Classify each of these reactions.
1) Ba(ClO3)2(s)--->BaCl2(s)+3O2(g)
2) 2NaCl(aq)+K2S(aq)--->Na2S(aq)+2KCl(aq)
3) CaO(s)+CO2(g)--->CaCO3(s)
4) KOH(aq)+AgCl(aq)---->KCl(aq)+AgOH(s)
5) Ba(OH)2(aq)+2HNO2(aq)--->Ba(NO2)2(aq)+2H2O(l)
Each classify reaction should be either one of this.
a. acid-base neutralization
b. precipitation
c. redox
d. none of the above

Answer:

Based on the given reaction, it is evident that the reaction is endothermic as indicated by a positive sign of enthalpy of reaction. Thus, it can be stated that the favoring of the forward reaction will take place by upsurging the temperature of the reaction mixture.  

Apart from this, based on Le Chatelier’s principle, any modification in the quantity of any species is performed at equilibrium and the reaction will move in such an orientation so that the effect of the change gets minimized. Therefore, a slight enhancement in the concentration of the reactant will accelerate the reaction in the forward direction and hence more formation of the product takes place.  

Answer: Thus the cell potential of an electrochemical cell is +0.28 V

Explanation:

The calculation of cell potential is done by :

[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

[tex]E^0_{[Fe^{2+}/Fe]}= -0.41V[/tex]

[tex]E^0_{[Pb^{2+}/Pb]}=-0.13V[/tex]

As Reduction takes place easily if the standard reduction potential is higher(positive) and oxidation takes place easily if the standard reduction potential is less(more negative). Thus iron acts as anode and lead acts as cathode.

[tex]E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Fe^{2+}/Fe]}[/tex]

[tex]E^0=-0.13- (-0.41V)=0.28V[/tex]

Thus the cell potential of an electrochemical cell is +0.28 V

Answer:

empirical formula = C3H7

molecular formula = C6H14

Answer:

C. They react strongly with water to produce an alkaline solution and hydrogen

Explanation:

All alkali metals react vigorously with cold water. In the reaction, hydrogen gas is given off and the metal hydroxide is produced.

Hope that helps.

Answer:

OH⁻ is the Bronsted-Lowry base.

Explanation:

A Bronsted-Lowry base is a substance that accepts protons. In the chemical equation, OH⁻ accepts a proton from H₂PO₄⁻ to become H₂O. H₂PO₄⁻ would be a Bronsted-Lowry acid because it donates a proton to OH⁻ and becomes HPO₄²⁻.

Hope that helps.

Answer:

Kc = [H₂CO₃] / [CO₂]

Explanation:

Equilibrium constant expression (Kc) of any reaction is defined as the ratio between molar concentrations in equilibrium of products over reactants.

Pure solids and liquids don't affect the equilibrium and you don't have to take its concentrations in the equilibrium.

Also, each specie must be powered to its reactant coefficient.

For example, for the reaction:

aA(s) + bB(aq) ⇄ cC(l) + nD(g) + xE(aq)

The equilibrium constant, kc is:

Kc = [D]ⁿ / [B]ᵇ[E]ˣ

You don't take A nor C species because are pure solids and liquids. b, n and x are the reactant coefficients of each substance. Ratio of products over reactants

Thus, for the reaction:

CO₂(aq) + H₂O(l) ⇄ H₂CO₃(aq)

The Kc is:

Answer:

0.1110 mol

Explanation:

Mass = 2g

Molar mass = 18.016 g/mol

moles = ?

These quantities are realted by the following equation;

Moles = Mass / Molar mass

Substituting the values of the quantities and solving for moles, we have;

Moles = 2 / 18.016 = 0.1110 mol

Answer:

The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.

Explanation:

The decomposition of the compound has an exponential behavior and process can be represented by this linear first-order differential equation:

[tex]\frac{dc}{dt} = -\frac{1}{\tau}\cdot c(t)[/tex]

Where:

[tex]\tau[/tex] - Time constant, measured in seconds.

[tex]c(t)[/tex] - Concentration of the compound as a function of time.

The solution of the differential equation is:

[tex]c(t) = c_{o} \cdot e^{-\frac{t}{\tau} }[/tex]

Where [tex]c_{o}[/tex] is the initial concentration of the compound.

The time is now cleared in the result obtained previously:

[tex]\ln \frac{c(t)}{c_{o}} = -\frac{t}{\tau}[/tex]

[tex]t = -\tau \cdot \ln \frac{c(t)}{c_{o}}[/tex]

Time constant as a function of half-life is:

[tex]\tau = \frac{t_{1/2}}{\ln 2}[/tex]

Where [tex]t_{1/2}[/tex] is the half-life of the composite decomposition, measured in seconds.

If [tex]t_{1/2} = 8\,s[/tex], then:

[tex]\tau = \frac{8\,s}{\ln 2}[/tex]

[tex]\tau \approx 11.542\,s[/tex]

And lastly, given that [tex]\frac{c(t)}{c_{o}} = \frac{1}{9}[/tex] and [tex]\tau \approx 11.542\,s[/tex], the time taken for the concentration to decrease to one-ninth of its initial value is:

[tex]t = -(11.542\,s)\cdot \ln\frac{1}{9}[/tex]

[tex]t \approx 25.360\,s[/tex]

The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.

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:

-471 Kj/mole acrylic acid

Explanation:

THIS IS THE COMPLETE QUESTION BELOW;

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In the first step, calcium carbide and water react to form acetylene and calcium hydroxide: CaC (s) + 2 H2O(g) - CH (9) + Ca(OH),(s) AH -414. kJ In the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6 C H (9) + 3 CO2(9) + 4H2O(g) - SCH,CHCO,H) AH-132. kJ Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ?

The two equations from the reaction can be written as;

a)CaC₂(s) + 2H₂O(l) ------->C₂H₂(g) + CaOH₂(s)

Δ H= -414Kj ........................ equation (a)

b)6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj ...................... equation (b)

In equation (b)acrylic acid was produced by the reaction between Acetylene carbon dioxide and water

Then we can multiply equation(a) by factor of 6 and the ΔH Then we have (6× -414Kj)= ΔH= -2484Kj.

6CaC₂(s) + 12H₂O(l) ------->6C₂H₂(g) + 6CaOH₂(s)

Δ H= -2484Kj.................. equation (c)

6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj

Then add equation (c) and equation(b) then we have

6CaC₂(s) + 16H₂O(l)+3CO₂(g)------> 5CH₂CHCO₂H(g) + 6CaOH₂(s) ΔH= -2352Kj

ΔH(net)= -2352Kj/5moles

=-471Kj/mole

therefore, net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ? is -471Kj/mole acrylic acid

Answer:

KNO₃ - Potassium nitrate

Fe(OH)₃ - Iron(III) hydroxide

H₂SO3 - Sulfurous acid

Al₂(SO₄)₃ - Aluminum sulfate

Hope that helps.

Answer:

[tex]m_{MgO}=0.239gMgO[/tex]

Explanation:

Hello,

In this case, the chemical reaction between magnesium and oxygen to yield magnesium oxide is:

[tex]2Mg+O_2\rightarrow 2MgO[/tex]

In such a way, for 0.144 g of magnesium reacting with sufficient oxygen, the mass of magnesium oxide, whose molar mass is 40.3 g/mol (2:2 mole ratio) turns out:

[tex]m_{MgO}=0.144gMg*\frac{1molMg}{24.3gMg} *\frac{2molMgO}{2molMg}* \frac{40.3gMgO}{1molMgO} \\\\m_{MgO}=0.239gMgO[/tex]

Best regards.

Explanation:

This reaction is in equilibrium and would hence obey lechatelier's principle. This principle states that whenever a system at equilibrium undergoes a change, it would react in way so as to annul that change.

Since it is an endothermic reaction, increasing the temperature would cause the reaction to shift towards the right.

This means that it favours product formation and more of the product would be formed.

Answer:

[tex][Cl^-]=232.3\frac{mgCl^-}{L}[/tex]

Explanation:

Hello,

In this case, we can represent the chemical reaction as:

[tex]Cl^-(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NO_3^-(aq)[/tex]

In such a way, since the mass of the obtained silver chloride is 93.9 mg, we can compute the chloride ions in the ground water by using the following stoichiometric procedure whereas the molar mass of chloride ions and silver chloride are 35.45 g/mol and 143.32 g/mol respectively:

[tex]m_{Cl^-}=93.3mgAgCl*\frac{1mmolAgCl}{143.32mgAgCl}*\frac{1mmolCl^-}{1mmolAgCl} *\frac{35.45mgCl^-}{1mmolCl^-} =23.23mgCl^-[/tex]

Finally, for the given volume of water in liters (0.100L), we compute the required concentration:

[tex][Cl^-]=\frac{23.2mgCl^-}{0.100L}\\[/tex]

[tex][Cl^-]=232.3\frac{mgCl^-}{L}[/tex]

Best regards.

The concentration of chloride ions in the groundwater sample is 230 mg/L.

Based on the given information,

Now the number of moles of AgCl precipitate can be calculated as,

n = Given mass/Molar mass

Now putting the values we get,

[tex]n = \frac{0.939 g}{143.32 g/mol} \\n = 6.5 * 10^{-4}[/tex]

Thus, 6.5 ×  10⁻⁴ moles of AgCl comprises 6.5 × 10⁻⁴ chloride ions. Therefore, 6.5 × 10⁻⁴ of chloride ions are present in the sample of 100 ml.

Now the molar mass of chloride ion is 35.453 g/mol, the mass of chloride ion will be,

Mass = Mole × Molar mass

Mass = 6.5 × 10⁻⁴ moles  × 35.453 g/mol

Mass = 0.0230 g or 23 mg

The volume of the groundwater sample is 100 ml or 0.1000 L.

Now the concentration of the chloride ions in the sample given is,

C = 23 mg/0.1000 L

C = 230 mg/L

Thus, the concentration of chloride ions in the groundwater sample is 230 grams per liter.

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Answer:

The correct answer is 8.10

Explanation:

Given:

A(g) + 2B(g) ↔ AB₂(g)   Kc = 59 ---- Eq. 1

A(g) + 3B(g) ↔ AB₃(g)   Kc = 478 ----- Eq. 2

We have to rearrange the chemical equations in order to obtain:

AB₂(g) + B(g) ↔ AB₃(g) Kc = ?

AB₂(g) is a reactant, so we have to use the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant is the same: Kc= 478.  The following is the sum of rearranged chemical equations, and the compounds in bold and italic are canceled:

 AB₂(g)       ↔   A(g) + 2B(g)          Kc₁= 1/59

A(g) + 3B(g) ↔   AB₃(g)                  Kc₂= 478

-----------------------------------------

AB₂(g) + B(g) ↔ AB₃(g)

If we add reactions at equilibrium, the equilibrium constants Kc are mutiplied as follows:

Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10

The value of the missing equilibrium constant is 8.10.

The value of the missing equilibrium constant is 8.10

Since

A(g) + 2B(g) ↔ AB₂(g)   Kc = 59 ---- Eq. 1

A(g) + 3B(g) ↔ AB₃(g)   Kc = 478 ----- Eq. 2

Now we have to rearrange the chemical equations in order to obtain:

AB₂(g) + B(g) ↔ AB₃(g) Kc = ?

Here AB₂(g) represents a reactant, so we have to applied the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant should be the same: Kc= 478.  

The following is the sum of rearranged chemical equations, and the compounds in bold and italic should be canceled:

AB₂(g)       ↔   A(g) + 2B(g)          Kc₁= 1/59

A(g) + 3B(g) ↔   AB₃(g)                  Kc₂= 478

-----------------------------------------

AB₂(g) + B(g) ↔ AB₃(g)

In the case when we add reactions at equilibrium, the equilibrium constants Kc are multiplied as follows:

Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10

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Answer:

The hydrogen can be gotten from the added Acid or water during "workup".

Explanation:

Basically we can say that the reaction describe in this question is a Reduction reaction because of the chemical compound called sodium borohydride. In the reaction described above we can see that there is a Reduction of ketone to alcohol by the compound; sodium borohydride.

For the reduction Reaction to occur the C-O bond must break so as to enable the formation of O-H bond and C-H bond.

So, "the reaction mixture is treated with water and H2SO4 to produce the desired alcohol", thus, the oxygen will definitely pick up the hydrogen from H2SO4 or H2O.

Answer:

A) [tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex].

B) [tex]Kc=0.0933[/tex].

C) 0.9 mol.

D) Increasing both temperature and pressure.

Explanation:

Hello,

In this case, given the information, we proceed as follows:

A)

[tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex]

B) For the calculation of Kc, we rate the equilibrium expression:

[tex]Kc=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]

Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent [tex]x[/tex], we have:

[tex][NH_3]=0.6M=2*x[/tex]

[tex]x=\frac{0.6M}{2}=0.3M[/tex]

Next, the concentrations of nitrogen and hydrogen at equilibrium are:

[tex][N_2]=\frac{1.5mol}{1.5L}-x=1M-0.3M=0.7M[/tex]

[tex][H_2]=\frac{4mol}{1.5L}-3*x=2.67M-0.9M=1.77M[/tex]

Therefore, the equilibrium constant is:

[tex]Kc=\frac{(0.6M)^2}{(0.7M)*(1.77M)^3}\\ \\Kc=0.0933[/tex]

C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.

D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.

Best regards.

The cell can obtain more sugar when the sugar undergo facilitated diffusion through a channel protein, and no energy is required.

FACILITATED DIFFUSION:

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Answer:

A. releases a large amount of heat

Explanation:

A reaction is said to be spontaneous if it can proceed on its own without the addition of external energy. A spontaneous reaction is not determined by the length of time, because some spontaneous reactions are completed after a long period of time. They are exothermic in nature. An example is the conversion of graphite to carbon which takes a long period of time to complete. Spontaneous reactions are known to increase entropy in a system. Entropy is the rate of disorder in a system.

In the combustion of fire, energy is released to the surroundings as there is a decrease in energy. This is an example of a spontaneous reaction because it is an exothermic reaction, which causes an increase in entropy and a decrease in energy.

Answer:

D- poor conductor

Explanation:

metallic properties decrease as we go on the right of the periodic table. Carbon is a non metal hence it is dull and a poor conductor.

it has a low density and is ductile.

Answer: Poor conductor

Explanation:

Answer:

[tex]m_{CO_2}=2.8gCO_2[/tex]

Explanation:

Hello,

In this case, the combustion of octane is chemically expressed by:

[tex]C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O[/tex]

In such a way, due to the 25/2:8 molar ratio between oxygen and carbon dioxide, we can compute the yielded grams of carbon dioxide (molar mass 44 g/mol) as shown below:

[tex]m_{CO_2}=3.2gO_2*\frac{1molO_2}{32gO_2} *\frac{8molCO_2}{\frac{25}{2}molO_2 } *\frac{44gCO_2}{1molCO_2}\\ \\m_{CO_2}=2.8gCO_2[/tex]

Best regards.

Answer:

Keq = 0.217

Explanation:

Let's determine the equilibrium reaction.

In gaseous state, water vapor can be decomposed to hydrogen and oxygen and this is a reversible reaction.

2H₂(g)  + O₂(g)  ⇄  2H₂O (g)         Keq

Let's make the expression for the equilibrium constant

Products / Reactants

We elevate the concentrations, to the stoichiometry coefficients.

Keq = [H₂O]² / [O₂] . [H₂]²

Keq = 0.250² / 0.8 . 0.6² =  0.217

Answer:

No, because sulfur does not typically form negative ions or oxidation states less than 2−. The binary compound formed by sulfur and sodium is Na2S

Explanation:

Sulphur is a member of group 16. The oxidation states expected for sulphur in group 16 are -1, -2, +1, +2,+3,+4,+5 or +6. The elements of group 16 usually form negative ions with oxidation number of -2. They do not typically form negative ions with oxidation state less than -2.

The implication of this is that we actually do not expect the existence of a compound in which sulphur forms an S^4- anion. In reality, such an anion does not exist. Rather a binary compound of sulphur and sodium will have the formula Na2S because it contains the S^2- anion.

The representative elements are the elements in the tall columns of the periodic table.

The representative elements that can also be referred to as the main group elements. They can be used to represent the chemistry of the group to which they belong.

Hence, the representative elements are the elements in the tall columns of the periodic table.

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Answer:

0.619 M to 3 significant figures.

Explanation:

1 mole of [tex]NiI_{2}[/tex] - 312.5 g

? mole of [tex]NiI_{2}[/tex] - 2.9 g

= 2.9/312.5

= 0.0928 moles.

Concentration = no. of moles/vol in litres = [tex]\frac{0.0928}{0.150L}[/tex]

= 0.619 M

Answer:

wavelength, λ =  486.6 nm

frequency, f = 6.16 * 10¹⁴ Hz

Explanation:

a. Wavelength

Using the wavelength equation; 1/λ = (1/hc) * 2.18 * 10⁻¹⁸ J * (1/nf² - 1/ni²)

Where nf is the final energy level; ni is the initial energy level; h is Planck's constant = 6.63 * 10⁻³⁴ J.s; c is velocity of light = 3 * 10⁸ m/s

1/λ = 1/(6.63 * 10⁻³⁴ J.s * 3 * 10⁸ m/s) * 2.18 * 10⁻¹⁸ J * (1/2² - 1/4²)

1/λ = 2.055 * 10⁶ m

λ = 4.866 * 10⁻⁷ m

wavelength, λ =  486.6 nm

b.  Frequency

Using f = c/λ

f = (3 * 10⁸ m/s) / 4.866 * 10⁻⁷ m

frequency, f = 6.16 * 10¹⁴ Hz

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:

Using energy.

Explanation:

Answer:

The change in the boiling point would be 2.1°C.

Explanation:

1 ) Let us first determine the molarity of this solution :

M = mol / kg,

M = 0.402 mol / 0.200 kg = 2.01 M NaCl

2 ) ΔT = i [tex]*[/tex] K [tex]*[/tex] m

ΔT = 2 [tex]*[/tex] 0.512C/m [tex]*[/tex] 2.01m

ΔT = 2.06C

As you can see, this is none of the answer choices. However the van't Hoff factor i in this case was taken to be 2, but this value is actually less than the predicted ideal solution. This is due to the ion pairing, causing i to be around 1.7 to 1.8. Therefore our solution is option b, 2.1°C.

Answer:

The heat absorbed by the sample of water is 3,294.9 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

Replacing:

Q= 4.184 [tex]\frac{J}{g*C}[/tex] * 45 g* 17.5 C

Solving:

Q=3,294.9 J

The heat absorbed by the sample of water is 3,294.9 J