Measurements show that the enthalpy of a mixture of gaseous reactants decreases by 162. kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -194. kJ of work is done on the mixture during the reaction. a.Calculate the change in energy of the gas mixture during the reaction. Round your answer to 3 significant digits b. Is the reaction exothermic or endothermic?

Answer:

0.122 moles

Explanation:

Final answer:

0.122 moles of KCl will be produced from the reaction of 15.0 g of KClO3, using stoichiometry and the molar mass of KClO3 which is 122.55 g/mol.

Explanation:

To determine the amount of KCl produced from 15.0 g of KClO3, we first need to know the molar mass of KClO3, which is 122.55 g/mol according to LibreTexts. By using stoichiometry, we can find out how many moles of KClO3 we have and, subsequently, how many moles of KCl will be produced according to the balanced chemical equation.

First, calculate the number of moles of KClO3:

15.0 g KClO3 × (1 mol KClO3 / 122.55 g) = 0.122 mol KClO3

From the balanced equation Î{2KClO3 -> 2KCl + 3O2}, we can see that 2 moles of KClO3 produce 2 moles of KCl. Therefore, the number of moles of KCl produced is equal to the number of moles of KClO3 available:

0.122 mol KClO3 × (2 mol KCl / 2 mol KClO3) = 0.122 mol KCl

Thus, 0.122 mol of KCl are produced from the reaction of 15.0 g of KClO3.

Answer:

CHC13

Explanation:

A slice or part of a homogeneous substance has the same density as the whole piece because the density is uniform throughout. In a heterogeneous substance, the density can vary and depends on the composition and structure of the part being measured. Density is found by dividing mass by volume and is useful for identifying substances.

The density of a substance is defined as its mass divided by its volume. In a homogeneous material, the density is consistent throughout, which means a slice or part of the substance would have the same density as the whole piece. For example, a solid iron bar is homogeneous, and therefore any part of it would have the same density as the entire bar. However, in a heterogeneous material, the density can vary from one part to another; an instance of this is Swiss cheese, which contains air pockets resulting in variable local densities.

To determine the density of a substance, you would divide the mass by the volume of the substance. This method applies to both whole objects and parts thereof. In the case of a cube, the volume is found by cubing the edge length. In practice, density is often used to help identify substances by comparison with known values, as different substances have characteristic densities.

Answer:

C

Explanation:

Well, we can see from the equation that hydrogen gas and ethane are in a 2:1 ratio based on their mole coefficients.

You need 2 mol H2 to make 1 mol C2H6. So that means you need 2*13.78 mol H2 to make 13.78 mol C2H6!

2*13.78 = 27.56 mol

C is correct.

Answer:

Check the explanation

Explanation:

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Answer: 0.225, x, 0.0788

Explanation:

Answer:

Find the given attachment

Option b (O=N−F and N=O−F) represents a pair of resonance structures.

The correct choice representing a pair of resonance structures is option b. O=N−F and N=O−F.

Resonance structures are different representations of a molecule or ion that show the delocalization of electrons. In option b, the double bond can be moved between the nitrogen and oxygen atoms, resulting in two resonance structures.

Resonance occurs when there are multiple valid Lewis structures that can be drawn for a molecule or ion. It indicates that the actual structure is a combination or hybrid of all the resonance structures.

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Answer : The new volume will be, 238.9 mL

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles of gas.

Mathematically,

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where,

[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.

[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.

We are given:

[tex]V_1=222mL\\T_1=30.0^oC=(30.0+273)K=303K\\V_2=?\\T_2=53.1^oC=(53.1+273)K=326.1K[/tex]

Putting values in above equation, we get:

[tex]\frac{222mL}{303K}=\frac{V_2}{326.1K}\\\\V_2=238.9mL[/tex]

Therefore, the new volume will be, 238.9 mL

Answer:

D. The experiment includes a good plan for what measurements to take.

Explanation:

Answer:

(a)

Rate of appearance of sucrose = - d[C12H22O11] / dt = - ( 0.808 - 1.002 ) / ( 60.0 - 0.0) = 0.00323 M/s

(b)

Rate of appearance of fructose = d[C6H12O6] / dt = (1.002 - 0.808) / (60.0 - 0.0) = 0.00323 M/s

(c)

k = (1 / t ) * ln[A]/[A]t

k = ( 1 / 60.0 ) * ln[1.002 / 0.808]

k = 0.00359 min-1

(d)

0.00359 = ( 1 / t ) * ln[1.002 / 0.212]

t = 432.6 min

(e)

Half life time = 0.693 / k = 0.693 / 0.00359 = 193 min

Explanation:

First-order reactions are defined as the chemical reactions in which rate of the reaction is linearly dependent on the concentration of only one reactant.

The answers can be explained as:

(a) Rate of appearance of the sucrose from the chemical reaction is:

Rate = [tex]\dfrac{\text d [\text C_{12}\text H_{22}\text O_{11}]}{\text {dt}}[/tex] = [tex]\dfrac{0.808 - 1.002}{60.0 -0.0}[/tex]

Rate = 0.00323 m/s

(b) Rate of appearance of Fructose from the given chemical reaction is:

Rate = [tex]\dfrac{\text d [\text C_{6}\text H_{12}\text O_{6}]}{\text {dt}}[/tex] = [tex]\dfrac{1.002 - 0.808 }{60.0 -0.0}[/tex]

Rate = 0.00323 m/s

(C) Rate constant for the reaction is:

[tex]\text k &= \dfrac{1}{\text t}\times \dfrac {\text{ln [A]}}{\text {[A]} \text t}[/tex]

[tex]\text k &= \dfrac{1}{60}\times \dfrac {\text{ln} (1.002)}{(0.808)}[/tex]

k = 0.00359 minute⁻¹

(d) Time required for the concentration of sucrose to drop from 1.002 to 0.212 M is:

[tex]0.00359 &= \dfrac{1}{t} \times {\text{ln}\dfrac{[1.002]}{[0.212]}[/tex]

t = 432.6 minutes

(e) The half-life of the decomposition of sucrose at 25°C is:

Half-life = [tex]\dfrac{0.693}{\text k} = \dfrac{0.693}{0.00359}[/tex]

Half-life = 193 minutes.

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Answer:

Check the explanation

Explanation:

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Answer:

Explanation:

The chemical equation for the reaction is :

[tex]C_6H_{12}O_6_{(s)} \to 2C_2H_5OH _{(l)} + 2CO_{2(g)}[/tex]

The standard enthalpy of formation [tex]\Delta H ^0_f[/tex]  of the above equation is as follows:

[tex]\Delta H^0_{f, C_6H_{12}O_6}[/tex] = -1274.4 kJ/mol

[tex]\Delta H ^0_{f, C_2H_{5}OH[/tex] = -277.7 kJ/mol

[tex]\Delta H ^0_{f, CO_2}[/tex] = -393.5 kJ/mol

[tex]\Delta H ^0_{rxn }= \sum n_p \Delta H ^0_{f,p} - \sum n_r \Delta H ^0_{f,r}[/tex]

where ;

[tex]n_p[/tex] = stochiometric coefficients of products

[tex]n_r=[/tex] stochiometric coefficients of reactants

[tex]\Delta H^0_{f.p}[/tex] = formation standard enthalpy of products

[tex]\Delta H^0_{f.r}[/tex] = formation standard enthalpy of reactants

[tex]\Delta H ^0_{rxn }= (2 \ mol* -2777.7 \ kJ/mol + 2 \ mol * - 393.5 \ kJ/mol) - (1\ mol *(-1274.4 \ kJ/mol)[/tex]

[tex]\Delta H ^0_{rxn }= -68 \ kJ[/tex]

For [tex]\Delta S ^0[/tex] ;

The standard enthalpy of formation of [tex]\Delta S ^0_f[/tex] of the reactant and the products are :

[tex]\Delta S ^0 _{f \ C_6H_{12}O_6} = 212 \ \ J/Kmol[/tex]

[tex]\Delta S ^0 _{f \ C_2H_5O_H} = 160.7 \ \ J/Kmol[/tex]

[tex]\Delta S ^0 _{f \ CO_2} = 213.63 \ \ J/Kmol[/tex]

The [tex]\Delta S ^0_{rxn}[/tex] is as follows:

[tex]\Delta S ^0_{rxn} = \sum n_p \Delta S^0_{f.p} - \sum n_r \Delta S^0_{f.r}[/tex]

[tex]\Delta S ^0_{rxn} = (2 \ mol *160.7 \ \ J/Kmol + 2 \ mol *213.63 \ \ J/Kmol) -(1*212 J/Kmol)[/tex]

[tex]\Delta S ^0_{rxn} =536 . 7 \ J/K[/tex]   (to kJ/K)

[tex]\Delta S ^0_{rxn} =536 . 7 \ J/K * \frac{1 \ kJ}{1000 \ J}[/tex]

[tex]\Delta S ^0_{rxn} =0.5367 \frac{kJ}{K}[/tex]

Given that;

at T = 25°C = ( 25 + 273) K = 298 K

[tex]\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}[/tex]

[tex]\Delta G^0 _{rxn} = -68 \ kJ - 298 * 0.5367 \frac{kJ}{K}[/tex]

[tex]\Delta G^0 _{rxn} = -227. 9 \ \ \ kJ[/tex]

As [tex]\Delta G^0 _{rxn}[/tex] is negative; the reaction is spontaneous

[tex]\Delta H^0 _{rxn}[/tex] = negative

[tex]\Delta S^0 _{rxn}[/tex] = positive; Therefore , the reaction is spontaneous at all temperature , We can then say that the spontaneity of this reaction [tex]\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}[/tex]  is dependent on temperature.

A scientific question may lead to a(n)

The answer is may lead to a hypothesis.

Explanation:

Answer: observations

Hypothesis

Experimentation

Explanation:in order and right on edge

Answer:

Check the explanation

Explanation:

Here, Nitrogen (N) undergoes oxidation and Chlorine (Cl) undergoes reduction.

To answer your question:

N is oxidized from an oxidation number of -3 to an oxidation number of -1.

Cl is reduced from oxidation number of +1 to an oxidation number of -1.

Now,

Borneol should have a lower Rf because of boiling point.

Answer:

0.0075 milliliters (7.5 microliters) will be taken from the stock solution and then diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.

Explanation:

This is a problem of dilution using the equation:

initial concentration x initial volume = final concentration x final volume.

The final volume to be prepared is 25 microliters.

The final concentration to be prepared is 3 M.

The initial volume to be taken is not known yet.

The initial concentration is 10 M.

Now, let's substitute these parameters into the the equation above.

10 x initial volume = 3 x 25

Initial volume = 3 x 25/10

     = 7.5 microliters

Note that: 1 microliter = 0.001 milliliters

Hence,

7.5 microliters = 0.0075 milliliters

This means that an initial volume of 0.0075 milliliters (7.5 microliters) will be taken from the stock solution. This amount will then be diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.

Answer:

A weakly acidic solution.

Explanation:

Final answer:

A solution with a Ka value much less than 1 refers to a weak acid, where only a small portion of the acid is ionized, resulting in relatively low concentrations of hydronium ions and the conjugate base.

Explanation:

The type of solution that would have a Ka value much less than 1 is a weak acid solution. This is because the Ka (acid dissociation constant) value indicates the strength of the acid in solution; the lower the Ka, the weaker the acid. For a Ka value significantly less than 1, this means that only a small fraction of the acid molecules donate protons (H+) to the water, resulting in few hydronium ions (H₋₃O⁺) and the conjugate base (A-).

In contrast, strong acids have a Ka value that approaches infinity since they are almost completely ionized in solution. Acetic acid, with a Ka value of 1.75 x 10⁻⁵, is an example of a weak acid, indicating that in equilibrium, the concentration of the undissociated acetic acid is much greater than the concentrations of the acetate and hydronium ions.

Absorption of 1.0 rad of gamma radiation would likely cause the greatest tissue damage due to its high energy and penetration ability.

The amount of tissue damage caused by radiation depends on various factors, including the type of radiation and its energy level.

Alpha particles are relatively heavy and highly ionizing, but they have low penetration power, so they are generally less damaging to tissues externally but can be more damaging if ingested or inhaled.

Beta particles have less mass and energy compared to alpha particles and can penetrate deeper into tissues, potentially causing more damage.

Gamma rays are electromagnetic radiation with high energy and penetration power. They can penetrate deeply into tissues, causing damage along their path.

Considering these factors, absorption of 1.0 rad of gamma radiation would likely result in the greatest tissue damage due to its high energy and penetration ability.

Answer:

False

False

True

False

False

False

Explanation:

The colour emitted by an excited atom depends on the emission Spectra of the atom rather than on the number of free electrons passing through the lamp.

When a free electrons hits an atom, the atom can be excited to various intermediate states other than the highest energy level depending on the amount of energy it absorbed.

Increase in voltage of the battery also increases the kinetic energy of the electron

The kinetic energy of the electron depends on the voltage difference not on the distance from the atom.

The Paschen, Brackett, and Pfund series of lines are due to transitions from higher-energy orbits to orbits with n = 3, 4, and 5, respectively; these transitions release substantially less energy, corresponding to infrared radiation.

Excited atoms can jump from a higher level to the ground state in a series of steps or directly to the ground state

Answer:

Work done on the door is = 0

Mechanical energy of the person is 100%

Mechanical energy of the door is 50%.

Explanation:

Mechanical energy, M.E., is the energy an object possesses due to its position and motion. When we talk of position, the body is at rest = potential energy. In terms of motion, the body is moving = kinetic energy.

This means that M.E = potential + kinetic

Also, Work done = force × distance

For the person pushing the door, the potential of the energy stored in the person and the kinetics of pushing the door handle are all present. Therefore,  M.E of person is 100%.

But despite the doors potential energy present, there is no motion from it, which means the M.E is only half or 50%.

Also, since distance moved by the door is 0, no work is done on the door.

Answer:

4.0 moles

Explanation:

The following data were obtained from the question:

Volume (V) = 12L

Pressure = 5.6 atm

Temperature (T) = 205K

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

Using the ideal gas equation: PV = nRT, the number of mole of the gas can be obtained as follow

PV = nRT

5.6 x 12 = n x 0.08206 x 205

Divide both side by 0.08206 x 205

n = (5.6 x 12)/(0.08206 x 205)

n = 4.0 moles

Therefore, the number of mole of the gas is 4.0 moles

Answer:

42050 J.

Explanation:

Data obtained from the question:

Mass (M) of ethanol = 50g

Heat of vaporisation (ΔHv) of ethanol = 841 J/g

Heat (Q) =.?

The heat required to boil 50g of ethanol can be obtained as follow:

Q = MΔHv

Q = 50 x 841

Q = 42050 J.

Therefore, the heat required to boil 50g of ethanol is 42050 J.

Answer:

See explaination

Explanation:

Please see the attached file for the structural diagram of the molecular formation.

It is detailed and properly presented at the attachment.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The pH is  [tex]pH = 4.94[/tex]

Explanation:

From the question we are told that

   The average concentration of NaOH is [tex][NaOH] = 0.101 M[/tex]

    The volume of NaOH is  [tex]V__{NaOH}} = 15.00 mL[/tex]

    The average concentration of Acetic acid is [tex][Acetic \ Acid] =0.497 \ M[/tex]

     The volume of Acetic acid is  [tex]V__{Acetic \ Acid}} = 5.00 \ mL[/tex]

The chemical equation for this reaction is  

         [tex]NaOH + CH_3COOH ---> CH_3 COONa + H_2 O[/tex]

The total volume of the solution is  

  [tex]V__{Total}} = V__{NaOH}} + V__{Acetic \ Acid}}[/tex]

Substituting values

      [tex]V__{Total}} = 15 + 5[/tex]

      [tex]V__{Total}} = 20mL = 20 *10^{-3} L[/tex]

The number of moles of NaOH is mathematically represented as

      [tex]n__{NaOH}} = [NaOH] * V__{NaOH}}[/tex]

substituting values

      [tex]n__{NaOH}} = 0.101 * 15*10^{-3}[/tex]

     [tex]n__{NaOH}} = 0.001515 \ moles[/tex]

The number of moles of Acetic acid is mathematically represented as

      [tex]n__{Acetic acid}} = [Acetic \ acid] * V__{Acetic acid}}[/tex]

substituting values

        [tex]n__{Acetic acid}} = 0.497 * 5*10^{-3}[/tex]

      [tex]n__{Acetic acid}} = 0.002485\ moles[/tex]

From the chemical equation

          1 mole of NaOH reacts with   1 mole of Acetic acid  to produce 1 mole of   [tex]CH_3 COONa[/tex] salt and 1 mole of  [tex]H_2 O[/tex]

So

  0.001515 moles  of NaOH reacts with  0.001515 moles of Acetic acid  to produce 0.001515 moles of   [tex]CH_3 COONa[/tex] salt and 0.001515 moles of  [tex]H_2 O[/tex]

This implies the number of moles of NaOH remaining after the react would be

      [tex]\Delta n__{NaOH}} = 0.001515 - 0.001515[/tex]

      [tex]\Delta n__{NaOH}} = 0 \ mole[/tex]

the number of moles of Acetic acid remaining after the react would be

    [tex]\Delta n__{Acetic acid}} = 0.002485 - 0.001515[/tex]

    [tex]\Delta n__{Acetic acid}} = 0.00097 \ moles[/tex]

the number of moles of [tex]CH_3 COONa \ salt[/tex] remaining after the react would be

    [tex]\Delta n__{CH_3 COONa \ salt}} = 0 + 0.001515[/tex]

    [tex]\Delta n__{CH_3 COONa \ salt}} = 0.001515 \ moles[/tex]

the number of moles of [tex]H_2 O[/tex] remaining after the react would be

   [tex]\Delta n__{H_2O}} = 0 + 0.001515[/tex]

   [tex]\Delta n__{H_2O}} = 0.001515 \ moles[/tex]

The expected pH is mathematically evaluated as

      [tex]pH = pK_a + log [\frac{[CH_3 COONa]}{[Acetic \ acid]} ][/tex]

Where [tex]pKa[/tex] is mathematically evaluated as

        [tex]pK_a = - log (K_a)[/tex]

The concentration of [tex]CH_3 COONa \ salt[/tex] is mathematically evaluated a s

[tex][CH_3 COONa] = \frac{\Delta n_CH_3 COONa \ salt }{V__{Total}}}[/tex]

substituting values

   [tex][CH_3 COONa] = \frac{0.001515}{20 *10^{-3}}[/tex]

    [tex][CH_3 COONa] = 0.07575M[/tex]

The concentration of  Acetic acid is mathematically evaluated as

          [tex][Acetic acid] = \frac{\Delta n__Acetic acid}{V__{Total}}}[/tex]

substituting values

        [tex][CH_3 COONa] = \frac{0.00097}{20 *10^{-3}}[/tex]

       [tex][CH_3 COONa] = 0.0485 M[/tex]

Substituting values into the equation for pH

          [tex]pH = - log (1 .8 *10^{-5}) + log [\frac{0.07575}{ 0.0485} ][/tex]

             [tex]pH = log [\frac{0.07575}{ 0.0485} ] - log (1 .8 *10^{-5})[/tex]

            [tex]pH = log [\frac{1.561856}{1.8*10^{-5}} ][/tex]

          [tex]pH = 4.94[/tex]

Answer: [H3O+]= 0.05 M

Explanation:

Message me for extra explanation.

snap- parkguy786

Answer:

1670 ml

Explanation:

molarity x Volume (Liters) = moles => Volume (Liters) = moles/Molarity

Volume needed = 2.50mol/1.50M = 1.67 Liters = 1670 ml.

Answer:

hes right

Explanation:

ik all

Answer:

Zn(s) + Fe^2+(aq) ---------> Zn^2+(aq) + Fe(s)

∆G= 61760J

K= 6.5 ×10^10

Explanation:

Zn(s) + Fe^2+(aq) ---------> Zn^2+(aq) + Fe(s)

E°reaction= E°Fe - E°Zn

E°reaction= (-0.44)-(-0.76)

E°reaction= 0.32V

We know that

∆G= -nFE°

Since n= 2 from the reaction equation and F= 96500C

∆G= -(2×96500×0.32)

∆G= 61760J

From

E°= 0.0592/n log K

0.32= 0.0592/2 log K

log K= 0.32/0.0296

log K= 10.81

K= Antilog(10.81)

K= 6.5 ×10^10

Answer:

Check the explanation

Explanation:

Kindly check the attached image below for the step by step explanation to the question above.

Answer:

-5.51 kJ/mol

Explanation:

Step 1: Calculate the heat required to heat the water.

We use the following expression.

[tex]Q = c \times m \times \Delta T[/tex]

where,

The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

[tex]Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ[/tex]

Step 2: Calculate the heat released by the methane

According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero

Qc + Qw = 0

Qc = -Qw = -22.0 kJ

Step 3: Calculate the molar heat of combustion of methane.

The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

[tex]\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol[/tex]

Answer: The molarity of [tex]H^+[/tex] in this solution is [tex]2.2\times 10^{-11}M[/tex]

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

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

Given : pH = 10.66

Putting in the values:

[tex]10.66=-\log[H^+][/tex]

[tex][H^+]=10^{-10.66}[/tex]

[tex][H^+]=2.2\times 10^{-11}[/tex]

Thus the molarity of [tex]H^+[/tex] in this solution is [tex]2.2\times 10^{-11}M[/tex]

The molarity of hydrogen ions, H⁺ in this solution is  2.2 * 10⁻¹¹

The molarity of a solution is a measure of the concentration in moles of a substance present in a liter of solution of that substance.

The pH of a solution is the negative logarithm in base 10 of the hydrogen ion concentration of the solution.

pH = -log[H⁺]

-pH = log[H⁺]

pH of solution = 10.66

-10.66 = log[H⁺]

taking antilogarithm of both sides

[H⁺] = 10⁻¹⁰°⁶⁶

[H⁺] = 2.2 * 10⁻¹¹

Therefore, the molarity of hydrogen ions, H⁺ in this solution is  2.2 * 10⁻¹¹

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