Be sure to answer all parts. What is the [H3O+] and the pH of a benzoic acid-benzoate buffer that consists of 0.22 M C6H5COOH and 0.41 M C6H5COONa? (Ka of benzoic acid = 6.3 × 10−5) Be sure to report your answer to the correct number of significant figures.

Answer:

There should react 1008.3 grams KOH

Explanation:

Step 1: Data given

yield = 20.85 %

actual yield of H2O = 67.5 grams

Molar mass of H2SO4 = 98.08 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass of KOH = 56.11 g/mol

Step 2: The balanced equation

H2SO4 + 2KOH ⇒ K2SO4 + 2H2O

Step 3: Calculate moles of H2O

Moles H2O = Mass H2O / Molar mass H2O

Moles H2O = 67.5 grams / 18.02 g/mol

Moles H2O = 3.746 moles

Step 4: Calculate theoretical  yield

% yield= (actual yield/theoretical yield)*100%

Theoretical yield= (Actual yield/Percent yield) * 100%

Theoretical yield = (3.746 moles /20.85) *100%

Theoretical yield = 17.97 moles of H2O

Step 5: Calculate moles of KOH

For 2 moles of H2O produced, we need 2 moles of KOH

For 17.97 moles of H2O produced, we need 17.97 moles of KOH

Step 6: Calculate mass of KOH

Mass KOH =moles KOH * Molar mass KOH

Mass KOH = 17.97 moles * 56.11 g/mol

Mass KOH = 1008.3 grams

There should react 1008.3 grams KOH

a) There must be added 4.465 grams of aspirin

b) The unknown has a molecular weight of 342.2 g/mol

Step 1: Data given

The vapor pressure of chloroform is 173.11 mm Hg at 25°C

a nonvolatile, nonelectrolyte (MW = 180.1 g/mol), must be added

Mass of chloroform = 244.0 grams

The vapor pressure is reduced to 171.05 mm Hg

chloroform = CHCl3 = 119.40 g/mol.

The boiling point of water, H2O, is 100.000 °C

Step 2: Calculate moles chloroform

244.0g CHCl3/ 119.40g/ mol  = 2.0436 mol CHCl3

step 3: Calculate moles aspirin

Moles aspirin = x grams / 180.1 g/mol

(p0 - p)/p0

⇒ p0 is the initial pressure = 173.11 mm Hg

⇒ p = the reduced pressure = 171.05 mm Hg

(p0 - p)/p0  = (173.11-171.05)/173.11  = (xg/180.1 g/mol) / (2.0436 mol + (xg/180.1g/mol))

0.0119 mol = (xg/180.1 g/mol) / (2.0436 mol + (xg/180.1g/mol))

0.02432 + (0.0119x/180.1) = (x/180.1)

0.02432 =(( x - 0.0119x)/180.1)

0.9981x = 4.38

x = 4.465 grams

There must be added 4.465 grams of aspirin

b)  What is the molecular weight they determined for this compound?

Step 1: Define molar mass of the compound:

Molar mass = mass / moles

Molar mass compound = 11.55 grams / moles

Step 2: Calculate ΔT

ΔT = Tsoln - Tpure

ΔT = 100.071 - 100.00 = 0.071 °C

Step 3: Calculate m

ΔT = m*Kb

0.071°C = m*0.512 °C/m

m = 0.071  °C / 0.512 °C/m

m = 0.13867 m  = 0.13867 mol/kg

Step 4: Calculate number of moles of the compound

Moles = mass of water * m

Moles = 0.2434 kg * 0.13867 mol/kg

Moles = 0.03375 moles

Step 5: Calculate molar mass

Molar mass  = mass / moles

Molar mass compound = 11.55 grams / 0.03375 moles

Molar mass compound = 342.2 g/mol

The unknown has a molecular weight of 342.2 g/mol (and can be sucrose. It has a molar mas of 342.2 g/mol, is nonvolatile and non-electrolyte.).

Answer:

The molarity is 0.85M

The mole fraction = 0.05

Explanation:

Step 1: Data given

Molarity of acetone = 1.15 m

Density of acetone = 0.788 g/cm³

Density of ethanol = 0.789 g/cm³

Molar mass of acetone = 58.08 g/mol

Step 2: Calculate number of moles acetone

Molality = moles solute / kg solvent = moles acetone / kg ethanol. (To make it easy, we will suppose we have 1 kg ethanol)

1.15 m = 1.15 moles acetone / 1 kg ethanol

Step 3: Calculate mass of acetone

1.15 moles acetone * (58.08 g/mol) = 66.792 g acetone

Step 4: Calculate volume of acetone

66.792 g acetone / 0.788 g/mL acetone = 84.76 mL acetone

Step 5: Calculate volume of ethanol

1000 g ethanol / 0.789 g/mL ethanol = 1267.43 mL ethanol

Step 6: Calculate total volume solution

Total solution volume = 84.76 + 1267.43 = 1352.2 mL  = 1.3522L

Step 7: Calculate molarity

Molarity of acetone = moles acetone /volume  solution = 1.15moles / 1.3522L

Molarity = 0.85 M

Step 8: Calculate moles ethanol

moles ethanol= mass/ molar mass = 1000g/ 46.0g/mol = 21.74 moles

Step 9: Calculate mole fraction

mole fraction acetone = (moles acetone / total moles) = (1.15 / (1.15 + 21.74)) = 0.05

Answer:

The minimum pressure should be 901.79 kPa

Explanation:

Step 1: Data given

Temperature = 25°C

Molarity of sodium chloride = 0.163 M

Molarity of magnesium sulfate = 0.019 M

Step 2: Calculate osmotic pressure

The formula for the osmotic pressure =

Π=MRT.

⇒ with M = the total molarity of all of the particles in the solution.

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

⇒ T = the temperature = 25 °C = 298 K

NaCl→ Na+ + Cl-

MgSO4 → Mg^2+ + SO4^2-

M = 2(0.163) + 2(0.019 M)

M = 0.364 M

Π = (0.364 M)(0.08206 atm-L/mol-K)(25 + 273 K)

Π = 8.90 atm

(8.90 atm)(101.325 kPa/atm) = 901.79 kPa

The minimum pressure should be 901.79 kPa

Answer: Option (B) is the correct answer.

Explanation:

According to the given reaction equation, formula to calculate [tex]\Delta n[/tex] is as follows.

   [tex]\Delta n[/tex] = coefficients of gaseous products - gaseous reactants                    

                  = 1 - 0

                  = 1

Also we know that,

        [tex]K_{p} = K_{c} \times (RT)^{\Delta n}[/tex]

         [tex]5.6 \times 10^{-3} = K_{c} \times (0.0821 \times 700)^{1}[/tex]

             [tex]K_{c} = 0.097 \times 10^{-3}[/tex]

For the equation, [tex]2HgO(s) \rightarrow 2Hg(l) + O_{2}(g)[/tex]

Activity of solid and liquid = 1

As,     [tex]K_{p} = \frac{P^{2}_{Hg} \times P_{O_{2}}}{P^{2}_{HgO}}[/tex]

          [tex]5.6 \times 10^{-3} = P_{O_{2}}[/tex]

Hence, [tex]P_{O_{2}}[/tex] = 0.0056 atm

Thus, we can conclude that partial pressure of oxygen gas at equilibrium is 0.0056 atm.

Final answer:

To calculate the partial pressures of the three gases at equilibrium, we can use the equilibrium constant equation. Plugging in the given initial partial pressures into the equation gives a Kp value of 0.0135.

Explanation:

To solve this problem, we can use the equation for calculating the equilibrium constant (Kp):

Kp = ([NO]²)/([N₂O][O₂])

We are given the initial partial pressures of each gas, so we can plug these values into the equation. The initial partial pressures are as follows: [NO] = 0.08 atm, [N₂O] = 0.62 atm, and [O₂] = 0.24 atm. Plugging these values into the equation:

Kp = (0.08²)/(0.62·0.24) = 0.0135

Therefore, the equilibrium constant (Kp) for this reaction at the given temperatures and pressures is 0.0135.

Answer:

(a) Chlorine atom

(b) 0

(c) 109.5°

(d) about the same

Explanation:

Perchlorate is a negatively charged molecule that is composed of one carbon atom and four oxygen atoms. In this molecule, the central chlorine atom that is present in +7 oxidation state, is covalently bonded to four oxygen atoms.

Thus the number of bond pairs is 4 and the number of lone pairs on the central chlorine atom is 0.

So according to the VSEPR theory, perchlorate ion has a tetrahedral geometry. Therefore, the ideal bond angle for the perchlorate ion is 109.5°.

The actual bond angle of perchlorate is about the same as ideal bond angle. This is because the perchlorate is resonance stabilized and thus all the chlorine-oxygen bonds are equivalent.

In the perchlorate anion, chlorine is the central atom with zero lone pairs. The ideal angle between the chlorine-oxygen bonds is 109.5 degrees, though the actual angle is likely slightly less due to repulsion between bond pairs.

Considering the perchlorate anion (ClO4−):

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Answer:

There are 29.4 grams of oxygen in the container

Explanation:

Step 1: Data given

Volume = 20.0 L

Pressure = 845 mmHg

Temperature = 22.0 °C

Molar mass of O2 = 32 g/mol

Step 2: Ideal gas law

p*V = n*R*T

⇒ p = the pressure of the gas = 845 mmHg = 1.11184

⇒ V = the volume of the gas = 20.0 L

⇒ n = the number of moles = TO BE DETERMINED

⇒R = the gasconstant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 22°C + 273 = 295 Kelvin

n = (p*V)/(R*T)

n = (1.11184*20.0)/(0.08206*295)

n = 0.9186 moles

Step 3: Calculate mass of NO2

Mass of O2 = Moles O2 * Molar mass O2

Mass of O2 = 0.9186 moles * 32 g/mol

Mass of O2 = 29.4 grams

There are 29.4 grams of oxygen in the container

Answer: + 1.40 V

Explanation:

The balanced chemical equation is:

[tex]3Cl_2(g)+2Fe(s)\rightarrow 6Cl^-(aq)+2Fe^{3+}(aq)[/tex]

Here Fe undergoes oxidation by loss of electrons, thus act as anode. Chlorine undergoes reduction by gain of electrons and thus act as cathode.

[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

[tex]E^0_{[Fe^{3+}/Fe]}=-0.04V[/tex]

[tex]E^0_{[Cl_2/Cl^-]}=+1.36V[/tex]

[tex]E^0=E^0_{[Cl_2/Cl^-]}- E^0_{[Fe^{3+}/Fe]}[/tex]

[tex]E^0=+1.36-(-0.04V)=+1.40V[/tex]

The standard cell potential for the reaction is +1.40 V

Answer:

9.39 × 10²² molecules

Explanation:

We can find the moles of gases (n) using the ideal gas equation.

P . V = n . R . T

where,

P is the pressure (standard pressure = 1 atm)

V is the volume

R is the ideal gas constant

T is the absolute temperature (standard temperature = 273.15 K)

[tex]n=\frac{P.V}{R.T} =\frac{1atm.3.50L}{(0.08206atm.L/mol.K).273.15K} =0.156mol[/tex]

There are 6.02 × 10²³ molecules in 1 mol (Avogadro's number). Then,

[tex]0.156mol.\frac{6.02\times10^{23}molecules}{mol} =9.39\times10^{22}molecules[/tex]

Final answer:

The number of gaseous molecules in the flask can be determined using Avogadro's number and the ideal gas law. For a 3.50 L flask at standard temperature and pressure, the number of molecules is approximately 2.40 x 10^24.

Explanation:

The number of gaseous molecules in the flask can be determined using Avogadro's number and the ideal gas law. Avogadro's number (6.022 x 10^23) represents the number of molecules in one mole of any substance. The ideal gas law equation (PV = nRT) relates the pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T) of a gas.

Since the air in the flask is at standard temperature and pressure, we can assume the temperature is 273 K and the pressure is 1 atm. Plugging in these values, along with the flask volume, into the ideal gas law equation allows us to calculate the number of moles of air in the flask. The moles can then be converted to the number of molecules by multiplying by Avogadro's number.

For a flask volume of 3.50 L, the number of gaseous molecules is approximately 2.40 x 10^24.

Answer:

So after reaction completion. 7.11 g of P₄ will be left over.  

Explanation:

So for number of moles of P₄ =    Given mass/ molar mass = 45.55/ 123.88

                                                                                      = 0.367 moles  

Number of moles of Cl₂ = Given mass/ molar mass = 131.9/70 = 1.884 moles  

We will divide the values by dividing excess of reactants with their coefficient for finding excess of reactants.  

P₄ = 0.367/ coefficient 1 = 0.367

Cl₂ =   1.884/ 6= 0.314

Phosphorus is thus the excess reactant.

We need to use the mole value of the limiting reactant in order to find the mass of P₄ used. Using dimensional analysis,

1.884 mol Cl₂  ( 1mole P₄/6 mol Cl₂  )  ( 123.88 g P₄/ 1 mol P₄)   = 38.44 g P₄

This is the amount which is used up. So by subtracting this amount from the Initial amount  

45.55 – 38.44 =  7.11 g of P₄

So after reaction completion. 7.11 g of P₄ will be left over.  

Final answer:

To find the mass of the excess reactant, we need to determine the limiting reactant first. By comparing the moles of each reactant, we can find that the excess reactant is chlorine (Cl2) and the mass left is 127.7 g.

Explanation:

To determine the mass of the excess reactant, we need to find the limiting reactant first. We can do this by comparing the moles of each reactant to their stoichiometric coefficients. Firstly, we need to convert the given masses of P4 and Cl2 to moles using their molar masses: 123.9 g/mol for P4 and 70.9 g/mol for Cl2. This gives us 0.368 mol of P4 and 1.86 mol of Cl2. From the balanced equation, we see that 1 mole of P4 reacts with 6 moles of Cl2, so we need 1/6th of the moles of Cl2 to react with all of the P4. This means that we need 0.368/6 = 0.061 mol of Cl2. The excess moles of Cl2 is then 1.86 - 0.061 = 1.799 mol. Finally, we convert the excess moles to grams using the molar mass of Cl2: 1.799 mol * 70.9 g/mol = 127.7 g. Therefore, the mass of the excess reactant (Cl2) is 127.7 g.

The detailed answer provides the balanced chemical equation for the oxidation of Cl2 by Cu3+ in an acidic solution, with states of matter indicated.

The balanced chemical equation describing the oxidation of Cl2 gas by Cu3+ to form chlorate ion (ClO3-) and Cu2+ in an acidic aqueous solution is:

3Cu^3+(aq) + 6Cl¯(aq) + 18H^+(aq) + Cl2(g) → 3Cu^2+(aq) + ClO3^-(aq) + 9H2O(l)

States of matter: (s) solid, (l) liquid, (g) gas, (aq) aqueous solution.

The oxidation of [tex]Cl_2[/tex] gas by [tex]Cu^{3+}[/tex] to form chlorate ion and [tex]Cu^{2+}[/tex] in an acidic aqueous solution is represented by the balanced chemical equation provided.

The given question concerns the concept of balancing chemical equations in chemistry and deriving correct chemical reactions for the mentioned reactants and products. Also, oxidation is mentioned which refers to addition of oxygen to an entity or removal of hydrogen from entity. The balanced chemical equation for the oxidation of [tex]Cl_2[/tex] gas by [tex]Cu^{3+}[/tex] in an acidic aqueous solution is:

[tex]2Cl_2(g) + 2Cu^{3+} (aq) + 4H^+ (aq) \rightarrow 2ClO^{3-} (aq) + 2Cu^{2+} (aq) + 2H_2O (l)[/tex]

This reaction results in the formation of chlorate ion ([tex]ClO^{3-}[/tex]) and [tex]Cu^{2+}[/tex] in the solution. The states of matter as gas, aqueous, and liquid are indicated in the equation.

Answer:

1.44%

Explanation:

Initially, there is 24.201g of the waste. Let this waste contain 'x' mols of  ammonia([tex]NH_{3}[/tex]). Now, this waste is dissolved in 72.053g of water, thus making the weight of the solution to 24.201+72.053 = 96.254g. Now, from this, 13.774g is taken. Note that, when the aqueous waste is dissolved in water, the ammonia particles are uniformly distributed, i.e, the 'x' mols of ammonia is uniformly present in the 96.254g of the solution. Hence, when we take 13.774g of the solution, only a fraction of the ammonia particles is taken. This fraction is equal to [tex]\frac{13.774}{96.254}[/tex] of x, which is equal to 0.1431 times 'x'.

For the HCl solution, 28.02mL of 0.1045M solution contains 28.02x0.1045 = 2.9281mmols of HCl is present in it.

The basic titration reaction that occurs is : [tex]HCl + NH_{3}[/tex]→[tex]NH_{4}Cl[/tex]  , i.e, one mol of ammonia requires one mol of HCl for neutralization. Therefore, for the above solution of HCl containing 2.9281mmols, same amount of ammonia, i.e, 2.9281mmols is required for complete neutralization.

Therefore, 0.1431x = 2.9281 mmols ⇒ x = 20.4619 mmols.

The molecular weight of ammonia is 17g/mol. Therefore, the weight of 20.4619mmols of ammonia has a weight(w) = 20.4619 x [tex]10^{-3}[/tex] x 17 = 0.3478 g.

Therefore the weight of ammonia in the initial aqueous waste is 0.3478 g. The total weight of the waste is 24.201g, hence, the percent weight of ammonia is given by [tex]\frac{0.3478}{24.201}[/tex]×100 = 1.44%

Final answer:

To calculate the weight percent of ammonia in the waste sample, the moles of HCl used in titration are converted to mass of NH3. This value is then adjusted for dilution and related to the original mass of the waste to find the weight percent of NH3, which is calculated to be 1.442%.

Explanation:

To calculate the weight percent of ammonia (NH3) in the aqueous waste sample from the fertilizer manufacturer, we follow a series of steps involving dilution and titration. The aliquot of the solution (13.774 g) titrated with 0.1045 M HCl required 28.02 mL to reach the endpoint, which indicates the amount of ammonia that was present in the aliquot.

We first need to calculate the number of moles of HCl that reacted with NH3 using the molarity of HCl and the volume used in the titration:

Since each mole of NH3 reacts with one mole of HCl, the moles of NH3 in the aliquot will also be 0.00293 mol. To find the mass of NH3, we multiply the moles by the molar mass of NH3 (17.031 g/mol).

To calculate the weight percent of NH3 in the original waste sample, we need to adjust the mass of NH3 found for the dilution and relate that to the original mass of the waste sample:

Therefore, the weight percent of ammonia in the initial aqueous waste sample is 1.442%.

Answer:

ΔH of solution is expected to be close to zero.

Explanation:

When we mix two non polar organic liquids like hexane and heptane,the resulting mixture formed is an ideal solution.An ideal solution is formed when the force of attraction between the molecules of the two liquids is equal to the force of attraction between the molecules of the same type.

For instance if liquids A and B are mixed,

[tex]F_{A-A}[/tex] = [tex]F_{B-B}[/tex] = [tex]F_{A-B}[/tex]

Hence the condition before and after mixing remains unchanged.

Since enthalpy change is associated with inter molecular force of attraction the enthalpy change for ideal solution is zero.

More examples of ideal solutions are:

1. Ethanol and Methanol

2. Benzene and Toluene

3. Ethyl bromide and Ethyl iodide

Answer:

1) The mass of water lost = 7.7892 grams

2) Z = 3: Cu(NO3)2*3H2O

Explanation:

Step 1: Data given

Mass of copper (II) nitrate hydrate, Cu(NO3)2•zH2O = 34.8243 grams

Mass of substance after heating = 27.0351 grams

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate mass of water

The mass of water is the mass lost after heating.

Mass water = 34.8243 - 27.0351 = 7.7892 grams of water

Step 3: Calculate moles of Cu(NO3)2

Moles Cu(NO3)2 = Mass Cu(NO3)2 / Molar mass Cu(NO3)2

Moles Cu(NO3)2 = 27.0351 grams / 187.56 g/mol

Moles Cu(NO3)2 = 0.144 moles

Step 4: Calculate moles of H2O

Moles H2O = 7.7892 grams / 18.02 g/mol

Moles H2O = 0.432 moles

Step 5: Calculate Z

z = moles H2O / moles Cu(NO3)2

Z = 0.432/0.144

Z = 3

This means we have 3 water molecules in the formula. This makes the formula ofthe hydrate: Cu(NO3)2*3H2O

Final answer:

To find the percent water in the hydrate, subtract the final mass after heating from the initial mass of the hydrate, then divide by the initial mass and multiply by 100. The mass of water lost is 7.7892g, and the percent water is 22.36%. To determine the value of z, calculate the moles of water and anhydrous Cu(NO3)2 to find z equals 3, making the formula Cu(NO3)2•3H2O.

Explanation:

The question involves determining the percent water in a hydrate, the mass of water lost during heating, and the value of z in the formula Cu(NO3)2•zH2O. Firstly, to calculate the mass of water lost, subtract the final mass after heating from the initial mass of the copper (II) nitrate hydrate. Then, to find the percent water, the mass of water lost is divided by the initial mass of the hydrate, multiplied by 100. Lastly, to determine the value of z, the moles of water lost are divided by the moles of anhydrous Cu(NO3)2.

To calculate the mass of water lost: 34.8243g (initial mass) - 27.0351g (final mass) = 7.7892g (mass of water lost).

To calculate the percent water in the hydrate: (7.7892g / 34.8243g) x 100 = 22.36%.

To find z, the number of waters, you need the molar mass of Cu(NO3)2 and H2O. Assume the molar mass of Cu(NO3)2 is approximately 187.55 g/mol. The mass of Cu(NO3)2 after heating is 27.0351g, which corresponds to 27.0351g / 187.55g/mol = 0.1441 mol. The moles of water lost is 7.7892g / 18.015g/mol = 0.4323 mol. Hence, z = 0.4323 mol / 0.1441 mol = 3, indicating the formula is Cu(NO3)2•3H2O.

Answer:

Ecell = 0.500 V

Explanation:

For a chemical cell, the cell potential (Ecell) is:

Ecell = E° - (0.0592/n)*logQ (Nernst equation)

Where n is the number of electrons involved in the redox reaction, and Q is the reaction coefficient.

Let's see a half-reaction:

2Co³⁺(aq) → 2Co²⁺(aq)

The charge goes from +6 (2*3) to +4 (2*2), so n = 6 - 4 = 2 electrons.

The reaction coefficient is the multiplication of the concentration of the products elevated by their coefficients, divided by the multiplication of the concentration of the reactants elevated by their coefficients:

Q = ([Cl₂]*[Co⁺²]²)/([Co⁺³]²*[Cl⁻]²)

The concentration of Cl₂ is the number of moles (n) divided by the volume(V), and can be calculated by the ideal gas law:

PV = nRT

n/V = P/RT  (P is the pressure, R is the gas constant, and T is the temperature)

P = 9.30 atm, R = 0.082 atm.L/mol.K, T = 25°C + 273 = 298 K

n/V = [Cl₂] = 9.30/(0.082*298)

[Cl₂] = 0.381 M

Q = [0.381*(0.369)²]/[(0.765)²*(0.486)²]

Q = 0.051877/0.138228

Q = 0.3753

Ecell = 0.483 - (0.0592/2)*log(0.3753)

Ecell = 0.483 - 0.0296*(-0.4256)

Ecell = 0.500 V

The cell potential of the electrochemical cell with the cobalt and chlorine electrodes has been 0.5 V.

The electrochemical cell has been the cell in which the chemical and electrical energy is converted by the charge transfer.

The cell potential ([tex]E_{cell}[/tex]) can be given as:

[tex]E^\cell=E^\circ-\dfrac{0.059}{n}\;\times\;log\; Q[/tex]

The standard cell potential  ([tex]E^\circ[/tex]) of the cell has been:

[tex]E^\circ=0.483\;\text V[/tex]

The charge transferred in the reaction ([tex]n[/tex]) can be given as 2.

The concentration of chlorine  has been:

[tex]\rm M=\dfrac{P}{RT} \\\\Cl_2=\dfrac{9.30}{0.0821\;\times\;298}\\\\ Cl_2=0.981\;M[/tex]

The value of Q is given as:

[tex]\rm Q=\dfrac{[Cl_2][CO_2]^2}{[CO_3]^2[Cl]^2}\\\\ Q=\dfrac{(0.981)(0.369)^2}{(0.765)^2(0.486)^2} \\\\Q=0.3753[/tex]

Substituting the values for the cell potential:

[tex]E_{cell}=0.483-\dfrac{0.059}{2}\;\times\;\text {log}\;0.3753 \\E_{cell}=0.5\;\rm V[/tex]

The cell potential is 0.5 V.

Learn more about cell potential, here:

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Answer:

The four quantum number for each electron will be:

[tex]1s^{2}[/tex]

[tex]n=1;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}[/tex]

[tex]2s^{2}[/tex]

[tex]n=2;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}[/tex]

[tex]2p^{6}[/tex]

[tex]n=1;l=1;\\m=+1,0.-1 \\s=+\frac{1}{2}/-\frac{1}{2}[/tex]

[tex]3s^{2}[/tex]

[tex]n=3;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}[/tex]

Explanation:

As the element is neutral, the number of protons will be equal to number of electrons which will be the atomic number of the element.

Number of electrons =12

Atomic number = 12

Element : Magnesium

The principal shell is represented by "n"

i) For "s" subshell the value of l =0 (azimuthal quantum number) thus m (magnetic quantum number)= 0

The two electrons in s subshell will have either plus half or minus half spin quantum number

ii) for "p" subshell the value for l =1

thus m = 0 or +1 or -1

The two electrons in each orbital will have either plus half or minus half spin quantum number

Final answer:

The complete set of quantum numbers for each electron in the given electron configuration, which corresponds to the element magnesium (Mg), is derived from the configuration 1s²2s²2p⁶3s², by stating the values for the principal, angular momentum, magnetic, and spin quantum numbers for each electron.

Explanation:

The electron configuration of a neutral atom is 1s22s22p63s2. To write a complete set of quantum numbers for each of the electrons, we can consider the principles of quantum mechanics. Each electron in an atom is described by four quantum numbers: the principal quantum number (n), angular momentum quantum number (l), magnetic quantum number (ml), and spin quantum number (ms).

For the 1s subshell (2 electrons), the quantum numbers for each electron are (1, 0, 0, +1/2) and (1, 0, 0, -1/2). For the 2s subshell (2 electrons), the quantum numbers are similar, with the principal quantum number being 2: (2, 0, 0, +1/2) and (2, 0, 0, -1/2). For the 2p subshell (6 electrons), there are three magnetic quantum numbers (-1, 0, +1), and two spin states. Thus, the six sets of quantum numbers are (2, 1, -1, +1/2), (2, 1, -1, -1/2), (2, 1, 0, +1/2), (2, 1, 0, -1/2), (2, 1, +1, +1/2), and (2, 1, +1, -1/2). For the 3s subshell (2 electrons), the quantum numbers are (3, 0, 0, +1/2) and (3, 0, 0, -1/2).

Based on this configuration, the neutral atom would be magnesium (Mg), which has an atomic number of 12, corresponding to the 12 electrons described.

Answer:

Heptane (d) > Hexane (b)  3,3 dimethyl pentane (c) > butane (a)

Explanation:

The boiling points of the alkanes are depends on the following factors:

The given molecules are as follows.

a) Butane

b) Hexane

c) 3,3 dimethyl pentane

d) Heptane.

The Highest to lowest boiling point order is as follows.

Heptane (d) > Hexane (b)  3,3 dimethyl pentane (c) > butane (a)

Explanation:

So, we need to break the bonds of liquid substance in order to convert it into vapor state. And, energy is absorbed for breaking of bonds which means that evaporation is an endothermic process.

Hence, the statement evaporation of water is an exothermic process is false.

A combustion reaction will always release heat energy. Hence, combustion reaction is exothermic in nature.

Hence, for an exothermic process value of [tex]\Delta H[/tex] is negative.

Thus, we can conclude that statements which are true are as follows.

Answer:

3.01 ·10↑22

Explanation:

First you want to convert the grams of Glucose to moles of Glucose.

[tex]\frac{1.5 grams of glucose}{180.15588grams/molesglucose} =.008326 Moles of Glucose[/tex]

Next find the formula units of glucose.

.008326Moles of Glucose · 6.022 · 10↑23Forumula Units*Moles↑-1 =

5.01 ·10↑21 Formula Units of Glucose

Now multiply the formula units of glucose by the amount of each element in the molecule.

So for Carbon:

6carbon · 5.01 · 10↑21 = 3.01 · 10↑22

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Answer:

Number of C-atoms =  3.01 *10^22

Explanation:

Step 1: Data given

Mass of glucose = 1.50 grams

Molar mass of Glucose = 180.156 g/mol

Molar mass of carbon = 12.01 g/mol

Step 2: Calculate number of moles glucose

Number of moles glucose = mass glucose / molar mass glucose

Moles glucose = 1.50 grams / 180.156 g/mol

Moles glucose = 0.00833 moles

Step 3: Calculate formula units of glucose

0.00833 moles glucose * 6.02 *10^23 = 5.01 *10^21

Step 4: Calculate number of carbon atoms

Glucose has 6 C'atoms in it's molecular formule

Number of C-atoms = 5.01 *10^21 * 6 = 3.01 *10^22

Answer:

[tex]\large \boxed{\text{B.) 2.8 atm}}[/tex]

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

[tex]\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}[/tex]

Data:

p₁ = 1520 Torr; T₁ =   27 °C

p₂ = ?;               T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = (  27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

[tex]\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\[/tex]

(c) Convert the pressure to atmospheres

[tex]p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}[/tex]

Answer:

ΔS = -0.1076 kJ /kg*K

Explanation:

Step 1: Data given

Initial state = 0.8 m³/kg and 25 °C = 298.15 K

Final state = 0. 3³/kg and 287 °C = 560.15 K

Cv = 0.686 kJ/kg*K

Step 2: Calculate the average temperature

The average temperature = (25°C + 287 °C)/2  =156 °C ( = 429 K)

Step 3: Calculate the ΔS

ΔS =(Cv, average) * ln(T2/T1) + R*ln(V2/V1)

ΔS = 0.686 * ln(560.15/298.15) + 0.2598*ln( 0.1/0.8)

ΔS = -0.1076 kJ /kg*K

Answer:

The pH of pure water at this temperature is 6.88.

Explanation:

Ionization constant of water = [tex]K_w=1.7 x 10^{-14}[/tex]

[tex]2H_2O(l)\rightleftharpoons H_3O^+(aq) + OH^-(aq)[/tex]

Water will dissociate into equal amount of hydronium and hydroxide ions

[tex][H_3O^+] = [OH^-]= x[/tex]

[tex]K_w=[H_3O^+]\times [OH^-][/tex]

[tex]K_w=1.7 x 10^{-14}=[H_3O^+]\times [OH^-][/tex]

[tex]1.7 x 10^{-14}=x^2[/tex]

[tex]x=1.304\times 10^{-7}[/tex]

The pH of the solution is a negative logarithm of hydronium ions concentration.

The pH of pure water at this temperature:

[tex]pH=-\log [H_3O^+][/tex]

[tex]pH=-\log (1.304\times 10^{-7})=6.88[/tex]

The pH of pure water is:

C)  6.88.

Ionization constant of water = [tex]1.7 * 10^{-14}[/tex]

Chemical reaction for ionization of water:

[tex]2H_2O(l)---- > H_3O^+(aq) + OH^-(aq)[/tex]

Water dissociates into equal amount of hydronium and hydroxide ions

[tex][H_3O^+]=[OH^-]=x\\\\\K_w=[H_3O^+]*[OH^-]\\\\1.7*10^{-14}=x^2\\\\x=1.307*10^{-7}[/tex]

The pH of the solution is a negative logarithm of hydronium ions concentration.

The pH of pure water at this temperature:

[tex]pH=-log[H_3O^+]\\\\pH=-log(1.304*10^{-7})\\\\pH=6.88[/tex]

Thus, correct option is C.

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Answer:

Detailed in explanation.

Explanation:

The pentose phosphate pathway is a two‑stage pathway that generates NADPH which is a reductant in many biosynthetic reactions and takes part in detoxifying reactive oxygen species, and ribose 5-phosphate which is a nucleotide (DNA and RNA) precursor. The substrate for the pentose phosphate pathway is glucose-6-phosphate.

The isomerization phase of the pentose phosphate pathway interconverts phosphorylated monosaccharides, some of which can feed into the glycolysis pathway.

Answer:

1) NADPH .

2) Ribose 5-phosphate .

3) G6P .

4) Nonoxidative .

5) glycolytic.

Explanation:

Hello,

In this case, each statement turn out into:

-The pentose phosphate pathway is a two‑stage pathway that generates NADPH (it is a donating electrons cofactor and a hydrogens provider to reactions catalyzed by enzymes) which is a reductant in many biosynthetic reactions and takes part in detoxifying reactive oxygen species, and ribose 5-phosphate (it is a fundamental constituent of the pyridine-based nucleotides such as nicotinamide adenine dinucleotide phosphate and nicotinamide adenine dinucleotide phosphate and the purine nucleotides adenosine diphosphate and ATP) which is a nucleotide (DNA and RNA) precursor.

-The substrate for the pentose phosphate pathway is G6P (glucose 6-phosphate, sometimes called the Robison ester, is a glucose based sugar phosphorylated at the hydroxyl group on carbon 6) .

-The nonoxidative  (not carrying out oxidation) phase of the pentose phosphate pathway interconverts phosphorylated monosaccharides, some of which can feed into the glycolytic (referred to glycolysis as the metabolic pathway that converts glucose into pyruvate and a hydrogen ion) pathway.

Best regards.

Answer:

The initial volume of the system: V₁ = 53.06 L

Explanation:

Given: heat absorbed by system: q =  50.5 J, Pressure: P = 0.491 atm, Final volume: V₂ = 56.2 L, The change in the internal energy: ΔE = -106.0 J

Initial volume: V₁ = ? L

According to the First Law of Thermodynamics:

ΔE = q - PΔV

⇒ PΔV = q - ΔE = 50.5 J - (-106.0 J) = 156.5 J

As, 1 L∙atm = 101.3 J  ⇒   1 J = (1 ÷ 101.3) L∙atm

⇒ PΔV = 156.5 J = (156.5 ÷ 101.3) L∙atm = 1.54 L∙atm

So,

ΔV = 1.54 L∙atm ÷ P = 1.54 L∙atm ÷ 0.491 atm = 3.14 L

∵ ΔV = V₂ - V₁ = 3.14 L

⇒ V₁ = V₂ -  3.14 L = 56.2 L -  3.14 L = 53.06 L

Therefore, the initial volume of the system: V₁ = 53.06 L

Answer:

Explanation:

2NO₂(g) → N₂O₄(g)   Kc=4.7

The definition of equilibrium is when the forward rate and the reverse rate are equal.

Because in the initial state there's only NO₂, there's no possibility for the reverse reaction (from N₂O₄ to NO₂). Thus the forward rate will be larger than the reverse rate.

Answer:

16.5 days

Explanation:

Given that:

Half life = 26.5 days

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

Where, k is rate constant

So,  

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

[tex]k=\frac {ln\ 2}{26.5}\ days^{-1}[/tex]

The rate constant, k = 0.02616 days⁻¹

Using integrated rate law for first order kinetics as:

[tex][A_t]=[A_0]e^{-kt}[/tex]

Where,  

[tex][A_t][/tex] is the concentration at time t

[tex][A_0][/tex] is the initial concentration

Given:

35.0 % is decomposed which means that 0.35 of [tex][A_0][/tex] is decomposed. So,

[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.35 = 0.65

t = 7.8 min

[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]

[tex]0.65=e^{-0.02616\times t}[/tex]

t = 16.5 days.

Answer:

1. 9.90 x 10^14 Hz

Explanation:

The work function is the minimum energy required to eject electron's from the surface of the metal. The frequency of photons that are capable of doing this is called the threshold frequency. Given the work function, the threshold frequency can be calculated in the following way

Work function = Planck's constant * frequency

6.56 x 10^-19 J = 6.62607004 × 10-34 m²-kg/s  * f

                      f = 9.90 x 10^14 Hz

Answer:

Here's what I get  

Explanation:

When a solid is converted to a liquid, heat is absorbed. TRUE.

Temperature remains constant while all of a solid is converted to a liquid. TRUE.

When heat is applied to a solid, the molecular motion decreases as the temperature increases. False. Molecular motion increases as the temperature increases.

The average kinetic energy of the system changes while all of a solid is converted to a liquid. False. The temperature remains constant during a phase change, so the average kinetic energy of the system does not change.

The temperature increases while all of a liquid is converted to a gas. False.

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

[tex]\Delta S=\frac{\Delta H_{freezing}}{T_f}[/tex]

where,

[tex]\Delta S[/tex] = change in entropy

[tex]\Delta H_{fus}[/tex] = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

[tex]\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol[/tex]

[tex]T_f[/tex] = freezing point temperature = [tex]-97.6^oC=273+(-97.6)=175.4K[/tex]

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

[tex]\Delta S=\frac{\Delta H_{freezing}}{T_m}[/tex]

[tex]\Delta S=\frac{-3170J/mol}{175.4K}[/tex]

[tex]\Delta S=-18.07J/mol.K[/tex]

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

Answer:

Entropy is defined as the measure of possible arrangement of the atoms or molecules. Entropy change for the the freezing process mole of 1 liquid methanol at -97.6 degree celcius and 1 atm will be -18.07 J/mol.K

Explanation:

The entropy change can be calculated by the given formula:

[tex]\begin{aligned} \Delta \text S &= \frac{\Delta \text H_{\text {freezing}}}{\text T_{f}} \end{aligned}[/tex]

in which,

[tex]\Delta \text S[/tex] = change in entropy

[tex]\Delta \text H_{\text {fusion}} &=[/tex] change in enthalpy of fusion that is -3.17 kJ/mol (as given).

By the formula of enthalpy change,

[tex]\begin{aligned} \Delta \text H_{\text {fusion}} &= - \Delta \text H_{\text {freezing}} &= -3.17 {\text {kJ/mol}} &= -3170 \text {J/mol}\end{aligned}[/tex]

Now,

[tex]{\text T_{f}}[/tex] = Freezing Point Temperature = [tex]-97.6^{o} \text C[/tex] = 273 + (-97.6) = -175.4 K

Substituting the values in above equation of entropy change, we get,

[tex]\begin{aligned} \Delta \text S &= \frac{\Delta \text H_{\text {freezing}}}{\text T_{f}} \end{aligned}[/tex]

[tex]\begin{aligned} \Delta \text S &= \frac{-3170 \text {J/mol}}{175.4 \text {K}} \\\\Delta \text S& = -18.07 \text {J/mol.K}\end{aligned}[/tex]

Thus, the value of change in entropy for freezing process of 1 mole of liquid methanol is -18.07 J/mol.k

For Further Reference:

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Explanation:

As it is given that both the given containers are at same temperature and pressure, therefore they have the same density.

So, mass of [tex]SF_{6}[/tex] in container- 1 is as follows.

    5.35 mol x molar mass of [tex]SF_{6}[/tex]

            = 7.61 mol x 146.06 g/mol

             = 1111.52 g

Therefore, density of [tex]SF_{6}[/tex] will be calculated as follows.

            Density = [tex]\frac{mass}{volume}[/tex]  

         density = [tex]\frac{1111.52 g}{2.09 L \times 1000 ml/L}[/tex]

                       = 0.532 g/mL

Now, mass of [tex]SF_{6}[/tex] in container- 2 is calculated as follows.

        4.46 L x 1000 mL/L x 0.532 g/mL

            = 2372.72 g

Hence, calculate the moles of moles [tex]SF_{6}[/tex] present in container 2 as follows.

  No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]  

                        = [tex]\frac{2372.72 g}{146.06 g/mol}[/tex]

                        = 16.24 mol

Since, 1 mol [tex]SF_{6}[/tex] contains 6 moles F atoms .

So, 16.24 mol  [tex]SF_{6}[/tex] contains following number of atoms.

                = [tex]16.24 mol \times 6[/tex]

                = 97.46 mol

Thus, we can conclude that moles of F atoms in container 2 are 97.46 mol.