In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 5 mL 4.0M acetone + 5 mL 1.0 M HCl + 5 mL 0.0050 M I2 + 10 mL H2O What was the molarity of acetone in the reaction mixture ? The volume of the mixture was 25 mL, and the number of moles of acetone was found to be 0.020 moles. MA = no. moles A / V of solution in liters

Answer:

[tex]K_{b}[/tex] is 0.000438 and [tex]pK_{b}[/tex] is 3.36

Explanation:

Methylamine is a monoprotic base.

For a monoprotic base, [tex]K_{b}=\frac{ca^{2}}{(1-a)}[/tex]

where, c is concentration of base in molarity and a is it's degree ionization

Here [tex]a=\frac{6.4}{100}=0.064[/tex] and c = 0.100 M

So, [tex]K_{b}=\frac{(0.100)\times (0.064)^{2}}{(1-0.064)}=0.000438[/tex]

We know, [tex]pK_{b}=-logK_{b}[/tex]

Hence, [tex]pK_{b}=-log(0.000438)=3.36[/tex]

Answer:

H2O<en<phen

Explanation:

The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.

The d-splitting for each ligand is as follows:

H2O (weak field ligand): smallest d-splitting

en (stronger field ligand): medium d-splitting

phen (strongest field ligand): largest d-splitting

The complexes from least to greatest d-splitting are as follows:

[Co(H2O)6]3+

[Co(en)3]3+

[Co(phen)3]3+

The reason for this ordering is that stronger field ligands cause a larger d-splitting. This is because stronger field ligands interact more strongly with the metal's d orbitals, which splits the d orbitals into two sets of orbitals with different energies.

The d-splitting of a transition metal complex is the energy difference between the two sets of d orbitals that are split by the ligand field. The magnitude of the d-splitting depends on the strength of the ligand field. Stronger field ligands cause a larger d-splitting.

The following table shows the spectrochemical series of ligands, which is a ranking of ligands from weakest to strongest field:

| Ligand | Spectrochemical series |

|---|---|---|

| H2O | Weak field |

| en | Medium field |

| phen | Strong field |

Based on the spectrochemical series, we can predict that the d-splitting for each ligand is as follows:

H2O (weak field ligand): smallest d-splitting

en (stronger field ligand): medium d-splitting

phen (strongest field ligand): largest d-splitting

The following table shows the expected d-splitting of the [Co(H2O)6]3+, [Co(en)3]3+, and [Co(phen)3]3+ complexes:

Complex Ligand D-splitting

[Co(H2O)6]3+ H2O Smallest

[Co(en)3]3+ en Medium

[Co(phen)3]3+ phen Largest

The d-splitting of a transition metal complex affects the color of the complex. Complexes with a larger d-splitting absorb higher energy light, which is in the visible region of the spectrum. This is why complexes with strong field ligands tend to be colored.

The following table shows the observed colors of the [Co(H2O)6]3+, [Co(en)3]3+, and [Co(phen)3]3+ complexes:

Complex Ligand Color

[Co(H2O)6]3+ H2O Purple

[Co(en)3]3+ en Purple-pink

[Co(phen)3]3+ phen Pink

The observed colors of the complexes are consistent with the predicted d-splitting. The [Co(phen)3]3+ complex has the largest d-splitting and absorbs the highest energy light, which is why it is pink. The [Co(H2O)6]3+ complex has the smallest d-splitting and absorbs the lowest energy light, which is why it is purple.

For more such information on:  ligand

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Answer:

The answer is 0.017 μmol of mercury iodide

Explanation:

To know the micromoles we must use the following conversion, 1 mmol = 1000μmol. we use a rule of three for the calculation of micromoles:

1mmol------------------1000μmol

1.7x10^-5mmol------- Xμmol

Clearing the X, we have:

X μmol = (1.7x10^-5x1000)/1 = 0.017μmol

Answer:

[tex]2.55*10^{11[/tex]

Explanation:

Equation for the heterogeneous system is given as:

[tex]2A_{(aq)} + 3 B_{(g)} + C_{(l)}[/tex]      ⇄      [tex]2D_{(s)}[/tex]    [tex]+[/tex]     [tex]3E_{(g)}[/tex]

The concentrations and pressures at equilibrium  are:

[tex][A] = 9.68*10^{-2}M[/tex]

[tex]P_B = 9.54*10^3Pa[/tex]

[tex][C]=14.64M[/tex]

[tex][D]=10.11M[/tex]

[tex]P_E=9.56*10^4torr[/tex]

If we convert both pressure into bar; we have:

[tex]P_B = 9.54*10^3Pa[/tex]

[tex]P_B = (9.54*10^3)*\frac{1}{10^5} bar[/tex]

[tex]P_B=9.54*10^{-2}bar[/tex]

[tex]P_E=9.56*10^4torr[/tex]

1 torr = 0.001333 bar

[tex]9.54*10^4 *0.001333 = 127.5 bar[/tex]

[tex]K=\frac{[P_E]^3}{[A]^2[P_B]^3}[/tex]

[tex]K=\frac{(127.5)^3}{(9.68*10^{-2})^2(9.54*10^{-2})^3}[/tex]

[tex]K=2.55*10^{11[/tex]

Answer: The equilibrium constant for the reaction is 2.47

Explanation:

We are given:

Initial moles of ammonia gas = 29.0 moles

Equilibrium moles of nitrogen gas = 13.0 moles

Volume of the tank = 7.50 L

Molarity is calculated by using the formula:

[tex]\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of tank}}[/tex]

[tex]\text{Initial molarity of ammonia}=\frac{29.0}{7.50}=3.87M[/tex]

[tex]\text{Equilibrium molarity of nitrogen gas}=\frac{13.0}{7.50}=1.73M[/tex]

The chemical equation for the decomposition of ammonia follows:

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

Initial:             3.87

At eqllm:       3.87-2x    x      3x

Evaluating the value of 'x'

[tex]\Rightarrow 3x=1.73\\\\x=\frac{1.73}{3}=0.577[/tex]

So, equilibrium concentration of ammonia = (3.87 - 2x) = [3.87 - 2(0.577)] = 2.716 M

Equilibrium concentration of nitrogen gas = x = 0.577 M

The expression of [tex]K_{eq}[/tex] for above equation follows:

[tex]K_{eq}=\frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]

Putting values in above equation, we get:

[tex]K_{eq}=\frac{(2.716)^2}{0.577\times (1.73)^3}\\\\K_{eq}=2.47[/tex]

Hence, the equilibrium constant for the reaction is 2.47

Answer: The volume of NaOH required is 402.9 mL

Explanation:

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

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]     .....(1)

Molarity of HCl solution = 0.315 M

Volume of solution = 503.4 mL = 0.5034 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

[tex]0.315M=\frac{\text{Moles of HCl}}{0.5034L}\\\\\text{Moles of HCl}=(0.315mol/L\times 0.5034L)=0.1586mol[/tex]

Molarity of sulfuric acid solution = 0.125 M

Volume of solution = 503.4 mL = 0.5034 L

Putting values in equation 1, we get:

[tex]0.125M=\frac{\text{Moles of }H_2SO_4}{0.5034L}\\\\\text{Moles of }H_2SO_4=(0.125mol/L\times 0.5034L)=0.0630mol[/tex]

As, all of the acid is neutralized, so moles of NaOH = [0.1586 + 0.0630] moles = 0.2216 moles

Molarity of NaOH solution = 0.55 M

Moles of NaOH = 0.2216 moles

Putting values in equation 1, we get:

[tex]0.55M=\frac{0.2216}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.2216}{0.55}=0.4029L=402.9mL[/tex]

Hence, the volume of NaOH required is 402.9 mL

Answer:

See explanation below

Explanation:

In order to solve this, we'll do it by parts, as this exercise takes some time to solve it, however, the procedure it's pretty easy to understand.

A. Molecular weight of the KHP

To do this, we need the atomic weight of each element of the KHP. These are the following:

H = 1 g/mol; K = 39 g/mol; C = 12 g/mol; O = 16 g/mol

Now, let's calculate the molecular weight. Remember to multiply the number of atoms by the atomic weight:

MM KHP = (1*5) + (39) + (4*16) + (12*8) = 204 g/mol

B. moles of KHP used

In this part, we already have the molecular weight, so, we can calculate the moles with the expression:

n = m/MM  (1)

The mass used of KHP is 0.5591 so the moles are:

n = 0.5591/204 = 2.74x10⁻³ moles

C. Concentration of NaOH

As the problem states, the KHP can be considered as a monoprotic acid, therefore, we can assume that the mole ratio between NaOH and KHP is 1:1 and we can use the following expression to calculate the concentration of the base:

MaVa = MbVb  (2)

But moles:

n = M*V  (3)

We have the moles of the KHP used, and the volume used to standarize the base, so we can solve for Molarity of the base:

Mb = na / Vb

Mb = 2.74x10⁻³ / 0.01339 = 0.2046 M

D. Molarity of vinegar solution

Using expression (2), we can calculate the vinegar solution, as we have the base volume used and volume of vinegar so:

MaVa = MbVb

Ma = MbVb/Va

Ma = 0.2046 * 8.38 / 5 = 0.3429 M

E. %W/V of vinegar

In this case, we use the following expression:

%W/V = mass solute / V solution * 100   (4)

The volume of solution would be the volume of the vinegar and volume of the base:

V solution = 8.38 + 5 = 13.38 mL

The mass of vinegar can be calculated, we have the concentration and volume, we can calculate the moles using expression (3):

n = 0.3429 * 0.005 = 1.71x10⁻³ moles

The mass of vinegar using the molecular weight of acetic acid (60 g/mol):

m = 1.71x10⁻³ * 60 = 0.1026 g

So the %:

%W/V = 0.1026/13.38 * 100 = 0.77%

F.  mg of ascorbic acid

We do the same thing as in part C and then, the mass of ascorbic acid can be calculated with the molecular weight:

MaVa = MbVb = na

na = 0.2046 * 0.01356 = 2,77x10⁻³  moles

m = 2.77x10⁻³ * 176.1271 = 0.4878 g or simply 487.8 mg

A) The Molecular Weight of KHP is 204.23 g/mol. B). Moles of KHP were used in the standardization of the NaOH solution .C) The concentration of NaOH is 0.2045 mol/L. D). The vinegar solution's molarity is 0.3426 mol/L, and E).it has a weight/volume percentage of 2.06%, while F). the Vitamin C tablet contains 487 mg of ascorbic acid.

Let's go step by step to find each required value:

A. The Molecular Weight of KHP (HKC₈H₄O₄) in g/mol :

The molecular formula of KHP (Potassium hydrogen phthalate) is HKC₈H₄O₄. Thus:

Adding these values gives us the Molecular Weight of KHP: 204.23 g/mol.

B. Moles of KHP were used in the standardization of the NaOH solution :

Moles of KHP = Mass of KHP / Molecular Weight of KHP

                       = 0.5591 g / 204.23 g/mol

                       = 0.002737 moles

                       = 2.74x10⁻³ moles

C. The concentration of NaOH solution in (mol/L) :

Molarity (M) = Moles of Solute / Volume of Solution in L
Volume of NaOH in L = 13.39 mL × (1 L / 1000 mL) = 0.01339 L
Thus, Molarity of NaOH = 0.002737 moles / 0.01339 L

                                       = 0.2045 mol/L

D. The molarity of the vinegar solution (mol/L) :

Firstly, we know the mole ratio between acetic acid (CH₃COOH) and NaOH is 1:1.
Moles of NaOH used for vinegar = Molarity of NaOH × Volume in L

                                                       = 0.2045 mol/L × 0.00838 L

                                                        = 0.001713 moles

Since it’s a 1:1 ratio, the moles of acetic acid (CH₃COOH) in the vinegar is 0.001713 moles
Molarity of vinegar = Moles of Acetic Acid / Volume of Vinegar Solution in L
Molarity of Vinegar = 0.001713 moles / 0.005 L

                                = 0.3426 mol/L

E. The weight/volume percentage of the vinegar solution (g/100 mL) :

Weight of Acetic Acid (CH₃COOH) in grams = Moles × Molecular Weight

                                     =  0.001713 moles × 60.05 g/mol = 0.1029 g
Since the vinegar sample was 5.00 mL, weight/volume %

                                     = (0.1029 g / 5.00 mL) × 100 = 2.06%

F) . The amount of ascorbic acid in the Vitamin C tablet in (mg) :

Moles of NaOH used = Molarity of NaOH × Volume in L

                                    = 0.2045 mol/L × 0.01356 L = 0.002774 moles
Since mole ratio between ascorbic acid and NaOH is 1:1, moles of ascorbic acid = 0.002774 moles
Mass of ascorbic acid = Moles × Molecular Weight = 0.002774 moles × 176.1271 g/mol = 0.487 grams
Thus, mass in mg = 0.487 g × 1000 = 487 mg

To address a conditional situation in Excel, you would use the IF function. In this case, the asked function is =IF(E9>=$B$7, 0, D9*$B$6/$B$5), wherein a condition is evaluated and if true, returns 0, if false, conducts a specified calculation.

The question asked is about using the IF statement in Excel, which falls under the Computers and Technology topic. Excel's IF function is used to create conditional statements, where different calculations can be performed depending on whether a particular condition is met.

For the given question, you are asked to insert the IF function in cell I9. The condition to test is if the value in cell E9 is greater than or equal to the value in cell B7. If this condition is true, the function will return 0. If the condition is not met, the function will execute a calculation: it multiplies the values in cells D9 and B6, then divides the result by the value in cell B5.

Your final function should look like this: =IF(E9>=$B$7,0,D9*$B$6/$B$5).

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Answer:

a), b), c) & f)

Explanation:

d) does not apply because Ms value can be either +½ or -½

e) does not apply because Ml - values range from -l to +l, hence l= 2 doesn't exist when l= 1

The four quantum numbers that govern the possible states of an electron in an atom are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (ml), and the spin quantum number (ms). Allowable combinations of quantum numbers follow specific rules.

The four quantum numbers that govern the possible states of an electron in an atom are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (ml), and the spin quantum number (ms).

Allowable combinations of quantum numbers for an electron can be determined according to the following rules:

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:

Rank from most acidic to most basic is

Explanation:

This question is missing options.

Options are

To rank these solutions first calculate either pH or pOH of these solutions.

We will use pH as an indicator to rank these.

Relation between pH and pOH is

    pH + pOH = 14    ................... Eq (A)

For pOH = 8.55

     use equation A

     pH + 8.55 = 14

     pH = 5.45

For pH = 5.45

     pH = 5.45

For 0.0023 M  HCl

HCl is a strong acid. It will ionize 100% in aqueous solution and produce Hydronium Ion. Formula to calculate pH of 0.0023 M  HCl is,

     pH = - log (Molarity)

     pH = - log (0.0023)

     pH = - (-2.64)

     pH = 2.64

For 0.0018 M  KOH

KOH is a base. It will produce Hydroxide Ion in aqueous solution. First calculate pOH and convert it into pH.

    pOH = - log (Molarity)

    pOH = - log (0.0018)

    pOH = 2.74

   use equation A

   pH + 2.74 = 14

   pH = 11.26

Lower the pH more acidic the solution is. Rank from most acidic to most basic is

The solutions, arranged from most acidic to most basic, are H₂SO₄, HCl, NH₄NO₃, NaCl, NaOH, and NaCN.

The arrangement of the given solutions in order of decreasing acidity is as follows:

Strong acids completely dissociate in water, producing a large number of hydronium ions (H₃O⁺) and making the solution highly acidic. Neutral salts, such as NH₄NO₃ and NaCl, do not affect the pH of the solution. Strong bases, like NaOH, ionize completely and produce a high concentration of hydroxide ions (OH⁻), resulting in a basic solution. Weak bases, such as NaCN, only partially ionize, resulting in a lower concentration of OH⁻ ions and a slightly basic solution.

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The complete question is here:

Arrange the following aqueous solutions, all at 25 ∘C, in order of decreasing acidity. Rank from most acidic to most basic. To rank items as equivalent, overlap them.

H₂SO₄

HCl

NH₄NO₃

NaCl

NaOH

NaCN

In a titration involving a diprotic acid, the equivalence points occur when the acid has lost its protons. If the first equivalence point occurs at 100 mL, the second should occur at 200 mL. resultant species in the solution will vary based on the volume of NaOH added.

A diprotic acid has two protons to donate in a reaction. During titration, protons are removed one at a time, thus presenting two titration curves or equivalence points. If the first equivalence point occurs at 100.0 mL, the second equivalence point typically occurs at twice that volume because the second proton is just as readily removed as the first. Therefore, the second equivalence point will be at 200.0 mL.

Different volumes of added NaOH will result in different major species present. For example, before the equivalence point is reached (0 mL < V< 25 mL), the pH increases gradually as the diprotic acid reacts with the added NaOH to form its conjugate base. At the equivalence point (V = 25 mL), pH increases abruptly as the reaction transitions from acidic to either neutral or basic, depending on whether the diprotic acid is strong or weak, respectively. After the equivalence point (V > 25 mL), the pH is determined by the amount of added NaOH.

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The vapor pressures of chloroform and acetone in the solution are 52.02 torr and 273.50 torr, respectively, and the total vapor pressure above the solution is 325.52 torr.

To calculate the vapor pressures of chloroform and acetone above the solution, we should make use of Raoult's law. According to Raoult's law, the partial pressure of each component of a mixture is the product of the vapor pressure of the pure component and its mole fraction in the mixture.

First, calculate the mole fractions of each substance. The molar mass of chloroform (CHCl3) is about 119.38 g/mol, and the molar mass of acetone (CH3COCH3) is about 58.08 g/mol. Therefore, the mole fractions of chloroform and acetone are 4.08 g CHCl3 * (1 mol / 119.38 g) = 0.0342 mol and 9.29 g CH3COCH3 * (1 mol / 58.08 g) = 0.1599 mol, respectively. The total moles are 0.0342 mol + 0.1599 mol = 0.1941 mol, so the mole fraction of chloroform is 0.0342/0.1941 = 0.176 and that of acetone is 0.1599/0.1941 = 0.824.

Next, apply Raoult's Law to find the partial pressure of each component in the mixture. The vapor pressure of chloroform is 0.176 * 295 torr = 52.02 torr, and the vapor pressure of acetone is 0.824 * 332 torr = 273.50 torr.

The total vapor pressure above the solution is the sum of the vapor pressures of chloroform and acetone, that is, 52.02 torr + 273.50 torr = 325.52 torr.

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Explanation:

Ethanol can be oxidized to ethanal or acetaldehyde which is further oxidized to acid that is acetic acid.

[tex]CH_3CH_2OH[/tex]→  [tex]CH_3CHO[/tex] [oxidation by loss of hydrogen]

Acetaldehyde can also be reduced back to ethanol again by adding hydrogen to it by using a reducing agent that is sodium tetrahydro borate, NaBH4.

     [tex]CH_3CHO[/tex]  →  [tex]CH_3CH_2COOH[/tex] [reduction by the gain of electrons]

Here, the oxidizing agent used is[tex]Cr_3O[/tex] in the presence of acetone.

To convert ethanol to the next more reduced form in the series, it will form an aldehyde called acetaldehyde.

To convert ethanol to the next more reduced form in the series, we need to consider the oxidation level of the molecules. Ethanol is an alcohol with the -OH group bonded to a carbon atom attached to two other carbon atoms, which means it will form an aldehyde upon oxidation. Acetaldehyde is the next more reduced form in the series. Acetaldehyde has a carbonyl group (C=O) instead of an -OH group.

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Answer:

1.63 g of H₂O is the theoretical yield

Explanation:

We determine the reactants for the reaction:

HCl, NaOH

We determine the products for the reaction:

H₂O, NaCl

The equation for this neutralization reaction is:

HCl (aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

We use both masses, from both reactants to determine the limiting.

First, we convert the mass to moles.

3.3 g . 1mol / 36.45 g = 0.090 moles of HCl

6.6 g . 1mol / 40 g = 0.165 moles of NaOH

Ratio is 1:1, so for 0.165 moles of hydroxide I need the same amount of acid. I have 0.090 HCl so the acid is the limiting reagent.

Let's work with stoichiometry. Ratio is 1:1, again.

1 mol of acid can produce 1 mol of water

Therefore, 0.090 moles of acid must produce 0.090 moles of H₂O

We convert the moles to mass, to define the theoretical yield

0.090 mol . 18g / 1 mol = 1.63 g

Answer:

The theoretical yield of H2O is 1.63 grams

Explanation:

Step 1: Data given

Mass of hydrochloric acid = 3.3 grams

Mass of sodium hydroxide = 6.6 grams

Molar mass hydrochloric acid (HCl) = 36.46 g/mol

Molar mass sodium hydroxide (NaOH) = 40.0 g/mol

Step 2: The balanced equation

HCl + NaOH → NaCl + H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles HCl = 3.3 grams / 36.46 g/mol

Moles HCl = 0.0905 moles

Moles NaOH = 6.6 grams / 40.0 g/mol

Moles NaOH = 0.165 moles

Step 4: Calculate the limiting reactant

HCl is the limiting reactant. There will react 0.0905 moles. NaOH is in excess. There will react 0.0905 moles moles. There will remain 0.165 - 0.0905 = 0.0745 moles

Step 5: Calculate moles H2O

For 1 mol HCl we need 1 mol NaOH to produce 1 mol NaCl and 1 mol H2O

For 0.0905 moles HCl we'll have 0.0905 moles H2O

Step 6: Calculate mass H2O

Mass H2O = moles * molar mass

Mass H2O = 0.0905 * 18.02 g/mol

Mass H2O = 1.63 grams

The theoretical yield of H2O is 1.63 grams

Answer: The theoretical yield of barium sulfate is 50.9 grams

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     .....(1)

Given mass of sodium sulfate = 32.4 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of sodium sulfate}=\frac{32.4g}{142g/mol}=0.228mol[/tex]

Given mass of barium chloride = 65.3 g

Molar mass of barium chloride = 208.23 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of barium chloride}=\frac{65.3g}{208.23g/mol}=0.314mol[/tex]

The chemical equation for the reaction of barium chloride and sodium sulfate follows:

[tex]Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl[/tex]

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of barium chloride

So, 0.228 moles of sodium sulfate will react with = [tex]\frac{1}{1}\times 0.228=0.228mol[/tex] of barium chloride

As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sodium sulfate produces 1 mole of barium sulfate.

So, 0.228 moles of sodium sulfate will produce = [tex]\frac{1}{1}\times 0.228=0.228moles[/tex] of barium sulfate

Now, calculating the mass of barium sulfate from equation 1, we get:

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.228 moles

Putting values in equation 1, we get:

[tex]0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol\times 223.4g/mol)=50.9g[/tex]

Hence, the theoretical yield of barium sulfate is 50.9 grams

Answer:

HF > H2O > NH3 > H2S > CBr4=CO2=0

Explanation:

Dipole moment is a vector quantity. Its a measure of polarity of a bond in a molecule and also a meaure of separation of positive and negative charge in a system.It occurs due to electronegativity difference between the atoms in a molecule.

In order for a molecule to have dipole moment, a molecule must exhibit high electronegativity difference and the shape of the moloecule must be asymmetry

HF has the highest electronegativity difference among all the molecules listed above hence its dipole moment is the greatest.

[tex]H_{2}O[/tex] has a bent structure. There are two O-H bonds hence more charge dipoles. The dipole moment is less than HF molecule because of the net dipole moments of two O-H bonds.

[tex]CO_{2}[/tex] is a linear molecule.However it has polar bonds.But because of the shape of the molecule, the two C-O bond dipoles cancel out each other hence the overall dipole moment will be zero.

Similarly in [tex]CBr_{4}[/tex], ,since the molecule is symmetry, the bond dipole cancels each other out hence the overall dipole moment will be zero.  

The increasing order of the dipole moment will be:

[tex]\rm CO_2[/tex] < [tex]\rm CBr_4[/tex] < [tex]\rm H_2S[/tex] < [tex]\rm NH_3[/tex] < [tex]\rm H_2O[/tex] < HF.

Dipole moment can be described as the measure of the polarity of the molecule. The higher the electronegativity difference, the more polar the bond.

In the given molecules,

The increasing order of the dipole moment will be:

[tex]\rm CO_2[/tex] < [tex]\rm CBr_4[/tex] < [tex]\rm H_2S[/tex] < [tex]\rm NH_3[/tex] < [tex]\rm H_2O[/tex] < HF.

The molecule with the highest dipole moment is HF.

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Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, what are the partial pressures of NO2 and N2O4 after equilibrium has been achieved at 45 degrees C?

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619=  [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373  

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10  = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

Answer:A. Dilute copper and ascorbic acid solutions should be disposed of in the waste container in the hood.

C. Metal and graphite electrodes should be rinsed with water, dried, and returned to their positions on the lab bench.

D. The contents of the half-cell module should be disposed of in the waste container in the hood.

Explanation: An electrochemical cell is a cell that has the capability of producing Electric energy from chemical reaction (voltaic cells) or using Electric energy to make chemical reactions to take place(electrolytic cell). For a proper or effective waste disposal in an electrochemical cell the following options are correct.

Dilute copper and ascorbic acid solutions should be disposed of in the waste container in the hood.

Metal and graphite electrodes should be rinsed with water, dried, and returned to their positions on the lab bench.

The contents of the half-cell module should be disposed of in the waste container in the hood.

In the Electrochemical Cells Experiment, the correct statements regarding waste disposal include disposing copper and ascorbic acid solutions in the waste container, flushing rinsings from the half-cell module down the sink, and rinsing and drying the electrodes.

The correct statements with respect to waste disposal in the Electrochemical Cells Experiment are:

Answer:

188 s

Explanation:

We are told the reaction is second order respect to A so we know the expression for the rate law is

rate = - Δ[A]/Δt = k[A]²

where the symbol Δ stands for change, [A] is the concentration of A, and k is the rate constant.

The integrated rate law for this equation from calculus is

1 / [A]t = kt + 1/[A]₀

where  [A]t is the concentration of A at time t, k is the rate constant, and [A]₀ is the initial concentration.

Since we have all the information required to solve this equation lets plug our values

1 / 0.280= 0.0130x t  + 1 / 0.890

( 1 / 0.280 - 1 / 0.890)M⁻¹ = 0.0130 M⁻¹ ·s⁻¹t

t = 188 s

Answer:

This shown on the second uploaded image

Explanation:

What is occurring in this reaction is the further deprotonation of the base and this would now react with HCl to give the final product

Final answer:

To calculate the entropy change for the surroundings during the reaction of N2(g) and O2(g) to form NO2(g), the standard enthalpy change of the reaction must be known. This value can be used with the equation ΔS° = -ΔH°/T to find the entropy change of the surroundings.

Explanation:

The student is asking to calculate the entropy change for the surroundings (ΔS°surroundings) when 1.90 moles of N2(g) react with O2(g) to form NO2(g) according to the reaction N2(g) + 2O2(g) → 2NO2(g) at standard conditions of 298 K.

To find this, we'll first need the standard enthalpy change (ΔH°) for the reaction, which can be obtained from standard thermodynamic tables. We then apply the equation ΔS° = -ΔH°/T, which relates the entropy change of the surroundings to the enthalpy change of the system at a constant temperature (T).

Given that the standard enthalpy change for the formation of NO2(g) is 33.2 kJ/mol, and the reaction produces 2 moles of NO2 for 1 mole of N2, the standard enthalpy change for the reaction when 1.90 moles of N2 react is (1.90 moles * 33.2 kJ/mol * 2). We'll convert kJ to J by multiplying by 1,000 and then calculate ΔS°surroundings.

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Final answer:

The standard entropy change for the reaction N2(g) + 2O2(g) → 2NO2(g) at 298K can be calculated using the equation: ΔS° = 2∙S°(NO2) - [S°(N2) + 2∙S°(O2)]. The entropy change is -198.3 J/mol K.

Explanation:

The standard entropy change for the reaction N2(g) + 2O2(g) → 2NO2(g) at 298K can be calculated using the equation: ΔS° = 2∙S°(NO2) - [S°(N2) + 2∙S°(O2)].

Using the standard entropy values at 298K from the reference table, we can substitute the values and calculate the entropy change:

ΔS° = 2∙192.5 - [191.5 + 2∙130.6]

ΔS° = -198.3 J/mol K

Answer:

a) The Venn diagram is presented in the attached image to this answer.

b) Check Explanation.

c) 0.3333

d) 0.1429

e) 0.6

Explanation:

Let the probability of a lake having high BOD be P(B) = 30% = 0.3

Probability of a lake having low pH = P(P) = 20% = 0.2

Probability that a lake has high BOD and low pH = P(B n P) = 10% = 0.1

Then, probability that a lake has normal BOD = P(B') = 1 - P(B) = 1 - 0.3 = 0.7

Probability that a lake has normal pH = P(P') = 1 - P(P) = 1 - 0.2 = 0.8

Total probability = P(U) = 100% = 1

a) The Venn diagram is presented in the attached image to this answer.

b) Two events are independent if and only if, P(A|B) = P(A) or P(B|A) = P(B).

For this question,

P(B|P) = P(B n P)/P(P) = 0.1/0.2 = 0.5 ≠ P(B) (which is 0.3)

And P(P|B) = P(B n P)/P(B) = 0.1/0.3 = 0.333 ≠ P(P) (which is 0.2).

It is evident that the two events aren't independent of each other.

c) If a stream has high BOD, what is the probability it will also have low pH?

This probability is given as P(P|B) meaning that, the probability of a lake having low pH given that it has high BOD.

Mathematically, this conditional probability is given by

P(P|B) = P(B n P)/P(B) = 0.1/0.3 = (1/3) = 0.3333

d) If a stream has normal levels of BOD, what is the probability it will also have low pH.

This probability is given as P(P|B'); that is, the probability of a lake having low pH given that it has normal BOD.

Mathematically,

P(P|B') = P(B' n P)/P(B')

P(B') = 0.7 (already found above)

But P(B' n P) = ?

Mathematically,

P(B' n P) = P(P) - P(B n P) = 0.2 - 0.1 = 0.1

P(P|B') = 0.1/0.7 = 0.1429

e) What is the probability that a stream will not exhibit either pollution indicator, i.e., will have normal BOD and pH levels?

This is given as P(B' n P')

Mathematically, this represents the region in the Venn diagram outside of the circles representing P(B) and P(P) and it's given mathematically as,

P(B' n P') = P(U) - [P(B n P') + P(B' n P) + P(B n P)] = 1 - (0.2 + 0.1 + 0.1) = 1 - 0.4 = 0.6 or 60%

When the stream has high BOD the probability of low pH is 0.33 and 0.142 when it has normal BOD. 0.6 is the probability that a stream will exhibit neither indicator.

BOD is the biological oxidation demand, that tells about the dissolved oxygen amount in the water body. BOD along with pH are the indicator of pollution.

The Venn diagram is attached in the image below.

The two indicators, high BOD and low pH are dependent on each other. It can be shown as:

[tex]\rm P(A|B) = P(A) \;or \;P(B|A) = P(B)[/tex]

But,

[tex]\begin{aligned}\rm P(B|P) &= \rm \dfrac{P(B \;n \;P)}{P(P)} \\\\&= \dfrac{0.1}{0.2} \\\\&= 0.5 \neq \rm P(B)\end{aligned}[/tex]

And  

[tex]\begin{aligned}\rm P(P|B) &=\rm \dfrac{P(B \;n \;P)}{P(B)} \\\\&= \dfrac{0.1}{0.3} \\\\&= 0.333 \neq \rm P(P) \end{aligned}[/tex]

Hence they are not independent of each other.

The probability of low pH at high BOD is given as P(P|B).

[tex]\begin{aligned}\rm P(P|B) &= \rm \dfrac{P(B \;n \;P)}{P(B)}\\\\ &= \dfrac{0.1}{0.3}\\\\&= 0.3333\end{aligned}[/tex]

Hence, 0.33 is the probability of low pH at high BOD.

The probability of low pH at normal BOD is given as P(P|B').

[tex]\begin{aligned}\rm P(P|B') &= \rm \dfrac{P(B' \;n\; P)}{P(B')}\\\\\rm P(P|B') &= \dfrac{0.1}{0.7} \\\\&= 0.1429\end{aligned}[/tex]

Hence,  0.1429 is the probability of low pH at normal BOD.

The probability that a stream will not exhibit any of the indicators is given by, P(B' n P').

[tex]\begin{aligned}\rm P(B' n P') &= \rm P(U) - [P(B n P') + P(B' n P) + P(B n P)]\\\\&= 1 - (0.2 + 0.1 + 0.1) \\\\&= 0.6\end{aligned}[/tex]

Hence, 0.6 is the probability that neither of the indicators will be expressed.

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The true statement for nuclear model for the atom is B: For a given element, the size of an atom is the same for all of the element’s isotopes is

Explanation:

The isotopes are the atoms of an element that have different number of neutrons in their nuclei but same number of electrons.

The number of protons remains the same in isotopes.

the atomic mass of the isotope differ because the atomic mass is number of proton+ number of neutrons.

The atomic number remains the same since number of protons equals the number of protons.

The atomic radii does not change because the arrangement or number of electrons do not differ among the isotopes. Since there is no change in atomic radii the size remains same in isotopes of an element. Although the size of nucleus increases due to more protons.

The correct option is B). The true statement is For a given element, the size of an atom is the same for all of the element’s isotopes.

In the nuclear model of the atom, the size of an atom is primarily determined by the number of protons in the nucleus, which is the same for all isotopes of a given element. This is because the number of protons determines the atomic number, and thus the number of electrons in a neutral atom.

Isotopes of an element have the same number of protons but different numbers of neutrons. While the addition of neutrons does increase the mass of the atom, it does not significantly affect the size of the atom. The electron cloud is attracted to the positively charged protons in the nucleus, and since the number of protons remains constant for all isotopes of an element, the size of the electron cloud, and therefore the atomic radius, remains relatively constant.

 The presence of additional neutrons can have a very slight effect on the atomic radius due to the increased mass of the nucleus, which might lead to a slightly stronger attraction between the nucleus and the electrons.

Answer:

0.8 x10^-1 mol/(L s)

Explanation:

2A + B -> 2C + D

For the reaction above, the differential rate is usually expressed as;

Rate = - (1 / 2) Δ[A] / Δt = - Δ[B] / Δt

The negative sign denotes disappearance.

Upon comparing ;  (1 / 2) Δ[A] / Δt = Δ[B] / Δt

If  initial rate of disappearance of A is 1.6x10^-1 mol/(L s);

That means

Δ[B] / Δt = 1.6x10^-1 mol/(L s) / 2

Δ[B] / Δt = 0.8 x10^-1 mol/(L s)

The initial rate of disappearance of B = 1.6x10^-1 mol/(L s)

Answer:

1. 0.45 mole

2. 49.95g

Explanation:

The following were obtained from the question:

Volume of solution = 300mL = 300/1000 = 0.3L

Molarity = 1.5 M

Mole of CaCl2 =?

1. We can obtain the mole of the solute as follow:

Molarity = mole of solute /Volume of solution

1.5 = mole of solute/0.3

Mole of solute = 1.5 x 0.3

Mole of solute = 0.45 mole

2. The grams in 0.45 mole of CaCl2 can be obtained as follow:

Molar Mass of CaCl2 = 40 + (35.5 x 2) = 40 + 71 = 111g/mol

Mole of CaCl2 = 0.45 mole

Mass of CaCl2 =?

Mass = number of mole x molar Mass

Mass of CaCl2 = 0.45 x 111

Mass of CaCl2 = 49.95g

Answer:

We have 0.45 moles CaCl2 in this 1.5 M solution

This 49.9 grams of CaCl2

Explanation:

Step 1: data given

Volume of the CaCl2 solution = 300 mL = 0.300 L

Molarity of the CaCl2 solution = 1.5 M

Molar mass CaCl2 = 110.98 g/mol

Step 2: Calculate number of moles in the solution

Moles CaCl2 = molarity solution * volume of solution

Moles CaCl2 = 1.5 M * 0.300 L

Moles CaCl2 = 0.45 moles

Step 3: Calculate mass CaCl2

Mass CaCl2 = moles CaCl2 * molar mass CaCl2

Mass CaCl2 = 0.45 moles * 110.98 g/mol

Mass CaCl2 = 49.9 grams CaCl2

We have 0.45 moles CaCl2 in this 1.5 M solution

This 49.9 grams of CaCl2

Answer:

The dissociation constant for the acid ( experimental ) is 1.45 lit/mol

Explanation:

The value of dissociation constant can be calculated as,

  [tex]K_{a}[/tex] = C × ∝²

Where, C = concentration of the solution = 0.329M

          ∝ = Degree of dissociation

again , Degree of dissociation can be obtained form :

                      [tex]p_{H}[/tex] = C × ∝

                         ∝ = [tex]\frac{p_{H} }{C}[/tex]

                        ∝ = [tex]\frac{2.327}{0.329}[/tex] = 7.072

So, now [tex]K_{a}[/tex] = C × ∝²

                     = 0.329 ×( 7.072)²

                     = 1.45 lit/ mol

Answer:

The Ka = 6.74 * 10^-5

Explanation:

Step 1: Data given

Concentration of benzoic acid = 0.329 M

pH = 2.327

Step 2: Calculate the Ka

pH = -log (√([HA]*Ka))

2.327 = -log (√(0.329*Ka))

10 ^ - 2.327 = √(0.329*Ka))

0.0047098 = √(0.329*Ka))

2.218 * 10^-5 = 0.329 * Ka

Ka = 6.74 * 10^-5

The Ka = 6.74 * 10^-5

Explanation:

Boiling point is the temperature at which vapor pressure of a liquid becomes equal to the atmospheric pressure.

So, more is the number of carbon atoms present in a compound that are linearly attached to each other more will be its boiling point. This is because then there are more intermolecular forces present in it and also it occupies more surface area.

Hence, in order to break these intermolecular forces more heat is required. As a result, boiling point of the compound increases.

Therefore, we can conclude that given compounds are ranked according to the increasing order in which you would expect them to begin boiling as follows.

    Hexane at high altitude < Octane at high altitude < Octane at sea level  < Octanol at sea level      

The solution described is saturated as it contains NaCl at a concentration equal to its solubility at 25°C, meaning water has dissolved the maximal amount of NaCl it can at that temperature.

The solution you've described is a saturated solution, as it contains NaCl at a concentration equal to its solubility at a given temperature, in this case 25°C. This means that the water has dissolved as much sodium chloride as it can at this temperature, and any additional NaCl would not dissolve.

A supersaturated solution, on the other hand, is a solution that contains more solute than is possible under stable equilibrium conditions. This state is achieved by adding excess solute, or by evaporation of the solvent. A supersaturated solution is unstable, and will return to its saturation point upon disturbance, causing the excess solute to precipitate out.

Lastly, an unsaturated solution contains less solute than the maximum amount that can be dissolved in the solvent at a given temperature. In this state, more solute can be added and dissolved.

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The solution containing 35 g of NaCl per 100 g of water at 25 °C is unsaturated.

To determine the saturation status of the solution, we need to compare the actual amount of solute (NaCl) dissolved in the solvent (water) with the maximum amount that can be dissolved at a given temperature, which is known as the solubility of the solute in the solvent.

The solubility of NaCl in water at 25 °C is approximately 36 g per 100 g of water. Since the given solution has 35 g of NaCl per 100 g of water, this is less than the solubility limit. Therefore, the solution can still dissolve more NaCl and is considered unsaturated.

 If the solution had exactly 36 g of NaCl per 100 g of water, it would be saturated, meaning it has reached its maximum capacity to dissolve NaCl at that temperature. If it had more than 36 g of NaCl per 100 g of water, it would be supersaturated, which is a state where the solution contains more solute than would normally be possible at equilibrium.

Answer:

c i believe, not sure

Explanation: