For the data set: 0.09, 0.10, 0.11, 0.13, 0.09, 0.11, 0.10, 0.07 To obtain information of the precision of the data set the standard deviation would be:

a. 0.018
b. 0.022
c. 0.0166
d. 0.01

Answer: the gas is lighter than CO2 and it is hydrogen

Explanation:Please see attachment for explanation

Answer:

88.8 minutes

Explanation:

Graham's law of diffusion relates rate of difusion by the following formula

Rate1 / rate 2 = √( Mass of argon / Mass of Neon)

Where rate = volume divided by time

Rate 1 = 10 ml / t1

Rate 2 = 10 ml / t2

Rate 1/ rate 2 = 10 ml / t1 ÷ 10 ml/ t2 = t2/ t1

t2/t1 = √(Mass of argon / mass of Neon) = √( 39.984/20.179)

125 / t1 = 1.4026

t1 = 125 / 1.4026 = 88.8 minutes

It will take approximately 95.3 seconds for 10.0 mL of Ne gas to effuse through the porous barrier.

To calculate the time it will take for 10.0 mL of Ne gas to effuse through a porous barrier, we can use the effusion rate ratio. In Example 9.21, it is stated that it takes 243 seconds for 4.46 × 10-5 mol Xe to effuse through a tiny hole. Using the effusion rate ratio, we can calculate that it will take 95.3 seconds for 4.46 × 10-5 mol Ne to effuse. Since Ne is lighter than Xe, its effusion rate will be larger, resulting in a smaller effusion time.

Explanation:

It is known that entropy is the measure of degree of randomness present in a substance due to the movement of its molecules.

More randomly the atoms are moving from one place to another more will be the entropy of system.

In order to calculate the entropy, we divide the amount of heat transferred by the temperature at which heat transfer occurs.

As the given temperature is 296.95 K and heat of vaporization is 24.8 kJ/mol. Therefore, calculate the molar entropy of vaporization as follows.

Molar entropy of vaporization = [tex]\frac{24.8 kJ/mol}{296.95 K}[/tex]

                                                  = 0.08351 kJ/mol K

Thus, we can conclude that molar entropy of vaporization of [tex]CC_{3}F[/tex] is 0.08351 kJ/mol K.

CFCs, including Trichlorofluoromethane, have been banned due to their destructive effect on the ozone layer. UV light causes them to release chlorine atoms, which react with and deplete ozone. The harmful side effect of this is increased solar radiation reaching Earth.

Chlorofluorocarbons, or CFCs, like Trichlorofluoromethane, were previously widely used in a variety of industries, such as refrigeration and aerosols. However, the stable nature of these compounds, combined with their ability to break down and release chlorine atoms under ultraviolet light, led to significant depletion of the ozone layer in the stratosphere. This process happens because as CFCs are broken down by UV light, they produce chlorine atoms. These chlorine atoms then react with ozone molecules, leading to a net decrease in stratospheric ozone. This ozone depletion had severe implications, such as increased risks from solar radiation, which prompted the worldwide effort to phase out the use of CFCs under the Montreal Protocol.

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Answer:

option b is correct

Explanation:

Taking the Van't Hoff equation

d ln Keq / dT = ΔH/(RT²)

then

Keq increases with increasing temperature (d ln Keq / dT>0) when ΔH>0 and decreases with increasing temperature (d ln Keq / dT<0) when ΔH<0

if Keq decreases (reactions shift toward reactants) from 7.9×10³ to 0.77 when temperature increases from 298 K to 713 K (d ln Keq / dT<0) → ΔH<0 ( exothermic reaction)

therefore option b is correct

The enthalpy of the reaction is predicted to be negative, indicating it is an exothermic reaction, because the equilibrium constant decreases with an increase in temperature, a characteristic of exothermic reactions. Therefore, the correct answer is b.

The question asks if the enthalpy of the reaction is positive or negative based on changes in the equilibrium constant at different temperatures. The equilibrium constant decreases from 7.9×10³ at 298 K to 0.77 at 713 K. According to the principle that for an endothermic reaction (ΔH° > 0), the magnitude of K increases with increasing temperature, and conversely, for an exothermic reaction (ΔH° < 0), the magnitude of K decreases with increasing temperature, we can deduce the nature of the reaction. As the equilibrium constant decreases with an increase in temperature, the reaction must be exothermic. Therefore, the correct answer is b. Negative, because exothermic reactions shift toward reactants at higher temperatures.

Answer:

None of the conditions will favor either the forward reaction or backward reaction , hence the answer is D

Explanation:

It's challenging to determine the precise initial amounts of gaseous elements X and Y in the reaction, XY(s) → X(g) + Y(g), with the given Kp and the conditions provided. Based on the data provided, the accurate answer would be 'none of the above'.

This question involves equilibrium in chemical reactions, specifically the concept of Kp, the equilibrium constant under constant pressure circumstances. In its simplest form, it represents a ratio of the concentrations of products and reactants, each raised to the power of their respective stoichiometric coefficients in the balanced chemical equation. In this context, if the reaction XY(s) → X(g) + Y(g) has a Kp = 4.1, it means that when equilibrium is reached, the ratio [X][Y]/[XY] is 4.1.

Given that one mole of gas at 0°C and 1 atmosphere of pressure occupies a volume of 22.4L (standard molar volume), we can establish the initial conditions of X and Y using the stoichiometric ratio. However, without additional specifics on the initial conditions of the solid XY such as its volume or weight, it's challenging to determine precise initial amounts of X and Y. Therefore, the accurate answer from the given choice would be (d) none of the above.

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Answer:

≅ 4.39 gm/cm^3

Explanation:

Given data:

width of block =3.2 cm

length = 17.1 cm

height = 5.0 cm

Mass = 1.2 kg

Therefore volume= width×length×height = 3.2×17.1×5= 273.6 cm^3

now, density = mass/volume

calculating density in gm/cm^3

= 1200/273.6 = 4.38596

≅ 4.39 gm/cm^3

Final answer:

The sulfur-containing product of the oxidation of two 2-methyl-1-propanethiol molecules is a disulfide, where the sulfur atoms form a bond between the two original thiol molecules, resulting in a dimer.

Explanation:

The student has asked for the sulfur-containing product of the oxidation reaction between two 2-methyl-1-propanethiol molecules. In the presence of mild oxidizing agents such as oxygen, sulfur atoms can be oxidized from the sulfhydryl (S-H) groups present in 2-methyl-1-propanethiol. When two such sulfur atoms, each with an unpaired electron, come into contact, they can form a sulfur-sulfur bond (disulfide bond), resulting in the dimerization of the two thiol molecules.

The chemical formula of 2-methyl-1-propanethiol is CH3CH2CH(SH)C(CH3)2. Upon oxidation, the sulfur atoms from two of these molecules will bond together to form a disulfide. The resulting molecule will be CH3CH2CH(S-S)CH2C(CH3)2.

The molecule C7H10NBr has an Index of Hydrogen Deficiency (IHD) of 4, indicating 4 degrees of unsaturation. These could be a combination of double bonds, triple bonds, or rings.

The Index of Hydrogen Deficiency (IHD) or Degree of Unsaturation for a molecule can be calculated using the formula IHD = 1/2(2C + 2 + N - X - H), where C is the number of Carbons, N the number of Nitrogens, X the number of Halogens, and H the total Hydrogen count. For the molecule C7H10NBr, applying the formula would yield:

IHD = 1/2(2*7 + 2 + 1 - 1 - 10) = 4

This means there are 4 degrees of unsaturation present in the molecule. The types of unsaturation could be double bonds, triple bonds, or rings. In this case, with a value of 4, it could be 4 double bonds, or 2 double bonds and 2 rings, or 2 rings and 1 triple bond, or 1 ring and 3 double bonds, or 2 triple bonds, and many more combinations.

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The Index of Hydrogen Deficiency for the compound C₇H₁₀N Br is 4, which indicates that the molecule can have a combination of 4 double bonds or 4 rings or a mix of both.

The Index of Hydrogen Deficiency (IHD) is a count of how many molecules of H2 are missing from an alkane having the same number of carbon atoms. It's used to identify the number of rings and/or double bonds in organic compounds. For the compoundC₇H₁₀N Br, consider that Br adds 1 to the hydrogen count, and N subtracts 1. Thus, the IHD calculation would be: IHD= (2*C + 2 + N - X - H)/2 = [2*7 + 2 +1 -0 -10]/2 = 4

Since it has an IHD of 4, it indicates that the molecule can have a combination of 4 double bonds or 4 rings or a mix of both.

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Answer:

85 cents/L is equal to 3.2176$/gallon.

gas wiuld be heaper to buy from the station A.

Explanation:

As 85 cents/L is more than $2.5/gallon therefore buying gas from station A would be cheaper than the other one.

Answer:

It will take 77.8 seconds for 95% of sample to react

Explanation:

As the given reaction obeys 1st order therefore-

                        [tex][N_{2}O_{5}]=[N_{2}O_{5}]_{0}\times (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}[/tex]

Where [tex][N_{2}O_{5}][/tex] is the concentration of [tex]N_{2}O_{5}[/tex] after "t" time, [tex][N_{2}O_{5}]_{0}[/tex] is the initial concentration of [tex]N_{2}O_{5}[/tex] and [tex]t_{\frac{1}{2}}[/tex] is half life

Here, [tex]\frac{[N_{2}O_{5}]}{[N_{2}O_{5}]_{0}}=\frac{100-95}{100}=0.05[/tex] and [tex]t_{\frac{1}{2}}=18.0 seconds[/tex]

So, [tex]0.05=(\frac{1}{2})^{\frac{t}{18.0seconds}}[/tex]

or, [tex]t=77.8 seconds[/tex]

So, it will take 77.8 seconds for 95% of sample to react

Answer:

ra to r+a

Explanation: In an enzyme catalyzed reaction, enzyme binds with the active side(3D) of the substrate substrate. It provides an alternative path for the reaction to take place with lower activation energy. in the reaction the kinetic energy of the molecules increases so reaction takes place at a higher rate. when the reaction is completed enzyme leaves the active side.

Final answer:

In the formation of Fe3O4, three atoms of iron (Fe) combine with four molecules of oxygen (O2), involving a total of three iron atoms and eight oxygen atoms to maintain the stoichiometric ratio as per the compound's chemical formula.

Explanation:

The question relates to a chemical reaction where three atoms of iron (Fe) combine with four molecules of oxygen (O2) to form the compound Fe3O4. This compound, known as magnetite, is an example of an iron oxide where the ratio of iron to oxygen atoms is maintained as per the stoichiometry of the compound's chemical formula. In this case, three atoms of iron will combine with oxygen to utilize four oxygen molecules (which equates to eight oxygen atoms), resulting in the formula Fe3O4. This shows the direct application of stoichiometry in understanding chemical formulas and the ratios in which elements combine to form compounds.

Answer: 13.47g of MgBr2

Explanation:

MM of MgBr2 = 24 + (2 x 80) = 24 + 160 = 184g/mol

Mass conc. Of MgBr2 = 0.183 x184 = 33.672g

33.672g of MgBr2 dissolves in 1000mL

Therefore Xg of MgBr2 will dissolve in 400mL

Xg of MgBr2 = ( 33.672 x 400)/1000

Xg of MgBr2 = 13.47g

Answer:

The bond order for C2 molecule is 2.

Explanation:

Bond order can be defined as the half of the difference between the number of electrons in the bonding orbital and the number of electrons in the antibonding orbitals. It can be represented mathematically by; .

Bond order,n= [number of electrons in the bonding molecular orbitals(BMO) - the number or electrons in the anti-bonding molecular orbitals(AMO) ] / 2.

The electronic configuration of the C2 molecule is given below;

C2 = (1sσ)^2 (1s^*σ)^2 (2sσ)^2 (2s^*σ)^2 (2pπ)^4.

The ones with the (*) are known as the Anti-bonding molecular orbitals while the ones without (*) are known as the bonding molecular orbitals. Hence, we have 8 Electrons from the bonding molecular orbitals and 4 Electrons from the anti-bonding molecular orbitals.

So, from the formula given above, the bond order of C2 molecule is;

===> 8-4/2= 4/2.

===> 2.

The bond order of the C-C bond in the C₂ molecule is 1. This is calculated using molecular orbital theory by identifying bonding and antibonding electrons and applying the bond order formula.

The bond order of the C-C bond in the C₂ molecule can be calculated using molecular orbital theory. First, we identify the number of bonding and antibonding electrons in the molecule. The molecular orbital configuration for C₂ is (σ2s)² (σ*2s)² (π2p)⁴. This gives us 4 bonding electrons (from π2p) and 2 antibonding electrons (from σ*2s).

To calculate the bond order, we use the formula:

Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2

Substituting the numbers, we get Bond order = (4 - 2) / 2 = 1.

Therefore, the bond order of the C-C bond in the C₂ molecule is 1.

Answer:

This is as a result of their property type

ΔG is extensive and E is Intensive. The explanation is as given below

Explanation:

Basically both ΔG and the cell potential or the electromotive force (E.M.F) has some disparity especially in their spontaneity, for spontaneous reaction ΔG = -ve while E = +ve and vice versa. But the most important disparity is their state function i.e one is intensive and the other is extensive property.

ΔG is an example of an extensive property, they are properties whose value is dependent on the volume or the size of the system. other examples are mass, volume etc.

E on the other hand is an intensive property, they are properties whose value is not dependent on the size of the system. As such, this differences explains why ΔG for a reaction scale with a reaction quantity and E does not.

The change in Gibbs free energy (ΔG) for a reaction scales with reaction quantity because it is an extensive property, while the standard electrode potential (E) does not scale with reaction quantity because it is an intensive property.

The change in Gibbs free energy (ΔG) for a reaction scales with reaction quantity because it is an extensive property, meaning it depends on the amount of substance involved in the reaction. On the other hand, the standard electrode potential (E) does not scale with reaction quantity because it is an intensive property, meaning it is independent of the amount of substance.

For example, in the combustion of hydrogen, the standard Gibbs free energy change (ΔG°) for the reaction is -237 kJ/mol for 1 mole of hydrogen and -474 kJ/mol for 2 moles of hydrogen. Since ΔG is an extensive property, it doubles when the quantity of hydrogen doubles. However, the standard electrode potential (E°) remains the same regardless of the quantity of hydrogen.

In summary, the scaling of ΔG with reaction quantity and the independence of E from reaction quantity are due to the differences in their nature as extensive and intensive properties, respectively.

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The enthalpy change for the sublimation of iodine, represented by I₂(s) → I₂(g), is 31 kJ/mol.

The enthalpy change for the sublimation of iodine, represented by I₂(s) → I₂(g), can be determined by comparing the enthalpy changes of the two given reactions. Since the equation ½ H₂(g) + ½ I₂(g) → HI(g) has a lower enthalpy change (-5.0 kJ/mol) compared to the equation ½ H₂(g) + ½ I₂(s) → HI(g) (26 kJ/mol), it means that the phase change from solid iodine to gaseous iodine requires an additional amount of energy. Thus, the enthalpy change for the sublimation of iodine is the difference between the two reaction enthalpies, which is 31 kJ/mol.

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Answer:

Ketone, alcohol and unsaturation

Explanation:

The functional groups are the groups that identify the organic function of the molecule. For example, the hydroxyl (-OH) represents alcohol, the carbonyl (represented in the figure below) can represent a ketone or an aldehyde, and so each function has its representation.

The cortisone molecule is represented below, and each group is marked in blue. They are:

a, c, and d = carbonyl of ketone;

b =  unsaturation (double bond) of alkene;

e, and f = hydroxyl of alcohol.

Answer: none of the above

Explanation:

It should be 2,1,1,2 to give a balanced chemical reaction

To balance the chemical equation between potassium iodide and lead (II) acetate, producing lead (II) iodide and potassium acetate, the coefficients must be 2 (for potassium iodide), 1 (for lead (II) acetate), 1 (for lead (II) iodide), and 2 (for potassium acetate).

The question is asking for the correct coefficients that balance the chemical equation between potassium iodide and lead (II) acetate to produce lead (II) iodide and potassium acetate. The balanced chemical equation is:

2KI + Pb(C2H3O2)2 → PbI2 + 2KC2H3O2

The resultant balanced equation shows that two molecules of potassium iodide react with one molecule of lead (II) acetate to yield one molecule of lead (II) iodide and two molecules of potassium acetate. Therefore, the coefficients for the balanced equation are 2, 1, 1, 2 respectively.

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Answer: The amount of water required to prepare given amount of salt is 398.4 mL

Explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

We are given:

Molarity of solution = 0.16 M

Given mass of manganese (II) nitrate tetrahydrate = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Putting values in above equation, we get:

[tex]0.16M=\frac{16\times 1000}{251\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{16\times 1000}{251\times 0.16}=398.4mL[/tex]

Volume of water = Volume of solution = 398.4 mL

Hence, the amount of water required to prepare given amount of salt is 398.4 mL

Answer: 17.6psi

Explanation:

V1 = 10.5 L

P1 = 14.3 psi

V2 = 8.55L

P2 =?

P1V1 = P2V2

14.3 x 10.5 = P2 x 8.55

P2 = (14.3 x 10.5) / 8.55

P2 = 17.6psi

Answer:

The final pressure is 17.56 psi

Explanation:

Step 1: Data given

Initial volume = 10.5 L

Initial pressure = 14.3 psi

Final volume = 8.55 L

Step 2: Calculate the new pressure

P1V1 = P2V2

⇒ P1 = The initial pressure = 14.3 psi

⇒ V1 = The initial volume = 10.5L

⇒ P2 = The final pressure = TO BE DETERMINED

⇒ V2 = The final volume = 8.55 L

14.3 * 10.5 = P2*8.55

P2 = 17.56 psi

The final pressure is 17.56 psi

Answer: C2H4S5

Explanation:

Since the total mass is 5.000g

Mass of sulphur = 5.000-(0.6357+0.1070)

Mass of sulphur = 4.2573g

Using Empirical relation

C= 0.6357 H= 0.1070 S= 4.2573

Divide through by their molar mass to obtain the smallest ratio

C= 0.6357/12 H=0.1070/1 S=4.2573/32

C= 0.053 H= 0.1070 S= 0.133

Divide through by the smallest ratio (0.053)

C=0.053/0.053 H=0.1070/0.053 S=0.133/0.053

C=1 H=2 S=2.5

1:2:2.5 ,multiply through by 2 ,to obtain whole numbera

2:4:5

Therefore the empirical formula is C2H4S5. Thus only gives the ratio

Molecular formula is the chemical formula .

(Empirical formula) n = molecular formula

(C2H4S5)n = molar mass

[(12×2) + ( 1×4) +(32×5)]n = 188.4

188n=188.4

n= 1

Molecular formula = (C2H4S5)×1

Therefore the chemical formula of

lenthionine is C2H4S5

Raoult's law accounts for the fact that the vapor pressure of a solvent will decrease as the mole fraction of the solvent is decreased. In considering the mole fraction, it is important to consider the total moles of dissolved particles. Remember: a particle can be a dissolved molecule or ion. Which aqueous solutions would have the lowest vapor pressure.

0.1 M [tex]NH_4NO_3(aq)[/tex] , 0.1 M [tex]NaF(aq)[/tex],  0.1 M [tex]LiNO_3(aq)[/tex],  0.1 M [tex]Na_3PO_4(aq)[/tex] and  0.1 M [tex]HC_2H_3O_2(aq)[/tex]

Answer: 0.1 M [tex]Na_3PO_4[/tex]

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]

where,

[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure

i = Van'T Hoff factor  

[tex]x_2[/tex] = mole fraction of solute  

1. For 0.1 M [tex]NH_4NO_3[/tex]

[tex]NH_4NO_3\rightarrow NH_4^{+}+NO_3^{-}[/tex]  

, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be [tex]1\times 0.1+1\times 0.1=0.2[/tex]

2. For 0.1 M [tex]NaF[/tex]

[tex]NaF\rightarrow Na^{+}+F^{-}[/tex]  

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be [tex]1\times 0.1+1\times 0.1=0.2[/tex]

3. For 0.1 M [tex]LiNO_3[/tex]

[tex]LiNO_3\rightarrow Li^{+}+NO_3^{-}[/tex]  

, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be [tex]1\times 0.1+1\times 0.1=0.2[/tex]

4. For 0.1 M [tex]Na_3PO_4[/tex]

[tex]Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}[/tex]  

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be [tex]3\times 0.1+1\times 0.1=0.4[/tex]

5. For 0.1 M [tex]HC_2H_3O_2(aq)[/tex]

[tex]HC_2H_3O_2(aq)\rightarrow CH_3COO^{-}+H^{+}[/tex] [/tex]

, i= 2 as it is a electrolyte and dissociate to give two ions, concentration of ions will be [tex]2\times 0.1=0.2[/tex]

Thus as concentration of solute is highest for 0.1 M [tex]Na_3PO_4[/tex] , the vapor pressure will be lowest.

Answer:

C

Explanation:

The chemical formula of salt is NaCl (sodium chloride).

Na is a metal and Cl is a non-metal.

This means that NaCl is an ionic compound.

Covalent compound is a compound made up of non-metal atoms only.

Salts are always ionic compounds. Therefore, option C is correct.

An ionic compound is a type of chemical compound that is composed of ions held together by electrostatic forces of attraction.

Ionic compounds play vital in various applications, such as in the formation of salts, the electrolysis of compounds, and the construction of solid-state electronic devices.

Salts are a specific type of ionic compound. They are compounds formed when an acid reacts with a base through a chemical reaction called neutralization.

Salts consist of cations (positively charged ions) and anions (negatively charged ions) held together by ionic bonds.

To learn more about the ionic compound, follow the link:

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The answer is a change in internal energy causes work to be done and heat to flow into the system.  

Explanation:

Boyle's law says, PV=RT

Answer:

297.0 K

Explanation:

Given data

We can find the boiling point of CFC-11 using the following expression.

ΔSvap = ΔHvap/Tb

Tb = ΔHvap/ΔSvap

Tb = (26.88 × 10³ J/mol)/(90.51 J/(mol ⋅ K))

Tb = 297.0 K

Answer: c. 0.200 m HOCH2CH2OH

Explanation:

The collective properties of solvents expressed that the freezing point of water is lowered when solute particles are dissolved in it.

When more particles are dissolved, the more the freezing point is deepened and lowered.

Here, all the molarities are equal, the deciding factor in measuring the lowering will be which substance produces the fewest particles in solution; then we relate it to lowering of the freezing point.

Ba(NO3)2 ---> Ba2+ + 2NO3(1-)

1 mol releases 3 mol of ions.

Mg(ClO4)2 ---> Mg2+ + 2ClO4(1-)

1 mol releases 3 mol of ions.

HOCH2CH2OH is covalent and doesn't ionize.

1 mol in water is just 1 mol of molecules.

Na3PO3 ---> 3Na+ + PO3(1-)

1 mol releases 4 mol ions.

The substance that produces the fewest particles in solution is the 0.200 m HOCH2CH2OH

The solution with the highest freezing point will be the one that forms the fewest particles when dissolved. In this case, that would be the ethylene glycol solution (HOCH2CH2OH), because it does not ionize in solution and therefore has a van't Hoff factor of 1.

In order to understand which solution has the highest freezing point, think about a concept called freezing point depression. The freezing point depression is due to the number of solute particles in a given volume of solvent; more particles will result in a lower freezing point. The key to the number of particles is the van't Hoff factor (i), which shows how many particles a solute breaks up into when dissolved.

Selected solution (a) Ba(NO3)2 is a salt which breaks up into three ions in solution: Ba2+ and 2NO3-, so i=3. Solution (b) Mg(ClO4)2 also breaks up into three ions: Mg2+ and 2ClO4-, so i=3. Solution (c) HOCH2CH2OH, ethylene glycol, is a nonelectrolyte and doesn't ionize in solution, so i=1. Solution (d) Na3PO3 breaks up into four ions: 3Na+ and PO33-, so i=4.

Therefore, the solution with the highest freezing point will be the one that forms the fewest particles when dissolved, which is 0.200 m HOCH2CH2OH.

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Answer:

a) pH = 2.573

b) pH = 4.347

Explanation:

a) weak acid: CH3COOH

∴ Ka = 1.8 E-5 = [H3O+][CH3COO-] / [CH3COOH]

∴ C CH3COOH = 0.40 M

mass balance:

⇒ 0.40 M = [CH3COO-] + [CH3COOH].........(1)

charge balance:

⇒ [H3O+] = [CH3COO-].........(2)

(2) in (1):

⇒ [CH3COOH] = 0.40 - [H3O+]

replacing in Ka:

⇒ Ka = 1.8 E-5 = [H3O+]² / ( 0.40 - [H3O+] )

⇒ [H3O+]² = 7.2 E-6 - 1.8 E-5[H3O+]

⇒ [H3O+]² + 1.8 E-5[H3O+] - 7.2 E-6 = 0

⇒ [H3O+] = 2.6743 E-3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.573

b) balanced reations:

∴ C CH3COOH = 0.40 M

∴ C CH3COONa = 0.20 M

mass balanced:

⇒ C CH3COOH + C CH3COONa = [CH3COO-] + [CH3COOH]

⇒ 0.60 = [CH3COO-] + [CH3COOH]......(1)

charge balanced:

⇒ [H3O+] + [Na+] = [CH3COO-]

∴ [Na+] = 0.20 M

⇒ [H3O+] + 0.20 M = [CH3COO-]........(2)

(2) in (1):

⇒ 0.60 M = ( [H3O+] + 0.20 ) + [CH3COOH]

⇒ [CH3COOH] = 0.40 - [H3O+]

replacing in Ka:

⇒ 1.8 E-5 = ([H3O+])([H3O+] + 0.20) / (0.40 - [H3O+])

⇒ 7.2 E-6  - 1.8 E-5[H3O+] = [H3O+]² + 0.20[H3O+]

⇒ [H3O+]² + 0.20[H3O+] - 7.2 E-6 = 0

⇒ [H3O+] = 4.499 E-5 M

⇒ pH = 4.347

The pH of the solution in (a) is 2.57 The pH of the solution in (b) is 4.4.

We have to set up the ICE table for the reaction;

           CH3CO2H + H2O ⇄ H3O^+   +   CH3CO2^-

I            0.40                            0                  0

C          -x                                  +x                 +x

E       0.40 - x                            x                   x

The pKa of  CH3CO2H  is 1.8 x 10-5

Hence,

1.8 x 10-5 = x^2/0.40 - x  

1.8 x 10-5 (0.40 - x ) = x^2

7.2  x 10-6 - 1.8 x 10-5x =  x^2

x^2 +  1.8 x 10^-5x - 7.2  x 10^-6 = 0

x = 0.00267 M

Hence;

pH = -log [0.00267 M] = 2.57

Using the Henderson Hasselbaclch equation;

pH = pKa + log [A-]/[HA]

pKa = - log Ka = -log[1.8 x 10^-5] = 4.7

Hence;

pH = 4.7 + log [0.20 M]/[0.40 M ]

pH = 4.4

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Answer: option E. None because in all the reactions O2 is in excess

Explanation:

In the above reactions, only Reaction A and Reaction B have oxygen (O2) as the limiting reactant. For the rest of the reactions, the other compound will be used up before all the O2 is spent.

In every reaction, you have 5.73 moles of O2 and 1.70 moles of the other reactants. You can determine which of the two reactants are limiting (which will be consumed first) by comparing the number of moles of that reactant divided by its stoichiometric coefficient (the numbers in front of the chemicals in the reaction) to the same calculation for the other reactant.

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Explanation:

Net ionic between HCl and the buffer equation

CH3COOH --> acetic acid, CH3COO- acetate , H+ proton from HCl

there is neutralization of the acetate, and an increase in acid

the ionic equation

CH3COO-(aq) + H+(aq) + Cl-(aq) --> CH3COOH(aq) + Cl-(aq)

Net ionic:

CH3COO-(aq) + H+(aq) --> CH3COOH(aq).

Final answer:

H+ ions from added HCl react with the acetate ions in a buffer to form acetic acid, preventing a significant pH change.

Explanation:

When a strong acid like HCl is added to a buffer made of acetic acid (HC₂H₃O₂) and sodium acetate (CH₃COONa), the hydrogen ions (H+) from HCl will react with the acetate ions (C₂H₃O₂-) from sodium acetate in the buffer. The net ionic reaction for this process is:

H*(aq) + C₂H₃O₂-(aq) → HC₂H₃O₂(aq)

The reaction shows that H+ ions are intercepted by the buffer's acetate anions to form acetic acid (HC₂H₃O₂), preventing a significant change in pH and maintaining the buffer's purpose.

Answer:

The pH of solution is 2.88 .

Explanation:

The reaction is :

[tex]CH_3COOH-->CH_3COO^-+H^+[/tex]

We know, [tex]K_a[/tex] for this reaction is = [tex]1.76\times 10^{-5}.[/tex]

Also, since volume of water is 1 L.

Therefore, molarity of solution is equal to number of moles.

Also, [tex]K_a=\dfrac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]

Let, amount of [tex]CH_3COO^- and\ H^+[/tex] produce is x.

So,

[tex]K_a=\dfrac{[x][x]}{[0.1]}\\1.76\times 10^{-5}=\dfrac{[x][x]}{[0.1]}[/tex]

[tex]x=0.0013\ mol.[/tex]

We know, [tex]pH=-log{x}=-log(0.0013)=2.88[/tex]

Therefore, pH of solution is 2.88 .

Hence, this is the required solution.

Final answer:

The final pH of a 1L solution containing 0.1 moles of acetic acid is determined by using the hydrogen ion concentration resulting from its dissociation. The pH is calculated to be approximately 2.879 following the calculation procedures outlined for weak acids.

Explanation:

Calculating pH of an Acetic Acid Solution

The final pH of a solution can be determined by calculating the hydrogen ion concentration ([H+]) resulting from the dissociation of acetic acid in water. Given that we have 0.1 moles of acetic acid added to water to make 1L of solution, the molarity of acetic acid is 0.100 M. Acetic acid (CH₃COOH) is a weak acid, and its dissociation in water is represented by the following equation:

CH₃COOH <--> CH₃COO⁻ + H⁺

We use the acid dissociation constant (Ka) to solve for [H+]. For acetic acid, Ka = 1.8 x 10⁻⁵ M. The calculation involves setting up an ICE table and using the quadratic formula, as outlined in the subject's textbook or reference. The calculation typically results in a hydrogen ion concentration of 1.32 x 10⁻³ M.

The pH is found by taking the negative logarithm of the hydrogen ion concentration:

pH = -log(1.32 x 10⁻³) = 2.879
Therefore, the pH of a 0.100 M solution of acetic acid is approximately 2.879.