Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate·x-hydrate (CuSO4·xH2O), where x is an integer. Part of their practical exam is to determine this integer x. They are working in pairs, though Hermione is doing most of the work. This should be discouraged! After dehydration they find that they are left with 0.96 g of the an-hydrate CuSO4. What is the unknown integer x. Round the answer to the nearest integer.

Answer:

The volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.

Explanation:

For the complete recrystalization,

the amount of hot water should be such that, the benzoic acid is completely soluble in it.

As we are given that the solubility of benzoic acid in hot water is 68.0 g/L. Now we have to determine the volume of water is needed to recrystallize 0.700 g of benzoic acid.

we conclude that,

As, 68.0 grams of benzoic acid soluble in 1 L of water.

So, 0.700 grams of benzoic acid soluble in  of water.

The volume of water needed = 0.01029 L = 10.29 mL

conversion used : (1 L = 1000 mL)

Therefore, the volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.

The question is incomplete, complete question is :

In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. This reaction is now the first step taken to make most of the world's fertilizer. Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 348 liters per second of dinitrogen are consumed when the reaction is run at 205°C and 0.72 atm. Calculate the rate at which ammonia is being produced.

Answer:

The rate of production of ammonia is 217.08 grams per second.

Explanation:

[tex]N_2+3H_2\rightarrow 2NH_3[/tex]

Volume of dinitrogen used in a second = 348 L

Temperature of the gas = T = 205°C = 205+273 K = 478 K

Pressure of the gas = P = 0.72 atm

Moles of dinitrogen = n

[tex]n=\frac{PV}{RT}=\frac{0.72 atm\times 348 L}{0.0821 atm L/mol K\times 478 K}=6.385 mol[/tex]

According to reaction, 1 mole of dinitriogen gives 2 mole of ammonia.Then 6.385 moles of dinitrogen will give:

[tex]\frac{2}{1}\times 6.385 mol=12.769 mol[/tex]

Mass of 12.769 moles of ammonia;

12.769 mol 17 g/mol = 217.08 g

217.08 grams of ammonia is produced per second.So, the rate of production of ammonia is 217.08 grams per second.

Answer:

0.66 mol

Explanation:

Zero Gauge pressure = 14.7 psi

Pressure read = 173 psi

Actual pressure = 173 psi - 14.7 psi = 158.3 psi

P (psi) = 1/14.696  P(atm)

So, Pressure = 10.77 atm

Given that:

Temperature = 20 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (20+ 273.15) K = 298.15 K

V = 1.50 L

Using ideal gas equation as:

[tex]PV=nRT[/tex]

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

10.77 atm × 1.50 L = n ×0.0821 L atm/ K mol  × 298.15 K

⇒n = 0.66 mol

The change in entropy when 95 g of ethyl acetate freezes at -84.0°C is -59.8 J/K, rounded to two significant digits.

The question involves calculating the change in entropy when 95 g of ethyl acetate freezes at -84.0℃. Given the heat of fusion ( Hf) for ethyl acetate is 10.5 kJ/mol, we first need to convert grams to moles using the molar mass of ethyl acetate (C4H8O2) which is 88.11 g/mol.

Next, we use the formula:

moles = mass (g) / molar mass (g/mol)

For 95 g ethyl acetate, moles = 95 g / 88.11 g/mol = 1.078 moles.

The change in entropy ( S) can be found using the formula:
S =  H / T

where T is the temperature in Kelvin. To convert -84.0℃ to Kelvin, we add 273.15:

T = -84.0 + 273.15 = 189.15 K

The entropy change then is:
S = 10.5 kJ/mol / 189.15 K
S per mole = 0.0555 kJ/K∙mol

Finally, for 1.078 moles, the total entropy change is:
S total = 0.0555 kJ/K∙mol × 1.078 mol
S total = 0.0598 kJ/K

Expressed in J/K (as 1 kJ = 1000 J),
S total = 59.8 J/K.

Since we are calculating the entropy change for freezing, the sign should be negative because entropy decreases during freezing.

Therefore, the change in entropy when 95 g of ethyl acetate freezes at -84.0℃ is -59.8 J/K, rounded to two significant digits.

Answer:

Ca(OH)2(s) + 2HCl(aq) →  CaCl2(aq) + 2H2O(l)

Explanation:

Step 1: Data given

Mass of Ca(OH)2 = 9.49 grams

Volume of a 0.490 M HCl solution = 30.0 mL = 0.030 L

Step 2: The unbalanced equation

Ca(OH)2 + HCl →  CaCl2 + H2O

On the left side we have 2x O (in Ca(OH)2 )  and on the right side we have 1x O (in H2O). To balance the amount of O we have to multiply H2O, on the right side , by 2.

Ca(OH)2 + HCl →  CaCl2 + 2H2O

On the right side we have 4x H (in 2H2O), and on the left side we have 3x H  (2x in Ca(OH)2 and 1x in HCl). To multiply the amount of H, we have to multiply HCl on the left side by 2.

Ca(OH)2(s) + 2HCl(aq) →  CaCl2(aq) + 2H2O(l)

Answer:

using higher concentration of the nucleophile

Explanation:

In SN2 reaction, the attack of the nucleophile on the substrate occurs simultaneously as the leaving group departs. The entering group normally attacks through the back side of the molecule. The reaction is concerted and bimolecular. This implies that the concentration of the nucleophile is important in the rate equation for the reaction. Hence increasing the concentration of the nucleophile will increase the rate of SN2 reaction.

Answer:

radius = 156 pm

Explanation:

The relation between radius and edge length of unit cell of BCC is

r=a[tex]\sqrt{3}[/tex]/4

Given

a = 360 pm

Therefore

r = r = radius = 360[tex]\sqrt{3}[/tex]/4= 155.88 pm

Or

156 pm

Answer: For the given reaction, the value of [tex]K_c[/tex] is greater than 1

Explanation:

For the given chemical equation:

[tex]CaCO3(s)\rightleftharpoons Ca^{2+}(aq.)+CO_3^{2-}(aq.)[/tex]

The expression of [tex]K_c[/tex] for above equation follows:

[tex]K_c=\frac{[Ca^{2+}]\times [CO_3^{2-}]}{[CaCO_3}]\\\\K_c=[Ca^{2+}]\times [CO_3^{2-}][/tex]

The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression

As, the denominator is missing and the numerator is the only part left in the expression. So, the value of [tex]K_c[/tex] will be greater than 1.

Hence, for the given reaction, the value of [tex]K_c[/tex] is greater than 1

Explanation:

Reaction equations for the given species is as follows.

   [tex]H_{3}PO_{4} + 2NaOH \righleftharpoons Na_{2}HPO_{4} + 2H_{2}O[/tex]

   [tex]NaH_{2}PO_{4} + NaOH \rightleftharpoons Na_{2}HPO_{4} + H_{2}O[/tex]

At the first equivalence point we need 2 × 10 mmol NaOH.

At the second equivalence point we need 5 mmol of NaOH.

Hence, total moles of NaOH required is as follows.

               (20 + 5) mmol = 25 mmol

We assume that volume of NaOH required is V.

      [tex]25 mmol NaOH \times \frac{10^{-3} mol NaOH}{1 mmol NaOH} \times \frac{1000 ml V}{0.10 \text{mol NaOH}}[/tex]

      = 250 ml V

Thus, we can conclude that 250 ml of 0.10 M NaOH must be added to reach second equivalence point of the titration of the [tex]H_{3}PO_{4}[/tex] and NaOH.

In the given case, 250 ml of 0.10 M NaOH must be added to attain the second equivalence point.

The reactions taking place in the given case are:

Based on the given information, a solution comprises 10 mmol of H₃PO₄ and 5 mmol of NaH₂PO₄.

For the first equivalence point, there is a need of 2 × 10 mmol NaOH.

For the second equivalence point, there is a need of 5 mmol NaOH.

Now the total moles of NaOH needed is,

= 20 + 5 mmol

= 25 mmol

Now let the volume of NaOH required be V. Now putting the values we get,

[tex]= 25 mmol NaOH * \frac{10^{-3} mole NaOH}{1 mmol NaOH} * \frac{1000 mLV}{0.10 mol NaOH} \\= 250 ml[/tex]

Thus, 250 ml of 0.10 M NaOH is need to be added to attain the second equivalence point of the titration.

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Answer:

1) 71.9 g/L

2) 0.221 m olal

3)  7.05% by mass

Explanation:

Step 1: Data given

Concentration of sucrose = 0.210 M

Molar weight of sucrose = 342.3 g/mol

Density of solution = 1.02 g/mL

Mass of water = 948.1 grams

Step 2: Convert this value into terms of g/L

(0.210 mol/L) * (342.3 g/mol) = 71.9 g/L

Calculate the molality

Step 1: Calculate mass water

Suppose we have a volume of 1.00L

Mass of the solution = 1000 mL * 1.02 g/mL = 1020 g solution

We know that there are 71.9 g of solute in a liter of solution from the first calculation. This means

(1020 grams solution) - (71.9 g solute) = 948.1 g = 0.9481 kg water

Step 2: Calculate molality

Molality = moles sucrose / mass water

(0.210 mol) / (0.9481 kg) = 0.221 mol/kg = 0.221 m olal

Mass %

% MAss = (mass solute / mass solution)*100%

(71.9 g) / (1020 g) *100% = 7.05% by mass

Explanation:

It is known that formula for the ionization energy of hydrogen atom is as follows.

               E = [tex]\frac{13.6 eV}{n^{2}}[/tex]

or,          n = [tex]\sqrt{\frac{13.6}{E}}[/tex]

The value of energy is given as 0.544 eV. Therefore, we will calculate the value of n as follows.

                   n = [tex]\sqrt{\frac{13.6}{E}}[/tex]

                      = [tex]\sqrt{\frac{13.6}{0.544 eV}}[/tex]

                      = 5

Thus, we can conclude that n equals to 5 for a hydrogen atom if 0.544 eV of energy can ionize it.

Final answer:

The principal quantum number n of a hydrogen atom that requires 0.850 eV of energy to ionize from an excited state is 4. This is calculated using the formula for the energy levels of hydrogen, En = -13.6 eV/n², and solving for n.

Explanation:

The question relates to the energy levels of a hydrogen atom and how much energy is required to ionize it when the electron is in an excited state. The ionization energy of hydrogen in the ground state (n = 1) is 13.6 eV. To find the principal quantum number n for a particular amount of energy required to ionize the hydrogen atom from an excited state, we use the formula: En = -13.6 eV/n². We can set this equal to the energy needed to ionize the atom, in this case, 0.850 eV, and solve for n.

First, we rearrange the formula so it looks like this:

-13.6 eV/n² = -0.850 eV
Then we solve for n:

n² = 13.6 eV / 0.850 eV
n² = 16
n = √16
n = 4

Thus, the hydrogen atom is in the n = 4 excited state when 0.850 eV of energy can ionize it.

Answer:

[H₂]  = 0.178 M

[I₂]    = 0.184 M

[HI]   = 1.415 M

Explanation:

For the equilibrium:

H₂(g) + I₂(g) ⇄ 2 HI(g)

the equilibrium constant is given by the equation:

Kc = [ HI]² / [H₂][I₂]

Lets use first the reaction quotient which has the same expression as the equilibrium constant to predict the direction the reaction will take, i.e towards reactants or product side.

Q =( 0.895)²/(0.438)(0.444) = 4.12

Q is less than Kc so the reaction will favor the product side.

We can set up the following table to account for all the species at equilibrium:

                                     H₂             I₂                HI

initial                        0.438        0.444          0.895

change                        -x               -x                +2x

equilibrium              0.438 - x    0.444 - x     0.895 + 2x

Now we are in position to express these concentrations  in terms of the equilibrium conctant, Kc

54.3 = (0.895 + 2x)² / (0.438 -x)(0.444 - x)

performing the calculatiopns will result in a quadratic equation:

0.801 + 3.580x +4x² = (0.194 - 0.882x + x²)x 54.3

Upon rearrangement and some algebra, we have

0.801 + 3.580 x + 4x² = 10.534 - 47.893x + 54.3 x²

0 = 9.733 - 51.473 x + 54.3 x²

This equation has two roots X₁ = 0.687 and X₂ = 0.26

The first is physically impossible since it will imply that more 0.687 will make the quantity at equilibrium for both H₂ and I₂ negative.

Therefore the concentrations at equilibrium of each  gas are:

[H₂] = (0.438 - 0.260)              = 0.178 M

[I₂]   = (0.444 - 0.260) M          = 0.184  M

[HI] = [0.895 + 2x(0.260)] M    = 1.415   M

Note if we plug these values into the equilibrium expression we get 61 which is due to the rounding errors propagating in the quadratic equation.

The equilibrium concentrations of H2, I2, and HI at 430 °C are calculated using an expression derived from the reaction quotient equation, plugged into the Kc equation, which is then solved for 'x'. The solutions found are the changes in molarities which applied to the initial molarities give the equilibrium concentrations

Let's denote the change in molarity of H2, I2, and HI as 'x'. At equilibrium, the molarities of H2, I2, and HI will be 0.438+x, 0.444+x, and 0.895-2x respectively. We know the equilibrium constant, Kc = 54.3. Thus, (0.895-2x)2/(0.438+x)(0.444+x) = 54.3. This is a quadratic equation in 'x' and needs to be solved to get the value of 'x'.

After finding 'x', put this value back into the equilibrium concentrations of the gases, i.e., 0.438+x for H2, 0.444+x for I2, and 0.895-2x for HI. These will give you the equilibrium concentrations of the gases at 430 °C.

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Answer:

≅ 16.81 kJ

Explanation:

Given that;

mass of acetone = 31.5 g

molar mass of acetone = 58.08 g/mol

heat of vaporization for acetone = 31.0 kJ/molkJ/mol.

Number of moles = [tex]\frac{mass}{molar mass}[/tex]

Number of moles of acetone = [tex]\frac{31.5}{58.08}[/tex]

Number of moles  of acetone = 0.5424 mole

The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;

Hence;

The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol

The heat required to vaporize 31.5 g of acetone = 16.8144 kJ

≅ 16.81 kJ

The balanced equation for the combustion of isooctane (C8H18) is C8H18 + 12.5O2 -> 8CO2 + 9H2O.

The balanced equation for the combustion of isooctane (C8H18) to produce carbon dioxide (CO2) and water (H2O) is:

C8H18 + 12.5O2 -> 8CO2 + 9H2O

In this equation, the coefficients have been adjusted to balance the number of atoms on both sides of the equation. There are 8 carbon atoms, 18 hydrogen atoms, and 25 oxygen atoms on both sides of the equation.

Final answer:

The balanced chemical equation for the complete combustion of isooctane (C₈H₁₈) is 2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g), comprising a step-by-step process of first balancing the carbon, then hydrogen, and finally the oxygen atoms.

Explanation:

The balanced chemical equation for the combustion of isooctane (C₈H₁₈) to produce carbon dioxide (CO₂) and water (H₂O) can be represented as follows:

2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g)

Here is a step-by-step explanation:

After balancing, you end up with the smallest possible integers that balance the equation.

For the given reaction between rubidium and Boron, only one electron will be transferred in the formation [tex]2 RbBr[/tex].

The potential difference between the two half-cells of an electrochemical cell is known as cell potential. The transfer of electrons from one-half cell to another produces the potential difference.

The given chemical reaction is:

[tex]2Rb_{(s)} + Br_{(l)}[/tex] ⇒ [tex]2RbBr_{(s)}[/tex]

The following electrochemical reaction can be divided into the reaction occurring at the cathode and the reaction occurring at the anode:

Oxidation reaction at the anode: [tex]2Rb_{(s)}[/tex] ⇒ [tex]2Rb^+_{(aq)} + 2e^-[/tex]

Reduction reaction at the cathode: [tex]Br_{2(l)} + 2e^-[/tex] ⇒ [tex]2Br^-_{(aq)}[/tex]

Since there are two units of RbBr in the aforementioned equation and two electrons were transported, only one electron is needed to generate one unit of RbBr.

Thus, only one electron is involved in the formation of RbBr.

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Final answer:

In the formation of one formula unit of RbBr, one electron is transferred from rubidium (Rb) to bromine (Br), as Rb loses one electron to form Rb+ and Br gains one electron to form Br-.

Explanation:

The question regards the number of electrons transferred in the formation of one formula unit of RbBr from the reaction 2 Rb + Br2 → 2 RbBr. In order to determine the electron transfer, we need to consider the oxidation states of the elements involved. Rubidium (Rb) is oxidized, meaning it loses electrons, and bromine (Br2) is reduced, meaning it gains electrons.

Rubidium starts with an oxidation state of 0 and becomes Rb+ in RbBr. This indicates that each Rb atom loses one electron. Bromine starts with an oxidation state of 0 as diatomic Br2 and each Br atom gains one electron to become Br- in RbBr. Therefore, for each RbBr formula unit produced, one electron is transferred from Rb to Br.

Answer:

Molten solder flows between the two base metals being soldered because of capillary action.

Explanation:

Capillary molten solder takes place when a metal and a melt is brought into contact with the tube and an accessory after heating. Due to the phenomenon of capillary action, the molten metal rises and extends in any direction, due to the small space that remains between the wall of the tube and that of the fitting; With this, when cooling, a completely hermetic union is achieved.

Molten solder flows between two base metals due to capillary action, which allows the solder to spread evenly and adhere to the bases, metal ensuring a strong bond.

Molten solder flows between the two base metals being soldered mainly because of a process known as capillary action. Capillary action is a natural occurrence where a liquid, in this case the molten solder, moves along a narrow space, such as between the two pieces of metal, against the force of gravity. The solder is drawn into the space and spread evenly between the two base metals because of the adhesive forces between the liquid and the surrounding materials are stronger than the cohesive forces within the liquid.

This phenomenon plays a fundamental role in the soldering process, as it allows the solder to spread evenly and adhere to the base metals, ensuring a strong and stable bond between them. This principle is used not only in soldering but also in various other fields such as plant biology, painting, and inkjet printing.

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Answer:

The activation energy is lowered by +85.35 KJ/mol

Explanation:

Rate constant is related to the activation energy and the temperature of the reaction through

K = Ae⁻ᴱᵃ/ᴿᵀ

In K = (In A) - (Ea/RT)

In K = (-Ea/RT) + In A

where k = the activity constant

Ea = activation energy

R = molar gas constant

T = absolute temperature in Kelvin

₁₂

At point 1, without the catalyst

In K₁ = (-Ea₁/RT) + In A (eqn 1)

At point 2 with the catalyst

In K₂ = (-Ea₂/RT) + In A (eqn 2)

Note that the molar gas constant, the absolute temperature in Kelvin and the activity constant are all the same for both points.

Subtract (eqn 1) from (eqn 2)

In K₂ - In K₁ = (-Ea₂/RT) - (-Ea₁/RT) + In A - In A

In (K₂/K₁) = (1/RT) (Ea₁ - Ea₂)

We were told that K₂/K₁ = 2.3 × 10¹⁴, then find the difference in Ea

R = 8.314 J/mol.K, T = 37°C = 310.15 K

In (2.3 × 10¹⁴) = (1/(8.314×310.15) (Ea₁ - Ea₂)

33.1 = (1/2578.5871) (Ea₁ - Ea₂)

(Ea₁ - Ea₂) = 33.1 × 2578.5871 = 85351.23 J/mol

Therefore, the activation energy is lowered by +85.35 KJ/mol

The activation energy is lowered by +85.35 KJ/mol

[tex]K = Ae^{\frac{-E_a}{RT} }\\\\In K = (In A) -( \frac{-E_a}{RT})\\\\In K = (\frac{-E_a}{RT}) + In A[/tex]

where

1. At point 1, without the catalyst

[tex]In K_1 = (-Ea_1/RT) + In A[/tex]............(i)

2. At point 2 with the catalyst

[tex]In K_ = (-Ea_2/RT) + In A[/tex]...........(ii)

On subtracting equation (i) from (ii)

[tex]In K_2 - In K_1 = (-Ea_2/RT) - (-Ea_1/RT) + In A - In A\\\\In (K_2/K_2) = (1/RT) (Ea_1 - Ea_2)[/tex]

Given:

On solving:

[tex]In (2.3 * 10^{14}) = (1/(8.314*310.15) (Ea_1 - Ea_2)\\\\33.1 = (1/2578.5871) (Ea_1 - Ea_2)\\\\(Ea_1 - Ea_2) = 33.1 * 2578.5871\\\\ (Ea_1 - Ea_2)= 85351.23 J/mol[/tex]

Thus, the activation energy is lowered by +85.35 KJ/mol.

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Answer:

kp  = 1.55

Explanation:

To get to this reaction, we just need to sum the other two reactions, and multiply coefficients if it's needed so:

C(s) + 2H₂O(g) ⇄CO₂(g) + 2H₂(g)   Kp1 = 3.23    (1)

H₂(g) + CO₂(g) ⇄ H₂O(g) + CO(g)    Kp2 = 0.601   (2)

we will multiply eqn 2 by 2 because we need to eliminate H₂

2H2(g) + 2CO2(g) ⇄ 2H2O(g) + 2CO(g)    Kp = (0.601)²

Now, add both equation

C(s) + 2H₂O(g) ⇄CO₂(g) + 2H₂(g)   Kp1 = 3.23  

2H2(g) + 2CO2(g) ⇄ 2H2O(g) + 2CO(g)    Kp = (0.601)²

we have,

C(s) + CO₂(g) ⇄ 2CO(g)  

Now the value of Kp will be:

kp = 3.23 × (0.693)²

kp  = 1.55

The molarity and molality of a solution of 14. g of benzoic acid dissolved in 425.mL of a solvent with a density of 0.92 g/mL are 0.26 M and 0.28 m respectively.

To calculate the molarity and molality of the solution, we'll first need to know how many moles of benzoic acid (C7H6O2) are present. Simply converting grams to moles using the molar mass of benzoic acid (122.12 g/mol), we get 14.g ÷ 122.12 g/mol = 0.11 mol.

Molarity is defined as moles of solute divided by liters of solution. Therefore, the molarity of the solution would be 0.11 mol ÷ 0.425 L = 0.26 M.

On the other hand, molality is calculated as moles of solute divided by kilograms of solvent. To obtain the mass of the solvent in kg, we need to multiply the volume by the density, 425.mL x 0.92 g/mL = 391 g = 0.391 kg. Consequently, the molality of the solution would be 0.11 mol ÷ 0.391 kg = 0.28 m.

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Answer:

A) i) 0.002 ppb (ii) 2ppt (iii) 3.7 x 10^(-6) μM

B) i)0.002 ppm. (ii) 2ppb

Explanation:

A) We know that 1 ppb = 1 μgram per liter, and so the concentration of Mirex in ppb would be 0.002 ppb.

1 ppt = 1 nanogram per liter of water, so the concentration of Mirex in ppt would be 2 ppt;

(0.002 μg/L) (100ng/μg) (1ppt/ng/L) = 2ppt.

Now, MW of Mirex = 540 g/mol ≡ μg/μmol

Thus, 1 μmole = 540 μgram,

Hence, the concentration of Mirex in μmoles would be;

(0.002 μg/L)/(540 μg/μmol) = 3.7 x 10^(-6) μM

B) i) 1 ppm = 1 μgram per gram.

Thus, the concentration of Mirex in ppm would be = 0.002 ppm.

ii) Now, 1 ppb = 1 nanogram per gram.

Thus, concentration of Mirex in ppb would be = (0.002 μg/g) (100ng/μg) (1ppb ng/g) = 2ppb

Final answer:

The concentration of mirex in water samples is 0.002 ppb, 2 ppt, and approximately 3.70 x 10⁻³ μM. In fish samples, the concentration of mirex is 0.002 ppm and 2 ppb.

Explanation:

The student is asking about converting the concentration of a pesticide, mirex, in water and fish samples to various units. The concentration in water samples is given as 0.002 μg/L. The conversions are as follows:

For the fish samples:

Final answer:

The molar mass of the unknown protein is 5.79 g/mol.

Explanation:

To find the molar mass of the unknown protein, we can use the formula for osmotic pressure:

π = MRT

Where π is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging the formula, we have:

M = π / (RT)

Substituting the given values:

M = 2.45 torr / (0.0821 L.atm/(mol.K) * 298 K)

M = 0.0996 mol/L

To convert the molarity to grams per liter, we need to multiply it by the molar mass

Molar mass = (0.0996 mol/L) * (10 mL/1 L) * (5.87 mg/1 mL)

Molar mass = 5.79 g/mol

Explanation:

The dissociation equations for the carbonic acid are as follows.

    [tex]H_{2}CO_{3}(aq) + H_{2}O(l) \rightarrow HCO^{-}_{3}(aq) + H_{3}O^{+}(aq)[/tex]

  [tex]HCO^{-}_{3}(aq) + H_{2}O(l) \rightarrow (CO^{2-}_{3})(aq) + H_{3}O^{+}(aq)[/tex]

The given buffer contains [tex]KHCO_{3}[/tex] and [tex]K_{2}CO_{3}[/tex]. This means that there will be [tex]HCO^{-}_{3}[/tex] ions and [tex]CO^{2-}_{3}[/tex] ions and since, both of them are present in the second step.

Therefore, [tex]K_{a2}[/tex] is more significant with respect to this reaction.

Thus, we can conclude that [tex]K_{a2}[/tex] is more important to this buffer.

A buffer resists changes in pH. When considering carbonic acid, which is a diprotic acid i.e., it can donate two protons, the first ionization is represented by Ka1, and the second by Ka2. In a buffer solution with physiological pH close to 7, Ka1 is more important because the first ionization of carbonic acid happens more readily at this pH.

Your question asks which Ka value is more important for this buffer, which contains 0.38 M KHCO3 and 0.71 M K2CO3. Carbonic acid is a diprotic acid with Ka1 = 4.5 × 10^−7 and Ka2 = 4.7 × 10^−11.

A buffer solution resists changes in pH by neutralizing added acids and bases. The effectiveness of a buffer is determined by the pKa of the acid component of the buffer, which is the negative logarithm of its Ka. Generally, the most effective buffers are those wherein the pKa is close to the desired pH.

In our case, since carbonic acid is diprotic it has two associated Ka values. The first dissociation (giving HCO3^-) has Ka1 and the second dissociation (giving CO3^2-) has Ka2. Typically in physiological circumstances such as in our blood, where pH is close to 7, the first ionization of carbonic acid matters more because it is the one that happens readily at physiological pH. Therefore, Ka1 would be more important in this buffer solution as it represents the first ionization of the acid.

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Answer:

a lipid molecule that contains at least one carbohydrate unit

Explanation:

A glycolipid -

It refers to the lipid which consists of the carbohyde group attached via a glycosidic bond , which are basically covalent bonds , is referred to as a glycolipid .

They are present on the surface of the eukaryotic cell membranes .

Glycolipids are important to connect from one tissue to another and facilitate cellular recognition .

Hence , from the given information of the question ,

The correct answer is a lipid molecule that contains at least one carbohydrate unit .

At a constant temperature of 50 °C, a volume of 1.0 L would contain approximately 0.045 moles of gas, whereas a volume of 3.0 L would contain approximately 0.134 moles of gas. These values are based on the standard molar volume of 22.4 L/mol at Standard Temperature and Pressure (STP).

In order to make this comparison, we can use Avogadro's Law which states that equal volumes of gases, at the same temperature and pressure, will contain an equal number of molecules or moles. We would also use the standard molar volume of an ideal gas at Standard Temperature and Pressure (STP), which is about 22.4 L/mol.

Given that our system's temperature is constant at 50 °C and comparing volumes of 1.0 L and 3.0 L, we simply calculate the number of moles in each volume. For 1.0 L, if we stay consistent with molar volume at STP, it equates to about 0.045 moles of gas (1 L / 22.4 L per mole). For 3.0 L, it would equate to about 0.134 moles of gas (3 L / 22.4 L per mole).

Please note that this approximation is based on the standard molar volume at STP, and accuracy may vary at temperatures significantly different from STP (such as 50 °C).

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Explanation:

In the given solvents -

Final answer:

Toluene is the best solvent choice for separating benzophenone, diphenyl methanol, and biphenyl using TLC on silica gel due to its balanced polarity, which will allow for better differentiation between the non-polar compounds.

Explanation:

In choosing a solvent for separating benzophenone, diphenyl methanol, and biphenyl using TLC on silica gel, toluene appears to be the most appropriate choice. Since phenyl groups are fairly non-polar, a non-polar or moderately polar solvent is needed to achieve good separation. Toluene is a good choice because it has a higher polarity than hexanes but is less polar than the other solvents listed, providing a balanced environment to distinguish the three compounds on the basis of their differing polarities. A solvent like methanol or water would be too polar, causing the compounds to stay near the baseline, while hexanes might be too non-polar, failing to separate the compounds efficiently. Therefore, toluene is the recommended solvent for TLC in this scenario.

Answer:

82.53 % protein in the milk.

Explanation:

Titration formula:

Conc (acid) × Vol (acid) = Conc (base) × Vol (base)

Beginning from back-titration:

0.5 M HCl × Vol  (HCl) = 0.3512 (conc of NaOH) × 34.50 mL (vol of NaOH)

=  [tex]\frac{12.1164}{0.5}[/tex] = 24.2328 mL

⇒ Vol of HCl used in initial titration is 24.2328 mL

So from Initial Titration:

Using same titration formula,

0.5 × 24.2328 = Conc. (base) × 100 ml

Conc (base) = 12.1164 ÷ 100

= 0.121164 M.

But concentration = mass ÷ molar mass

0.121164 = [tex]\frac{100 g}{ molar mass}[/tex]

= 825.3277 g/mol of protein

⇒ [tex]\frac{825.32765}{1000}[/tex] × 100

= 82.53 % protein in the milk.

Answer:

152 mL is the volume of KOH required to reach the equivalence point.

Explanation:

[tex]HNO_3(aq)+KOH(aq)\rightarrow KNO_3(aq)+H_2O(l)[/tex]

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HNO_3[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.

We are given:

[tex]n_1=1\\M_1=0.590 M\\V_1=90.0 mL\\n_2=1\\M_2=0.350 M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]1\times 0.590 M\times 90.00=1\times 0.350 M\times V_2[/tex]

[tex]V_2=\frac{1\times 0.590 M\times 90.0 mL}{1\times 0.350 M}=151 .7 mL\approx 152 mL[/tex]

152 mL is the volume of KOH required to reach the equivalence point.

Answer:

152 ml.

Explanation:

Given:

Volume of HNO3 = 90 ml

Molar concentration of HNO3 = 0.59 M

Molar concentration of KOH = 0.35 M

Equation of the reaction

KOH + HNO3 --> KNO3 + H2O

Number of moles of HNO3 = molar concentration × volume

= 0.59 × 0.09

= 0.0531 moles.

By stoichiometry, 1 mole of HNO3 reacts with 1 mole of KOH. Therefore,

Number of moles of KOH = 0.0531 moles.

Volume = 0.0531 ÷ 0.350

= 0.152 l

= 152 ml.

Explanation:

Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.

a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement  a is wrong.

b. Expression for second order reaction is as follows:

[tex]\frac{1}{[A]} =\frac{1}{[A]_0} +kt[/tex]

the above equation can be written in the form of Y = mx + C

so, the plot between 1/[A] and t is linear. So the statement b is true.

c.

Expression for half life is as follows:

[tex]t_{1/2}=\frac{1}{k[A]_0}[/tex]

As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.

d.

Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong

Answer:

Kp = 0.16

Explanation:

Step 1: Data given

Initial pressure of Cl2 = 4.92 atm

Initial pressure of Br2 = 4.65 atm

The equilibrium pressure of BrCl is found to be 1.597 atm

Step 2: The balanced equation

Cl2(g) + Br2(g) ⇔ 2 BrCl(g

Step 3: The initial pressures

pCl2 = 4.92 atm

pBr2 = 4.65 atm

pBrCl = 0 atm

Step 4: The pressure at the equilibrium

For 1 mol Cl2 we need 1 mol Br2 to produce 2 moles BrCl

pCl2 = 4.92 - X atm

pBr2 = 4.65 - Xatm

pBrCl = 2X atm = 1.597 atm

X = 1.597/2 = 0.7985 atm

pCl2 = 4.92 - X atm = 4.92 - 0.7985 = 4.1215 atm

pBr2 = 4.65 - Xatm = 3.8515 atm

Step 5: Calculate Kp

Kp = (BrCl)² / (Cl2)*(Br2)

Kp = 1.597² / (4.1215*3.8515)

Kp = 0.16

Final answer:

To calculate Kp for the given reaction, start with the initial pressures of Cl2 and Br2, calculate the change in pressure due to the formation of BrCl, and apply these values in the Kp equation. The result is Kp = 0.084.

Explanation:

The student asked how to calculate Kp for the reaction Cl2(g) + Br2(g) ⇌ 2 BrCl(g) at a given temperature, given initial pressures for Cl2 and Br2, and the equilibrium pressure of BrCl. To solve for Kp, we utilize the change in concentration according to the reaction's stoichiometry and apply it to the equation for the equilibrium constant in terms of pressure (Kp).

Initial pressures: Cl2 = 4.92 atm, Br2 = 4.65 atm. At equilibrium, BrCl = 1.597 atm. The change in pressure for Cl2 and Br2 to form 2 BrCl is equal to the pressure of BrCl divided by 2, since the stoichiometry of the reaction dictates twice the amount of BrCl for each reactant consumed. Thus, the change (δ) is 1.597 atm / 2 = 0.7985 atm. Therefore, the pressures of Cl2 and Br2 at equilibrium are 4.92 - 0.7985 atm and 4.65 - 0.7985 atm, respectively.

To find Kp, the equation is Kp = (PBrCl)2 / (PCl2 × PBr2). Substituting the equilibrium pressures into this equation gives Kp = (1.5972) / ((4.92 - 0.7985) × (4.65 - 0.7985)). Solving this yields Kp = 0.084 (to two decimal places).

Answer: Option (c) is the correct answer.

Explanation:

It is known that like dissolves like. As cyclohexane is a hydrocarbon and we know that hydrocarbons are non-polar in nature. So, a non-polar compound will be soluble in a hydrocarbon.

For example, NaBr is an ionic compound and when it i added to liquid cyclohexane then it will not dissolve. Similarly, [tex]CH_{2}Cl_{2}[/tex] and HI are polar covalent compounds and they will not dissolve in liquid cyclohexane.

But [tex]CH_3CH_2CH_2CH_2CH_3[/tex] is a non-polar compound. So, when it is added to liquid cyclohexane then it will readily dissolve.

Thus, we can conclude that pentane ([tex]CH_3CH_2CH_2CH_2CH_3[/tex]) will most likely dissolve in cyclohexane to form a solution.