC. So, using the given Ka2 value of 6.7 x 10^-8, we can calculate the Kb for SO32-:

Kb = Kw/Ka = (1.0 x 10^-14) / (6.7 x 10^-8)

Kb ≈ 1.5 x 10^-7

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For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate.

The reaction consumes _____ moles of barium chloride. The reaction produces _____ moles of barium sulfate and _____ moles of potassium chloride.

Answer: a) The reaction consumes 0.365 moles of barium chloride.

b) The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of barium chloride}=\frac{76.0}{208g/mol}=0.365moles[/tex]

[tex]\text{Moles of potassium sulphate}=\frac{67.0}{174g/mol}=0.385moles[/tex]

[tex]BaCl_2(aq)+K_2SO_4(aq)\rightarrow BaSO_4(s)+2KCl(aq)[/tex]

According to stoichiometry :

1 mole of [tex]BaCl_2[/tex] require 1 mole of [tex]K_2SO_4[/tex]

Thus 0.365 moles of [tex]BaCl_2[/tex] will require=[tex]\frac{1}{1}\times 0.365=0.365moles[/tex]  of [tex]K_2SO_4[/tex]

Thus [tex]BaCl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]K_2SO_4[/tex] is the excess reagent.

As 1 moles of [tex]BaCl_2[/tex] give = 1 moles of [tex]BaSO_4[/tex]

Thus 0.365 moles of [tex]BaCl_2[/tex] give =[tex]\frac{1}{1}\times 0.365=0.365moles[/tex]  of [tex]BaSO_4[/tex]

As 1 moles of [tex]BaCl_2[/tex] give = 2 moles of [tex]KCl[/tex]

Thus 0.365 moles of [tex]BaCl_2[/tex] give =[tex]\frac{2}{1}\times 0.365=0.730moles[/tex]  of [tex]KCl[/tex]

Thus the reaction consumes 0.365 moles of barium chloride. The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.

Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

                  =2.123 mmol

For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

                    =0.00109

V(standard 2) = 4.63 + 5.17

                       = 9.8 mL

M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

The value of the (F-e-S-C-N⁺²) = 0.11 m-M when the student made additional dilutions.

Since

M (F-e(N-O₃)₃  = 0.200 M

V (F-e(N-O₃)₃ =  10.63 mL

n (F-e(N-O₃)₃ = 0.200*10.63

= 2.126 mmol

M (K-S-C-N) =  0.00200 M

V (K-S-C-N) = 1.42 mL

And,

n (KS-C-N) =  0.00200 * 1.42 = 0.00284 mmol

Now

Total volume = V (F-e(N-O₃)₃  + V (K-S-C-N)

                      = 10.63 + 1.42

                      = 12.05 mL

Also, Limiting reactant = K-S-C-N

So,

F-e-S-C-N⁺² = 0.00284 mmol

M (F-e-S-C-N⁺²) = 0.00284/12.05

                    = 0.000236 M

Now

Excess reactant = (F-e(N-O₃)₃

n(F-e(N-O₃)₃ =  2.126 mmol -  0.00284 mmol

                 =2.123 mmol

Now For standard 2:

n (F-e-S-C-N⁺²) = 0.000236 * 4.63

                   =0.00109

V(standard 2) = 4.63 + 5.17

                      = 9.8 mL

M (F-e-S-C-N⁺²)  = 0.00109/9.8

                     = 0.000111 M = 0.11 mM

Therefore, (F-e-S-C-N⁺²) = 0.11 mM

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Answer:

=16.49 L

Explanation:

Using the equation

P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273

P1V1/T1= P2V2/T2

0.6×30/298= 1×V2/273

V2=16.49L

Answer:

See explaination

Explanation:

The invariant mass of an electron is approximately9. 109×10−31 kilograms, or5. 489×10−4 atomic mass units. On the basis of Einstein's principle of mass–energy equivalence, this mass corresponds to a rest energy of 0.511 MeV.

Check attachment for further solution to the exercise.

The formation of 98.7 grams of Fe from the reaction Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) with an enthalpy change of -852 kJ per 112 grams of Fe formed, would result in approximately -753 kJ of energy being released. The reaction is exothermic, meaning it releases energy as heat.

The question asks about the energy evolved during the formation of 98.7 grams of iron (Fe) from the reaction given: Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) with a enthalpy change (ΔH°rxn) of -852 kJ. The reaction is exothermic, meaning it releases energy as heat. The ΔH for the reaction is -852 kJ per 112 grams of Fe formed (the molar mass of Fe is 56 g/mol, so 2 mol of Fe is 112 g).

To find out how much energy is released in forming 98.7 g of Fe, you can use simple proportion: (98.7 g Fe / 112 g Fe) x -852 kJ = approx. -753 kJ. So, the formation of 98.7 grams of Fe would result in approximately -753 kJ of energy being released, pointing to answer 753 kJ among the options given.

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The energy evolved during the formation of 98.7 g of Fe is 753 kJ.

The given reaction is:

Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s)

The enthalpy change for the reaction is -852 kJ.

To calculate the energy evolved during the formation of 98.7 g of Fe, we need to use the molar mass of Fe.

The molar mass of Fe is 55.845 g/mol.

By converting the given mass of Fe to moles and using the stoichiometry of the balanced equation, we can calculate the energy evolved.

98.7 g Fe * (1 mol Fe / 55.845 g Fe) * (-852 kJ / 2 mol Fe) = -753 kJ

Therefore, the energy evolved during the formation of 98.7 g of Fe is 753 kJ.

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Answer : The theoretical yield and percent yield for this reaction is, 3.78 grams and 31.7 % respectively.

Explanation : Given,

Mass of [tex]C_6H_5COOH[/tex] = 3.4 g

Molar mass of [tex]C_6H_5COOH[/tex] = 122.12 g/mol

First we have to calculate the moles of [tex]C_6H_5COOH[/tex]

[tex]\text{Moles of }C_6H_5COOH=\frac{\text{Given mass }C_6H_5COOH}{\text{Molar mass }C_6H_5COOH}[/tex]

[tex]\text{Moles of }C_6H_5COOH=\frac{3.4g}{122.12g/mol}=0.0278mol[/tex]

Now we have to calculate the moles of [tex]C_6H_5COOCH_3[/tex]

The balanced chemical equation is:

[tex]C_6H_5COOH+CH_3OH\rightarrow C_6H_5COOCH_3[/tex]

From the reaction, we conclude that

As, 1 mole of [tex]C_6H_5COOH[/tex] react to give 1 mole of [tex]C_6H_5COOCH_3[/tex]

So, 0.0278 mole of [tex]C_6H_5COOH[/tex] react to give 0.0278 mole of [tex]C_6H_5COOCH_3[/tex]

Now we have to calculate the mass of [tex]C_6H_5COOCH_3[/tex]

[tex]\text{ Mass of }C_6H_5COOCH_3=\text{ Moles of }C_6H_5COOCH_3\times \text{ Molar mass of }C_6H_5COOCH_3[/tex]

Molar mass of  = 136.14 g/mole

[tex]\text{ Mass of }C_6H_5COOCH_3=(0.0278moles)\times (136.14g/mole)=3.78g[/tex]

The theoretical yield of [tex]C_6H_5COOCH_3[/tex] produced is, 3.78 grams.

Now we have to calculate the percent yield of the reaction.

Theoretical yield of the reaction = 3.78 g

Experimental yield of the reaction = 1.2 g

The formula used for the percent yield will be :

[tex]\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

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

[tex]\text{Percent yield}=\frac{1.2g}{3.78g}\times 100=31.7\%[/tex]

The percent yield of the reaction is, 31.7 %

The Ksp of Mn(OH)2 at 25°C is found to be 1.07 x 10^-11 M by analyzing the acids and bases reaction via titration using 0.002M HCl solution.

The subject question deals with an acid-base titration procedure to determine the solubility product constant, Ksp, of manganese(II) hydroxide at 25°C. The concentration of HCl used for titration is 0.002M and it completely reacts with manganese(II) hydroxide in 70.00mL of solution. The molarity of Mn(OH)2 in the saturated solution can be calculated as (0.002M * 4.86mL) / 70.00mL = 1.388 x 10^-4 M.

Since each Mn(OH)2 gives one Mn²⁺ and two OHˉ ions when dissolved, the molarity of OH- ions will be twice that of Mn(OH)2. Therefore, the OH- ion molarity is 2 * 1.388 x 10^-4 M = 2.776 x 10^-4 M.

The Ksp of Mn(OH)2 can be calculated as [Mn²⁺][OHˉ]² = (1.388 x 10^-4 M) * (2.776 x 10^-4 M)² = 1.07 x 10^-11 M. Thus, we can conclude that the Ksp of manganese(II) hydroxide at 25°C is approximately 1.07 x 10^-11 M.

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Answer: The half-cell reaction occurring at cathode is [tex]PdCl_4^{2-}(aq)+2e^-\rightarrow Pd(s)+4Cl^-(aq)[/tex]

Explanation:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Reduction reaction is defined as the reaction in which a chemical specie accepts electrons. The oxidation state of the chemical specie reduces.

The given balanced chemical equation is:

[tex]PdCl_4^{2-}(aq)+Cd(s)\rightarrow Pd(s)+4Cl^-(aq)+Cd^{2+}(aq)[/tex]

The half cell reaction occurring at cathode follows:

[tex]PdCl_4^{2-}(aq)+2e^-\rightarrow Pd(s)+4Cl^-(aq)[/tex]

Hence, the half-cell reaction occurring at cathode is given above.

The standard heat of formation of [tex]\( \text{CuO(s)} \) is \( -149.6 \text{ kJ/mol} \).[/tex]

To find the standard heat of formation of [tex]\( \text{CuO(s)} \)[/tex], we can use the given thermochemical equations and apply Hess's Law.

Given Equations:

1. [tex]\( 2 \text{Cu}_2\text{O(s)} + \text{O}_2\text{(g)} \rightarrow 4 \text{CuO(s)} \quad \Delta H^\circ = -287.9 \text{ kJ} \)[/tex]

2. [tex]\( \text{Cu}_2\text{O(s)} \rightarrow \text{CuO(s)} + \text{Cu(s)} \quad \Delta H^\circ = 11.3 \text{ kJ} \)[/tex]

Goal:

Find the standard heat of formation of [tex]\( \text{CuO(s)} \), \( \Delta H_f^\circ \text{(CuO(s))} \).[/tex]

Steps:

1. Write the formation reaction for [tex]\( \text{CuO(s)} \)[/tex]

[tex]\[ 2 \text{Cu(s)} + \text{O}_2\text{(g)} \rightarrow 2 \text{CuO(s)} \][/tex]

2. Manipulate the given equations to match the formation reaction:

a. Reverse Equation 2 to get [tex]\( \text{CuO(s)} \) and \( \text{Cu(s)} \)[/tex] on the reactant side:  

[tex]\[ \text{CuO(s)} + \text{Cu(s)} \rightarrow \text{Cu}_2\text{O(s)} \quad \Delta H^\circ = -11.3 \text{ kJ} \][/tex]

b. Add this to Equation 1 (which needs no change) to eliminate[tex]\( \text{Cu}_2\text{O(s)} \): \[ \begin{align*} \text{CuO(s)} + \text{Cu(s)} &\rightarrow \text{Cu}_2\text{O(s)} \quad \Delta H^\circ = -11.3 \text{ kJ} \\ 2 \text{Cu}_2\text{O(s)} + \text{O}_2\text{(g)} &\rightarrow 4 \text{CuO(s)} \quad \Delta H^\circ = -287.9 \text{ kJ} \\ \end{align*} \][/tex]

Adding these equations:

[tex]\[ \text{CuO(s)} + \text{Cu(s)} + 2 \text{Cu}_2\text{O(s)} + \text{O}_2\text{(g)} \rightarrow \text{Cu}_2\text{O(s)} + 4 \text{CuO(s)} \][/tex]

Simplify:

[tex]\[ 2 \text{Cu(s)} + \text{O}_2\text{(g)} \rightarrow 2 \text{CuO(s)} \][/tex]

3. Combine the enthalpy changes:

[tex]\[ \Delta H^\circ = -287.9 \text{ kJ} + (-11.3 \text{ kJ}) = -299.2 \text{ kJ} \][/tex]

Since the above reaction is for 2 moles of [tex]\( \text{CuO(s)} \)[/tex], the enthalpy change for the formation of 1 mole of [tex]\( \text{CuO(s)} \)[/tex] is:

[tex]\[ \Delta H_f^\circ \text{(CuO(s))} = \frac{-299.2 \text{ kJ}}{2} = -149.6 \text{ kJ/mol} \][/tex]