Problem 19.24a Using the reagents below, list in order (by letter, no period) those necessary to convert the starting material into the given product: Note: More than 1 mole of the starting material may be used. Not all spaces provided may be needed. Type "na" in any space where you have no reagent. a. heat, -CO2 b. NaOEt c. (CH3CH2)2CuLi d. CH2Cl2, PCC e. C3H7C(O)CH(C2H5)C(O)C2H5 f. CH3CH2Li g. CH3C(O)Cl, AlCl3 h. NBS, ROOR i. H2NC(O)NH2 j. HN(CH3)2 k. OH-, H2O, heat then H3O l. H3O

Answer:

Moles = 0.375

Explanation:

Moles= m/M

= 12/32 = 0.375mol

There are approximately 0.375 moles of O₂ in 12.0 grams of O₂.

To find the number of moles in a given mass of a substance, one can use the formula:

[tex]\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \][/tex]

The molar mass of oxygen (O₂) is approximately 32.0 g/mol, since one oxygen atom has an atomic mass of approximately 16.0 g/mol and O₂ has two oxygen atoms.

Given the mass of O₂ is 12.0 g, we can calculate the moles as follows:

[tex]\[ \text{moles of \ } {O_2} = \frac{12.0 \text{ g}}{32.0 \text{ g/mol}} \][/tex]

[tex]\[ \text{moles of} \ {O_2} = 0.375 \text{ mol} \][/tex]

Therefore, there are 0.375 moles of O₂ in 12.0 grams of O₂.

Answer:

Problem 1 => 8778 joules

Problem 2 => 14,630 joules

Explanation:

Reference the Heating Curge of Water Problem posted earlier. These are temperature change fragments of that type problem. As you read the problem note the 'temperature change' phrase in the problem. This should signal you to use the q = m·c·ΔT expression as opposed to the q = m·ΔH expression where no temperature change is noted; i.e., melting or evaporation/boiling.

For the listed problems...

Problem 1: Amount of heat needed to heat 150 grams water from 21.0 to 35.0 Celcius.

Note the temperature change in the problem context => use q = m·c·ΔT ...

q = (150g)(4.18j/g·°C)(35.00°C - 21.0°C) = 8778 joules (4 sig. figs. based on the 150.0g value of water)

Problem 2: Amount of heat needed to heat 250.0 grams water from 31.0 to 45.0 Celcius.

Same type problem. Note temperature change in text of problem => use q = m·c·ΔT ...

q = (250.0g)(4.18j/g·°C)(45.0°C - 31.0°C) = 14,630 joules

Hope this helps. Doc

Answer:

B. CH3COOH pH > 4.7 (4.8)

Explanation:

∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol

⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol

⇒ C CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)

⇒ C CH3COOH = 0.0905 M

∴ mol CH3COONa = (0.05 L )*(0.20 mol/L) = 0.01 mol

⇒ C CH3COONa =  (0.01 mol + 5 E-4 mol) / (0.105 L )

⇒ C CH3COONa = 0.1 M

∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5

⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])

⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]

⇒ [H3O+]² + 0.1000175[H3O+] - 1.584 E-6 = 0

⇒ [H3O+] = 1.5835 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = - Log (1.5835 E-5)

⇒ pH = 4.8004 > 4.7

Final answer:

After adding NaOH to the acetate buffer, the acetate ion concentration increases and the pH of the solution will be greater than the pKa of acetic acid (4.7). Option D, with C2H3O2- and pH > 4.7, is correct.

Explanation:

The question involves calculating the pH of an acetate buffer solution after the addition of sodium hydroxide, NaOH. The buffer consists of acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2). The pKa of acetic acid is 4.7, and equal concentrations of the buffer components are typically ideal. When NaOH is added, it neutralizes some of the acetic acid, converting it to acetate and increasing the pH slightly.

Before the addition of NaOH, the buffer has equal concentrations of acetic acid and acetate, so its pH is close to the pKa of acetic acid, which is 4.7. After adding NaOH, more acetic acid is converted to acetate, but the pH will still be maintained close to the pKa since the buffer action resists changes in pH. Therefore, the correct pairing is that the acetate ion (C2H3O2-) will be present in greater concentration and the pH of the solution will be greater than 4.7 after NaOH is added, making option D correct.

Answer:

The peak at mass 100 with a 8% relative abundance is the molecular ion peak

Explanation:

Molecular ion peak has the highest charge to mass ratio,

Mass of 100 is same as mass to charge ratio =100

Answer:

We need 1136.4 kJ energy

Explanation:

Step 1: Data given

The mass of water = 369 grams

the initial temperature = -10°C

The finaltemperature = 125 °C

Cm (ice)=36.57 J/(mol⋅∘C)

Cm (water)=75.40 J/(mol⋅∘C)

Cm (steam)=36.04 J/(mol⋅∘C)

ΔHfus=+6010 J/mol

ΔHvap=+40670 J/mol

Step 2: :Calculate the energy needed to heat ice from -10 °C to 0°C

Q = n*C*ΔT

⇒Q = the energy needed to heat ice to 0°C

⇒with n = the moles of ice = 369 grams / 18.02 g/mol = 20.48 moles

⇒with C = the specific heat of ice = 36.57 / J/mol°C

⇒ ΔT = the change of temperature = 10°C

Q = 20.48 moles * 36.57 J/mol°C * 10°C

Q = 7489.5 J = 7.490 kJ

Step 3: calculate the energy needed to melt ice to water at 0°C

Q = n* ΔHfus

Q = 20.48 moles * 6010 J/mol

Q = 123084.8 J = 123.08 kJ

Step 4: Calculate energy needed to heat water from 0°C to 100 °C

Q = n*C*ΔT

⇒Q = the energy needed to heat water from 0°C to 100 °C

⇒with n = the moles of water = 369 grams / 18.02 g/mol = 20.48 moles

⇒with C = the specific heat of water  =75.40 J/(mol⋅∘C)

⇒ ΔT = the change of temperature = 100°C

Q = 20.48 moles * 75.40 J/mol°C* 100°C

Q = 154419.2 J = 154.419 kJ

Step 5: Calculate energy needed to vapourize water to steam at 100°C

Q = 20.48 moles * 40670 J/mol

Q = 832921.6 J = 832.922 kJ

Step 6: Calculate the energy to heat steam from 100 °C to 125 °C

⇒Q = the energy needed to heat steam from 100 °C to 125 °C

⇒with n = the moles of steam = 369 grams / 18.02 g/mol = 20.48 moles

⇒with C = the specific heat of steam  = 36.04 J/(mol⋅∘C)

⇒ ΔT = the change of temperature = 25 °C

Q = 20.48 moles * 36.04 J/mol*°C * 25°C

Q = 18452.5 J = 18.453 kJ

Step 7: Calculate total energy needed

Q = 7489.5 J + 123084.8 J + 154419.2 J +  832921.6 J + 18452.5 J

Q = 1136367.6 J

Q = 1136.4 kJ

We need 1136.4 kJ energy

The amount of energy needed is [tex]\( \\1120 \text{ kJ}} \).[/tex]

To calculate the amount of energy required to change 369 g of water from ice at[tex]\(-10^\circ \text{C}\)[/tex] to steam at [tex]\(125^\circ \text{C}\),[/tex] we need to consider the different phases of water and the energy required for each phase change and temperature change.

Let's break down the calculation step by step.

1. Energy to heat ice from [tex]\(-10^\circ \text{C}\) to \(0^\circ \text{C}\).[/tex]

  First, calculate the energy required to heat the ice initially at[tex]\(-10^\circ \text{C}\)[/tex] to its melting point [tex]( \(0^\circ \text{C}\) ).[/tex]

  Specific heat capacity of ice [tex](\(C_m\)): \(36.57 \text{ J/(mol} \cdot \text{°C)}\).[/tex]

  Molar mass of water (H₂O). [tex]\(18.015 \text{ g/mol}\).[/tex]

  Amount of ice in moles:

[tex]\[ \text{moles of ice} = \frac{369 \text{ g}}{18.015 \text{ g/mol}} \approx 20.49 \text{ mol} \][/tex]

  Energy required to heat ice.

[tex]\[ Q_1 = n \times C_m \times \Delta T = 20.49 \text{ mol} \times 36.57 \text{ J/(mol} \cdot \text{°C)} \times (0^\circ \text{C} - (-10^\circ \text{C})) = 7485.39 \text{ J} \][/tex]

[tex]\( Q_1 = 7.48539 \text{ kJ} \)[/tex]

2. Energy to melt the ice at [tex]\(0^\circ \text{C}\).[/tex]

Latent heat of fusion [tex](\(\Delta H_{\text{fus}}\)) for ice: \(6.01 \text{ kJ/mol}\)[/tex].

Energy required to melt ice:

[tex]\[ Q_2 = \text{moles of ice} \times \Delta H_{\text{fus}} = 20.49 \text{ mol} \times 6.01 \text{ kJ/mol} = 123.05 \text{ kJ} \][/tex]

[tex]\( Q_2 = 123.05 \text{ kJ} \)[/tex]

3.Energy to heat water from[tex]\(0^\circ \text{C}\) to \(100^\circ \text{C}\)[/tex].

Specific heat capacity of water[tex](\(C_m\))[/tex]: [tex]\(75.40 \text{ J/(mol} \cdot \text{°C)}\)[/tex].

Energy required to heat water:

[tex]\[ Q_3 = \text{moles of water} \times C_m \times \Delta T = 20.49 \text{ mol} \times 75.40 \text{ J/(mol} \cdot \text{°C)} \times (100^\circ \text{C} - 0^\circ \text{C}) = 155,297.80 \text{ J} \][/tex] [tex]\( Q_3 = 155.2978 \text{ kJ} \)[/tex]

4. Energy to vaporize water at [tex]\(100^\circ \text{C}\)[/tex]:

 Latent heat of vaporization [tex](\(\Delta H_{\text{vap}}\)) for water: \(40.67 \text{ kJ/mol}\).[/tex]

  Energy required to vaporize water:

[tex]\[ Q_4 = \text{moles of water} \times \Delta H_{\text{vap}} = 20.49 \text{ mol} \times 40.67 \text{ kJ/mol} = 833.34 \text{ kJ} \][/tex]

 [tex]\( Q_4 = 833.34 \text{ kJ} \)[/tex]

5. Energy to heat steam from [tex]\(100^\circ \text{C}\) to \(125^\circ \text{C}\):[/tex]

  Specific heat capacity of steam [tex](\(C_m\)): \(36.04 \text{ J/(mol} \cdot \text{°C)}\).[/tex]

  Energy required to heat steam:

[tex]\[ Q_5 = \text{moles of steam} \times C_m \times \Delta T = 20.49 \text{ mol} \times 36.04 \text{ J/(mol} \cdot \text{°C)} \times (125^\circ \text{C} - 100^\circ \text{C}) = 1831.79 \text{ J} \][/tex]

[tex]\( Q_5 = 1.83179 \text{ kJ} \)[/tex]

6. Total energy required.

  Summing up all the energy contributions.[tex]\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 = 7.48539 \text{ kJ} + 123.05 \text{ kJ} + 155.2978 \text{ kJ} + 833.34 \text{ kJ} + 1.83179 \text{ kJ} = 1120.00598 \text{ kJ} \][/tex]

Answer:

See explanation below

Explanation:

In this reaction we have the ethyl acetoacetate which is reacting with 2 eq of sodium etoxide. The sodium etoxide is a base and it usually behaves as a nucleophyle of many reactions. Therefore, it will atract all the acidics protons in a molecule.

In the case of the ethyl acetoacetate, the protons that are in the methylene group (CH3 - CO - CH2 - COOCH2CH3) are the more acidic protons, therefore the etoxide will substract these protons instead of the protons of the methyl groups. This is because those hydrogens (in the methylene group) are between two carbonile groups, which make them more available and acidic for any reaction. As we have 2 equivalents of etoxide, means that it will substract both of the hydrogen atoms there, and then, reacts with the Br - CH2CH2 - Br and form a product of an aldolic condensation.

The mechanism of this reaction to reach X is shown in the attached picture.

Answer: The value of [tex]E^{o}[/tex] for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          [tex]AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)[/tex]

Its solubility product will be as follows.

       [tex]K_{sp} = [Ag^{+}][Cl^{-}][/tex]

Cell reaction for this equation is as follows.

     [tex]Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)[/tex]

Reduction half-reaction: [tex]Ag^{+} + 1e^{-} \rightarrow Ag(s)[/tex],  [tex]E^{o}_{Ag^{+}/Ag} = 0.799 V[/tex]

Oxidation half-reaction: [tex]Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}[/tex],   [tex]E^{o}_{AgCl/Ag}[/tex] = ?

Cell reaction: [tex]Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)[/tex]

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, [tex]K_{sp} = [Ag^{+}][Cl^{-}][/tex]

                                               = [tex]1.77 \times 10^{-10}[/tex]

Therefore, according to the Nernst equation

           [tex]E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}[/tex]

At equilibrium, [tex]E_{cell}[/tex] = 0.00 V

Putting the given values into the above formula as follows.

         [tex]E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}[/tex]

        [tex]0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}[/tex]    

       [tex]E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}[/tex]

                  = [tex]0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}[/tex]

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

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

       [tex]0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}[/tex]

       [tex]0.577 V = 0.799 V - E^{o}_{AgCl/Ag}[/tex]

       [tex]E^{o}_{AgCl/Ag}[/tex] = 0.222 V

Thus, we can conclude that value of [tex]E^{o}[/tex] for the half-cell reaction is 0.222 V.

The E° for the half-reaction can be calculated by using the Nernst equation, solubility product constant (Ksp) and the standard reduction potential for Ag+(aq) + e− ⇀ ↽ − Ag(s) which is +0.799 V.

The E° for the half-reaction, AgCl (s) + e− ⇀ ↽ − Ag (s) + Cl− (aq) can be calculated using the Nernst equation, considering the relationship between standard reduction potential and solubility product constant.

First, we can rewrite the half-reaction for the dissolution of AgCl: AgCl(s) ⇀ ↽ − Ag+(aq) + Cl−(aq). The solubility product constant (Ksp) for this reaction is 1.77 × 10^−10.

We also know that the standard reduction potential of Ag+(aq) + e− ⇀ ↽ − Ag(s) is +0.799 V.

We can relate this data through the Nernst equation: E=E°−(RT/nF)lnQ. Here, we can set Q as Ksp and adjust the equation to solve for our wanted E.

Using these values in Nernst equation would yield the E for the half-reaction.

https://brainly.com/question/31593791

#SPJ3

The half-life of the radioactive isotope is 8 days because after 16 days (which is two half-lives), the sample is reduced from 20 x 10^10 atoms to 5 x 10^10 atoms.

The question deals with the concept of the half-life of a radioactive isotope, which is the time required for half of the radioactive atoms in a sample to decay. According to the problem statement, an initial sample containing 20 x 1010 atoms decays to 5 x 1010 atoms in 16 days. To find the half-life (t1/2), we can use the fact that after one half-life, the number of atoms would be halved. Starting with 20 x 1010 atoms, after one half-life, there would be 10 x 1010 atoms, and after two half-lives, there would be 5 x 1010. This indicates that two half-lives have passed in 16 days, making the half-life 8 days.

The correct answer is 8 days, option E.

Answer: pH of the solution after the addition of 600.0 ml of LiOH is 7.

Explanation:

The given data is as follows.

    Volume of [tex]HClO_{4}[/tex] = 900.0 ml = 0.9 L,

   Molarity of [tex]HClO_{4}[/tex] = 0.18 M,

Hence, we will calculate the number of moles of [tex]HClO_{4}[/tex] as follows.

       No. of moles = Molarity × Volume

                             = 0.18 M × 0.9 L

                             = 0.162 moles

Volume of NaOH = 600.0 ml = 0.6 L

Molarity of NaOH = 0.27 M

No. of moles of NaOH = Molarity × Volume

                                     = 0.27 M × 0.6 L

                                     = 0.162 moles

This shows that the number of moles of [tex]HClO_{4}[/tex] is equal to the number of moles of NaOH.

Also we know that,

         [tex]HClO_{4} + NaOH \rightarrow NaClO_{4} + H_{2}O[/tex]

As 1 mole of [tex]HClO_{4}[/tex] reacts with 1 mole of NaOH then all the hydrogen ions and hydroxide ions will be consumed.

This means that pH = 7.

Thus, we can conclude that pH of the solution after the addition of 600.0 ml of LiOH is 7.

Answer:

35.6 g

Explanation:

Step 1: Calculate the [H⁺]

We use the following expression

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -13.90 = 1.259 × 10⁻¹⁴ M

Step 2: Calculate the [OH⁻]

We use the ionic product of water (Kw).

Kw = 10⁻¹⁴ = [H⁺] × [OH⁻]

[OH⁻] = 0.7943 M

Step 3: Calculate the moles of OH⁻

[tex]\frac{0.7943mol}{L} \times 0.800 L = 0.635 mol[/tex]

Step 4: Calculate the required moles of KOH

KOH is a strong base that dissociates according to the following equation.

KOH → K⁺ + OH⁻

The molar ratio of KOH to OH⁻ is 1:1. Then, the required moles of KOH are 0.635 moles.

Step 5: Calculate the mass of KOH

The molar mass of KOH is 56.11 g/mol.

[tex]0.635 mol \times \frac{56.11g}{mol} =35.6 g[/tex]

Answer:

Check the explanation

Explanation:

kindly check the attached image below to see the answer to question A and B

(c) The energy balance equation if the pressure is constant (ΔE=0) is,

ΔE= ΔU + Δ[tex]E_{k}[/tex]

From the above relation, it is clear that some heat energy used to raises the temperature of the air.

Hence, the internal energy is not equal to zero. Therefore, the rate of transfer of heat to air is not equal to the rate of thane in kinetic energy of the air.

That is ΔE ≠ Δ[tex]E_{k}[/tex]

Answer:

a) [tex]\delta E_k[/tex] = 112.164 W

b) [tex]\delta E_k[/tex]  = 42.567 W

c) From what we've explained in part b;  increase in temperature of the system is caused by rate of heat transfer . Therefore, not all heat is used to  increase kinetic energy. Hence, since not all the heat is used to increase the kinetic energy . It is not valid and it is incorrect to say that rate of heat transfer is equal to the change in kinetic energy.

Explanation:

The given data include:

Inlet diameter [tex](d_1)[/tex] = 7 cm = 0.7 m

Inlet velocity [tex](v_1)[/tex] = 42 m/s

Inlet pressure [tex](P_1)[/tex] = 130 KPa

Inlet temoerature [tex](T_1)[/tex] = 300°C = (300 + 273.15) = 573.15 K

a)  Assuming Ideal gas behaviour

Inlet Volumetric flowrate [tex](V_1)[/tex] = Inlet velocity [tex](v_1)[/tex] × area of the tube

[tex]= v_1*(\frac{\pi}{4})0.07^2\\\\= 0.161635 \ m^3/s[/tex]

Using Ideal gas law at Inlet 1

[tex]P_1V_1 =nRT_1[/tex]

where ; n = molar flow rate of steam

making n the subject of the formula; we have:

[tex]n = \frac{P_1V_1}{RT_1}[/tex]

[tex]n = \frac{130*42}{8.314*573.15}[/tex]

[tex]n= 4.4096*10^{-3} \ Kmol/s[/tex]

Moleular weight of air = 28.84 g/mol

The mass flow rate = molar flowrate × molecular weight of air

[tex]= 4.4096*10^{-3} \ * 28.84\\= 0.12717 \ kg/s[/tex]

Finally: the kinetic  energy at Inlet [tex](\delta E_k) = \frac{1}{2}*mass \ flowrate *v_1^2[/tex]

= [tex]0.5*0.12717*42^2[/tex]

= 112.164 W

b) If the air is heated to 400°C;

Then temperature at 400°C = (400 + 273.15)K = 673.15 K

Thee pressure is also said to be constant ;

i.e [tex]P_1= P_2[/tex] = 130 KPa

Therefore; the mass flow rate is also the same ; so as the molar flow rate:

Thus; [tex]n= 4.4096*10^{-3} \ Kmol/s[/tex]

Using Ideal gas law at Inlet 2

[tex]P_2V_2 = nRT_2[/tex]

making [tex]V_2[/tex] the subject of the formula; we have:

[tex]V_2 = \frac{nRT_2}{P_2}[/tex]

[tex]V_2 = \frac{4.4906*10^{-3}*8.314*673.15}{130}[/tex]

[tex]V_2 = 0.189836 \ m^3/s[/tex]

Assuming that the diameter is constant

[tex]d_1 = d_2 = 0.07 \ cm[/tex]

Now; the velocity at outlet = [tex]\frac{V_2}{\frac{\pi}{4}d_2^2}[/tex]

= [tex]\frac{0.0189836}{\frac{\pi}{4}(0.07)^2}[/tex]

= 49.33 m/s

Change in kinetic energy [tex]\delta E_k[/tex]  = [tex]\frac{1}{2}*mass \ flowrate * \delta V[/tex]

= [tex]0.5*0.12717 *(49.33^2 -42^2)[/tex]

= 42.567 W

c).

From what we've explained in part b; increase in temperature of the system is caused by rate of heat transfer . Therefore, not all heat is used to  increase kinetic energy. Hence, since not all the heat is used to increase the kinetic energy . It is not valid and it is incorrect to say that rate of heat transfer is equal to the change in kinetic energy.

Answer:

a. Copper (II) oxide is the limiting reactant.

b. [tex]m_{N_2}=5.30g[/tex]

c. [tex]Y=87\%[/tex]

d. [tex]m_{NH_3}=2.60gNH_3[/tex]

Explanation:

Hello,

In this case, for the given reaction:

[tex]2 NH_3(g) + 3 CuO(s) \rightarrow N_2(g) + 3 Cu(s) + 3 H_2O(l)[/tex]

a. The limiting reactant is identified by computing the available moles of ammonia and the moles of ammonia that react with 45.2 g of copper (I) oxide as shown below:

[tex]n_{NH_3}^{Available}=9.05gNH_3*\frac{1molNH_3}{17gNH_3}=0.532molNH_3\\n_{NH_3}^{Reacted}=45.2gCuO*\frac{1molCuO}{79.545gCuO}*\frac{2molCuO}{3molCuO} =0.379molNH_3[/tex]

In such a way, as there more ammonia available than that is reacted, we say it is in excess and the copper (II) oxide the limiting reactant.

b. Here, with the reacting moles of ammonia, we compute the yielded grams of nitrogen:

[tex]m_{N_2}=0.379molNH_3*\frac{1molN_2}{2molNH_3}*\frac{28gN_2}{1molN_2}\\m_{N_2}=5.30g[/tex]

c. Now, since the 5.30 g of nitrogen are the expected grams of it, the percent yield of nitrogen is compute by dividing the real obtained mass over the theoretical previously computed mass:

[tex]Y=\frac{4.61g}{5.30g} *100\%\\Y=87\%[/tex]

d. Finally, as 0.532 moles of ammonia are available, but just 0.379 moles react, the unreacted moles are:

[tex]n_{NH_3}=0.532molNH_3-0.379molNH_3=0.153molNH_3[/tex]

That in grams are:

[tex]m_{NH_3}=0.153molNH_3*\frac{17gNH_3}{1molNH_3}=2.60gNH_3[/tex]

Best regards.

The limiting reactant is CuO, the percent yield of nitrogen gas is 29%.

The equation of the reaction is;

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)

Number of moles of NH3 =  9.05 g/17 g/mol = 0.53 moles

Number of moles of CuO = 45.2 g/80 g/mol = 0.565 moles

Since 2 moles of NH3 reacts with 3 moles of CuO

0.53 moles of NH3 reacts with 0.53 moles × 3 moles/ 2 moles

= 0.795 moles

We can see that there is not enough CuO in the system hence it is the limiting reactant.

Number of moles of N2 produced = 0.565 moles × 28 g/mol =15.82 g of N2

Percent yield = Actual yield/Theoretical yield × 100/1

Percent yield = 4.61 g/15.82 g  × 100/1

Percent yield = 29%

If 2 moles of NH3 reacts with 3 moles of CuO

x moles of NH3 reacts with 0.565 moles of CuO

x =  2 moles × 0.565 moles/3 moles

x = 0.38 moles

Number of moles of excess reactant left over = 0.53 moles - 0.38 moles

= 0.15 moles

Mass of excess reactant left over = 0.15 moles × 17 g/mol = 2.55 g

Learn more: https://brainly.com/question/11897796

Answer:

Reaction Quotient, Kq = {[a-ketoglutarate]x[L-alanine]}/{[L-glutamate]x[pyruvate]}

or, Kq = {(1.6x10-2)x(6.25x10∧-3)}/{(3x10∧-5)x(3.3x10-4)} = 1.01 x 10∧4

Since Kq > Keqb ; therefore the reaction will proceed in the backward direction, in order words the reaction will not occur in forward direction. i.e formation of reactants will be favored.

Explanation:

Answer:

ΔS = -114.296 J/K

Explanation:

Step 1: Data given

Temperature = 298 K

Number of moles N2 = 2.19 moles

S°(N2) = 191.6 J/K*mol

S°(O2)= 161.1 J/k*mol

S°(N2O) = 219.96 J/K*mol

Step 2: The balanced equation

2N2(g) + O2(g) ⇆ 2N2O(g)

Step 3: Calculate ΔSrxn

ΔSrxn = ∑S°products -  ∑S°reactants

ΔSrxn = (2*219.96) - (161.1 + 2*191.6) J/K*mol

ΔSrxn = -104.38 J/K

Step 4: Calculate ΔS for 2.19 moles

The reaction is for 2 moles N2

ΔS = -114.296 J/K

Final answer:

The entropy change for a chemical reaction can be determined using standard absolute entropies. In this case, the given reaction of N2(g) and O2(g) to form N2O(g) can be analyzed using these principles.

Explanation:

The entropy change for the given reaction can be calculated using the standard absolute entropies at 298K. The formula to calculate the entropy change is ΔS = ΣnS(products) - ΣnS(reactants). In this case, for the reaction 2N2(g) + O2(g) → 2N2O(g), the entropy change can be calculated using the standard absolute entropies of each component.

Step 1: Data given

Temperature = 298 K

Number of moles N2 = 2.19 moles

S°(N2) = 191.6 J/K*mol

S°(O2)= 161.1 J/k*mol

S°(N2O) = 219.96 J/K*mol

Step 2: The balanced equation

2N2(g) + O2(g) ⇆ 2N2O(g)

Step 3: Calculate ΔSrxn

ΔSrxn = ∑S°products -  ∑S°reactants

ΔSrxn = (2*219.96) - (161.1 + 2*191.6) J/K*mol

ΔSrxn = -104.38 J/K

Step 4: Calculate ΔS for 2.19 moles

The reaction is for 2 moles N2

ΔS = -114.296 J/K

3.47 x [tex]10^{23}[/tex] atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x [tex]10^{23}[/tex]

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X [tex]10^{23}[/tex]

from the relation,

1 mole of element contains 6.022 x [tex]10^{23}[/tex] atoms.

so no of moles of gold given = [tex]\frac{3.47 X 10^{23} }{6.022 X 10^{23} }[/tex]

0.57 moles of gold.

from the relation:

number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x [tex]10^{23}[/tex] atoms of gold have mass of 113.44 grams

The mass of 3.47 x 10^23 gold atoms calculated using Avogadro's number and the molar mass of gold equals approximately 113.52 grams. To calculate this, the given number of atoms was first converted into moles and the number of moles were then multiplied by the molar mass.

In chemistry, to calculate the mass of a certain number of atoms, we use the concept of a mole and Avogadro's number (6.022 x 1023). For gold atoms, the molar mass is 197 g/mol. Given that the number of gold atoms is 3.47 x 1023, we can use the relationship that 1 mole of gold atoms = 6.022 x 1023 gold atoms. Therefore, the mass of 3.47 x 1023 gold atoms can be calculated as follows:

First, we convert the number of atoms to moles using Avogadro's number: 3.47 x 1023 atoms / 6.022 x 1023 atoms/mol = 0.576 moles of gold.

Then, we multiply this moles by the molar mass of gold to get the mass in grams: 0.576 moles * 197 g/mol = 113.52 g.

https://brainly.com/question/16034692

#SPJ3

Answer:

In a Grignard reagent, the carbon bonded to the magnesium has a partial negative charge, because carbon is more electronegative than magnesium. This makes this carbon of the Grignard nucleophilic.

Explanation:

Organometallic compounds have covalent bonds between carbon atoms and metallic atoms. Organometallic reagents are useful because they have nucleophilic carbon atoms, in contrast to the electrophilic carbon atoms of the alkyl halides. The majority of metals are more electropositive than carbon, and the bond is biased with a positive partial charge on the metal and a negative partial charge on the carbon.

Answer:

In a Grignard reagent, the carbon bonded to the magnesium has a partial negative charge, because carbon is more electronegative than magnesium. This makes this carbon of the Grignard nucleophilic In a Grignard reagent, the carbon bonded to the magnesium has a partial positive charge, because carbon is more electronegative than magnesium. This makes this carbon of the Grignard less electrophilic.

Explanation:

The Grignard reagent is a highly reactive organomagnesium compound formed by the reaction of a haloalkane with magnesium in an ether solvent. The carbon atom of a Grignard reagent has a partial negative charge. Hence a Grignard reagent will have a general formula RMgX. Where R must contain a carbanion.

Grignard reagents are versatile reagents used in many applications in organic chemistry.

Answer:

The maximum volume is 122.73 mL

Explanation:

125 mL of butter that is 0.360 M in CH₃COOH and CH₃COO⁻.

pKa of acetic acid = 4.76

Using Henderson Hasselbalch equation.

pH =pKa + log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex]

= 4.76 + [tex]log \frac{0.3600}{0.3600}[/tex]

= 4.76

Assuming the pH of the buffer changes by a unit of 1 , then it will lose its' buffering capacity.

When HCl is added , it reacts with  CH₃COO⁻ to give CH₃COOH.

However, [CH₃COOH] increases and the log term  results in a negative value.

Let  assume, the new pH is less than 4.76

Let say 3.76; calculating the concentration when the pH is 3.76; we have

3.76 = 4.76 + log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex]

log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 4.76 - 3.76

log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 1.00

[tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 0.1

Let number of  moles of acid be x   (i.e change in moles be x);  &

moles of acetic acid and conjugate base present be = Molarity × Volume

= 0.360 M × 125 mL

= 45 mmol

replacing initial concentrations and change in the above expression; we have;

[tex]\frac{[CH_3COO^-]}{[CH_3COOH]} = 0.1[/tex]

[tex]\frac{45-x}{45+x} =0.1[/tex]

0.1(45 + x) = 45 - x

4.5 + 0.1 x = 45 -x

0.1 x + x = 45 -4.5

1.1 x = 40.5

x = 40.5/1.1

x = 36.82

So, moles of acid added = 36.82 mmol

Molarity = 0.300 M

So, volume of acid = [tex]\frac{moles}{molarity }[/tex] =  [tex]\frac{36.82 \ mmol}{0.300}[/tex]

= 122.73 mL

Answer:

The structure in the first image file attached below shows the arrangement of atoms in hexagonal close packing

we need to show that the ratio between the height of the unit cell divided by its edge length is 1.633

In the structure, the two atoms are shown apart. But in fact the two atoms are touching. Therefore, the edge length is the sum of the radius of two atoms. If we assume r as radius then the expression for edge length will be as follows.

a = 2r    ..............(1)

other attached files show additional solutions in steps

The ideal c/a ratio in an HCP unit cell is 1.633, depicting the most compact arrangement of spheres. However, real HCP metals, influenced by bonding characteristics and atomic sizes, display varying c/a ratios. For Mg with known atomic radius in HCP arrangement, the lattice constants, c/a ratio, and theoretical density can be calculated.

The HCP unit cell (hexagonal-close-packed) is one of the simplest lattice types in crystal structures. It includes two types of atoms, one at the corners (and center) of a hexagonal prism and the other at the top and bottom of the prism. The ideal ratio of c/a (height of the unit cell c divided by edge length a) corresponds to the most compact possible arrangement of spheres, and is theoretically calculated to be √(8/3), or approximately 1.633.

In real-world applications, various HCP metals display c/a ratios varying slightly, such as 1.58 for Be and 1.89 for Cd. This depends on bonding characteristics and varying atomic sizes in the metal crystal structure. For example, for Mg (magnesium) which has a known atomic radius of 0.1605 nm in HCP arrangement, we can calculate the lattice constants, c and a, and thus the c/a ratio. Theoretical density can be determined using the mass of atoms in a unit cell, Avogadro's number, and the volume of unit cell.

Learn more about HCP Unit Cell Geometry here:

https://brainly.com/question/31874975

#SPJ3

Answer:

(a)  Distillate (D) = 0.597 kg

    Bottom product (B) = 0.403 kg

(b) Compositions of distillate  product:

     Benzene = 0.995

     Toluene = 0.05

   Compositions of bottom product:

    Benzene = 0.0149

    Naphthalene = 0.1241

    Toluene = 0.8610

(c) Fraction of toluene fed that is recovered in the bottom product = 99.14%

Explanation:

See the attached file for the calculation

The question is incomplete, the table of the question is given below

Answer:

I) xA= 0.34, yA= 0.55

ii) 76.2 mole % vapor

iii) Percentage of vapor volume = 98%

Explanation:

i) xA= 0.34, yA= 0.55

 xA= 0.34, yA= 0.55

ii)      0.50 = 0.55 nv + 0.34 nL

     Therefore, nV =    0.762 mol vapor and nL = 0.238 mol liquid

This shows 76.2 mole % vapor

iii)  ρA= 0.791 g/cm3 and,  ρE = 0.789 g/cm3

Therefore, ρ = 0.790 g/cm3

Now, we have:

MA = 58.08 g/mol and ME= 46.07 g/mol

So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol

1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor

Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3

Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3

Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %

Find the figure in the attachment

Answer:

1. The balanced equation is given below:

CaCO3 —> CaO + CO2

2. The coefficients are: 1, 1, 1

Explanation:

Step 1:

The word equation is given below:

calcium carbonate decomposes to produce calcium oxide and carbon dioxide.

Step 2:

The elemental equation. This is illustrated below:

Calcium carbonate => CaCO3

calcium oxide => CaO

carbon dioxide => CO2

CaCO3 —> CaO + CO2

A careful observation of the above equation shows that the equation is balanced.

The coefficients are: 1, 1, 1

Final answer:

The balanced equation for the decomposition reaction of calcium carbonate is CaCO3(s) → CaO(s) + CO2(g), representing the formation of calcium oxide and carbon dioxide gas from calcium carbonate.

Explanation:

The decomposition reaction of calcium carbonate forming calcium oxide and carbon dioxide can be represented by the following balanced chemical equation:

CaCO3(s) → CaO(s) + CO2(g)

This equation indicates that one mole of calcium carbonate (CaCO3) decomposes to form one mole of calcium oxide (CaO) and one mole of carbon dioxide (CO2) gas upon heating. The reactants and products are represented using their smallest possible integer coefficients, reflecting the principle of conservation of mass and the stoichiometry of the decomposition reaction.

Answer:

copper

Explanation:

These are usually copper wires with no insulation. They make the path through which the electricity flows. One piece of the wire connects the current from the power source (cell) to the load

Answer: The Molar mass of Cro2 (or Chromium (IV) oxide) is 83.9949 g/mol.

Explanation: The atomic mass of Chromium is 51.9961 and the atomic mass of Oxygen is 15.9994. You then add both numbers and multiply by 2. Put it in the calculator like this: 51.9961 + 15.9994 x 2 and you will get 83.9949 g/mol.

Answer:

B

Explanation:

B. There are two atoms of Nitrogen and two atoms of Hydrogen combined to make Ammonia.

6.81% is the percent yield when 12 g of CO2 are formed experimentally from the reaction of 0.5 moles of C8H18 reacting with excess oxygen.

Explanation:

Data given:

mass of carbon dioxide formed = 12 grams (actual yield)

atomic mass of CO2 = 44.01 grams/mole

moles of [tex]C_{8} H_{18}[/tex] = 0.5

Balanced chemical reaction:

2 [tex]C_{8} H_{18}[/tex] + 25 [tex]O_{2}[/tex] ⇒ 16 C[tex]O_{2}[/tex] + 18 [tex]H_{2} _O{}[/tex]

number of moles of carbon dioxide given is

number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]

number of moles= [tex]\frac{12}{44}[/tex]

number of moles of carbon dioxide gas = 0.27 moles

from the reaction 2 moles of [tex]C_{8} H_{18}[/tex] reacts to produce 16 moles of C[tex]O_{2}[/tex]

So, when 0.5 moles reacted it produces x moles

[tex]\frac{16}{2}[/tex] = [tex]\frac{x}{0.5}[/tex]

x = 4

4 moles of carbon dioxide formed, mass from it will give theoretical yield.

mass  = number of moles x molar mass

mass = 4 x 44.01

        = 176.04 grams

percent yield =[tex]\frac{actual yield}{theoretical yield}[/tex] x 100

percent yield = [tex]\frac{12}{176.04}[/tex] x 100

     percent yield  = 6.81 %

Answer:

Explanation:

check the attachment below for correct explanations.

The enolate ion is formed from a carbonyl function via a strong base, leading to resonance structures between keto and enolate forms. The resonance involves shifting of electron pairs, with the enolate form being more prevalent under basic conditions.

An enolate ion is formed when a strong base, B, abstracts a proton α to a carbonyl function. The negatively charged oxygen donates a pair of electrons, pushing the electrons from the double bond into the alpha carbon to form the enolate form. The curved arrow represents the movement of a pair of electrons.

The resonance of the keto form to the enolate form involves shifting of electron pairs, with the electron pair on the carbon moving to form a C=O bond and the π bond between carbon and oxygen moving to the oxygen, forming a p-orbital with a lone pair of electrons and making the oxygen negatively charged.

By this mechanism, the keto and enolate forms are interconvertible with the enolate form being more prevalent under basic conditions. The full resonance structures for both the keto and enolate forms would consist of all their bonds, charges, and nonbonding electron pairs.

https://brainly.com/question/36343750

#SPJ3

To construct a simulated 1H NMR spectrum for 1-chloropropane, you need to consider the chemical shifts and splitting patterns of the hydrogen atoms. Follow these steps: identify the different types of hydrogen atoms, determine the chemical shifts, assign integration values, and drag and drop splitting patterns.

To construct a simulated 1H NMR spectrum for 1-chloropropane, you need to consider the chemical shifts and splitting patterns of the hydrogen atoms. Here's a step-by-step guide:

https://brainly.com/question/9812005

#SPJ6