A common recipe is to make 3%(wt/vol) HCl in ethanol. HCl has a formula weight of 36.46 grams per mole. If the stock solution of HCl is 1 moles per 1000 mL, how many mL of HCl need to be added to achieve a final volume of 250 mL of acid alcohol solution? Report your answer to two decimal places.

Answer:

The specific heat of copper is  [tex]C= 392 J/kg\cdot ^o K[/tex]

Explanation:

From the question we are told that

The amount of energy contributed by each oscillating lattice site  is  [tex]E =3 kT[/tex]

       The atomic mass of copper  is  [tex]M = 63.6 g/mol[/tex]

        The atomic mass of aluminum is  [tex]m_a = 27.0g/mol[/tex]

        The specific heat of aluminum is  [tex]c_a = 900 J/kg-K[/tex]

 The objective of this solution is to obtain the specific heat of copper

       Now specific heat can be  defined as the heat required to raise the temperature of  1 kg of a substance by  [tex]1 ^o K[/tex]

  The general equation for specific heat is  

                    [tex]C = \frac{dU}{dT}[/tex]

Where [tex]dT[/tex] is the change in temperature

             [tex]dU[/tex] is the change in internal energy

The internal energy is mathematically evaluated as

                       [tex]U = 3nk_BT[/tex]

      Where  [tex]k_B[/tex] is the Boltzmann constant with a value of [tex]1.38*10^{-23} kg \cdot m^2 /s^2 \cdot ^o K[/tex]

                    T is the room temperature

                      n is the number of atoms in a substance

Generally number of  atoms in mass of an element can be obtained using the mathematical operation

                      [tex]n = \frac{m}{M} * N_A[/tex]

Where [tex]N_A[/tex] is the Avogadro's number with a constant value of  [tex]6.022*10^{23} / mol[/tex]

          M is the atomic mass of the element

           m actual mass of the element

  So the number of atoms in 1 kg of copper is evaluated as  

             [tex]m = 1 kg = 1 kg * \frac{10000 g}{1kg } = 1000g[/tex]

The number of atom is  

                       [tex]n = \frac{1000}{63.6} * (6.0*0^{23})[/tex]

                          [tex]= 9.46*10^{24} \ atoms[/tex]

Now substituting the equation for internal energy into the equation for specific heat

          [tex]C = \frac{d}{dT} (3 n k_B T)[/tex]

              [tex]=3nk_B[/tex]

Substituting values

         [tex]C = 3 (9.46*10^{24} )(1.38 *10^{-23})[/tex]

            [tex]C= 392 J/kg\cdot ^o K[/tex]

Higher concentration make a reaction faster due to faster collisions. The correct answer is "There are more collisions per second."

When the concentration of reactants is increased, it leads to a higher number of particles or molecules in the reaction mixture. As a result, there are more frequent collisions between the reactant particles per unit time.

In a chemical reaction, the reactant particles must collide with sufficient energy and proper orientation for a successful reaction to occur. By increasing the concentration, the chances of successful collisions increase because there are more particles available to collide with each other.

However, it's important to note that while a higher concentration increases the frequency of collisions, it does not necessarily guarantee that all collisions will result in a reaction. The energy of the collisions and proper orientation are also essential factors for a successful reaction.

Learn more about collisions from the link given below.

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Answer:

C HI = 0.0968 M

Explanation:

∴ a: HI(g)

∴ order (α) = 1

∴ rate constant (K) = 7.21/s

∴ Initial concenttration HI(g) (Cao) = 0.440 M

∴ t = 0.210 s ⇒ Ca = ?

⇒ - δCa/δt = KCa

⇒ - ∫δCa/Ca = K*∫δt

⇒ Ln(Cao/Ca) = K*t

⇒ Cao/Ca = e∧(K*t)

⇒ Cao/e∧(K*t) = Ca

⇒ Ca = (0.440)/e∧((7.21*0,21))

⇒ Ca = 0.440/4.54533

⇒ Ca = 0.0968 M

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

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:

742.2 L

Explanation:

First we must find the number of moles of nitroglycerine reacted.

Molar mass of nitroglycerine= 227.0865 g/mol

Mass of nitroglycerine involved = 1×10^3 g

Number of moles of nitroglycerine= 1×10^3g/227.0865 g/mol

n= 4.40361 moles

T= 1985°C + 273= 2258K

P= 1.100atm

R= 0.082atmLmol-1K-1

Using the ideal gas equation:

PV= nRT

V= nRT/P

V= 4.40361× 0.082× 2258/1.1

V= 742 L

Considering the reaction stoichiometry and ideal gas law, the volume of gas produced is 5375.626 L.

The balanced reaction is:

4 C₃H₅N₃O₉(s) → 12 CO₂(g) + 10 H₂O(g) + 6 N₂(g) + O₂(g)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Then 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas [12 moles of CO₂(g) + 10 moles of H₂O(g) + 6 moles of N₂(g) + 1 mole O₂(g)]

Being the molar mass of nitroglycerine C₃H₅N₃O₉(s) 227 g/mole, then the amount of moles that 1 kg (1000 g) of the compound contains can be calculated as:

[tex]1000 gramsx\frac{1 mole}{227 grams}= 4.405 moles[/tex]

Then you can apply the following rule of three: if by stoichiometry 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas, 4.405 moles of C₃H₅N₃O₉(s) will produce how many moles of gas?

[tex]amount of moles of gas=\frac{4.405 moles of nitroglycerinex29 moles of gas}{4 moles of nitroglycerine}[/tex]

amount of moles of gas= 31.93625 moles

On the other hand, an ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P× V = n× R× T

In this case, you know:

Replacing:

1.1 atm× V= 31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K

Solving:

V= (31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K) ÷ 1.1 atm

V= 5375.625 L

Finally, the volume of gas produced is 5375.626 L.

Learn more:

Answer: a) 0.070 moles of oxygen were produced.

b) New pressure due to the oxygen gas is 2.4 atm

Explanation:

According to ideal gas equation:

[tex]PV=nRT[/tex]

P = pressure of gas = 2.7 atm

V = Volume of gas = 700 ml = 0.7 L

n = number of moles = ?

R = gas constant =[tex]0.0821Latm/Kmol[/tex]

T =temperature = 329 K

[tex]n=\frac{PV}{RT}[/tex]

[tex]n=\frac{2.7atm\times 0.7L}{0.0821 L atm/K mol\times 329K}=0.070moles[/tex]

Thus 0.070 moles of oxygen were produced.

When the 700 mL flask cools to a temperature of 293K.

[tex]PV=nRT[/tex]

[tex]P=\frac{nRT}{V}[/tex]

[tex]P=\frac{0.070\times 0.0821\times 293}{0.7}[/tex]

[tex]P=2.4atm[/tex]

The new pressure due to the oxygen gas is 2.4 atm

Answer:

Explanation:

check the attachment below for correct explanations.

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.

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 %

The balanced chemical equation for the overall chemical reaction is 2NO2(g) + F2(g) → 2NO2F(g). The experimentally observable rate law for the overall chemical reaction is Rate = k [NO2]^2 [F2]. The rate constant k for the overall chemical reaction can be expressed in terms of k1, k2, k-1, and k-2 as k = k1 (k2/k-1).

The balanced chemical equation for the overall chemical reaction is:

2NO2(g) + F2(g) → 2NO2F(g)

The experimentally observable rate law for the overall chemical reaction is:

Rate = k [NO2]2 [F2]

The rate constant k for the overall chemical reaction can be expressed in terms of k1, k2, k-1, and k-2 as:

k = k1 (k2/k-1)

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.

The most likely substrates for the oligo-(α1→4)-(α1→4) glucanotransferase catalytic center of glycogen debranching enzyme are maltosyl and maltotriosyl residues.

The glycogen debranching enzyme has two activities; it acts as a glucosidase and a glucanotransferase. The (α‑6) glucosidase activity hydrolyzes α-1,6 glycosidic bonds, whereas the oligo-(α1→4)-(α1→4) glucanotransferase activity shifts α-1,4-linked glucose chains from one branch to another, often creating a substrate that the glucosidase can act upon. Based on the mechanism of glycogenolysis, the enzyme glycogen phosphorylase releases glucose units from the linear chain until a few are left near the branching point, and it is here that the glucan transferase action is relevant. The glucan transferase shifts the remaining α-1,4 linked glucose units, leaving a single α-1,6 linked glucose that the glucosidase can then release. Hence, the most likely substrates for the oligo-(α1→4)-(α1→4) glucanotransferase catalytic center are maltosyl and maltotriosyl residues, as these are the short α-1,4 linked glucose chains that are left after phosphorylase action.

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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:

15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.

Explanation:

A → B + C

The rate law of the reaction will be :

[tex]R=k[A]^x[/tex]

Initial rate of the reaction when concentration of the substrate was 0.4 M:

[tex]0.183 M/s=k[0.4 M]^x[/tex]..[1]

Initial rate of the reaction when concentration of the substrate was 0.8 M:

[tex]0.183 M/s=k[0.8 M]^x[/tex]...[2]

[1] ÷ [2] :

[tex]\frac{0.183 M/s}{0.183 M/s}=\frac{k[0.4 M]^x}{k[0.8 M]^x}[/tex]

x = 0

The order of the reaction is zero.

For the value of rate constant ,k:

[tex]0.183 M/s=k[0.4 M]^x[/tex]..[1]

x = 0

[tex]0.183 M/s=k[0.4 M]^0[/tex]

k= 0.183 M/s

The half life of the zero order kinetics is given by :

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

Where:

[tex][A_o][/tex] = Initial concentration of A

k = Rate constant of the reaction

So, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M:

[tex]t_{1/2}=\frac{2.77 M}{0.183 M/s}=15.1 s[/tex]

15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.

The half-life for the decomposition of substrate 1 when the initial concentration is 2.77 M is approximately 18 minutes.

The half-life of a reaction is the time required for one-half of a given amount of reactant to be consumed. In this case, we have initial rate data for the decomposition of substrate 1 with varying concentrations. To determine the half-life of substrate 1, we need to find the time it takes for the concentration to decrease to half its initial value.

Based on the given data, we can see that the initial rate of the reaction is constant at 0.183 M/s for different initial concentrations of substrate 1 (0.4 M, 0.8 M, and 2 M). Since the initial concentration of substrate 1 is given as 2.77 M, it will take the same amount of time as the other initial concentrations to reach half its initial concentration.

Therefore, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M is the same as the half-life observed when the initial concentration is 0.4 M, 0.8 M, or 2 M, which is approximately 18 minutes.

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Answer:

The phosphorylation of glucose to glucose-6-phosphate is endergonic reaction that is coupled to the exergonic hydrolysis of ATP.

Explanation:

In glycosis, the first reaction that takes place is the phosphorylation of glucose to glucose-6-phosphate by the enzyme hexokinase. This is an exergenic reaction. This is a coupled reaction in which phosphorylation of glucose is coupled to ATP hydrolysis. The free energy of ATP hydrolysis fuels glucose phosphorylation.

Answer:

2-Tert-butyl-cyclohexanone, 2,6-Di-Tert-Butylcyclohexanone and Di-isopropyl-amine.  

Explanation:

Answer:

B

Explanation:

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

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:

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n = 0.0989 moles

n = PV / RT

P = 2.09atm

V = 1.13L

R = 0.08206

T = 291K

Plug the numbers in the equation.

n = (2.09atm)(1.13L) / (0.08206)(291K)

n = 0.0989 moles

Final answer:

Using the ideal gas law PV = nRT, and rearranging for n (moles), we plug in the given values of pressure, volume, and temperature alongside the ideal gas constant to calculate the number of moles of gas present. There are approximately 0.0988 moles of gas present in 1.13 L of gas at [tex]\(2.09 \, \text{atm}\) and \(291 \, \text{K}\).[/tex]

Explanation:

To calculate the number of moles of gas present, we can use the ideal gas law equation:

[tex]\[ PV = nRT \][/tex]

First, we need to convert the given pressure to atm and the volume to liters if they are not already in those units.

Given:

- [tex]\( P = 2.09 \, \text{atm} \)[/tex]

-[tex]\( V = 1.13 \, \text{L} \[/tex]

- [tex]\( T = 291 \, \text{K} \)[/tex]

Now, we can rearrange the ideal gas law equation to solve for \( n \):

[tex]\[ n = \frac{PV}{RT} \][/tex]

[tex]\[ n = \frac{(2.09 \, \text{atm})(1.13 \, \text{L})}{(0.0821 \, \text{atm} \cdot \text{L} / \text{mol} \cdot \text{K})(291 \, \text{K})} \][/tex]

[tex]\[ n = \frac{2.3617}{23.9211} \][/tex]

[tex]\[ n \approx 0.0988 \, \text{mol} \][/tex]

Therefore, There are approximately 0.0988 moles of gas present in 1.13 L of gas at [tex]\(2.09 \, \text{atm}\) and \(291 \, \text{K}\).[/tex]

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:

[tex]T = 34.493\,^{\textdegree}C[/tex]

Explanation:

The equilibrium temperature of the gas-container system is:

[tex]Q_{Cu} = -Q_{g}[/tex]

[tex](0.3\,kg)\cdot\left(386\,\frac{J}{kg\cdot ^{\textdegree}C} \right)\cdot (T-20\,^{\textdegree}C) = (1.5\,mol)\cdot \left(12.5\,\frac{J}{mol\cdot ^{\textdegree}C} \right)\cdot (124\,^{\textdegree}C-T)[/tex]

[tex]\left(115.8\,\frac{J}{^{\textdegree}C} \right)\cdot (T-20\,^{\textdegree}C) = \left(18.75\,\frac{J}{^{\textdegree}C} \right)\cdot (124\,^{\textdegree}C-T)[/tex]

[tex]134.55\cdot T = 4641[/tex]

[tex]T = 34.493\,^{\textdegree}C[/tex]

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.

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The final pressure of the ammonia gas, given that the the volume decreased by 50.0%, is 11 atm

How to calculate the final pressure f the ammonia gas?

To solve this question in the most simplified way, we shall begin by listing out the given data from the question. This is shown below:

The final pressure of the ammonia gas can be calculated as shown below:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\\\frac{4.6\ \times\ V}{252} = \frac{P_2\ \times\ 0.5V}{296}\\\\252\ \times\ P_2\ \times\ 0.5V = 4.6\ \times\ V\ \times\ 296\\\\P_2 = \frac{4.6\ \times\ V\ \times\ 296}{252\ \times\ 0.5V} \\\\P_2 = 11\ atm[/tex]

Thus, the final pressure is 11 atm

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:

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:

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.