Heterocyclic aromatic compounds undergo electrophilic aromatic substitution in a similar fashion to that undergone by benzene with the formation of a resonance-stabilized intermediate. Draw all of the resonance contributors expected when the above compound undergoes bromination

Answer:

Molar solubility is 1.12x10⁻⁴M

Explanation:

The dissolution of magnesium hydroxide is:

Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻

The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:

Mg(OH)₂(s) ⇄ X + 2X

Where X is solubility.

If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:

2X = 2.24x10⁻⁴M

X = 2.24x10⁻⁴M/2

X =1.12x10⁻⁴M

Answer:

See explanation

Explanation:

The reaction of chlorine with the pictured compound will occur via free radical mechanism. The stability of the free radical formed will depend on its structure.

The order of stability of free radicals is methyl < primary < secondary < tertiary. Hence a tertiary carbon free radical is the most stable.

Looking at the compound, the radical will form at the position shown in the image attached since it will lead to a secondary free radical which is more stable.

The structure that should be drawn is shown below.

It should be within the pictured compound that will arise via a free radical mechanism. The stability should be based on the structure. The stability of the order of free radicals should be methyl < primary < secondary < tertiary. Thus, a tertiary carbon free radical should be most stable.

Here look at the compound, the radical should form at the position that should be shown in the image that resulted in the secondary free radical i.e. more stable.

Answer:

Atomic radius less than 15%, similar structure and same valency.

Explanation:

The conditions that are favorable for extensive solid solubility of one element in another are the following.

The atomic radius of the solute and solvent atoms must be less than 15%. The structure of both solute and solvent are similar. Solubility completes when both have same valency. Valency means number of electrons in the outermost shell. If both solute and solvent has same number of electrons so it will be completely soluble in each other.

The conditions that are favorable for extensive solid solubility of one element in another is the same size, electrongativity and valency.

Hume - Rothery rules are the sets of some important rules which gives idea about the desired condition for the formation of solid solution.

Following main points are described in this rule:

Hence size, electronegativity and valency are the conditions.

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Answer:

1.05 V

Explanation:

Since;

E°cell= E°cathode- E°anode

E°cathode= -0.40 V

E°anode= -1.45 V

E°cell= -0.40-(-1.45) = 1.05 V

Equation of the process;

2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)

n= 8 electrons transferred

From Nernst's equation;

Ecell = E°cell - 0.0592/n log Q

Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]

Since log 1=0

Ecell= E°cell= 1.05 V

Answer:

D

Explanation:

Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.

The action that destroys the buffer is option c. adding 0.050 moles of HCl.

It is a solution of a weak acid and salt.

Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.

The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this,  there will be only acid in the solution.

Since

moles of HC2H3O2 = 1*0.250 = 0.250

moles of NaC2H3O2 = 1*0.050 = 0.050.

moles of HCl is added = 0.050

Now

The reaction between HCl and NaC2H3O2

[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]

Now

BCA table is

            NaC2H3O2  HCl       HC2H3O2

Before 0.050 0.050 0.250

Change -0.050 -0.050 +0.050

After 0 0 0.300

Now, the solution contains the acid (HC2H3O2 ) only.

Therefore addition of 0.050 moles of HCl will destroy the buffer.

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Answer:

Explanation:

Sr(OH)2 (aq) ⇔ Sr+2 (aq) + 2OH- (aq)

Answer:

In 5 US liquid gallons, there are 80 cups.

Explanation:

To get from gallons to cups, just multiply the amount of gallons you have by 16.

Answer:

[tex]SF_4[/tex]

Explanation:

The first thing would be to calculate the number of moles of each element in the compound.

No of moles of sulfur (S) = mass/molar mass

                                        = 29.68/32.065 = 0.9256

No of moles of fluorine (F) = mass/molar mass

                                       = 70.32/18.998 = 3.7014

Then, let us find the atomic ratio of each of the element in the compound by dividing by the no of moles by the smallest no of mole:

            S            :             F

      [tex]\frac{0.9256}{0.9256}[/tex] = 1       :       [tex]\frac{3.7014}{0.9256}[/tex] = 4

Therefore, the empirical formula for the compound is [tex]SF_4[/tex]

Answer:

D. Pure gold

Explanation:

Hello,

In this case, since gold, as a heavy metal, is said to be less reactive than hydrogen, when it undergoes electrolysis process when forming a salt, due to the action of the electric current, we can appreciate the formation of a layer of gold on the surface of the cathode via a reduction half-reaction from gold (I) to metallic gold:

[tex]Au^++1e^-\rightarrow Au^0[/tex]

Thereby, D. Pure gold is formed as the product of the reaction.

In contrast, more reactive metals than hydrogen such as sodium or potassium, will remain in solution so the hydrogen converted to hydrogen gas.

Best regards-

Answer:

b

Explanation:

Provided the experiment is about the transfer of heat from one medium to another, any residual hot water on the metal would have added to the heat transferred from the metal to the cold water system.

The residual hot water on the metal posses its own heat and when such is transferred along with the metal into the cold water, the heat from the residual hot water will interfere with the measurement of the actual heat transferred to the cold water by the metal. Hence, the accuracy of the result would be impacted.

The correct option is b.

Answer:

A divergent plate boundary  

Explanation:

At a divergent boundary, the plates pull away from each other and generate new crust.

Answer:

Because the ocean becomes larger, this is a divergent plate boundary. Divergent plates cause the ocean floor to expand, making the ocean larger.

Explanation:

PLATO ANSWER

Answer:

[tex]m_{PtBr_4}=0.306gPtBr_4[/tex]

Explanation:

Hello,

In this case, since the solubility product of platinum (IV) bromide is 8.21x10⁻⁹, and the dissociation is:

[tex]PtBr_4(s)\rightleftharpoons Pt^{4+}(aq)+4Br^-(aq)[/tex]

The equilibrium expression is:

[tex]Ksp=[Pt^{4+}][Br^-]^4[/tex]

Thus, since the salt is added to a solution initially containing 1.00 grams of potassium bromide, there is an initial concentration of bromide ions:

[tex][Br^-]_0=\frac{1.00gKBr*\frac{1molKBr}{119gKBr}*\frac{1molBr^-}{1molKBr} }{0.250L}=0.0336M[/tex]

Hence, in terms of the molar solubility [tex]x[/tex], we can write:

[tex]8.21x10^{-9}=(x)(0.0336+4x)^4[/tex]

In such a way, solving for [tex]x[/tex], we obtain:

[tex]x=0.00238M[/tex]

Which is the molar solubility of platinum (IV) bromide. Then, since its molar mass is 514.7 g/mol, we can compute the grams that get dissolved in the 250.0-mL solution:

[tex]m_{PtBr_4}=0.00238\frac{molPtBr_4}{1L}*0.250L *\frac{514.7gPtBr_4}{1molPtBr_4} \\\\m_{PtBr_4}=0.306gPtBr_4[/tex]

Best regards.

Answer:

See the answer below

Explanation:

The complete question can be seen in the attached image.

Phenolphthalein is an indicator that is often utilized in an acid-base reaction to indicate the endpoints of such reactions due to its ability to change color from pink/colorless to colorless/pink depending on if the final solution is acidic or basic.

Phenolphthalein is usually colorless in acidic solutions and appears pink in basic solutions. The more basic or alkaline a solution is, the stronger the pink color of phenolphthalein. Hence;

1. Ammonia with a pH of 11 is basic, phenolphthalein will turn pink.

2. Battery acid with a pH of 1 is acidic, it will remain colorless.

3. Lime juice with a pH of 2 is acidic, it will remain colorless.

4. Mashed avocado with a pH of 6.5 is acidic, it will remain colorless.

5. Seawater with a pH of 8.5 is basic, it will turn pink.

6. Tap water with a pH of 7 is neutral, it will remain colorless

Phenolphthalein is a chemical compound with the formula[tex]C_{20}H_{14}O_4[/tex]. Phenolphthalein is often used as an indicator in acid-base titrations. For this application, it turns colorless in acidic solutions and pink in basic solutions

Phenolphthalein works as in:-

According to the question, There are 5 solutions having different ph and the indication only turns basic solution to pink

The indicator only turn the basic solution pink and these solutions are as follows,

Hence, these are the answer.

For more information, refer to the link:-

Answer:

Each chlorine atom shares a pair of electrons with the sulfur atom. ... Each chlorine atom shares all its valence electrons with the sulfur atom

Answer:

[tex]V_{HCl}=0.208L=208mL[/tex]

Explanation:

Hello,

In this case, since the chemical reaction is:

[tex]2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O[/tex]

We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:

[tex]n_{HCl}=2*n_{Mg(OH)_2}[/tex]

In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:

[tex]n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g} =0.00858mol[/tex]

Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:

[tex][H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M[/tex]

Then, since the concentration and the volume define the moles, we can write:

[tex][HCl]*V_{HCl}=2*n_{Mg(OH)_2}[/tex]

Therefore, the neutralized volume turns out:

[tex]V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL[/tex]

Best regards.

Answer and Explanation:

a. The equation of K of this reaction is shown below:-

3 A + 5 B + 4 C↔5 D + 7 E + F

[tex]K = \frac{(D)^5 (E)^7 (F)}{(A)^3 (B)^5 (C)^4}[/tex]

b. The moles of compound F is shown below:-

3 A + 5 B + 4 C↔5 D + 7 E + F

               2 moles

Now, the mole of produced is

[tex]= \frac{1}{4} \times \ moles\ of\ c[/tex]

Now, we will the value of c by using the above equation

[tex]= \frac{1}{4} \times 2[/tex]

After solving the above equation we will get

0.5 moles

Answer:

550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M

Explanation:

It is possible to find the pH of a buffer by using H-H equation:

pH = pKa + log [A⁻]/[HA]

For the formic buffer (HCOOH/HCOONa):

pH = 3.74 + log [HCOONa]/[HCOOH]

As you need a buffer of pH 3.65:

pH = 3.74 + log [HCOONa]/[HCOOH]

3.65 = 3.74 + log [HCOONa]/[HCOOH]

Where  [HCOONa]/[HCOOH] can be taken as the moles of each specie.

As molarity of both solutions is 0.10M (0.10mol / L) and you need 1L of solution, total moles of the buffer are:

Replacing (2) in (1):

0.81283 = 0.10 - [HCOOH]  /[HCOOH]

0.81283[HCOOH] = 0.10 - [HCOOH]

1.81283[HCOOH] = 0.10

And moles of HCOONa are:

[HCOONa] = 0.1 mol - 0.055mol =

As concentration of the solutions is 0.1M, the volume you need to add of both solutions is:

HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M

HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M

The number should be considered like 550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M.

Here we used the H-H equation:

pH = pKa + log [A⁻]/[HA]

Now

For the formic buffer (HCOOH/HCOONa):

So,

pH = 3.74 + log [HCOONa]/[HCOOH]

Now

need a buffer of pH 3.65:

So,

pH = 3.74 + log [HCOONa]/[HCOOH]

3.65 = 3.74 + log [HCOONa]/[HCOOH]

0.81283 = [HCOONa]/[HCOOH] (1)

here  [HCOONa]/[HCOOH] can be considered as the moles of each specie.

Now the total moles should be

0.10 moles = [HCOONa] + [HCOOH] (2)

Now

0.81283 = 0.10 - [HCOOH]  /[HCOOH]

0.81283[HCOOH] = 0.10 - [HCOOH]

1.81283[HCOOH] = 0.10

[HCOOH] = 0.055 moles

And moles of HCOONa should be

[HCOONa] = 0.1 mol - 0.055mol =

[HCOONa] = 0.045 moles

Now

HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M

HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M

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Answer:

68%

Explanation:

Since we need a percentage we can use any number we want for our initial value.

5(1/2)^900/1620 = 3.40

(3.40 / 5)*100 = 68%

To make sure lets use a different initial amount

1(1/2)^900/1620 = 0.68

(0.68/1) * 100 = 68%

To solve this question, we'll assume the initial amount of radium-226 to be 1.

Now, we shall proceed to obtaining the percentage of radium-226 that will after 900 years. This can be obtained as illustrated below:

Determination of the number of half-lives that has elapsed.

Half-life (t½) = 1620 years

Time (t) = 900 years

[tex]n = \frac{t}{t_{1/2}}\\\\n = \frac{900}{1620}\\\\n = \frac{5}{9}[/tex]

Determination of the amount remaining

Initial amount (N₀) = 1

Number of half-lives (n) = 5/9

[tex]N = \frac{N_{0} }{2^{n}}\\\\N = \frac{1}{2^{5/9}}[/tex]

Determination of the percentage remaining.

Initial amount (N₀) = 1

Amount remaining (N) = 0.68

Percentage remaining = N/N₀ × 100

Percentage remaining = 0.68/1 × 100

Therefore, the percentage amount of radium-226 that remains after 900 years is 68%

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Answer:

The correct answer is 0.25 atm.

Explanation:

As mentioned in the given question that the chamber comprises equal molar concentrations of He, Ne, Ar and Kr gas. So, let us assume that the moles of all the gases will be x then the total number of moles will be 4x.  

The formula for calculating mole fraction is,  

Mole fraction = mole of the substance/total moles  

The mole fraction of Kr = x/4x = 0.25

The total pressure given in the chamber is 1 atm. Therefore, the partial pressure will be,  

Partial pressure = mole fraction * Total pressure

Partial pressure = 0.25 * 1 = 0.25 atm.  

The partial pressure of Kr is 0.25 atm.

We are told that there is an equal molar amounts of He, Ne, Ar, and Kr. If we decide to say, let the molar amount of each gas be x, the total number of moles  of all the gases will now be; x + x+ x + x = 4x

Let us recall that partial pressure of a gas can be obtained form the formula;

Partial pressure = mole fraction × total pressure

Mole fraction of Kr = x/4x = 1/4

Total pressure = 1 atm

Partial pressure of Kr = 1/4 × 1 = 0.25 atm

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Answer:

1. The balanced molecular equation is given below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

2. The net ionic equation is given below:

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

Explanation:

1. The balanced molecular equation

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

The above equation can be balance as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NaNO3 as shown below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

Now, the equation is balanced.

2. The bal net ionic equation.

This can be obtained as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —>

In solution, Na2SO4 and Ba(NO3)2 will dissociate as follow:

Na2SO4(aq) —> 2Na^+(aq) + SO4^2-(aq)

Ba(NO3)2(aq) —> Ba^2+(aq) + 2NO3^-(aq)

Na2SO4(aq) + Ba(NO3)2(aq) —>

2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) —> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)

Cancel the spectator ions i.e Na^+ and NO3^- to obtain the net ionic equation.

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

Answer:

1)Ni=1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d10 called full-filled

2)Cr=1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5 called half-filled

Answer:

Exceed the buffer capacity and Raise the pH by several units

Explanation:

Options are:

Raise the pH slightly

Lower the pH slightly

Raise the pH by several units

Lower the pH by several units

Not change the pH

Exceed the buffer capacity

The hypochlorous acid, HClO, is in equilibrium with Hypochlorite ion (From potassium hypochlorite, ClO⁻) producing a buffer. Using H-H equation, pH of initial buffer is:

pH = pKa + log [ClO⁻] / [HClO]

pKa for hypochlorous acid is 7.53

pH = 7.53 + log [0.581M] / [0.436M]

pH = 7.65

Barium hydroxide reacts with HClO producing more ClO⁻, thus:

Ba(OH)₂ + 2HClO →  2ClO⁻ + 2H₂O

As 0.479 moles of Barium hdroxide are added. For a complete reaction you require 0.479mol * 2 = 0.958 moles of HClO

As you have just 0.436 moles (Volume = 1L),

The addition will:

The Ba(OH)₂ that reacts is:

0.436 moles HClO * (1mole (Ba(OH)₂ / 2 mol HClO) = 0.218 moles Ba(OH)₂ and will remain:

0.479 mol - 0.218 mol = 0.261 moles Ba(OH)₂

As 1 mole of Ba(OH)₂ contains 2 moles of OH⁻, moles of OH⁻ and molarity is:

0.261 moles* 2 = 0.522 moles OH⁻ = [OH⁻]

pOH = -log [OH⁻]

pOH = 0.28

And pH = 14 - pOH:

pH = 13.72

Thus, after the addition the pH change from 7.65 to 13.62:

Answer:

Explanation:

Answer:

1, 1, 1, 1  

Explanation:

potassium hydrogen sulfate + potassium hydroxide ⟶ potassium sulfate + water(l)

                 KHSO₄                   +               KOH              ⟶            K₂SO₄   + H₂O

1. Put a 1 in front of the most complicated-looking formula (K₂SO₄?):

KHSO₄  + KOH ⟶ 1K₂SO₄ + H₂O

2. Balance S:

We have fixed 1 S on the right. We need 1 S on the left. Put a 1 in front of KHSO₄ to fix it.

1KHSO₄  + KOH ⟶ 1K₂SO₄ + H₂O

3. Balance K:

We have fixed 2 K on the right and 1 K on the left. We need 1 more K on the left. Put a 1 in front of KOH.

1KHSO₄  + 1KOH ⟶ 1K₂SO₄ + H₂O

4. Balance O

We have fixed 4 O on the right and 5 O on the left. We need 1 more O on the right. Put a 1 in front of H₂O.

1KHSO₄  + 1KOH ⟶ 1K₂SO₄ + 1H₂O

Every formula has a coefficient. The equation should be balanced.

5. Check that atoms balance:

[tex]\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{K} & 2 &2\\\text{H} & 2 & 2\\\text{S} & 1 & 1\\\text{O}&5&5\\\end{array}[/tex]

It checks.

The coefficients are 1, 1, 1, 1.

Answer:

[tex]0.033g[/tex]

Explanation:

Hello,

In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:

[tex]11mg\longrightarrow 1kg\\\\x\ \ \ \ \ \ \longrightarrow 3kg[/tex]

Thus, we obtain:

[tex]x=\frac{3kg*11mg}{1kg}\\ \\x=33mg[/tex]

That in grams is:

[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]

Regards.

Answer:

Cu2 + 2Mg-> 2Cu+ Mg2

Explanation:

Balance the equation and make sure both the reactant and the products are the same

Hope it will be helpful

[tex]Cu^{+2} + 2Mg[/tex]  -> [tex]2Cu + Mg^+2[/tex]  is the correct half-reactions.

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.

[tex]Cu^{+2} + 2Mg[/tex]  -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.

Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.

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Answer:

B. 4.52 X10-9 M

Explanation:

Our goal for this question is to calculate the concentration of hydronium ions [tex]H^+[/tex] produced by water in a vessel with a concentration of hydroxide ions of [tex]2.21X10^-^6~M[/tex]. So, our first approach can be the ionization reaction of water:

[tex]H_2O_(_l_)~->~H^+~_(_a_q_)~+~OH^-~_(_a_q_)[/tex]

If we write the Keq expression for this reaction we will have:

[tex]Keq=[H^+][OH^-][/tex]

Now, water is the universal solvent, so, Keq has a special name. In the equilibrium problems for water we have to use "Kw" instead of "Keq":

[tex]Kw=[H^+][OH^-][/tex]

From this equation, we know the Kw value () and the concentration of the hydroxide ions ([2.21X10^-^6~M]). If we replace these values into the equation we can solve for [tex][H^+][/tex]:

[tex]1.0X10^-^1^4=[H^+][2.21X10^-^6~M][/tex]

[tex][H^+]=\frac{1.0X10^-^1^4}{2.21X10^-^6}=4.52X^-^9[/tex]

I hope it helps!

Answer:

B. 4.52 * 10^-9M

Explanation:

did the test

Answer:

[tex]m_{NaNH_2}=30.42gNaNH_2[/tex]

[tex]m_{H_2}=0.783gH_2[/tex]

Explanation:

Hello,

In this case, the reaction between sodium and ammonia is:

[tex]2Na+2NH_3\rightarrow 2NaNH_2+H_2[/tex]

Thus, as we know the initial masses of both sodium and ammonia, we should first identify the limiting reactant, for which we firstly compute the available moles of sodium:

[tex]n_{Na}=18.0gNa*\frac{1molNa}{23.0gNa}=0.783molNa[/tex]

And the moles of sodium consumed by 21.4 g of ammonia (2:2 mole ratio):

[tex]n_{Na}^{\ consumed}=21.4gNH_3*\frac{1molNH_3}{17gNH_3} *\frac{2molNa}{2molNH_3} =1.26molNa[/tex]

In such a way, since less moles of sodium are available than consumed by ammonia, we can say, sodium is the limiting reactant. Furthermore, the mass of both sodium amide (39 g/mol) and hydrogen gas (2 g/mol) that are produced turn out:

[tex]m_{NaNH_2}=0.783molNa*\frac{2molNaNH_2}{2molNa}*\frac{39gNaNH_2}{1molNaNH_2}=30.42gNaNH_2[/tex]

[tex]m_{H_2}=0.783molNa*\frac{1molH_2}{2molNa}*\frac{2gH_2}{1molH_2}=0.783gH_2[/tex]

Best regards.

Answer:

2.8x10⁻¹⁵ M.

Explanation:

Hello,

In this case, the dissociation reaction for silver phosphate is:

[tex]Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4(aq)[/tex]

Therefore, the equilibrium expression is:

[tex]Ksp=[Ag^+]^3[PO_4^-][/tex]

In such a way, since the initial solution contains an initial concentration of silver ions (from silver nitrate) of 0.10M, we can write the equilibrium expression in terms of the reaction extent [tex]x[/tex]:

[tex]2.8x10^{-18}=(0.10+3x)^3*(x)[/tex]

Thus, solving for [tex]x[/tex] we have:

[tex]x=2.8x10^{-15}M[/tex]

Thus, the molar solubility of silver phosphate is 2.8x10⁻¹⁵ M.

Regards.

Answer:

pH of the buffer is 7.33

Explanation:

The mixture of the ions H₂PO₄⁻ and HPO₄²⁻ produce a buffer (The mixture of a weak acid, H₂PO₄⁻, with its conjugate base, HPO₄²⁻).

To find pH of a buffer we use H-H equation:

pH = pka + log [A⁻] / [HA]

Where A⁻ is conjugate base and HA weak acid.

For the H₂PO₄⁻ and HPO₄²⁻ buffer:

pH = pka + log [HPO₄²⁻] / [H₂PO₄⁻]

Computing values of the problem:

pH =7.21 + log [0.125M] / [0.095M]

pH = 7.33

Answer:

CH3C-NH+> CH3CH=NH2 >CH3CH2NH3+

Explanation:

The acid strength has to do with the ease of the loss of hydrogen ion from the cationic specie. Hydrogen ion will easily be lost from any specie which contains an atom, group of atoms or bond which withdraws electrons along the chain of the N-H bond.

The pi bond system is known to be highly electronegative and withdraws electrons along the chain hence a withdrawal of electron density along the chain which makes the hydrogen ion to be easily lost from a system which contains a pi bond along the chain. A triple bond is more electronegative than a double bond, hence the answer above.