A compound has two elements, A and B with electronegativities of 0.8 and 3.0 respectively. If the compound has a formula of A2B, which statement describes the bonding best? Group of answer choices The compound is largely ionic with A as the anion. The AB bond is largely covalent with a δ- on A. The compound is largely ionic with A as the cation. The AB bond is largely covalent with a δ+ on A.

Answer:

2 moles of HNO3

Explanation:

The equation seems to be balanced correctly. The problem is we done know what you started with. We will assume it is 3 moles of NO2.

If that is the case then 2 moles of HNO3 will be produced.

Answer:

Hey there!

That would be the alkaline earth metals.

Hope this helps :)

Answer: alkaline earth metals

Explanation:

Answer:

Hydroxyl

Explanation:

A hydroxyl group is a functional group that attaches to some molecules containing an oxygen and hydrogen atom, bonded together. Also spelled hydroxy, this functional group provides important functions to both alcohols and carboxylic acids.

The functional groups are the part of the organic chemistry that confers the characteristic feature of a molecule. The molecule has a hydroxyl group in its structure. Thus, option C is correct.

Hydroxyl functional groups are the atoms or molecules that provide a distinctive property to a compound. It has a chemical formula of -OH that has oxygen covalently bonded to the hydrogen atom.

The hydroxyl group is called the alcohol group that is seen in methanol, ethanol, propanol, etc. The presence of hydrogen allows the compound to form a water bond with other molecules and makes them soluble and polar.

Therefore, option C. the molecule has a hydroxyl or alcoholic functional group attached to its carbon atom.

Learn more about the hydroxyl functional group here:

https://brainly.com/question/4682253

#SPJ5

Answer:

1) positive

2) carbocation

3) most stable

4) faster

Explanation:

A common test for the presence of alcohols can be achieved using the Lucas reagent. Lucas reagent is a mixture of concentrated hydrochloric acid and zinc chloride.

The reaction of Lucas reagent reacts with alcohols leading to the formation of an alkyl chloride. Since the reaction proceeds via a carbocation mechanism, tertiary alcohols give an immediate reaction. Once a tertiary alcohol is mixed with Lucas reagent, the solution turns cloudy almost immediately indicating an instant positive reaction.

Secondary alcohols may turn cloudy within five minutes of mixing the solutions. Primary alcohols do not significantly react with Lucas reagent obviously because they do not form stable carbocations.

Therefore we can use the Lucas reagent to distinguish between primary, secondary and tertiary alcohols.

Answer:

[tex]P=6.10atm[/tex]

Explanation:

Hello,

In this case, we can study the ideal gas equation that relates temperature, volume, pressure and moles as shown below:

[tex]PV=nRT[/tex]

Thus, since we are asked to compute the pressure y simply solve for it as follows:

[tex]P=\frac{nRT}{V}=\frac{11.1mol*0.082\frac{atm*L}{mol*K}*300K}{44.8L}\\ \\P=6.10atm[/tex]

Best regards.

Answer:

NH4Br + AgNO3 —> AgBr + NH4NO3

Explanation:

When ammonium bromide and silver(I) nitrate react, the following are obtained as shown below:

NH4Br(aq) + AgNO3(aq) —>

In solution, NH4Br(aq) and AgNO3(aq) will dissociate as follow:

NH4Br(aq) —> NH4+(aq) + Br-(aq)

AgNO3(aq) —> Ag+(aq) + NO3-(aq)

The double displacement reaction will occur as follow:

NH4+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) —> Ag+(aq) + Br-(aq) + NH4+(aq) + NO3-(aq)

NH4Br(aq) + AgNO3(aq) —> AgBr(s) + NH4NO3(aq)

Answer:

171 mL of HCl

Explanation:

The first thing we want to do is consider the reaction between Al(OH)3 and water - as that is the expected reaction that is taking place,

Al(OH)3 + 3HCl → AlCl3 + 3H2O

Knowing this, let's identify the mass of Al(OH)3. Aluminum = 27 g / mol, Oxygen( 3 ) = 16 [tex]*[/tex] 3 = 48, Hydrogen ( 3 ) = 1 [tex]*[/tex] 3 = 3 - 27 + 48 + 3 = 78 g / mol. This value is approximated however ( 78 g / mol ), as the molar mass of each substance is rounded as well. Another key thing we need to do here is to convert 340 mg → grams, considering that that unit is a necessity with respect to moles, as you might know - 340 mg = 0.340 g.

Now we can calculate how much moles of HCl will be present in solution, provided we have sufficient information for that,

(0.340 g Al(OH)3) / (78.0036 g / mol Al(OH)3) [tex]*[/tex] (3 mol HCl / 1 mol Al(OH)3)

⇒ (.004358773185 g^2 / mol Al(OH)3) [tex]*[/tex] (3 HCl / Al(OH)3 )

⇒ .01307632 mol HCl

We can apply this same concept on the reaction of Mg(OH)2 and water, receiving the number of moles of HCl when that takes place. Then we can add the two ( moles of HCl ) and divide by the value " 0.18 mol / L " given to us.

" Mg(OH)2 + 2HCl → MgCl2 + 2H2O "

Molar mass of Mg(OH)2 = 58.3197 g / mol,

516 mg = 0.516 g

(0.516 g Mg(OH)2) / (58.3197 g / mol Mg(OH)2) [tex]*[/tex] (2 mol HCl / 1 mol Mg(OH)2)

= .017695564 mol HCL

___________

( .01307632 + .017695564 ) / ( 0.18 M HCl )

= 0.170954911 L

= 171 mL of HCl

Answer: The energy of a Br–F bond is 110 kJ/mol

Explanation:

The balanced chemical reaction is,

[tex]Br_2(g)+3F_2(g)\rightarrow 2BrF_3(g)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)][/tex]

[tex]\Delta H=[(n_{Br_2}\times B.E_{Br_2})+(n_{F_2}\times B.E_{F_2}) ]-[(n_{BrF_3}\times B.E_{BrF_3})][/tex]

[tex]\Delta H=[(n_{Br_2}\times B.E_{Br-Br})+(n_{F_2}\times B.E_{F_F}) ]-[(n_{BrF_3}\times 3\times B.E_{Br-F})][/tex]

where,

n = number of moles

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

[tex]\Delta H=[(1\times 193)+(3\times 155)]-[(2\times 3\times B.E_{Br-F})][/tex]

[tex]B.E_{Br-F}=110kJ/mol[/tex]

Thus the energy, in kJ/mol, of a Br–F bond is 110

Answer:

Ionic Crystal

Explanation:

Answer:

The Empirical Formular is given as; Ti₆Al₄V

Explanation:

The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium.

Elements                        Titanium            Aluminium        Vanadium

Percentage                    64.39                 24.19                   11.42

Divide all through by their molar mass

                                     64.39 / 47.87      24.19 / 27               11.42 / 50.94

                                       =  1.345                = 0.896                 = 0.224

Divide all though  by the smallest number (0.224)

                                     1.345 / 0.224        0.896 / 0.224       0.224 / 0.224

                                     = 6                         = 4                             = 1

The Empirical Formular is given as; Ti₆Al₄V

Using the stepwise procedure for obtaining the empirical formula of a compound, the empirical formula is [tex] T_{6}Al_{4}V[/tex]

Titanium :

Divide by Molar mass : = 64.39/47.87 = 1.345

Aluminum :

Divide by Molar mass : = 24.19/27 = 0.896

Vanadium :

Divide by Molar mass : = 11.42/50.94 = 0.224

Divide by the smallest :

Titanium = 1.345 / 0.224 = 6.00

Aluminum = 0.896 / 0.224 = 4

Vanadium = 0.224 / 0.224 = 1

Hence, the empirical formula is [tex] T_{6}Al_{4}V[/tex]

Learn more : https://brainly.com/question/17091379

Answer:

a. Strong acid, pH = 1.69

b. Strong base, pH = 10

c. Strong base, pH = 11

d. Weak acid, pH = 4.90

e. Strong base, pH ≅ 7 (pH should be higher than 7, but the base is so diluted)

f. Weak base, pH = 10.96

g. Acidic salt, pH = 5.12

h. Basic salt, pH = 8.38

i. Neutral salt, pH = 7

j. Acidic salt, pH < 7

Explanation:

a. HBr →  H⁺  +  Br⁻

Hydrobromic acid is a strong acid.

pH = - log [H⁺]

- log 0.02 = 1.69

b. NaOH → Na⁺  +  OH⁻

Sodium hydroxide is a strong base.

pH = 14 - pOH

pOH = - log [OH⁻]

pH = 14 - (-log 0.0001) = 10

c. Ba(OH)₂ → Ba²⁺  +  2OH⁻

Barium hydroxide is a strong base

[OH⁻] = 2 . 0.0015 = 0.003M

pH = 14 - (-log 0.003) = 11

d. HCN + H₂O ⇄  H₃O⁺  + CN⁻

This is a weak acid, it reacts in water to make an equilibrium between the given protons and cyanide anion.

To calculate the [H₃O⁺] we must apply, the Ka

Ka = [H₃O⁺] . [CN⁻] / [HCN]

6.2×10⁻¹⁰ = x² / 0.25-x

As Ka is really small, we can not consider the x in the divisor, so we avoid the quadratic formula.

[H₃O⁺] = √(6.2×10⁻¹⁰ . 0.25) = 1.24×10⁻⁵

-log 1.24×10⁻⁵ = 4.90 → pH

e.  KOH →  K⁺  +  OH⁻

2×10⁻¹⁰ M → It is a very diluted concentration, so we must consider the OH⁻ which are given, by water.

In this case, we propose the mass and charges balances equations.

Analytic concentration of base = 2×10⁻¹⁰ M = K⁺

[OH⁻] = K⁺ + H⁺ → Charges balance

The solution's hydroxides are given by water and the strong base.

Remember that Kw = H⁺ . OH⁻, so H⁺ = Kw/OH⁻

[OH⁻] = K⁺ + Kw/OH⁻. Let's solve the quadratic equation.

[OH⁻] = 2×10⁻¹⁰ + 1×10⁻¹⁴ /OH⁻

OH⁻² = 2×10⁻¹⁰. OH⁻ + 1×10⁻¹⁴

2×10⁻¹⁰. OH⁻ + 1×10⁻¹⁴ - OH⁻²

We finally arrived at the answer [OH⁻] = 1.001ₓ10⁻⁷

pH = 14 - (- log1.001ₓ10⁻⁷) = 7

The strong base is soo diluted, that water makes the pH be a neutral value.

Be careful, if you determine the [OH⁻] as - log 2×10⁻¹⁰, because you will obtain as pOH 9.69, so the pH would be 4.31. It is not possible, KOH is a strong base and 4.30 is an acid pH.

f. Ammonia is a weak base.

NH₃ +  H₂O  ⇄  NH₄⁺  + OH⁻

Kb = OH⁻  .  NH₄⁺  /  NH₃

1.74×10⁻⁵ = x² / 0.05 - x

We can avoid the x from the divisor, so:

[OH⁻] = √(1.74×10⁻⁵ . 0.05) = 9.32×10⁻⁴

pH = 14 - (-log 9.32×10⁻⁴ ) = 10.96

g. NH₄Cl, an acid salt. We dissociate the compound:

NH₄Cl →  NH₄⁺  +  Cl⁻.  We analyse the ions:

Cl⁻ does not make hydrolisis to water. In the opposide, the ammonium can react given OH⁻ to medium, that's why the salt is acid, and pH sould be lower than 7

NH₄⁺  +  H₂O  ⇄  NH₃  +  H₃O⁺   Ka

Ka = NH₃  .   H₃O⁺ / NH₄⁺

5.70×10⁻¹⁰ = x² / 0.1 -x

[H₃O⁺] = √ (5.70×10⁻¹⁰  . 0.1) = 7.55×10⁻⁶

pH = - log 7.55×10⁻⁶ = 5.12

As Ka is so small, we avoid the x from the divisor.

h. CaF₂  →  Ca²⁺  +  2F⁻

This is a basic salt.

The Ca²⁺ does not react to water. F⁻ can make hydrolisis because, the anion is the strong conjugate base, of a weak acid.

F⁻  +  H₂O  ⇄  HF  +  OH⁻          Kb

Kb = x² / 2 . 0.2 - x

Remember that, in the original salt we have an stoichiometry of 1:2, so 1 mol of calcium flouride may have 2 moles of flourides.

As Kb is small, we avoid the x, so:

[OH⁻] = √(1.47×10⁻¹¹ . 2 . 0.2) = 2.42×10⁻⁵

14 - (-log 2.42×10⁻⁵) = pH → 8.38

i . Neutral salt

BaNO₃₂  →   Ba²⁺  +  2NO₃⁻

Ba²⁺ comes from a strong base, so it is the conjugate weak acid and it does not react to water. The same situation to the nitrate anion. (The conjugate weak base, from a strong acid, HNO₃)

pH = 7

j.  Al(NO₃)₃, this is an acid salt.

Al(NO₃)₃  →  Al³⁺  +  3NO₃⁻

The nitrate anion is the conjugate weak base, from a strong acid, HNO₃ so it does not make hydrolisis. The Al³⁺ comes from the Al(OH)₃ which is an amphoterous compound (it can react as an acid or a base) but the cation has an acidic power.

Al·(H₂O)₆³⁺  + H₂O ⇄  Al·(H₂O)₅(OH)²⁺  + H₃O⁺

Answer:

Rubidium

Explanation:

Answer:

Molecular compounds are inorganic compounds that take the form of discrete (covalent) molecules. Examples include such familiar substances as water (H2O) and carbon dioxide (CO2).

Answer:

The voltage is  [tex]E_{cell} = 0.944 \ V[/tex]

Explanation:

Generally the half reaction for Zn, Zn2 half-cell is mathematically represented as

    [tex]Zn_{(s)}[/tex]   ⇔   [tex]Zn^{2+}_{ (aq)} + 2e^-[/tex]    (reference study academy)

and the electric potential for this is a constant value

     [tex]E_{zn } = -0.7618 \ V[/tex]

Generally the half reaction for Al, Al3 half-cell is mathematically represented as

      [tex]Al^{3+} _{(aq)} + 3e^-[/tex]  ⇔  [tex]Al_{(s)}[/tex]

and the electric potential for this is constant value  

       [tex]E_{Al } = -1.662 \ V[/tex]

Therefore the cell potential for an electrochemical cell  is mathematically represented as

          [tex]E_{cell} = E_{zn } - E_{Al }[/tex]

substituting values

         [tex]E_{cell} = -0.718 - (-1.662)[/tex]

        [tex]E_{cell} = 0.944 \ V[/tex]

Answer:

This means the the sign of q for the reaction was _NEGATIVE _____ and the reaction was _EXOTHERMIC_____.

Explanation:

In calorimetry, when heat is absorbed by the solution, the q-value of the solution will have a positive value. This means that the reaction will produce heat for the solution to absorb and thus the q-value for the reaction will be negative. This is an exothermic reaction.

Whereas, when heat is absorbed from the solution, the q-value for the solution will have a negative value. This means that the reaction will absorb heat from the solution and so the reaction is endothermic, and q value for the reaction is positive.

So, from the question, since the q-value of water is positive, it means that heat is absorbed by the solution and the reaction will produce a negative value of q and it's an exothermic reaction because the reaction produces heat for the solution.

Answer:

The correct answer is -8.80 kJ.

Explanation:

The ΔH° can be determined by using the formula,  

ΔH°rxn = ΔH°f (products) - ΔH°f(reactants)

Based on the given information, the ΔH°f of H2S(g) is -20.6, for Ag2S (s) is -32.6 and for H2O (l) is -285.8 kJ/mole.  

Now putting the values we get,  

= [2 molΔH°f (Ag2S) + 2 molΔH°f (H2O)] - [4 molΔH°f(Ag) + 2 molΔH°f(H2S) + 1 molΔH°f(O2)]

=[2 mol (-32.6 kJ/mol) + 2 mol(-285.8 kJ/mol)] - [4 mol(0.00 kJ/mol) + 2 mol (-20.6 kJ/mol) + 1 mol (0.00 kJ/mol)

= [(-65.2 kJ) + (-571.6 kJ)] - [(-41.2 kJ)]

= -595.6 kJ

Thus, the enthalpy change of -595.6 kJ takes place when 4 mol of Ag reacts by the equation mentioned.  

The mass of Ag given is 6.38 grams, the molecular mass of Ag is 107.9 g/mol. The formula for calculating moles is,  

Moles = mass/molar mass

= 6.38 g / 107.9 g/mol

= 0.0591 mol

Now the change in enthalpy when 0.0591 mol of Ag reacts by the given reaction is (-595.6 kJ/4 mol) × 0.0591 mol = -8.80 kJ

The negative sign indicates that the heat is released in the process. Therefore, the -8.80 kJ of heat is released by 6.38 grams of Ag in the given case.  

Answer:

Empirical formula: C₅H₅O

Molecular formula: C₁₀H₁₀O₂

Explanation:

When a compound containing C, H and O elements is combusted, the general reaction is:

CₐHₓOₙ + O₂ → a CO₂ + X/2 H₂O

Thus, you can find moles of carbon and hydrogen knowing moles of CO₂ and H₂O that are produced.

Moles CO₂ = Moles C = 0.0877g × (1mol / 44g) =

2.0x10⁻³ moles of CO₂ = moles C

Moles H₂O = 1/2 Moles H = 0.018g × (1mol / 18g) =

1x10⁻³ moles of H₂O; 2.0x10⁻³ moles H

The mass of the moles of C and H are:

2x10⁻³ moles C ₓ (12g / mol) = 0.024g C

2x10⁻³ moles H ₓ (1g / mol) = 0.002g H

Thus, mass of Oxygen is 32.3mg - 24mg C - 2mg O = 6.3mg O

Moles are:

0.0063g O ₓ (1mol / 16g) = 4x10⁻⁴ moles O

Empirical formula is the simplest ratio of atoms in a compound. Dividing each amount of moles for each atom in the 4x10⁻⁴ moles of oxygen (The lower moles), you will obtain:

C: 2.0x10⁻³ / 4x10⁻⁴ = 5

H: 2.0x10⁻³ / 4x10⁻⁴ = 5

O:  4x10⁻⁴ / 4x10⁻⁴ = 1

Thus, empirical formula is:

The molar mass of the empirical formula is:

12×5 + 1×5 + 16×1 = 81g/mol

As molar mass of the compound is 162g/mol, molecular formula is twice empirical formula:

Answer:

[tex]2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)[/tex]

[tex]Entalpy=-2861.9~KJ[/tex]

Explanation:

In this case, we have to start with the reagents:

[tex]Al~+~NH_4NO_3[/tex]

The compounds given by the problem are:

-) Nitrogen gas =  [tex]N_2[/tex]

-) Water vapor  =  [tex]H_2O[/tex]

-) Aluminum oxide =  [tex]Al_2O_3[/tex]

Now, we can put the products in the reaction:

[tex]Al_(_S_) ~+~NH_4NO_3_(_aq_) ->N_2_(_g_) ~+~H_2O_(_g_) ~+~Al_2O_3_(_S_) [/tex]

When we balance the reaction we will obtain:

[tex]2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)[/tex]

Now, for the enthalpy change, we have to find the standard enthalpy values:

[tex]Al_(_S_)=0~KJ/mol[/tex]

[tex]NH_4NO_3_(_a_q_)=-132.0~KJ/mol[/tex]

[tex]N_2_(_g_)=0~KJ/mol[/tex]

[tex]H_2O_(_g_)=~-~241.8~KJ/mol[/tex]

[tex]Al_2O_3_(_S_)=~-~1675.7~KJ/mol[/tex]

With this in mind, if we multiply the number of moles (in the balanced reaction) by the standard enthalpy value,  we can calculate the energy of the reagents:

[tex](0*2)~+~(-132*3)=~-396~KJ[/tex]

And the products:

[tex](0*3)~+~(-241.8*6)~+~(-1675.7*1)=-3125.9~KJ[/tex]

Finally, for the total enthalpy we have to subtract products by reagents :

[tex](-3125.9~KJ)-(-396~KJ)=-2729.9~KJ[/tex]

I hope it helps!

Answer:

6.9 Kg/mol•dL

Explanation:

To convert 6.9×10⁴ g/mol•L to kg/mol•dL,

First, we shall convert to kg/mol•L.

This can be achieved by doing the following:

Recall: 1 g = 1×10¯³ Kg

1 g/mol•L = 1×10¯³ Kg/mol•L.

Therefore,

6.9×10⁴ g/mol•L = 6.9×10⁴× 1×10¯³

6.9×10⁴ g/mol•L = 69 Kg/mol•L

Finally, we shall convert 69 Kg/mol•L to Kg/mol•dL.

This is illustrated below:

Recall: 1 L = 10 dL

1 Kg/mol•L = 1×10¯¹ Kg/mol•dL

Therefore,

69 Kg/mol•L = 69 × 1×10¯¹

69 Kg/mol•L = 6.9 Kg/mol•dL

Therefore, 6.9×10⁴ g/mol•L is equivalent to 6.9 Kg/mol•dL.

The correct answer is C. There is no oxygen in space, so there can be no combustion.

Explanation:

Fire and flames are the result of a chemical process known as combustion. Moreover, for combustion to occur there are two essential elements. The first one is a fuel or a substance that releases energy and ignites, and the second one is an oxidant, which accepts electrons. This mix and reaction causes high temperatures and release of heat in the form of fire and flames.

This implies, that for fireballs or any other form of fire to exist there must be oxygen or any substance that replaces it. This does not occur in space because the levels of oxygen are extremely low, this means, at least oxygen is added fireballs are not possible in this context as there is no oxygen, and therefore no combustion (Option C).

Answer:

C.) There is no oxygen in space, so there can be no combustion.

Explanation:

I got it correct on founders edtell

Answer:18.4 ML

Explanation:

easy add

Answer:

Silver ion - Lewis acid, Ammonia -  Lewis base

Explanation:

The reaction is given as;

Ag+(aq) + 2NH3(aq) ⇌ [Ag(NH3)]2+(aq)

A lewis acid is an electron pair acceptor. While a lewis base is any substance that that can donate a pair of nonbonding electrons.

This reaction however is a complexation reaction, where ammonia is reacting with the silver ion.

Silver ion accepts electrons in this reaction, hence it is the lewis acid. The ammonia on the other hand donates the electrons used in bonding so it is the lewis base.

Answer:

THE HEAT OF COMBUSTION IS 4995.69 kJ/mol OF OCTANE.

Explanation:

Heat capacity = 6.18 kJ/C

Temperature change = 41.5 C - 22.0 C = 19.5 C

Heat required to raise the temperature by 19.5 °C is:

Heat = heat capacity * temperature change

Heat = 6.18 kJ/ C * 19.5 C

heat = 120.51 kJ of heat

120.51 kJ of heat is required to raise the temperature of 2.75 g sample of  a liquid octane.

Molar mass of octane = ( 12* 8 + 1 * 18) = 114 g/mol

So therefore, the heat of the reaction per mole of octane will be:

120.51 kJ of heat is required for 2.75 g of octane

x J of heat will be required for 114 g of octane

x J = 120.51kJ * 114 / 2.75

x = 4995.69 kJ of heat per mole.

In conclusion, the heat of the combustion reaction in kJ / mole of octane is 4995.69 kJ/mol

Answer:

Explanation:

Hello,

Among the options given on the attached document, since phenolic functional group is characterized by a benzene ring bonded with a hydroxyl group (C₆H₅OH) we can see that the first option correctly points out such description. Thus, answer is on the second attached picture. Other options are related with other sections found in eugenol that are not phenolic.

Best regards.

The first option identified the phenol in eugenol.

According to the attached image, since the phenolic functional group should be characterized by a benzene ring bonded along with a hydroxyl group (C₆H₅OH) so here we can see that the first option correctly points out such description. However, other options are related to other sections found in eugenol that are not phenolic.

learn more about molecule here: https://brainly.com/question/13127022

Answer:

Bleaching Powder's chemical formula is CaOCl2 and is called Calcium Oxychloride. It is prepared on dry slaked lime by chlorine gas. 2. ... It gives calcium chloride, chlorine and water when bleaching powder reacts with hydrochloric acid.

Explanation:

Answer:

C

Explanation:

Turning liquid to a solid is like freezing water to ice and requires the water to LOSE (release) heat causing an exothermic reaction.

Answer:

Explanation:

CH₃COOH + NaOH = CH₃COONa + H₂O .

.02M

CH₃COOH  = CH₃COO⁻ + H⁺

C                       xC             xC

Ka = xC . xC / C = x² C

1.8 x 10⁻⁵ = x² . .02

x² = 9 x 10⁻⁴

x = 3 x 10⁻²

= .03

concentration of H⁺ = xC = .03 . .02

= 6 x 10⁻⁴ M , volume =  45 x 10⁻³ L

moles of H⁺  = 6 X 10⁻⁴  x 45 x 10⁻³

= 270 x 10⁻⁷ moles

= 2.7 x 10⁻⁵ moles

concentration of NaOH = .0200 M , volume = 35 x 10⁻³ L

moles of Na OH = 2 X 10⁻²  x 35 x 10⁻³

= 70 x 10⁻⁵ moles

=  

NaOH is a strong base so it will dissociate fully .

there will be neutralisation reaction between the two .

Net NaOH remaining = (70 - 2.7 ) x 10⁻⁵ moles

= 67.3 x 10⁻⁵ moles of NaOH

Total volume = 45 + 35 = 80 x 10⁻³

concentration of NaOH after neutralisation.= 67.3  x 10⁻⁵ / 80 x 10⁻³ moles / L

= 8.4125  x 10⁻³ moles / L

OH⁻ = 8.4125  x 10⁻³

H⁺ = 10⁻¹⁴ / 8.4125  x 10⁻³

= 1.1887 x 10⁻¹²

pH = - log (  1.1887 x 10⁻¹² )

= 12 - log 1.1887

= 12 - .075

= 11.925 .