A 10.1 g sample of NaOH is dissolved in 250.0 g of water in a coffee-cup calorimeter. The temperature increases from 23.0 °C to ________°C. Specific heat of liquid water is 4.18 J/g-K and ΔH for the dissolution of sodium hydroxide in water is 44.4 kJ/mol.

Answer : The heat of this reaction of AgCI formed will be, 66.88 KJ

Explanation :

First we have to calculate the heat of the reaction.

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

where,

q = amount of heat = ?

[tex]c[/tex] = specific heat capacity = [tex]4.18J/g.K[/tex]

m = mass of substance = 120 g

[tex]T_{final}[/tex] = final temperature = [tex]22^oC=273+22=295K[/tex]

[tex]T_{initial}[/tex] = initial temperature = [tex]22.8^oC=273+22.8=295.8K[/tex]

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

[tex]q=120g\times 4.18J/g.K\times (295.8-295)K[/tex]

[tex]q=401.28J[/tex]

Now we have to calculate the number of moles of [tex]AgNO_3[/tex] and [tex]HCl[/tex].

[tex]\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }AgNO_3=0.20mole/L\times 0.03L=0.006mole[/tex]

[tex]\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }HCl=0.10mole/L\times 0.1L=0.01mole[/tex]

Now we have to calculate the limiting reactant.

The balanced chemical reaction will be,

[tex]AgNO_3+HCl\rightarrow AgCl+HNO_3[/tex]

As, 1 mole of [tex]AgNO_3[/tex] react with 1 mole of HCl

So, 0.006 mole of [tex]AgNO_3[/tex] react with 0.006 mole of HCl

From this we conclude that, [tex]HCl[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]AgNO_3[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgCl.

The given balanced reaction is,

[tex]Ag^++Cl^-\rightarrow AgCl[/tex]

From this we conclude that,

1 mole of [tex]Ag^+[/tex] react with 1 mole [tex]Cl^-[/tex] to produce 1 mole of [tex]AgCl[/tex]

0.006 mole of [tex]Ag^+[/tex] react with 0.006 mole [tex]Cl^-[/tex] to produce 0.006 mole of [tex]AgCl[/tex]

Now we have to calculate the heat of this reaction of AgCI formed.

As, 0.006 mole of AgCl produced the heat = 401.28 J

So, 1 mole of AgCl produced the heat = [tex]\frac{401.28}{0.006}=66880J=66.88KJ[/tex]

Therefore, the heat of this reaction of AgCI formed will be, 66.88 KJ

Answer:

the percentage by mass of Nickel(II) iodide = 23.58%

Explanation:

% by mass of solute = (mass of solute/mass of solution) x 100%

% by mass of NiI2 = (mass of NiI2/mass of solution) x 100%

% by mass of NiI2 = (5.47 grams/23.2 grams) x 100% = 23.58% m/m

Answer:

50 Joule

Explanation:

Diatomic gas

Q = Heat given = 70 J

n = number of moles

Cp = specific heat at constant pressure

ΔT = Change in temperature

R = Gas constant  

Change in energy

ΔE = Q-w

⇒ΔE = n(Cp)ΔT-nRΔT

As it is a diatomic gas Cp = (7/2)R

Putting the value of Cp in the above equation we get

Q = (7/2)RΔT

ΔE = (5/2)RΔT

Dividing the equations we get

ΔE/Q = 5/7

⇒ΔE = (5/7)Q

⇒ΔE = (5/7)×70

⇒ΔE = 50 J

∴ The internal energy change is 50 Joule

To find the % yield of SO3, we calculate the theoretical yield based on the molar mass and stoichiometry of the reaction. The theoretical yield is shown to be 30.02 g, and with an actual yield of 7.9 g, the % yield is calculated to be 26.32%.

To calculate the % yield of SO3, we first need to determine the theoretical yield based on the reaction stoichiometry. The balanced equation is:

2SO2 + O2 → 2SO3

Given that excess sulfur (S) is present, the limiting reactant is O2. Using the molar mass of O2 (32.00 g/mol) and SO3 (80.06 g/mol), we can calculate the theoretical yield:

Calculate moles of O2: moles = mass / molar mass = 6.0 g / 32.00 g/mol = 0.1875 mol.

Stoichiometry tells us that 1 mol of O2 produces 2 mol of SO3, so the expected moles of SO3 is 2 × 0.1875 mol = 0.375 mol.

Calculate the theoretical yield of SO3: 0.375 mol × 80.06 g/mol = 30.02 g.

Next, we compare the theoretical yield to the actual yield to determine the percent yield:

Percent yield = (actual yield / theoretical yield) × 100 = (7.9 g / 30.02 g) × 100 = 26.32%

The percent yield of SO3 in this experiment is 26.32%.

The percent yield of SO3 in this experiment is 26.32%.

To find the % yield of SO3, we calculate the theoretical yield based on the molar mass and stoichiometry of the reaction. The theoretical yield is shown to be 30.02 g, and with an actual yield of 7.9 g, the % yield is calculated to be 26.32%.

To calculate the % yield of SO3, we first need to determine the theoretical yield based on the reaction stoichiometry. The balanced equation is:

2SO2 + O2 → 2SO3

Given that excess sulfur (S) is present, the limiting reactant is O2. Using the molar mass of O2 (32.00 g/mol) and SO3 (80.06 g/mol), we can calculate the theoretical yield:

Calculate moles of O2: moles = mass / molar mass = 6.0 g / 32.00 g/mol = 0.1875 mol.

Stoichiometry tells us that 1 mol of O2 produces 2 mol of SO3, so the expected moles of SO3 is 2 × 0.1875 mol = 0.375 mol.

Calculate the theoretical yield of SO3: 0.375 mol × 80.06 g/mol = 30.02 g.

Next, we compare the theoretical yield to the actual yield to determine the percent yield:

Percent yield = (actual yield / theoretical yield) × 100 = (7.9 g / 30.02 g) × 100 = 26.32%

Answer: Copper is getting oxidized and is a reducing agent. Silver is getting reduced and is oxidizing agent.

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

[tex]X\rightarrow X^{n+}+ne^-[/tex]

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.

[tex]X^{n+}+ne^-\rightarrow X[/tex]

Oxidizing agents are defined as the agents which oxidize other substance and itself gets reduced. These agents undergoes reduction reactions.

Reducing agents are defined as the agents which reduces the other substance and itself gets oxidized. These agents undergoes reduction reactions.

For the given chemical reaction:

[tex]Cu(s)+2AgNO_3(aq.)\rightarrow 2Ag(s)+Cu(NO_3)_2(aq.)[/tex]

The half reactions for the above reaction are:

Oxidation half reaction:  [tex]Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-[/tex]

Reduction half reaction:  [tex]2Ag^+(aq.)+2e^-\rightarrow 2Ag(s)[/tex]

From the above reactions, copper is loosing its electrons. Thus, it is getting oxidized and is considered as a reducing agent.

Silver is gaining electrons and thus is getting reduced and is considered as an oxidizing agent.

In the given redox reaction,

The oxidized substance is Cu(s)

The reduced substance is AgNO₃

The oxidizing agent is AgNO₃

The reducing agent is Cu(s)

From the question,

The given redox reaction is

Cu(s)+2AgNO3(aq)⟶2Ag(s)+Cu(NO3)2(aq)

The equation for the reaction can be written properly as

Cu(s) + 2AgNO₃(aq)⟶2Ag(s) + Cu(NO₃)₂(aq)

In order to identify the oxidized substance, the reduced substance, the oxidizing agent, and the reducing agent in the redox reaction,

First, we will define some of terms

Oxidizing agent

An oxidizing agent is substance which oxidizes something else. It gains electrons and is reduced in a chemical reaction. Therefore, an oxidizing agent is the reduced substance in a given chemical reaction.

Reducing agent

A reducing agent reduces other substances and loses electrons. It is oxidized in a chemical reaction. Therefore, a reducing agent is the oxidized substance in a given chemical reaction.

Oxidation

Oxidation can be defined as gain of oxygen. It can also be defined as an increase in oxidation state

Reduction

Reduction can be defines as loss of oxygen. It can also be defines as a decrease in oxidation state.

From the given reaction

The oxidation state of Cu increased from 0 to +2. This means Cu was oxidized.

(NOTE: The oxidation state of Cu in Cu(NO₃)₂ is +2)

∴ Cu(s) was oxidized and is the reducing agent

Also,

The oxidation state of Ag reduced from +1 to 0. This means Ag was reduced.

(NOTE: The oxidation state of Ag in AgNO₃ is +1)\

∴ AgNO₃ was reduced and it is the oxidizing agent

Hence, in the given redox reaction

The oxidized substance is Cu(s)

The reduced substance is AgNO₃

The oxidizing agent is AgNO₃

The reducing agent is Cu(s)

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Answer : The temperature decreases for an adiabatic expansion process.

Explanation :

As per first law of thermodynamic,

[tex]\Delta U=q+w[/tex]

where,

[tex]\Delta U[/tex] = internal energy

q = heat

w = work done

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

For expansion of a gas, the work is to be done by the system. So, 'w' will be negative.

[tex]\Delta U=-w[/tex]

and,

[tex]\Delta U[/tex] will be also negative. That means,

[tex]\Delta U=U_2-U_1<0[/tex]

Or, [tex]U_2<U_1[/tex]

From the above we conclude that, the final internal energy will be lesser than the initial internal energy and as we know that, the internal energy is depend on the temperature.

That means, the temperature of the final state will be less than that of the initial state. So, the temperature decreases for an adiabatic expansion process.

Hence, the temperature decreases for an adiabatic expansion process.

In an adiabatic expansion, the temperature of a gas decreases as the gas does work on its surroundings without any heat transfer.

When a gas expands adiabatically, the process occurs without heat transfer into or out of the system. Instead, changes to the system take place due to the work done on or by the system itself. Therefore, in an adiabatic expansion, the gas does work on its surroundings, which causes it to lose energy. Loss of energy, in this context, translates to a decrease in internal temperature of the gas. So, in adiabatic expansion, the temperature of a gas decreases.

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Answer : The percent yield is, 32.79 %

Explanation :  

First we have to calculate the moles of [tex]CH_3CH_2OH[/tex].

[tex]\text{Moles of }CH_3CH_2OH=\frac{\text{Mass of }CH_3CH_2OH}{\text{Molar mass of }CH_3CH_2OH}=\frac{65.2g}{46.07g/mole}=1.415mole[/tex]

Now we have to calculate the moles of [tex]CH_3CH_2OCH_2CH_3[/tex]

The balanced chemical reaction will be,

[tex]2CH_3CH_2OH(l)\rightarrow CH_3CH_2OCH_2CH_3(l)+H_2O(l)[/tex]

From the balanced reaction, we conclude that

As, 2 moles of [tex]CH_3CH_2OH[/tex] react to give 1 mole of [tex]CH_3CH_2OCH_2CH_3[/tex]

So, 1.415 moles of [tex]CH_3CH_2OH[/tex] react to give [tex]\frac{1.415}{2}=0.7075[/tex] mole of [tex]CH_3CH_2OCH_2CH_3[/tex]

Now we have to calculate the mass of [tex]CH_3CH_2OCH_2CH_3[/tex]

[tex]\text{Mass of ether}=\text{Moles of ether}\times \text{Molar mass of ether}[/tex]

[tex]\text{Mass of }ether=(0.7075mole)\times (74.12g/mole)=52.44g[/tex]

The theoretical yield of ether, [tex]CH_3CH_2OCH_2CH_3[/tex]  = 52.44 g

Now we have to calculate the percent yield of [tex]CH_3CH_2OCH_2CH_3[/tex]

[tex]\%\text{ yield of ether}=\frac{\text{Actual yield of ether}}{\text{Theoretical yield of ether}}\times 100=\frac{17.2g}{52.44g}\times 100=32.79\%[/tex]

Therefore, the percent yield is, 32.79 %

Answer:

The answer is lithium(LI)

Explanation:

A)  [tex]HCl(aq)+NaOH(aq)\rightarrow H_2O(l)+NaCl(aq)[/tex]

Mass of sodium hydroxide= 2.7 g

Moles of base = [tex]\frac{2.7 g}{40 g/mol}=0.0675 mol[/tex]

According to reaction , 1 mol of NaOH neutralizes with 1 mol of HCl.

Then 0.0675 mol of base will neutralize:

[tex]\frac{1}{1}\times 0.0675 mol=0.0675 mol[/tex] of HCl.

Mass of 0.0675 mol of HCl =  0.0675 mol × 35.5 g/mol = 2.396 g

2.396 grams of acid will completely react with and neutralize 2.7 g of the sodium hydroxide.

B)  [tex]2HNO_3(aq)+Ca(OH)_2(aq)\rightarrow 2H_2O(l)+Ca(NO_3)_2(aq)[/tex]

Mass of calcium hydroxide= 2.7 g

Moles of base = [tex]\frac{2.7 g}{57 g/mol}=0.04736 mol[/tex]

According to reaction , 2 mol of [tex]HNO_3[/tex] neutralizes with 1 mol of [tex]Ca(OH)_2[/tex].

Then 0.04736 mol of base will neutralize:

[tex]\frac{2}{1}\times 0.04736 mol=0.09472 mol[/tex] of [tex]HNO_3[/tex]

Mass of 0.09472 mol of [tex]HNO_3[/tex] :

0.09472 mol × 63g/mol = 5.9673 g

5.9673 grams of acid will completely react with and neutralize 2.7 g of the calcium hydroxide.

C)  [tex]H_2SO_4(aq)+2KOH(aq)\rightarrow 2H_2O(l)+K_2SO_4(aq)[/tex]

Mass of potassium hydroxide= 2.7 g

Moles of base = [tex]\frac{2.7 g}{56 g/mol}=0.04821 mol[/tex]

According to reaction , 1 mol of [tex]H_2SO_4[/tex] neutralizes with 2 mol of [tex]KOH[/tex].

Then 0.04821 mol of base will neutralize:

[tex]\frac{1}{2}\times 0.04821 mol=0.02410 mol[/tex] of [tex]H_2SO_4[/tex]

Mass of 0.02410 mol of [tex]H_2SO_4[/tex] :

0.02410 mol × 98 g/mol = 2.3618 g

2.3618 grams of acid will completely react with and neutralize 2.7 g of the potassium hydroxide.

The amount, in grams, of each acid that would be needed for each of the reactions represented by the equations respectively, would be 2.46 grams, 4.59 grams, and 2.36 grams.

From the first equation: HCl(aq)+NaOH(aq)→H2O(l)+NaCl(aq)

The mole ratio of HCl to NaOH is 1:1.

Mole of 2.7 g NaOH = 2.7/40 = 0.0675 moles

Equivalent mole of HCl = 0.0675 moles

Mass of 0.0675 mole HCl = 0.0675 x 36.458 = 2.46 grams

For the second equation: 2HNO3(aq)+Ca(OH)2(aq)→2H2O(l)+Ca(NO3)2(aq)

Mole ratio of base to acid = 1:2

Mole of 2.7 grams Ca(OH)2 = 2.7/74.093 = 0.0364 moles

Equivalent mole of HNO3 = 0.0364 x 2 = 0.0728 moles

Mass of 0.0728 mole HNO3 = 0.0728 x 63.01 = 4.59 grams

For the third equation: H2SO4(aq)+2KOH(aq)→2H2O(l)+K2SO4(aq)

Mole ratio of acid to base = 1;2

Mole of 2.7 grams KOH = 2.7/56.1 = 0.0481 moles

Equivalent mole of H2SO4 = 0.0481/2 = 0.024 moles

Mass of 0.024 mole H2SO4 = 0.024 x 98.079 = 2.36 grams

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Hey there!:

Answer : D

A water molecule can donate a proton to another water molecule, forming H3O⁺and OH⁻ in solution.

Hope this helps!

The correct option is d) A water molecule can donate a proton to another water molecule, forming H₃O⁺ and OH⁻ in solution because water autoionizes means a water molecule can donate a proton to another water molecule, forming H₃O⁺ and OH⁻ in solution.

When we say that water autoionizes, it refers to the process where two water molecules react with each other to form hydronium ions (H₃O⁺) and hydroxide ions (OH⁻). This is a natural process that occurs in pure water and is a fundamental concept in understanding the nature of acids and bases. The reaction can be represented as follows:

[tex]\[ 2 \text{H}_2\text{O} \leftrightarrow \text{H}_3\text{O}^+ + \text{OH}^- \][/tex]

This means that one water molecule acts as an acid (proton donor) and another acts as a base (proton acceptor), leading to the formation of these ions. This process is essential for understanding the pH of water and its behavior as both an acid and a base in chemical reactions.

The complete question is:

What does it mean to say that water autoionizes?

a) Water is polar, like an ion is polar. Therefore it is autoionic.

b) Water will cause many ionic compounds to dissociate without additional input of energy.

c) Water is a good conductor of electricity, and when it conducts, it ionizes.

d) A water molecule can donate a proton to another water molecule, forming H3O+ and OH? in solution.

Answer:

so the reaction rate increases by a factor 6.

Explanation:

For the given equation the reaction is first order with respect to both ester and sodium hydroxide

So we can say that the rate law is

[tex]Rate(initial)=K[NaOH][CH_{3}COOC_{2}H_{5}][/tex]

now as per given conditions the concentration of ester is increased by half it means that the new concentration is 1.5 times of old concentration

The concentration of NaOH is quadrupled means the new concentration is 4 times of old concentration.

The new rate law is

[tex]Rate(final)=K[1.5XNaOH][4XCH_{3}COOC_{2}H_{5}][/tex]

the final rate = 6 X initial rate

so the reaction rate increases by a factor 6.

The reaction rate of ethyl acetate with sodium hydroxide would increase by a factor of 6 if the concentration of ethyl acetate is increased by half and the concentration of NaOH is quadrupled, as it is first order in both reactants.

The reaction of ethyl acetate with sodium hydroxide is:

This reaction is first order concerning both CH₃COOC₂H₅ and NaOH. The rate law for this reaction can be written as:

If the concentration of CH₃COOC₂H₅ is increased by half, its new concentration becomes 1.5 times its initial concentration. If the concentration of NaOH is quadrupled, its new concentration becomes 4 times its initial concentration. Therefore, the rate of the reaction increases by a factor of:

So, the reaction rate would increase by a factor of 6.

Under standard conditions :

E(cell) = E(cathode) - E(anode)

Note : cathode has the larger numeric value and anode has the smaller. Therefore

E(cell) = +1.36V - ( -3.04V)

= 1.36 + 3.04

= +4.40V

Answer :

(a) The value of [tex]K_b[/tex] of the lactate ion is, [tex]7.14\times 10^{-39}[/tex]

(b) The value of [tex]K_b[/tex] of the conjugate base of pyruvic acid is, [tex]3.53\times 10^{-38}[/tex]

Explanation :

Solution for (a) :

As we are given : [tex]K_a=1.4\times 10^{24}[/tex]

As we know that,

[tex]K_a\times K_b=K_w[/tex]

where,

[tex]K_a[/tex] = dissociation constant of an acid = [tex]1.4\times 10^{24}[/tex]

[tex]K_b[/tex] = dissociation constant of a base = ?

[tex]K_w[/tex] = dissociation constant of water = [tex]1\times 10^{-14}[/tex]

Now put all the given values in the above expression, we get the dissociation constant of a base (lactate ion).

[tex]1.4\times 10^{24}\times K_b=1\times 10^{-14}[/tex]

[tex]K_b=7.14\times 10^{-39}[/tex]

Therefore, the value of [tex]K_b[/tex] of the lactate ion is, [tex]7.14\times 10^{-39}[/tex]

Solution for (b) :

As we are given : [tex]K_a=2.83\times 10^{23}[/tex]

As we know that,

[tex]K_a\times K_b=K_w[/tex]

where,

[tex]K_a[/tex] = dissociation constant of an acid = [tex]2.83\times 10^{23}[/tex]

[tex]K_b[/tex] = dissociation constant of a base = ?

[tex]K_w[/tex] = dissociation constant of water = [tex]1\times 10^{-14}[/tex]

Now put all the given values in the above expression, we get the dissociation constant of a base (conjugate base of pyruvic acid).

[tex]2.83\times 10^{23}\times K_b=1\times 10^{-14}[/tex]

[tex]K_b=3.53\times 10^{-38}[/tex]

Therefore, the value of [tex]K_b[/tex] of the conjugate base of pyruvic acid is, [tex]3.53\times 10^{-38}[/tex]

The added KI does not have any impact  

The reaction invovles Titration of vitaminc ( Ascorbic acid)

ascorbic acid + I₂ → 2 I⁻  +  dehydroascorbic acid

the excess iodine is free reacts with the starch  indicator, forming the blue-black starch-iodine complex.  

This is the endpoint of the titration. since alreay excess KI is added ( the source of Iodine), it does not have an influence.

Answer B

Hope this helps!

Final answer:

Adding excess potassium iodide (KI) to a vitamin C sample for titration with N-bromosuccinimide (NBS) results in increased production of iodine, which requires more sodium thiosulfate (Na₂S₂O₃) for back titration. Therefore, the amount of NBS needed to titrate the sample will increase.

Explanation:

When a student mistakenly adds a potassium iodide (KI) solution twice while preparing a vitamin C sample for titration with N-bromosuccinimide (NBS), this will lead to the addition of excess KI in the solution. The presence of additional KI will reduce the titrand more, producing a stoichiometric amount of iodine (I₂). This increased amount of I₂ will then be determined by back titration using sodium thiosulfate (Na₂S₂O₃) as a reducing titrant. Adding excess KI does not change the amount of vitamin C in the sample, but it does result in an increased production of I₂ which needs to be titrated with sodium thiosulfate. Therefore, the larger quantity of KI will increase the amount of NBS needed to titrate the sample since there is more I₂ produced that needs to be accounted for during the titration process.

Answer:

Choice D) The nebula is moving away from the observer.

Explanation:

Is the emission here a result of electron transition or thermal radiation?

The red-pink emission here is as a line rather than a continuous spectrum. In other words, the red-pink line observed is a result of electron transition. The energy difference will be constant. That should be the same case on the earth as it is in space at the nebula.

Also, this energy difference does not depend on the temperature of the hydrogen. Only that at higher temperature, low-energy radiations will be less prominent. The wavelength will still be 656 nm when the light was emitted from the nebula.

The wavelength observed on the earth is longer than the wavelength emitted. The Doppler's effect is likely to be responsible. As the star moves away from the earth, the distance that light from the star needs to travel keep increasing. Consider two consecutive peaks from the star. When compared with the first peak, the second peak will need to travel a few more kilometers and will need a few more fractions of a second to get to the earth. It would appear to an observer on the earth that the frequency of the light is lower than it actually is. Accordingly, the wavelength will appear to be longer than it was when emitted from the star.

Conversely, the wavelength will appear shorter if the source is moving toward to observer. For this star, the wavelength appears to be longer than it really is. In other words, the star is moving away from the earth.

The ratio between the speed at which the star moves away from the earth and the speed of the light can be found using the equation: (Source: AstronomyOnline)

[tex]\displaystyle \frac{v}{c} = \frac{\Delta \lambda}{\lambda_0} \approx 0.009[/tex].

The gas cloud is moving away from Earth at about 1% the speed of light due to the Doppler effect.So,option D is correct.

The gas cloud is moving away from Earth at about 1% the speed of light. This can be inferred from the observed shift in the wavelength of the hydrogen emission line from 656.3 nm to 656.6 nm.

The shift in wavelength is due to the Doppler effect, indicating the motion of the source relative to the observer.

Answer:

(a) forward direction

(b) forward direction

(c) backward direction.

Explanation:

Given , the chemical reaction in equilibrium is,

SO₂(g)  + Cl₂(g)  ⇄  SO₂Cl₂ (g)

The direction of the reaction by changing the concentration can be determined by Le Chatelier's principle,

It states that ,

When a reaction is at equlibrium , Changing the concentration , pressure,  temperature disturbs the equilibrium , and the reaction again tries to attain equilibrium by counteracting the changes.

(a)

For the reaction , Cl₂ is added to the system , i.e. , increasing the concentration of Cl₂ ,Now, according to Le Chatelier , The reaction will move in forward direction , to reduce the increased amount of Cl₂.

Hence, reaction will go in forward direction.

(b)

Removing SO₂Cl₂ from the system ,i.e. , decreasing the concentration of SO₂Cl₂  , according to Le Chatelier , the reaction will move in forward direction , to increase the amount of reduced SO₂Cl₂.

Hence, reaction will go in forward direction.

(c)

Removing SO₂ from the system , i.e. decreasing the concentration of SO₂ ,  according to Le Chatelier , the reaction will move in backward direction , to increase the amount of reduced SO₂.

Hence, reaction will go in backward direction.

Answer : The Lewis structures for the two molecules are shown below.

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 52.2 g

Mass of H = 13.1 g

Mass of O = 34.7 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{52.2g}{12g/mole}=4.35moles[/tex]

Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{13.1g}{1g/mole}=13.1moles[/tex]

Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{34.7g}{16g/mole}=2.17moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{4.35}{2.17}=2.00\approx 2[/tex]

For H = [tex]\frac{13.1}{2.17}=6.03\approx 6[/tex]

For O = [tex]\frac{2.17}{2.17}=1[/tex]

The ratio of C : H : O = 2 : 6 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = [tex]C_2H_6O_1[/tex]  = [tex]C_2H_6O[/tex]

The empirical formula weight = 12(2) + 6(1) + 1(16) = 46 gram/eq

Now we have to calculate the value of 'n'.

Formula used :

[tex]n=\frac{\text{Molecular formula weight}}{\text{Empirical formula weight}}=\frac{45g/mole}{46g/eq}=0.9\approx 1[/tex]

Molecular formula = [tex](C_2H_6O)_n=(C_2H_6O)_1=C_2H_6O[/tex]

So, there are two possibilities for the arrangements of atoms. That means, it will be an ethanol [tex](H_3C-CH_2-OH)[/tex] or dimethyl ether [tex](H_3C-O-CH_3)[/tex].

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

As we know that carbon has '4' valence electrons, oxygen has '6' valence electrons and hydrogen has '1' valence electron.

Therefore, the total number of valence electrons in [tex]C_2H_6O[/tex] = 2(4) + 6(1) + 6 = 20

According to Lewis-dot structure, there are 16 number of bonding electrons and 4 number of non-bonding electrons.

Thus, the Lewis structures for the two molecules are shown below.

Hey there!:

Answer : LiF < MgO

Explanation

Lattice energy depends on the charge of the constituent ions. In MgO, the charge of cation and anion are +2 and -2 respectively.In LiF, the charge of cation and anion are +1 and -1. Similarly, if the electronegativity of the anion is higher, the higher will be the lattice energy. Also, an ionic compound with smaller cation size will have higher lattice energy.

Here, the correct order is,

LiF > LiI    

NaCl > KCl    

MgO > CsCl  

CaO > BaCl2

LiF < MgO

Hope this helps!

The lattice energy of a compound increases as the ion charges increase and the ion sizes decrease. Therefore, the correct order of increasing lattice energy provided in the question is 'LiF < MgO' due to higher ionic charges in MgO.

The lattice energy of a compound increases with the increasing charge of its ions and decreasing ionic size. For example, the lattice energy of LiF, with both ions having a charge of 1, is 1023 kJ/mol, but for MgO, where both ions have a charge of 2, the lattice energy increases significantly to 3900 kJ/mol. Thus, the correct order of increasing (more exothermic) lattice energy from your given pairs would be 'LiF < MgO' as MgO's higher ionic charges make its lattice energy more exothermic than that of LiF.

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Answer:Cell reaction is going forward.

Explanation:

For any chemical reaction to be spontaneous or to move in forward direction the ΔG ,the Gibbs free energy must be negative.

The cell potential of a battery is positive for a spontaneous reaction, so for a battery to give output its cell potential must be positive.

Thermodynamics and electro-chemistry are related in the following manner:

ΔG=-nFE

n=number of electrons involved

F=Faradays constant

E=cell pottential of battery

so from the above equation ΔG would only be negative when E cell that is the cell potential is positive.

For a battery which is being used its cell potential is positive and hence the ΔG would be negative. So the cell reaction occurring would be in forward direction as ΔG is negative.

when the cell potential Ecell is 0 then ΔG is also zero then the reaction occurring in battery would be at equilibrium.

When the cell potential Ecell is - then ΔG is positive and the reaction would be occurring backwards.

Answer : The [tex]\Delta G[/tex] for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,  

[tex]Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)[/tex]

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : [tex]Cu\rightarrow Cu^{2+}+2e^-[/tex]

Reduction : [tex]2Ag^++2e^-\rightarrow 2Ag[/tex]

Now we have to calculate the Gibbs free energy.

Formula used :

[tex]\Delta G^o=-nFE^o[/tex]

where,

[tex]\Delta G^o[/tex] = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

[tex]E^o[/tex] = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

[tex]\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole[/tex]

Therefore, the [tex]\Delta G[/tex] for this reaction is, -88780 J/mole.

"The change in Gibbs free energy ΔG for the reaction is -94.48 kJ/mol.

To calculate ΔG for the reaction involving the reduction of silver ions with elemental copper, we can use the standard cell potential E° and the following relationship:

[tex]\[ \Delta G = -nFE\° \][/tex]

where:

- [tex]\( \Delta G \)[/tex] is the change in Gibbs free energy in joules (J).

-  n  is the number of moles of electrons transferred in the reaction.

-  F  is the Faraday constant, which is approximately 96485 J/(V·mol).

- E°  is the standard cell potential in volts (V).

The balanced chemical equation for the reaction is:

[tex]\[ \text{Cu}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s) \][/tex]

From the equation, we can see that two moles of electrons (n = 2) are transferred when one mole of copper is oxidized to form one mole of copper(II) ions.

Given that the standard cell potential  E° is 0.46 V, we can now plug in the values:

[tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \[ \Delta G = -(2 \text{ mol})(96485 \text{ J/(V\·mol)})(0.46 \text{ V}) \] \[ \Delta G = -(2)(96485)(0.46) \] \[ \Delta G = -186177 \text{ J} \][/tex]

To convert joules to kilojoules, we divide by 1000:

[tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \[ \Delta G = -\frac{186177}{1000} \text{ kJ} \] \[ \Delta G = -186.177 \text{ kJ/mol} \] \[ \Delta G \approx -186 \text{ kJ/mol} \][/tex]

However, there seems to be a discrepancy in the significant figures used in the calculation. The standard cell potential was given to two decimal places (0.46 V), so the final answer should be rounded accordingly:

Upon re-evaluating the calculation with the correct rounding:

[tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \[ \Delta G = -(2)(96485)(0.46) \] \[ \Delta G = -94480 \text{ J/mol} \] \[ \Delta G = -94.48 \text{ kJ/mol} \][/tex]

Therefore, the correct change in Gibbs free energy for the reaction is -94.48 kJ/mol."

Answer : The mass of [tex]CS_2[/tex] is, 555.028 grams

Explanation :

First er have to calculate the concentration of [tex]S_2[/tex].

[tex]\text{Concentration of }S_2=\frac{\text{Moles of }S_2}{\text{Volume of solution}}=\frac{8.08mole}{5.35L}=1.51mole/L[/tex]

Now we have to calculate the concentration of [tex]CS_2[/tex].

The given balanced chemical reaction is,

                          [tex]S_2(g)+C(s)\rightleftharpoons CS_2(g)[/tex]

Initial conc.         1.51       0         0

At eqm. conc.   (1.51-x)  (x)       (x)

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[CS_2]}{[S_2]}[/tex]

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

[tex]9.40=\frac{x}{(1.51-x)}[/tex]

By solving the term 'x', we get :

x = 1.365 M

Concentration of [tex]CS_2[/tex] = x M = 1.365 M

Now we have to calculate the moles of [tex]CS_2[/tex].

[tex]\text{Moles of }CS_2=\text{Concentration of }CS_2}\times \text{Volume of solution}=1.365mole/L\times 5.35L=7.303mole[/tex]

Now we have to calculate the mass of [tex]CS_2[/tex].

Molar mass of [tex]CS_2[/tex] = 76 g/mole

[tex]\text{Mass of }CS_2=\text{Moles of }CS_2}\times \text{molar mass of }CS_2}=7.303mole\times 76g/mole=555.028g[/tex]

Therefore, the mass of [tex]CS_2[/tex] is, 555.028 grams

The amount of carbon disulfide CS2 at equilibrium in the given reaction with sulfur S2 and carbon C can be determined by using the provided equilibrium constant Kc and the initial concentration of S2. The equilibrium concentration of CS2 corresponds to 14.19 M. To determine its mass in grams, this concentration is multiplied by the volume of the reaction vessel and the molar mass of CS2.

In the given chemical reaction S2 (g) + C(s) <-> CS2 (g) in which sulfur S2 (g) and carbon C(s) combine to form carbon disulfide CS2 (g), the equilibrium constant Kc at 900K is given as 9.40.

This value of the equilibrium constant represents the ratio of the concentration of the product (CS2) to the concentration of the reactants (S2 and C). Since the reaction involves heating sulfur S2 and carbon C in excess, we can solve for the equilibrium concentration of CS2 based on the initial amount of S2 used (8.08 mol) and the volume of the reaction vessel (5.35 L).

First we calculate the initial concentration of S2 as [S2] = 8.08 mol / 5.35 L = 1.51 M. Given the reaction stoichiometry, each mol of S2 produces one mol of CS2, so at equilibrium [CS2] equals to the equilibrium concentration of S2.

Substituting these into the equilibrium expression Kc = [CS2]/[S2] and solving, we find [CS2] = Kc * [S2] = 9.40 * 1.51 M = 14.19 M, which represents the equilibrium concentration of CS2. The amount in grams of CS2 can then be found by multiplying this concentration by the volume of the vessel and the molar mass of CS2 to give the total mass of CS2 in the reaction vessel at equilibrium.

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Answer:

It is due to the nature of the reactants

Explanation:

To ignite a solid, we require more heat component compared to liquids and gases. For ignition to occur, oxygen gas combines with a reactant in most cases.

Some factors affect the rate rate at which a chemical proceeds. One of the factors is the nature of reactants.

The solid phase is very slow while the gaseous phase is rapid and fast.

            solid phase < liquid phase <  gas phase

Gases are free and the molecules move in all direction. They easily combine and react very fast.

It generally takes more enthalpy to ignite a solid than a gas or liquid because solids consist of tightly bound molecules with strong intermolecular forces. More energy is required to break these connections and induce a phase change. In contrast, gas and liquid states have weaker intermolecular forces, requiring less energy for ignition.

The reason it generally takes more enthalpy to ignite a solid than a gas or liquid comes down to the state of matter and its molecular structure. Solids are composed of tightly bound molecules with strong intermolecular forces, thus, require more energy, or enthalpy, to break these connections and ignite.

In contrast, gas and liquid states of matter have weaker intermolecular forces. In gases, the molecules are separated by large distances, making it easier to ignite as less energy is required to break the bonds. In liquids, the attractive forces between molecules are weaker than in solids, therefore, less enthalpy is needed for ignition.

The phase of a substance - solid, liquid, or gas - plays a significant role in how much enthalpy is needed for ignition. The amount of energy needed for a phase change - such as from solid to liquid (melting) or from liquid to gas (vaporization) – depends on the strength and the number of bonds to be broken, which is generally higher in solids.

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An ionic bond, typically stronger than most covalent bonds, is formed through the transfer of electrons. Once ionic compounds are dissolved in water, they can conduct electricity. Therefore statement c is true.

The statement that is TRUE among the following is c. An ionic bond is generally stronger than most covalent bonds. Ionic bonds are formed through the transfer, not sharing, of electrons between atoms and once these ionic compounds are dissolved in water, they actually can conduct electricity as the ions are free to move.

This is why ionic compounds in their aqueous or molten states can conduct electricity, and not typically in a solid, room-temperature state. The movement of these charged particles allows for the flow of electric current.

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Final answer:

The true statement concerning ionic bonds and compounds is related to their electrical conductivity when dissolved in water, not the comparative strength of ionic and covalent bonds or their inherent ability to conduct electricity in solid form.

Explanation:

The question refers to properties and characteristics of ionic bonds and ionic compounds. Ionic bonds are formed through the transfer, not sharing, of electrons between a positively charged ion (cation) and a negatively charged ion (anion), resulting in a strong electrostatic attraction. Unlike covalent bonds, where electrons are shared, ionic bonds involve a clear donor and acceptor of electrons, creating components with opposite charges that are attracted to each other. This bond is relatively strong but generally weaker than the strongest covalent bonds due to the difference in electron sharing and exchange mechanisms.

One of the most critical properties of ionic compounds is their ability to conduct electricity when dissolved in water or melted. In solid state, ionic compounds form a crystal lattice structure that does not conduct electricity due to the lack of free-moving charged particles. However, once dissolved in water, the ions become free to move, allowing the solution to conduct electricity. Therefore, statement b, suggesting that ionic compounds rarely conduct electricity when dissolved in water, is incorrect.

To summarize, the correct statement among the given options is that an ionic bond is much stronger than most covalent bonds is false on two fronts: It misrepresents the nature of ionic and covalent bond strengths and overlooks the electrical conductivity of dissolved ionic compounds, which is a defining characteristic of such compounds.

Answer: The concentration of radon after the given time is [tex]3.83\times 10^{-30}mol/L[/tex]

Explanation:

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

[tex]t_{1/2}=\frac{0.693}{k}[/tex]

We are given:

[tex]t_{1/2}=3.82days[/tex]

Putting values in above equation, we get:

[tex]k=\frac{0.693}{3.82}=0.181days^{-1}[/tex]

Rate law expression for first order kinetics is given by the equation:

[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]

where,  

k = rate constant = [tex]0.181days^{-1}[/tex]

t = time taken for decay process = 3.00 days

[tex][A_o][/tex] = initial amount of the reactant = [tex]1.45\times 10^{-6}mol/L[/tex]

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

[tex]0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}[/tex]

[tex][A]=3.83\times 10^{-30}mol/L[/tex]

Hence, the concentration of radon after the given time is [tex]3.83\times 10^{-30}mol/L[/tex]

Explanation:

a) Manganese (II) phosphite :[tex]Mn_3(PO_3)_2[/tex]

Number of phosphite ions in manganese (II) phosphite = [tex]7.23\times 10^{22}[/tex]

In one molecule of manganese (II) phosphite there are 2 ions of phosphite.

Then [tex]7.23\times 10^{22}[/tex] ions of phosphite will be ions will be in:

[tex]\frac{1}{2}\times 7.23\times 10^{22}=1.446\times 10^{23} molecules[/tex]

Moles of manganese (II) phosphite;= [tex]\frac{1.446\times 10^{23}}{6.022\times 10^{23}}= 0.2401 mol[/tex]

Mass of 0.2401 mol of Manganese (II) phosphite :

0.2401 mol × 322.75 g/mol = 77.49 g

77.49 is the mass in grams of a sample of manganese (II) phosphite.

b) Molar mass of ammonium chromate =252.07 g/mol

Percentage of Nitrogen:

[tex]\frac{2\times 14g/mol}{252.07 g/mol}\times 100=11.10\%[/tex]

Percentage of hydrogen:

[tex]\frac{8\times 1g/mol}{252.07 g/mol}\times 100=3.17\%[/tex]

Percentage of chromium:

[tex]\frac{2\times 52 g/mol}{252.07 g/mol}\times 100=41.25\%[/tex]

Percentage of oxygen:

[tex]\frac{7\times 16g/mol}{252.07 g/mol}\times 100=44.44\%[/tex]

c) Molar mass of the substance = 202.23 g/mol

Percentage of the hydrogen = 4.98 %

Let the molecular formula be [tex]C_xH_y[/tex]

Percentage of the carbon = 95.02 %

[tex]\frac{12 g/mol\times x}{202.23 g/mol}\times 100=95.02\%[/tex]

x= 16.01 ≈ 16

Percentage of the hydrogen= 4.98 %

[tex]\frac{1 g/mol\times y}{202.23 g/mol}\times 100=4.98\%[/tex]

y= 10.07 ≈ 10

The molecular formula of the substance is [tex]C_{16}H_{10}[/tex].

The empirical formula of the substance is [tex]C_{\frac{16}{2}}H_{\frac{10}{2}}=C_8H_5[/tex].

Answer: 2.5 grams

Explanation:

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}=\frac{2g}{4g/mol}=0.5moles[/tex]

Avogadro's Law: This law states that volume is directly proportional to the number of moles of the gas at constant pressure and temperature.

[tex]V\propto n[/tex]   (At constant temperature and pressure)  

[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas = 2.00 L

[tex]V_2[/tex] = final volume of gas = 4.50 L

[tex]n_1[/tex] = initial number of moles = 0.5 moles

[tex]n_2[/tex] = final number of moles = ?

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]\frac{2.00}{0.5}=\frac{4.50}{n_2}[/tex]

[tex]n_2=1.125moles[/tex]

Thus moles of helium added to the cylinder = (1.125-0.5)= 0.625 moles

Mass of helium added =[tex]moles\times {\text {Molar mass}}=0.625\times 4=2.5grams[/tex]

2.5 grams of helium were added to the cylinder if the volume was changed from 2.00 L to 4.50 L.

To find the amount of helium added to the cylinder, we can utilize the ideal gas law. We found that each mole of helium fills a 4-L volume. When the volume was increased to 4.5 L, it corresponded to 1.125 moles of helium. Subtracting the initial 0.5 moles, we found that 0.625 moles of helium were added, which is equivalent to 2.50 g.

This problem can be solved by understanding the concept of ideal gases and using the ideal gas law, which states that the volume of a gas is directly proportional to the number of gas molecules when pressure and temperature are constant. In the case of the helium cylinder, pressure, temperature, and the type of gas (helium) remain constant. The only variables changing are the volume and the amount of gas.

Initially, we know that 2.00 g of He fills a 2.00 L volume. It is known that 1 mol of an ideal gas at room temperature fills 22.4 L. Helium, He, has a molar mass of approximately 4 g/mol. The initial 2 g of He corresponds to 0.5 mol. This means that each mole of He fills (2.00 L/0.5 mol) = 4.00 L.

When you increased the volume to 4.5 L while keeping pressure and temperature constant, you essentially allowed for more moles of He. Using the proportion established earlier, 4.5 L of He corresponds to 4.5 L *(1 mol/4.00 L) = 1.125 mol of He.

The difference between the final and initial amounts of gas indicates the amount of He added: 1.125 mol - 0.5 mol = 0.625 mol. This corresponds to 0.625 mol * 4 g/mol = 2.50 g of helium, which is the amount added to the cylinder.

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Answer:

31.9178 °C is the final temperature of the water

Explanation:

[tex]2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ[/tex]

Mass of benzene burned = 6.200 g

Moles of benzene burned = [tex]\frac{6.200 g}{78 g/mol}=0.0794 mol[/tex]

According to reaction , 2 moles of benzene gives 6542 kJ of energy on combustion.

Then 0.0794 mol of benzene on combustion will give:

[tex]\frac{6542 kJ}{2}\times 0.0794 kJ=259.7174 kJ=Q[/tex]

Mass of water in which Q heat is added = m = 5691 g

Initial temperature = [tex]T_i=21^oC[/tex]

Final temperature = [tex]T_f[/tex]

Specific heat of water = c = 4.18 J/g°C

Change in temperature of water = [tex]T_f-T_i[/tex]

[tex]Q=mc\Delta t=mc(T_f-T_i)[/tex]

[tex]259,717.4 J=5691 g\times 4.18 J/g^oC\times (T_f-21^oC)[/tex]

[tex]T_f=31.91 ^oC[/tex]

31.9178 °C is the final temperature of the water

The final temperature of the water : 31.916 °C

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released

Q in = Q out

Heat can be calculated using the formula:

Q = heat, J

m = mass, g

c = specific heat, joules / g ° C

∆T = temperature difference, ° C / K

From reaction:

2C₆H₆ (l) + 15O₂ (g) ⟶12CO₂ (g) + 6H₂O (l) +6542 kJ, heat released by +6542 kJ to burn 2 moles of C₆H₆

If there are 6,200 g of C₆H₆ then the number of moles:

mol = mass: molar mass C₆H₆

mol = 6.2: 78

mol C6H6 = 0.0795

so the heat released in combustion 0.0795 mol C6H6:

[tex]\rm Q=heat=\dfrac{0.0795}{2}\times 6542\:kJ\\\\Q=260.0445\:kJ[/tex]

the heat produced from the burning is added to 5691 g of water at 21 ∘ C

So :

Q = m . c . ∆T  (specific heat of water = 4,186 joules / gram ° C)

260044.5 = 5691 . 4.186.∆T

[tex]\rm \Delta T=\dfrac{260044.5}{5691\times 4.186}\\\\\Delta T=10.916\\\\\Delta T=T(final)-Ti(initial)\\\\10.916=T-21\\\\T=31.916\:C[/tex]

the difference between temperature and heat  

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Specific heat  

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relationships among temperature, heat, and thermal energy.  

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Answer: The balanced chemical equation is given below.

Explanation:

Double-displacement reaction is defined as the reaction in which exchange of ions takes place. The general equation representing this reaction follows:

[tex]AB+CD\rightarrow CB+AD[/tex]

The balanced chemical equation for the reaction of barium hydroxide and perchloric acid follows:

[tex]Ba(OH)_2(aq.)+2HClO_4(aq.)\rightarrow Ba(ClO_4)_2(aq.)+2H_2O(l)[/tex]

By Stoichiometry of the reaction:

1 mole of barium hydroxide reacts with 2 moles of perchloric acid to produce 1 mole of barium chlorate and 2 moles of water molecule.

Hence, the balanced chemical equation is given above.

When barium hydroxide (Ba(OH)2) reacts with perchloric acid (HClO4), it forms barium perchlorate (Ba(ClO4)2) and water (H2O) through a neutralization reaction. This process is a double-replacement reaction where the cations of the reactants exchange places.

In the provided question, a neutralization reaction takes place when an aqueous solution of barium hydroxide (Ba(OH)2) reacts with perchloric acid (HClO4). During this reaction, we get the products - water (H2O) and a salt which is barium perchlorate (Ba(ClO4)2). The balanced equation for this double-replacement neutralization reaction becomes: Ba(OH)2 (aq) + 2HClO4(aq) → Ba(ClO4)2(aq) + 2H2O(l).

As part of understanding the concept, in a double-replacement reaction, the cation of one compound exchanges places with the cation of the other compound to form two new compounds. In a neutralization reaction, an acid reacts with a base to produce a salt and water.

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Answer : The final concentration of the seawater is, 2.909 mole/L

Explanation :

Formula used for osmotic pressure :

[tex]\pi=CRT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure  = 70.0 bar = 70 atm

R = solution constant  = 0.0821 Latm/moleK

T= temperature of solution = [tex]20^oC=273+20=293K[/tex]

C = final concentration of seawater = ?

Now put all the given values in the above formula, we get the concentration of seawater.

[tex]70atm=C\times 0.0821Latm/moleK\times 293K[/tex]

[tex]C=2.909mole/L[/tex]

Therefore, the final concentration of the seawater is, 2.909 mole/L

To determine the final concentration of seawater when reverse osmosis stops at a pressure of 70.0 bar and 20 °C, additional data such as the initial osmotic pressure of the seawater is required. Without this data, the final concentration cannot be calculated.

In reverse osmosis, water purification occurs by forcing water from a more concentrated solution to a less concentrated one by applying pressure greater than the osmotic pressure. When a pressure of 70.0 bar is applied to seawater at 20 °C, the process will continue until the osmotic pressure of the sea water is equal to the applied pressure. Since the question asks for the final concentration at which the reverse osmosis stops, we would need information about the initial osmotic pressure of seawater to calculate the final concentration. Typically, however, this value can vary, and since the necessary data to perform the calculation is not provided, we cannot accurately provide the final concentration of the seawater.

Reverse osmosis systems, such as those used in desalination plants, continuously introduce seawater under pressure and collect pure water, hence the process carries on indefinitely and the actual concentration in the plants would be constantly changing based on the amount of seawater processed and pure water extracted.

Chemical equations are symbolic representations of chemical reactions where the reactant entity is given on the left side and the product entity is on the right side. The coefficient next to the symbol and entity formula is the absolute value of the stoichiometric number.

The arrow symbol that is usually used is used to distinguish between different types of reactions. To show the type of reaction.

The chemical reaction formula shows the process of how one thing becomes another. Most often, this is written in the format:

Reactants → Products

The right arrow is the most common arrow in a chemical reaction formula. The direction shows the direction of the reaction.

Double arrows indicate reversible reactions. Reactants become products and products can become reactants again using the same process.

Two arrows with a single thorn pointing in the opposite direction show a reversible reaction if the reaction is in equilibrium.

This arrow is used to show the equilibrium reaction where again the arrow points to the stronger side of the reaction.

The reaction above shows a product that is stronger than the reactants. The lower reaction shows the reactants are preferred over the product.

Single Double Arrows are used to indicate resonance between two molecules.

Typically, the reactants will be the resonant isomers of the product.

Curved arrows with single spines on arrows indicate the path of electrons in the reaction. Electrons move from tail to head.

Curved arrows are usually displayed on individual atoms in a skeletal structure to indicate where the electron will be moved from within the product molecule.

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Grade:  College

Subject:  Chemistry

keywords: Chemical equations

Final answer:

When examining a chemical equation that describes a molecule's behavior in water, you can determine whether it is a strong electrolyte, weak electrolyte, or nonelectrolyte. Strong electrolytes dissociate completely into ions when dissolved, weak electrolytes only partially ionize, and nonelectrolytes do not yield ions.

Explanation:

A simple way to estimate whether a molecule is a strong electrolyte, a weak electrolyte, or a nonelectrolyte is to examine the chemical equation that accurately describes its behavior in water. The direction and type of reaction arrow in the equation can provide valuable information. A chemical equation reported in the literature may include a forward reaction arrow (→), equilibrium reaction arrow (⇌), or reverse reaction arrow (←). By determining whether ions are likely to be present and in what quantities, you can classify the molecule.

Strong electrolytes are substances that dissociate completely into ions when dissolved, resulting in a larger number of dissolved particles and high conductivity. Examples include ionic compounds like sodium chloride (NaCl).

Weak electrolytes only partially ionize when dissolved, resulting in a smaller fraction of ions and lower conductivity. Examples include weak acids like acetic acid (CH3COOH).

Nonelectrolytes do not yield ions when dissolved in water and do not conduct electricity. Examples include molecular compounds like glucose (C6H12O6).