A saturated solution of lead(II) iodide, PbI2 has an iodide concentration of 3.0 x 10^-3 mol/L.
a) What is the molar solubility of PbI2?
b) Determine the solubility constant, Ksp, for lead(II) iodide.
c) Does the molar solubility of lead (II) iodide increase, decrease, or remain unchanged with the addition of potassium iodide to the solution? EXPLAIN.

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 three gases, in the three identical containers, will all have the same number of molecules

Explanation:

If these three gases (Helium He, Neon Ne, and Oxygen [tex]O_{2}[/tex]) are all contained in separate identical containers with the same volume. And they are all stored at the same temperature, and pressure. Then, they'll all contain the same number of molecules. This is in line with Avogadro's law which states that "Equal volume of all gases, at the same temperature and pressure, have the same number of molecules."

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.

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:

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:

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:

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:

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) The diode voltage,  [tex]V_D =[/tex]  0.776 V

(b) The diode current, [tex]I_D =[/tex] 3.81 x 10⁻²⁰ A

Explanation:

Given;

saturation current in diode, [tex]I_s[/tex] = 10⁻¹⁷ A

nonideality factor, n = 1.05

(a) the diode voltage

Given diode current, [tex]I_D[/tex] = 70 μA = 7 x 10⁻⁶ A

Diode voltage is calculated as;

[tex]V_D = nV_Tln(1+ \frac{I_D}{I_S} )[/tex]

Where;

[tex]V_T[/tex] is thermal voltage at 25°C = 0.025

[tex]V_D = 1.05 * 0.025 ln(1+ \frac{70*10^{-6}}{1*10^{-17}})\\\\V_D = 0.02625ln(1+ 7*10^{12})\\\\V_D = 0.776 \ V[/tex]

b) the diode current for VD = 0.1 mV

[tex]V_D = nV_Tln(1 +\frac{I_D}{I_S} )\\\\ln(1 +\frac{I_D}{I_S} ) = \frac{V_D}{nV_T} \\\\ln(1 +\frac{I_D}{I_S} ) = \frac{0.1*10^{-3}}{1.05*0.025} \\\\ln(1 +\frac{I_D}{I_S} ) = 0.00381\\\\1 +\frac{I_D}{I_S} = e^{0.00381}\\\\1+ \frac{I_D}{I_S}= 1.00381\\\\ \frac{I_D}{I_S}=1.00381 - 1\\\\ \frac{I_D}{I_S}= 0.00381\\\\I_D = 0.00381(I_S)\\\\I_D = 0.00381(10^{-17})\\\\I_D = 3.81*10^{-20} \ A[/tex]

Answer:

[tex]Kp=0.0386[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2NO+O_2\rightleftharpoons 2NO_2[/tex]

For which the equilibrium expression is:

[tex]Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}[/tex]

Whereas, at equilibrium, each pressure is computed in terms of the initial pressure and the reaction extent via:

[tex]p_{NO_2}=2x\\p_{NO}=522-2x\\p_{O_2}=421-x[/tex]

And the total pressure:

[tex]p_{eq}=p_{NO_2}+p_{NO}+p_{O_2}\\\\p_{eq}=2x+522-2x+421-x\\\\p_{eq}=943-x[/tex]

Yet it is 748 torr, for which the extent is:

[tex]x=943-p_{eq}=943-748\\\\x=195torr[/tex]

Therefore, Kp turns out:

[tex]Kp=\frac{(2x)^2}{(522-2x)^2(421-x)}\\\\Kp=\frac{(2*195)^2}{(522-2*195)^2(421-195)}\\\\Kp=0.0386[/tex]

Best regards.

Answer:

Hey there!

That would be the alkaline earth metals.

Hope this helps :)

Answer: alkaline earth metals

Explanation:

Answer:

Your answer will either be 22.9897 or 22.990 !!

Explanation:

Answer:

Ionic Crystal

Explanation:

Complete question:

Write the condensed formula from left to right, starting with (CH3)x where x is a number.

See attached image for the structure formula of the compound

Answer:

(CH₃)₂CHC(CH₃)₃ named as 2,2,3-Trimethylbutane

Explanation:

If we number the longest chain of the carbon starting from the left, we will observe that there are four carbons in the straight chain as shown in the image.

Starting from first carbon from the left of the carbon chain, at carbon number number 2, there two alkyl group, that is two methyl (CH3 is two). Also at carbon number 3, there are three alkyl group, that is three methyl (CH3 is three).

The condensed formula will be written as;

(CH₃)₂CHC(CH₃)₃

This compound is named as 2,2,3-Trimethylbutane, an isomer of Heptane

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 .

Answer:

Rubidium

Explanation:

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:

A and C

I hope this helps you:)

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:

Option A. 2096.1 K

Explanation:

The following data were obtained from the question:

Al2O3(s) + 3C(s) —> 2Al(s) + 3CO(g)? Enthalpy (H) = +1287 kJ mol¯¹

Entropy (S) = +614 JK¯¹ mol¯¹

Temperature (T) =...?

Entropy, enthalphy and temperature are related by the following equation:

Change in Entropy (ΔS) = Change in Enthalphy (ΔH) /Temperature (T)

ΔS = ΔH/T

With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:

Enthalpy (H) = +1287 kJ mol¯¹ = 1287000 Jmol¯¹

Entropy (S) = +614 JK¯¹mol¯¹

Temperature (T) =...?

ΔS = ΔH/T

614 = 1287000/ T

Cross multiply

614 x T = 1287000

Divide both side by 614

T = 1287000/614

T = 2096.1 K

Therefore, the temperature at which the reaction will be feasible is 2096.1 K.

Answer:

Option (b) Hydrocyanic acid, 4.9×10^-10

Explanation:

Data obtained from the question include:

Ka of Hydrofluoric acid = 3.5×10^-4

Ka of Hydrocyanic acid = 4.9×10^-10

Ka of Nitrous acid = 4.6×10^-4

To know which acid is least acidic, we shall determine the the pKa value for each acid.

This is illustrated below:

For Hydrofluoric acid

Ka = 3.5×10^-4

pKa =..?

pKa = –Log Ka

pKa = –Log 3.5×10^-4

pKa = 3.5

For Hydrocyanic acid

Ka = 4.9×10^-10

pKa =..?

pKa = –Log Ka

pKa = –Log 4.9×10^-10

pKa = 9.3

For Nitrous acid

Ka = 4.6×10^-4

pKa =..?

pKa = –Log Ka

pKa = –Log 4.6×10^-4

pKa = 3.3

Summary:

Acid >>>>>>>>>>>>> Ka >>>>>>>> pKa

Hydrofluoric acid >> 3.5×10^-4 >> 3.5

Hydrocyanic acid >> 4.9×10^-10 > 9.3

Nitrous acid >>>>>>> 4.6×10^-4 >> 3.3

NB: The smaller the pKa value, the more acidic the compound is and the larger the pKa value, the less acidic the compound will be.

From the above calculations, Hydrocyanic acid has the highest pKa value.

Therefore, Hydrocyanic acid is the least acidic compound

Answer:

Approximately [tex]1.854\; \rm mol\cdot L^{-1}[/tex].

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

Look up the relative atomic mass of [tex]\rm Sr[/tex], [tex]\rm O[/tex], and [tex]\rm H[/tex] on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

Calculate the formula mass of [tex]\rm Sr(OH)_2[/tex]:

[tex]M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}[/tex].

[tex]M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}[/tex] means that each mole of [tex]\rm Sr(OH)_2[/tex] formula units have a mass of [tex]121.634\; \rm g[/tex].

The question states that there are [tex]10.60\; \rm g[/tex] of [tex]\rm Sr(OH)_2[/tex] in this solution.

How many moles of [tex]\rm Sr(OH)_2[/tex] formula units would that be?

[tex]\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}[/tex].

There are [tex]8.71467\times 10^{-2}\; \rm mol[/tex] of [tex]\rm Sr(OH)_2[/tex] formula units in this [tex]47\; \rm mL[/tex] solution. Convert the unit of volume to liter:

[tex]V = 47\; \rm mL = 0.047\; \rm L[/tex].

The molarity of a solution measures its molar concentration. For this solution:

[tex]\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

(Rounded to four significant figures.)

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:

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:

Partial pressure (Ar) = 316.1mmHg

Explanation:

In the mixture of Ar and H₂ you can find the total moles of both gases using general gas law and with the mass of the sample and molar weight of each gas find the mole fraction of Argon and thus, its partial pressure.

Moles of gases:

PV = nRT

P = 910.0mmHg ₓ (1atm / 760mmHg) = 1.1974atm

V = 1.66L

n = Moles gases

R = 0.082atmL/molK

T = 54.9°C + 273.15K = 328.05K

PV = nRT

1.1974atm*1.66L = n*0.082atmL/molK*328.05K

Knowing H₂ = 2.016g/mol and Ar = 39.948g/mol you can write:

1.13g = 2.016X + 39.948Y (1)

Where X = moles of hydrogen and Y = moles of Argon.

Also we can write:

0.0739moles = X + Y (2)

Total moles of the sample are moles of hydrogen + moles Argon

Replacing 2 in 1:

1.13g = 2.016(0.0739-Y) + 39.948Y

1.13 = 0.1564 - 2.016Y + 39.948Y

0.9736 = 37.932Y

0.02567 = Y = moles of Argon

As total moles are 0.0739moles, mole fraction of Ar in the sample are:

XAr = 0.02567mol / 0.0739mol

X Ar = 0.347

Last, partial pressure of Ar = X Ar * total pressure.

Partial pressure (Ar) = 0.347*910.0mmHg

Answer:

The correct answer is 2.75 grams of HCl.

Explanation:

The given balanced equation is:  

CaCO₃ (s) + 2HCl (aq) ⇒ CaCl₂ (aq) + H₂O (l) + CO₂ (g)

Based on the given information, one mole of calcium carbonate is reacting with two moles of HCl. The molecular mass of HCl is 36.5 grams, thus, the mass of 2 moles of HCl will be, 36.5 × 2 = 73 grams

The molecular mass of CaCO₃ is 100 gram per mole, that is, the mass of 1 mole of CaCO₃ is 100 grams, therefore, the mass of HCl required for reacting with 3.75 grams of CaCO₃ will be,  

= 3.75 × 2 × 36.5 / 100 = 2.74 grams of HCl.