Identify the Brønsted-Lowry acid, the Brønsted-Lowry base, the conjugate acid, and the conjugate base in each reaction:

(a) C5H5N(aq)+H2O(l)→C5H5NH+(aq)+OH−(aq)
(b) HNO3(aq)+H2O(l)→H3O+(aq)+NO3−(aq)

Answer:

Explanation:

1) Balanced chemical equation (given):

2) Mole ratio:

3) Starting ratio (given)

Since 0.69 > 0.67 means that you start with more Al atoms (4.5 moles) than what are needed to react with the given CuCl units (6.5 mol); i.e. when the 6.5 moles of CuCl are consumed, there will be an excess of Al, and the reaction will stop, because the CuCl is over, meaning the latter is the limiting reactant (it limits the yield).

Answer:

option b

Explanation:

When the energy is released the process is called exothermic reaction. This happens when the bonds are broken in the reactants and the system release energy.

Answer:

a. forming bonds

Explanation:

The energy required to break a bond is endothermic that is energy is absorbed to break a bond.

The energy is released in the formation of a bond that is energy is released when a bond is formed.

The formula to find the ∆H of the reaction is  

∆H (reaction) = ∆H (bonds Broken) - ∆H (bonds formed)

For example  

[tex]N_2+3H_2< >2NH_3[/tex]

[tex]N_2[/tex] contains one N≡N triple bond (Bond breaking 946 KJ per mol)

[tex]H_2[/tex] contains a single H-H bond (bond breaking 436 KJ per mol)

[tex]NH_3[/tex] contains 3, N - H single bonds(389 KJ per mol)

So ∆H (bonds broken) = 946 + (3 × 436) = 2254KJ

∆H (Bonds formed ) = (2 × 3 × 389) = 2334KJ

So

∆H (reaction) = 2254 KJ - 2334 KJ

= - 80KJ and the reaction is Exothermic

In this example we see energy required to break the bond is lesser than energy released in forming the bond.

The solute that dissolves in ethanol depends on the polarity of the solute and the solvent. If both the solute and the solvent are polar, then the solute will dissolve in the solvent. Similarly, if both the solute and the solvent are nonpolar, then the solute will also dissolve in the solvent. However, if the solute and the solvent have different polarities, they will not dissolve in one another.

A substance can dissolve in a solvent, and form a solution, if the solute and solvent are attracted to each other. For example, water molecules that are held together by hydrogen bonding will dissolve solutes that can also hydrogen bond, like ethanol (CH3CH₂OH). The new hydrogen bonds between the water and the ethanol molecules (solvent-solute attractions) are nearly as strong as the hydrogen bonds in water (solvent-solvent) and ethanol (solute-solute) alone, making the process of solution formation (also called dissolution or dissolving) favorable.

When water mixes with other polar substances, like ethanol, some of the hydrogen bonding between water molecules replace with similar hydrogen bonding with ethanol molecules. Since the electrostatic potential energy is similar, the natural tendency to go towards more dispersion drives the dispersion of ethanol molecules uniformly in water resulting in the solution.

Nonpolar compounds do not dissolve in water. The attractive forces that operate between the particles in a nonpolar compound are weak dispersion forces. However, the nonpolar molecules are more attracted to themselves than they are to the polar water molecules. When a nonpolar liquid such as oil is mixed with water, two separate layers form because the liquids will not dissolve into each other. When another polar liquid such as ethanol is mixed with water, they completely blend and dissolve into one another. Liquids that dissolve in one another in all proportions are said to be miscible. Liquids that do not dissolve in one another are deemed immiscible. The general rule for deciding if one substance is capable of dissolving another is 'like dissolves like'. A nonpolar solid such as iodine will dissolve in nonpolar lighter fluid, but will not dissolve in polar water.

A main goal of most environmental scientists is to achieve biodiversity.

Answer:

Acetylcholine causes an end-plate potential by triggering the opening of sodium channels.

Acetylcholine (ACh), a neurotransmitter released by motor neurons, causes an end-plate potential by binding to receptors in the motor end plate and triggering a depolarization process that eventually leads to an action potential.

Acetylcholine (ACh) is a neurotransmitter that plays a pivotal role in triggering an end-plate potential within the neuromuscular system. The process starts when an action potential travels down the motor neuron's axon, triggering the release of these neurotransmitters. The ACh then binds to receptors in the motor end plate and initiates a series of events that lead to changes in ion permeability, the influx of sodium ions into the muscles cells, and ultimately a reduction in the voltage difference between the inside and outside of the cell - a process called depolarization. When ACh binds at the motor end plate, this depolarization is called an end-plate potential.

This depolarization then spreads along the muscle fiber membrane, the sarcolemma, creating an action potential, which moves across the entire cell, creating a wave of depolarization. Therefore, the acetylcholine indeed causes the end-plate potential by triggering the release of these neurotransmitters and the subsequent sequence of events.

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Answer:

The amount of ionization in solution.

Explanation:

Strong acids ionize fully in solution to release a large number of hydrogen ions responsible for the acidic properties. On the contrary weak acids ionize partially in solution.

In strong acids less energy is required to break the hydrogen- anion bonds.

Answer:

the amount of ionization that occurs in solution

Explanation:

An acid is a substance that interacts with water to produce excess hydroxonium ions, H₃0⁺ in an aqueous solution.

The ionization of acids in solution determines the strength of the acid.

A strong acid is one that ionizes almost completely in solutions and a weak acid is one that ionizes slightly and sets up an equilibrium.

             HCl + H₂0 → H₃0⁺ + Cl⁻ this is the ionization of strong acid

            CH₃COOH + H₂O ⇄  H₃0⁺ + CH₃COO⁻ ionization of a weak acid

Final answer:

The epicenter of an earthquake can be located using the arrival times of S- and P-waves at three or more seismographic stations, allowing for triangulation and precise determination of the distance to the earthquake source.

Explanation:

To locate the epicenter of an earthquake, seismographic stations use the arrival times of S- and P-waves. The S-wave is slower than the P-wave, so the difference in arrival times at three or more seismographic stations can be used to determine the distance to the epicenter. By triangulating the distances from multiple stations, the epicenter can be pinpointed.

For example, if the S-wave and P-wave travel at speeds of 4.00 km/s and 7.20 km/s respectively, the difference in arrival times can be used to calculate the distance. The precision of the seismographs in measuring the arrival times allows for a precise determination of the distance to the earthquake source.

It is important to note that uncertainties in the propagation speeds of the waves can introduce greater uncertainty in determining the distance to the epicenter.

Answer:

d. Occurs when there is equilibrium between solute going into solution and solute coming out of solution.

Explanation:

This is the definition of a saturated solution.

It is also the reason why options a, b, and c are wrong.

Answer: The equilibrium concentration of [tex]PCl_5[/tex] is 1.285 M.

Explanation:

The chemical equation for the decomposition of phosphorus pentachloride follows:

[tex]PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)[/tex]

The expression for equilibrium constant is given as:

[tex]K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]

We are given:

[tex]K_c=0.042[/tex]

[tex][PCl_3]=0.18M[/tex]

[tex][Cl_2]=0.30M[/tex]

The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.

Putting values in above equation, we get:

[tex]0.042=\frac{0.18\times 0.30}{[PCl_5]}[/tex]

[tex][PCl_5]=1.285[/tex]

Hence, the equilibrium concentration of [tex]PCl_5[/tex] is 1.285 M.

The concentration of PCl5 is determined to be 1.29 M.

The student is asking about the concentration of PCl5 at equilibrium in the reaction PCl5(g) ⇌ PC13(g) + Cl2(g) given the equilibrium constant (K) and the concentrations of the products PCl3 and Cl2.

To solve for the concentration of PCl5, we use the equilibrium constant expression:

K = [PCl3][Cl2] / [PCl5]

Given that K = 0.042, [PCl3] = 0.18 M, and [Cl2] = 0.30 M, we can rearrange the expression to solve for [PCl5]:

[PCl5] = [PCl3][Cl2] / K

[PCl5] = (0.18 M * 0.30 M) / 0.042

After carrying out the calculation, the concentration of PCl5 at equilibrium is found to be 1.29 M.

Answer:

I think it's the Zeroth law of thermodynamics.

Answer:

the second law of thermodynamics.

Explanation:

Answer:

D) blood glucose levels are low

Explanation:

When glucose levels are low, glucagon levels increase. The glucagon will then bind to receptors on the adipose cells' surface, thus setting off further reactions which culminates in fatty acids being released from the cells. They will then flow through the circulatory system, where they will either bind to a protein in blood or be used by muscle tissue for energy.

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

[tex]\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole[/tex]

[tex]\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole[/tex]

The balanced chemical reaction will be,

[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = [tex]137ml+137ml=274ml[/tex]

[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g[/tex]

Now we have to calculate the heat absorbed during the reaction.

[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]

where,

q = heat absorbed = ?

[tex]c[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

m = mass of water = 274 g

[tex]T_{final}[/tex] = final temperature of water = 325.8 K

[tex]T_{initial}[/tex] = initial temperature of metal = 298 K

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

[tex]q=274g\times 4.18J/g^oC\times (325.8-298)K[/tex]

[tex]q=31839.896J=31.84KJ[/tex]

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

[tex]\Delta H=\frac{q}{n}[/tex]

where,

[tex]\Delta H[/tex] = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

[tex]\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole[/tex]

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

The enthalpy of neutralization for the reaction of HCl and NaOH, given the provided conditions, is calculated to be 89.39 kJ/mol. The heat absorbed by the reaction solution is computed using the specific heat capacity, the mass of the solution (equivalent to its volume due to the density of water), and the temperature change.

To calculate the enthalpy of neutralization of HCl and NaOH, we first need to determine the heat produced during the reaction. The reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is:

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

We know the following:

First, we calculate the amount of heat (q) absorbed:

q = mass of solution × c × ΔT

Since the density of the solution is the same as water, we use the volume as mass (assuming 1 g/cm³ density), so:

mass = volume of HCl + volume of NaOH = 137 g + 137 g = 274 g

q = 274 g × 4.18 J/g·K × 27.8 K = 31822.32 J

Since the reaction involves equal volumes and concentrations of HCl and NaOH, the number of moles of HCl reacting will be the same as the number of moles of NaOH:

moles HCl = moles NaOH = volume × concentration = 0.137 dm³ × 2.6 mol/dm³ = 0.3562 mol

The enthalpy of neutralization (ΔH_neut) is the heat divided by the number of moles of either HCl or NaOH:

ΔH_neut = q / moles = 31822.32 J / 0.3562 mol = 89384.67 J/mol = 89.39 kJ/mol

Therefore, the enthalpy of neutralization of HCl and NaOH is 89.39 kJ/mol, which is option D.

Answer:

they are the same process, just reversed

Explanation:

Electrolysis is the process that converts electrical energy into chemical energy. The process involves the decomposition of an ionic compound by means of electric current passed through its solution.

A battery is an electrochemical cell in which chemical energy is converted to electrical energy. Here chemical reactions usually redox produces electric current.

Answer:

pH = 3.74

Explanation:

Given:

Initial volume of CH3COOH, V1 = 25.00 ml

Initial concentration of CH3COOH, M1 = 0.10 M

Initial volume of CH3COONa, V1 = 25.00 ml

Initial concentration of CH3COONa, M2 = 0.010 M

Ka (CH3COOH) = 1.8*10^-5

To determine:

pH of the solution

Calculation:

The given solution of CH3COOH/CH3COONa is in fact a buffer whose pH is given by the Henderson-Hasselbalch equation where:

[tex]pH = pKa + log\frac{[A-]}{[HA]} ----(1)[/tex]

where A- = concentration of conjugate base = [CH3COONa]

HA = weak acid = [CH3COOH]

Step 1: Calculate the final concentration of CH3COONa

V1 = 25.00 ml

V(final) = Total volume = 25.00 + 25.00 = 50.00 ml

M1 = 0.010 M

[tex]M1V1 = M2V2\\\\M2 = \frac{M1V1}{V2} = \frac{0.010 M * 25.00 ml}{50.00ml} =0.005M[/tex]

Step 2: Calculate the final concentration of CH3COOH

V1 = 25.00 ml

V(final) = Total volume = 25.00 + 25.00 = 50.00 ml

M1 = 0.10 M

[tex]M1V1 = M2V2\\\\M2 = \frac{M1V1}{V2} = \frac{0.10 M * 25.00 ml}{50.00ml} =0.05M[/tex]

Step 3: Calculate the pH

Based on equation (1)

[tex]pH = pKa + log\frac{[CH3COONa]}{[CH3COOH]} ----(1)[/tex]

pKa = -log Ka = -log(1.8*10^-5) = 4.74

[tex]pH = -logKa + log\frac{[CH3COONa]}{[CH3COOH]}[/tex]

[tex]pH = 4.74 + log\frac{[0.005]}{[0.05]} [/tex]

pH = 3.74

The addition of complexes to the solution changes the final concentration. The pH of the solution with the mixing of the two different solutions is 3.74.

The pH has been the hydrogen ion concentration in the solution. It can be given with the acid dissociation ability of the compound, or the ability of a compound to release hydrogen ions.

The addition of 25 ml solutions resulted in the final volume of 50 ml. The final concentration of the solutions is given as:

[tex]\rm Initial\;concentration\;\times\;Initial\;volume=Final\;concentration\;\times\;Final\;Volume[/tex]

The Final concentration of [tex]\rm CH_3COONa[/tex] salt is:

[tex]\rm 0.01\;M\;\times\;25\;mL=50\;mL\;\times\;[CH_3COONa]\\\\CH_3COONa=\dfrac{0.01\;M\;\times\;25\;mL}{50\;mL}\\\\ CH_3COONa=0.005\;M[/tex]

The final concentration of [tex]\rm CH_3COOH[/tex] acid is :

[tex]\rm 0.1\;M\;\times\;25\;mL=50\;mL\;\times\;[CH_3COOH]\\\\CH_3COOH=\dfrac{0.1\;M\;\times\;25\;mL}{50\;mL}\\\\ CH_3COOH=0.05\;M[/tex]

The pH of the solution is given as:

[tex]\rm pH=log\;Ka\;+\;log\;\dfrac{salt}{acid}[/tex]

Substituting the values in the equation:

[tex]\rm pH=-log\;1.08\;\times\;10^-^5\;+\;log\;\dfrac{0.005}{0.05} \\pH=4.74\;+\;(-1)\\pH=3.74[/tex]

The pH of the solution of sodium acetate and acetic acid is 3.74.

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Answer : The percent composition by volume of mixture of [tex]CO[/tex] and [tex]CO_2[/tex] are, 18.94 % and 81.06 % respectively.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

[tex]R\propto \sqrt{\frac{1}{M}}[/tex]

And the relation between the rate of effusion and volume is :

[tex]R=\frac{V}{t}[/tex]

or, from the above we conclude that,

[tex](\frac{V_1}{V_2})^2=\frac{M_2}{M_1}[/tex]            ..........(1)

where,

[tex]V_1[/tex] = volume of helium gas = 29.7 ml

[tex]V_2[/tex] = volume of mixture = 9.28 ml

[tex]M_1[/tex] = molar mass of helium gas  = 4 g/mole

[tex]M_2[/tex] = molar mass of mixture = ?

Now put all the given values in the above formula 1, we get the molar mass of mixture.

[tex](\frac{29.8ml}{9.28ml})^2=\frac{M_2}{4g/mole}[/tex]

[tex]M_2=40.97g/mole[/tex]

The average molar mass of mixture = 40.97 g/mole

Now we have to calculate the percent composition by volume of the mixture.

Let the mole fraction of [tex]CO[/tex] be, 'x' and the mole fraction of [tex]CO_2[/tex] will be, (1 - x).

As we know that,

[tex]\text{Average molar mass of mixture}=\text{Mole fraction of }CO[/tex]

[tex]\text{Average molar mass of mixture}=(\text{Mole fraction of }CO\times \text{Molar mass of } CO)+(\text{Mole fraction of }CO_2\times \text{Molar mass of } CO_2)[/tex]

Now put all the given values in this expression, we get:

[tex]40.94g/mole=((x)\times 28g/mole)+((1-x)\times 44g/mole)[/tex]

[tex]x=0.1894[/tex]

The mole fraction of [tex]CO[/tex] = x = 0.1894

The mole fraction of [tex]CO_2[/tex] = 1 - x = 1 - 0.1894 = 0.8106

The percent composition by volume of mixture of [tex]CO[/tex] = [tex]0.1894\times 100=18.94\%[/tex]

The percent composition by volume of mixture of [tex]CO_2[/tex] = [tex]0.8106\times 100=81.06\%[/tex]

Therefore, the percent composition by volume of mixture of [tex]CO[/tex] and [tex]CO_2[/tex] are, 18.94 % and 81.06 % respectively.

Answer:

Explanation:

When a chemical reaction happens, the net change of enthalpy of the products and the reactants is the heat of reaction:

Then, when ∑ ΔHproducts > ∑ ΔA reactants, ΔH rxn > 0 and heat is released. This is what is called an exothermic reaction.

In an exothermic reaction, heat is released, ΔH > 0, and the surroundings will get hotter.

In and endothermic reaction heat is absorved, ΔH < 0, and the surroundings will get cooler.

Answer:

Joules

Explanation:

The another ones are units of tempeture (B and D) and unit of electricity that relatione energy and charge. In chemistry the energy es measured in Joules, because the energy is  work done on an object when a force of one newton acts on that object in the direction of the force's motion through a distance of one metre. In other words, J=Nm

Answer:

Option A is true

Explanation:

When you are a scientist conducting an experiment on energy transfer .

The reaction in which you measure a large amount of heat energy transfer.

We have to find the units which you should record them in

Energy: It is defined as the capability of doing the work .

When current I is flowing in ampere A  V is potential in volt  v applied  in the experiment  and R be resistance  in ohm used in experiment for time t in seconds

Then, heat energy =[tex]VI[/tex]=A-volt=Joule

Heat energy=[tex]I^2Rt[/tex]

Heat energy=[tex]A^2ohm sec[/tex]=Joule

The S.I unit of heat energy is Joules.

Hence, option A is true.

Answer:

b. the rate of destruction is equal to the rate of carbonate input.

Explanation:

At the Calcite Compensation Depth(CCD) the rate of addition of calcite and dissolution of the mineral is the same. Below this depth, all calcite minerals are dissolved.

The CCD occurs at a depth of 3000-4000m. It varies from places to places within the ocean. Other factors play important roles in determining the CCD. Some of the factors are temperature and pressure.

Answer:

Explanation:

1) Data:

2) Formula:

Combined law of gases:

3) Solution:

T₂ = P₂ V₂ T₁ / (P₁ V₁)

T₂ = 872.15 mmHg × 7.63 liter × 261.2 K / ( 424.9 mmHg × 0.66 liter)

T₂ = 6198 K

Answer: pH = 4.11

Explanation: pH of the buffer solution is calculated using Handerson equation:

[tex]pH=pKa+log(\frac{base}{acid})[/tex]

pKa is calculated from the given Ka value as:

pKa = - log Ka

[tex]pKa=-log1.34*10^-^5[/tex]

pKa = 4.87

pH of the solution before adding HCl to it:

[tex]pH=4.87+log(\frac{0.112}{0.322})[/tex]

pH = 4.87 - 0.46

pH = 4.41

Now, 0.069 moles of HCl are added to the buffer solution. This added HCl react with base(sodium propanoate) to produce acid(propanoic acid).

Initial moles of acid = 0.322*1.44 = 0.464

initial moles of base = 0.112*1.44 = 0.161

moles of base after reacting with HCl = 0.161 - 0.069 = 0.092

moles of acid after addition of HCl = 0.464 + 0.069 = 0.533

Let's plug in the values in Handerson equation to calculate the pH:

[tex]pH=4.87+log(\frac{0.092}{0.533})[/tex]

pH = 4.87 - 0.76

pH = 4.11

So, the original pH of the buffer solution is 4.41 and after addition of HCl the pH is 4.11 .

The pH of solution following the addition of 0.069 moles of HCl is 4.11.

The pH of a buffer solution can be calculated using Handerson equation as follows:

pH = pka + log (base/acid)

pKa of the acid is calculated from the given Ka value as follows:

pKa = - log Ka

pKa = - log 1.34 × 10-⁵

pKa = 4.87

The pH of the solution before adding HCl to it is as follows:

pH = 4.87 + log(0.112/0.322)

pH = 4.87 - 0.46 = 4.41

According to this question, 0.069 moles of HCl are added to the buffer solution.

Therefore, the pH of the buffer after adding HCl is:

pH = 4.87 + log(0.092/0.533)

pH = 4.87 - 0.76 = 4.11

Therefore, the pH of solution following the addition of 0.069 moles of HCl is 4.11.

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Final answer:

The reaction 2N2O(g) → 2N2(g) + O2(g) indicates a positive entropy change (ΔS) since the number of gas molecules increases, but the enthalpy change (ΔH) cannot be determined without more information.

Explanation:

For the reaction 2N2O(g) → 2N2(g) + O2(g), we're interested in determining whether the enthalpy change (ΔH) and the entropy change (ΔS) are positive or negative. Enthalpy is a measure of the heat change during a reaction, while entropy measures the disorder or randomness.

Generally, the formation of a gas from non-gaseous reactants would indicate an increase in entropy due to the increase in randomness. However, since both the reactants and products in this reaction are gases, we look at the change in the number of moles of gas. The reaction goes from two moles of N2O (g) to a total of three moles of gases (two moles of N2 and one mole of O2), which means there is an increase in the number of gaseous particles, suggesting an increase in disorder (a positive ΔS).

Without specific data on the heat exchanged during this reaction (enthalpy change), the students would not be able to definitively determine ΔH with the information provided in the question. They would only be able to infer that ΔS is positive because the number of gas molecules increases.

Final answer:

For the reaction 2N2O(g) → 2N2(g) + O2(g), it is likely that ΔH is negative (exothermic decomposition) and ΔS is positive (increase in gas molecules, indicating higher entropy). Hence, the answer is (C) a negative ΔH and a positive ΔS.

Explanation:

The reaction 2N2O(g) → 2N2(g) + O2(g) involves the decomposition of dinitrogen oxide into nitrogen and oxygen gases. To determine the signs of ΔH (enthalpy change) and ΔS (entropy change), we look at the reactants and products. In the given reaction, we're starting with two molecules of dinitrogen oxide and producing four molecules (two of nitrogen and one of oxygen). Entropy (ΔS) is a measure of disorder, so an increase in the number of gas molecules typically indicates an increase in entropy (ΔS > 0), since gases have more disorder than solids or liquids.

Regarding enthalpy (ΔH), without specific data, it's not definitive what the enthalpy change is. However, we can sometimes infer whether a reaction is exothermic or endothermic based on the type of reaction. Decomposition reactions, especially those that break down compounds into their elemental states, are often exothermic. Hence, it's probable (but not certain) that ΔH < 0. Therefore, the most likely answer is that the decomposition of dinitrogen oxide has a negative ΔH and a positive ΔS, which corresponds to option (C).

Answer : The value of [tex]\Delta E[/tex] of the reaction is, 479.958 KJ/mole

Explanation :

The relation between the internal energy and enthalpy of reaction is:

[tex]\Delta E=\Delta H-\Delta n_g\times RT[/tex]

where,

[tex]\Delta E[/tex] = internal energy of the reaction = ?

[tex]\Delta H[/tex] = enthalpy of the reaction = 483.6 KJ/mole = 483600 J/mole

From the balanced reaction we conclude that,

[tex]\Delta n_g[/tex] = change in the moles of the reaction = Moles of product - Moles of reactant = 3 - 2 = 1 mole

R = gas constant = 8.314 J/mole.K

T = temperature = [tex]165^oC=273+165=438K[/tex]

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

[tex]\Delta E=483600J/mole-(1mole\times 8.314J/mole.K\times 438K)[/tex]

[tex]\Delta E=479958.468J/mole[/tex]

[tex]\Delta E=479.958KJ/mole[/tex]

Therefore, the value of [tex]\Delta E[/tex] of the reaction is, 479.958 KJ/mole

Final answer:

To calculate the change in internal energy for the given reaction, we can use the equation ∆U = q - P∆V, where q is the heat transferred and P∆V is the work done on or by the system. Given the enthalpy change for the reaction, we can calculate the heat transferred and therefore determine the change in internal energy.

Explanation:

The question asks for the change in internal energy (∆U) for the reaction 2H2O(g) → 2H2(g) + O2(g) at a pressure of 1.0 atm and a temperature of 165°C. To find ∆U, we need to use the equation ∆U = q - P∆V, where q is the heat transferred and P∆V is the work done on or by the system. Since the pressure is constant, P∆V is zero, so we only need to calculate the heat transferred, q.

Given that the enthalpy change (ΔH) for the reaction is 483.6 kJ/mol, we can use the equation q = nΔH, where n is the number of moles. Since 2.0 moles of H2O(g) are being converted, the heat transferred can be calculated as q = 2.0 mol × 483.6 kJ/mol = 967.2 kJ.

Therefore, the change in internal energy for this reaction, ∆U, is 967.2 kJ.

Answers:

A. Disrupts hydrogen bonding in proteins, linearizing the protein

B. Provides an overall negative charge on proteins, making the migration distance on gel a function of only protein size

Final answer:

SDS in gel electrophoresis A. linearizes proteins into a rod-like shape and coats them with a uniform negative charge, allowing separation by molecular weight when an electric current is applied.

Explanation:

The sodium dodecyl sulfate (SDS) in gel electrophoresis serves primarily two functions:

SDS binds to proteins roughly in proportion to the number of amino acids, which correlates with the protein's mass. Thus, SDS-PAGE allows for the separation of proteins primarily by their molecular weight once an electric current is applied. The molecular weight of the proteins is estimated by comparing their migration distance to that of known standards.

Answer:  Hypothesis

Explanation:  Hypothesis is a kind of idea that has been put forward prior to experimentation and whose results are not out yet.

The laws makes the theory then there are some rules which are needed to be followed in order to make the most out of the experiments. Then prior to the experiments , there comes the hypothesis whis the explanation of the work and which would be tested through the experiments.

Answer:  The equilibrium concentration of H2O(g) is 12.52  x [tex]10^{-3}[/tex]

Explanation:   The given equilibrium reaction is -

[tex]C_{2}H_{4}(g) + H_{2}O(g) \rightleftharpoons  C_{2}H_{5}OH(g)[/tex]

Equilibrium constant is the ratio of concentration of the products to the reactants.

Mathematically, it can be written as-

 Kc =  [tex][C_{2}H_{5}OH(g)][/tex]  /  [tex][C_{2}H_{4}][/tex]   [tex] [H_{2}O][/tex]  

Given values are -

Kc = 9.0 x [tex]10^{3}[/tex]

[tex][C_{2}H_{5}OH(g)][/tex] = 1.69M

[tex][C_{2}H_{4}][/tex] = 0.015M

Susbtituting these values in the equation we get

9.0 x [tex]10^{3}[/tex] = 1.69M / 0.015M [tex] [H_{2}O][/tex]  

[tex] [H_{2}O][/tex]   = 1.69M / 0.015M x  9.0 x [tex]10^{3}[/tex]

[tex] [H_{2}O][/tex]  = 12.52  x [tex]10^{-3}[/tex]

Answer: Option (c) is the correct answer.

Explanation:

Activation energy or free energy of a transition state is defined as the  minimum amount of energy required to by reactant molecules to undergo a chemical reaction.

So, when activation energy is decreased then molecules with lesser amount of energy can also participate in the reaction. This leads to an increase in rate of reaction.  

Also, increase in temperature will help in increasing the rate of reaction.

Whereas at a given temperature, every molecule will have different energy because every molecule travels at different speed.

Hence, we can conclude that out of the given options false statement is that at a given temperature and time all molecules in a solution or a sample will have the same energy.

Answer:

b. 9.29 L is the new volume of the gas if the pressure is constant.

Explanation:

As per Charles’s law  

At constant pressure for a given amount of a gas,

Volume is directly proportional to its temperature.

Thus the expression is [tex]V \propto T[/tex]

[tex]\frac {V}{T} = k[/tex] where k is a constant  

When there is a change in the volume and temperature the expression will be

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

where [tex]V_1[/tex] and [tex]T_1[/tex] are the initial volume and initial temperature and [tex]V_2[/tex] and [tex]T_2[/tex] are the final volume and temperature.

Plugging in the values given

[tex]\frac {(9.3L)}{(279.5K)}=\frac {V_2}{(279.1K)}[/tex]

[tex]V_2=\frac {(9.3L\times279.1K)}{279.5K} \\\\=9.29L[/tex]

(Answer)

Answer:

4.083 * 10^20 atoms.

Explanation:

One Mole of phosphorus  contains 6.022 * 10^23 atoms (Avogadros number)'

Since 1 mole of Phosphorus  has a mass of  30.974 grams, 21 milligrams has

6.022 * 10^23  * 0.021 / 30.974

= 0.004083 * 10^23

= 4.083 * 10^20

To calculate the standard enthalpy of formation of carbon disulfide (CS2), we can use Hess's Law and the given enthalpy changes for the reactions involving carbon, sulfur, and oxygen. The standard enthalpy of formation of CS2 from its elements is -310.3 kJ/mol.

To calculate the standard enthalpy of formation of carbon disulfide (CS2) from its elements, we can use Hess's Law and the given enthalpy changes for the reactions involving carbon (C), sulfur (S), and oxygen (O).

First, we can use the given reaction for the formation of carbon dioxide (CO2) to find the enthalpy change for the reaction involving carbon:

C(graphite) + O2(g) → CO2(g) ΔH o rxn = −393.5 kJ/mol

Next, we can use the given reaction for the formation of sulfur dioxide (SO2) to find the enthalpy change for the reaction involving sulfur:

S(rhombic) + O2(g) → SO2(g) ΔH o rxn = −296.4 kJ/mol

Now, we can use the given reaction for the formation of carbon disulfide (CS2) to find the enthalpy change for this reaction:

CS2 + 3O2(g) → CO2(g) + 2SO2(g) ΔH o rxn = −1073.6 kJ/mol

By rearranging these equations and manipulating the enthalpy changes, we can find the standard enthalpy of formation of carbon disulfide:

CS2 = CO2 - C(graphite) - 2SO2 = -1073.6 kJ/mol - (-393.5 kJ/mol) - 2(-296.4 kJ/mol) = -1073.6 kJ/mol + 393.5 kJ/mol - 2(-296.4 kJ/mol) = -310.3 kJ/mol

Therefore, the standard enthalpy of formation of carbon disulfide (CS2) from its elements is -310.3 kJ/mol.

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The standard enthalpy of formation of carbon disulfide (CS2) is calculated using Hess's Law and given reactions. After rearranging and combining reactions to form CS2 from its elements, the standard enthalpy of formation for CS2 is found to be -87.3 kJ/mol.

To calculate the standard enthalpy of formation (ΔHfo) of carbon disulfide (CS2), we use Hess's Law and the given reactions:

The standard enthalpy of formation of CS2 is calculated by rearranging the reactions to derive the formation reaction for CS2 from its elements:

In doing so, we get:

C(graphite) + 2S(rhombic) + O2(g) → CS2(l), ΔHfo = -1073.6 kJ/mol + (1 x 393.5 kJ/mol) + (2 x 296.4 kJ/mol)

We sum the enthalpy changes of these steps to find the enthalpy of formation for CS2 which is:

ΔHfo(CS2) = -1073.6 kJ/mol + 393.5 kJ/mol + 592.8 kJ/mol = -87.3 kJ/mol

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Using Hess's law, the enthalpy change for the reaction can be calculated by manipulating and adding together the enthalpy changes of the given reactions to match the sought reaction.

The enthalpy change for the reaction: 2 C(graphite) + 3H₂(g) → C₂H₆(g) can be calculated using Hess's law. According to

, the enthalpy change for a reaction is equivalent to the sum of the enthalpy changes for each step that makes up that reaction.

Looking at the given equations, we can manipulate them to establish a path to our desired equation. We can reverse the first equation and multiply it by 2, keep the second equation as it is and also multiply it by 3, and then reverse and divide the third equation by 2. Once these are added together, they should form our sought reaction. The overall enthalpy change for the reaction is then found by summing the enthalpy changes of each of these steps multiplied by their corresponding coefficients.

This approach utilizes the fact that enthalpy is a state function, meaning it only depends on the initial and final states of the system, not the path taken.

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The enthalpy change for the reaction 2 C(graphite) + 3H2(g) → C2H6(g) can be calculated using Hess's Law and the given enthalpy changes for other reactions. The reaction is broken down into steps and the total enthalpy change is the sum of the changes for each step. The calculated enthalpy change for the reaction is -941.8 kJ/mol.

To calculate the enthalpy change for the reaction 2 C(graphite) + 3H2(g) → C2H6(g), we use the values given for three known reactions and apply Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction process, regardless of the number of steps.

First, reverse the equation C(graphite) + O2(g) → CO2(g)ΔH o rxn = −393.5 kJ/mol and multiply it by 2 to get 2C(graphite) + 2O2(g) → 2CO2(g), ΔH = 2*(-393.5) kJ/mol = -787 kJ/mol.

Second, multiply the equation H2(g) + 1/2 O2(g) → H2O(l)ΔH o rxn = −285.8 kJ/mol by 6 to get 6H2(g) + 3 O2(g) → 6H2O(l), ΔH = 6*(-285.8) kJ/mol = -1714.8 kJ/mol.

Finally, add the values obtained to the equation for the formation of ethane, 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)ΔH o rxn = −3119.6 kJ/mol, but reverse it to get C2H6(g) → 2C(graphite) + 3H2(g), ΔH = -(-3119.6/2) kJ/mol = 1560 kJ/mol.

The total enthalpy change for the reaction is then found by adding these values together: (-787) + (-1714.8) + 1560 = -941.8 kJ/mol. Thus, the enthalpy change for the reaction 2 C(graphite) + 3H2(g) → C2H6(g) is -941.8 kJ/mol.

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