alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decimal places.

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:

0.17 lb

Explanation:

78 g * (1 lb/454 g)=0.17 lb

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Answer:

7.23 J

Explanation:

Step 1: Given data

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)

Q = 7.23 J

Answer:

A (Atomic mass number or Nucleon number)

Explanation:

The mass number is the total number of protons and nucleons in an atomic nucleus.

Hope this helps.

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Answer:

The pH of the solution is 9.06.

Explanation:

The reaction of the dissociation of NH₃ in water is:

NH₃(aq) + H₂O(l)  ⇄  NH₄⁺(aq) + OH⁻(aq)     (1)

[NH₃] - x                     [NH₄⁺] + x     x  

The concentration of NH₃ and NH₄⁺ is:

[tex] [NH_{3}] = \frac{n_{NH_{3}}}{V_{T}} = \frac{C_{i}_{(NH_{3})}*Vi_{NH_{3}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.12 M*0.2 L}{0.2 L + 0.25 L} = 0.053 M [/tex]

[tex] [NH_{4}^{+}] = \frac{C_{i}_{(NH_{4}^{+})*V_{NH_{4}^{+}}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.15 M*0.25 L}{0.2 L + 0.25 L} = 0.083 M [/tex]

From equation (1) we have:

[tex]Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}[/tex]

[tex] 1.8 \cdot 10^{-5} = \frac{(0.083 + x)*x}{0.053 - x} [/tex]

[tex] 1.8 \cdot 10^{-5}(0.053 - x) - (0.083 + x)*x = 0 [/tex]

By solving the above equation for x we have:

x =  1.15x10⁻⁵ = [OH⁻]

The pH of the solution is:

[tex] pOH = -log([OH^{-}]) = -log(1.15 \cdot 10^{-5}) = 4.94 [/tex]

[tex] pH = 14 - pOH = 14 - 4.94 = 9.06 [/tex]

Therefore, the pH of the solution is 9.06.

I hope it helps you!

Answer:

0.032 moles

Explanation:

no of moles =

[tex] \frac{mass \: in \: grams}{relative \: molecular \: mass} [/tex]

=

[tex] \frac{2.4}{40 + 32 + 2} [/tex]

= 0.032

Calcium hydroxide reacted with nitric acid the total number of moles will be 0.032 moles.

A mole is Avogadro's number of particles, which is exactly 6.02214076×1023.

The mole is widely used in chemistry as a convenient way to express amounts of reactants and products of chemical reactions. For example, the chemical equation 2H2 + O2 → 2H2O can be interpreted to mean that for each 2 mol dihydrogen (H2) and 1 mol dioxygen (O2) that react 2 mol of water (H2O) form.

Number of moles = Mass of substance / Mass of one mole Number of moles

mass of substance = 2.4g

molar mass of calcium hydroxide is  (1 ×40.078g/mol Ca) +(2 × 15.999g/mol O) + (2 × 1.008g/mol H) = 74.092 g/mol Ca (OH)2

substituting the value,

number of moles = 2.4 / 74.029

                          = 0.032 moles

Therefore, moles of calcium hydroxide will be 0.032 moles

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Answer:

CO3^2-

Explanation:

In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.

When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.

Answer:

D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice

Explanation:

Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).

All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.

Answer:

a) Monomers: monomers are unit molecules, that can react together with other monomers, to form a long chain molecule called a polymer. Th polymer formed can also be in a three dimensional network. The process of this conversion of monomers to polymers is called polymerization.

b) Repeating unit: A repeating unit is a unit of the polymer formed, whose repetition would produce a long complete polymer chain. A polymer is made up of these repeating links of molecules that form a long chain of molecules.

c) Condensation polymerization: This is a form of condensation reaction, that involves the combination of molecules into polymers with the loss of small molecules such as water or methanol as by products.

d) Cross-linked polymer: This is a polymer formed from a type of bonding of molecules. The bonding is usually in the form of covalent bonds or ionic bonds and the polymers can be either synthetic polymers or natural polymers.  The cross-links leads to an alteration in the physical properties of the polymer.

The definition of following terms are :

a) Monomers:

The monomers are unit atoms, that can respond in conjunction with other monomers, to create a long chain molecule called a polymer.

The polymer shaped can too be in a three dimensional arrange.

b) Repeating unit:

A rehashing unit may be a unit of the polymer shaped, whose reiteration would produce a long total polymer chain.

A polymer is made up of these rehashing joins of atoms that shape a long chain of molecules.

c) Condensation polymerization:

This is often a frame of condensation response, that includes the combination of particles into polymers with the misfortune of little particles such as water or methanol as by products.

d) Cross-linked polymer:

This can be a polymer shaped from a sort of holding of particles.

The cross-links leads to an modification within the physical properties.

Definition is a rhetorical style that uses various techniques to impress upon the reader the meaning of a term, idea, or concept.

Definition may be used for an entire essay but is often used as a rhetorical style within an essay that may mix rhetorical styles.

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Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.

hydro = water

Answer:

Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!

Explanation:

The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)

Answer:

5000g

Explanation:

mass= density × volume

Since the unit of density here is g/mL, we need to convert the volume to mL.

1L= 1000mL

4L= 4 ×1000 = 4000 mL

Mass of object

= 1.25 ×4000

= 5000g

Answer:

5,000 grams

Explanation:

The mass of an object can be found by multiplying the volume by the density.

mass= volume * density

The density is 1.25 g/mL and the volume is 4 L.

First, we must convert the volume to mL. The density is given in grams per milliliter, but the volume is given in liters.

There are 1,000 mL per L. The volume is 4 L. Therefore, we can multiply 4 and 1,000.

4 * 1,000 = 4,000

The volume is 4,000 mL.

Now, find the mass of the object.

mass= volume * density

volume = 4,000

density= 1.25

mass= 4,000 * 1.25 = 5,000

Add the appropriate units for mass, in this case, grams, or g.

mass= 5,000 g

The mass of the object is 5,000 grams.

Answer:

I. dipole-dipole

III. dispersion

IV. hydrogen bonding

Explanation:

Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.

London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.

Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.

Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.

Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.

Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.

Answer:

The intermolecular forces present in CH_3NH_2 includes

Explanation:

The intermolecular forces present in CH_3NH_2 includes II. (ion-dipole) and IV. (hydrogen bonding)

It is a polar molecule due to NH polar bond and it can form Hydrogen bond also due to NH bond.

Interaction will be dipole- dipole and Hydrogen dispersion forces can always be taken into account.

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Answer:

ΔHrxn = -635.14kJ/mol

Explanation:

We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:

(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol

(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole

(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole

Reaction (1) - (3) produce:

Ca(s) + H2O(l) → H2(g) + CaO(s)

ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol

Now this reaction + 1/2(2):

Ca(s) + ½ O2(g) → CaO(s)

ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)

Answer:

The correct answer is 1332 KJ.

Explanation:

Based on the given information,  

ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.  

Now the balanced equation is:  

C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)

ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2

ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)

ΔH°rxn = -1411.9 KJ

Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.  

Now based on the balanced chemical reaction,  

ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2

ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)

ΔS°rxn = -267.68 J/K or -0.26768 KJ/K

T = 25 °C or 298 K

Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get

ΔG = -1411.9 - 298.0 × (-0.2677)

ΔG = -1332 KJ.  

Thus, the maximum work, which can obtained is 1332 kJ.  

Answer:

8.05 moles

Explanation:

5.50 / 2.954 = x / 4.325

x = 8.05

According to ideal gas equation, if the temperature and pressure stay constant during the process 0.520 moles have been added  so that the balloon expands to 4.325 L.

The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.The law was proposed by Benoit Paul Emile Clapeyron in 1834.

In the given example if pressure and temperature are constant then V=nR substituting V=4.325 l and R=8.314  so n=V/R=4.325/8.314=0.520 moles.

Thus, 0.520 moles of helium are added if the temperature and pressure stay constant during this process.

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Answer:

NSAIDs are steroidal anti-inflammatories, their action is on the phospholipase A2 enzyme, this enzyme is responsible for breaking down the phospholipids of the membrane to trigger an inflammatory response. This is how steroidal anti-inflammatory drugs inhibit ALL inflammatory pathways (not like NSAIDs that they only inhibit the COX pathway).

These corticosteroid drugs cannot exceed the systemic mineralocorticoid value 1 in the body, since this corticosteroid hormone is also released by the adrenal cortex.

The NSAIDs generate: sporadic peaks in blood glucose, hypertension, fluid retention, increase in body fat mass, possible suppression of the adrenal cortex over time, inhibiting endogenous synthesis of corticosteroids.

On the other hand, naproxen and ibuprofen are NSAIDs, that is, non-steroidal anti-inflammatory drugs that do not influence both routes of inflammation, but only COX, this enzyme is abbreviated as COX but is called cyclooxygenase, and is responsible for a single route of inflammation.

NSAIDs such as naproxen and ibuprofen can cause gastric disorders such as ulcers or gastritis if they are consumed in a very repetitive manner.

In addition, both drugs are anti-inflammatory, analgesic and antipyretic. Although its two main functions are the first two, it was shown to have an effect in lowering body temperature.

That they are anti-inflammatory means that they inhibit the path of inflammation and analgesics the path of pain.

Explanation:

Both types of drugs generate the same effect but by different mechanisms.

Some are steroids and others are not, although steroids are considered to have a greater risk of benefit that is why they are administered against more systematically compromised instances such as anaphylactic shock.

NSAIDs such as naproxen and ibuprofen are the most prescribed today, since they have few risks and very good benefits, meaning that their adverse effects are not lethal or highly relevant and have a good effect on symptoms.

Both must be administered with care when treating a diabetic patient since corticosteroids generate glycemic peaks or increase in blood glucose, and NSAIDs compete for plasma protein with oral hypoglycemic agents, thus generating that these are in higher free concentrations. high diffusing better through the tissues and increases the potency of the adverse effects of these.

Answer:

When equation for neutralization of HBr by Ca(OH)₂ is correctly balanced, 1.2046*10²⁴ molecules of water will be formed

Explanation:

A neutralization reaction is one in which an acid (or acidic oxide) reacts with a base (or basic oxide). In the reaction a salt is formed and in most cases water is formed. A Salt is an ionic compound formed by the union of ions and cations through ionic bonds.

In the reactions of a strong acid (those substances that completely dissociate) with a strong base (they dissociate completely, giving up all their OH-), the complete neutralization of the species is carried out:

2 HBr (aq) + Ca(OH)₂ (s) → CaBr₂ (aq) + 2 H₂O (l)

The reaction is already balanced, complying with the law of conservation of matter. This law states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reactants must be equal to the number of atoms present in the products.

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), 2 moles of water H₂O are formed.

On the other hand, Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023 * 10²³ particles per mole. Avogadro's number applies to any substance.

Then you can apply the following rule of three: if 1 mole of H₂O contains 6.023*10²³ molecules, 2 moles of H₂O, how many molecules does it contain?

[tex]amount of molecules=\frac{2moles*6.023*10^{23}molecules }{1 mole}[/tex]

amount of molecules= 1.2046*10²⁴ molecules

When equation for neutralization of HBr by Ca(OH)₂ is correctly balanced, 1.2046*10²⁴ molecules of water will be formed

Answer:

Atomic mass is calculated by adding protons and neutrons.

Explanation:

Atomic mass is the sum of protons and neutrons in an atomic nucleus. For example, the element Oxygen has 8 protons (derived from the atomic number) and 8 neutrons (derived from subtracting the amount of protons from the atomic mass).

We can craft an equation to show the relationship between these variables.

M - N = P, where M = Mass, N = Neutrons, and P = Protons

This equation can be rearranged to show the relationship between the neutrons and protons leading to the atomic mass. Simply add N to both sides of the equation.

M = N + P

This shows that atomic mass is equivalent to the sum of protons and neutrons in an atom's nucleus.

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:

PLEASE LOOK INN TO THE FILE YOU WILL GET ANSWER AND ALSO SUMMARY THANKS FOR ASKING QUESTION.

Explanation:

Answer:

1. metals

2. atom

3. homogeneous

4. compounds

5. lustrous

6. saturated

7. colloidal

8. homogeneous

Explanation:

Answer:

there is one atom of oxygen and two atoms of hydrogen

Explanation:

Answer:

Explanation:

we know that

ΔH=m C ΔT

where ΔH is the change in enthalpy (j)

m is the mass of the given substance which is water in this case

ΔT IS the change in temperature and c is the specific heat constant  

we know that given mass=2.9 g

ΔT=T2-T1 =98.9 °C-23.9°C=75°C

specific heat constant for water is 4.18 j/g°C

therefore ΔH=2.9 g*4.18 j/g°C*75°C

ΔH=909.15 j

Answer:

0.0583g

Explanation:

The equation of the reaction is;

2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)

From the question, number of moles of HNO3 reacted= concentration × volume

Concentration of HNO3= 0.100 M

Volume of HNO3 = 20.00mL

Number of moles of HNO3= 0.100 × 20/1000

Number of moles of HNO3 = 2×10^-3 moles

From the reaction equation;

2 moles of HNO3 reacts with 1 mole of Mg(OH)2

2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2

But

n= m/M

Where;

n= number of moles of Mg(OH)2

m= mass of Mg(OH)2

M= molar mass of Mg(OH)2

m= n×M

m= 1×10^-3 moles × 58.3 gmol-1

m = 0.0583g

Answer:

A precipitate will form.

[Ag⁺] = 2.8x10⁻⁵M

Explanation:

When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:

Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)

Ksp is defined as:

Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are in equilibrium

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are the actual concentrations

If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,

The actual concentrations are -Where 500mL is the total volume of the solution-:

[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M

[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M

And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴

As Q > Ksp; a precipitate will form

In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:

[Ag⁺] = 0.06M - 2X

[CrO₄²⁻] = 0.165M - X

Where X is defined as the reaction coordinate

Replacing in Ksp expression:

1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]

Solving for X:

X = 0.165M → False solution. Produce negative concentrations.

X = 0.0299986M

Replacing, equilibrium concentrations are:

[Ag⁺] = 0.06M - 2(0.0299986M)

[CrO₄²⁻] = 0.165M - 0.0299986M

[CrO₄²⁻] = 0.135M

Answer:

Half-life at 629K = 252.4min

Explanation:

Using Arrhenius equation:

[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]

And as Half-life in a first order reaction is:

[tex]t_{1/2}=\frac{ln2}{K}[/tex]

We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:

[tex]58.0min=\frac{ln2}{K}[/tex]

K = 0.01195min⁻¹ = K₁

[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]

[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]

[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]

K₂ = 2.75x10⁻³ min⁻¹

And, replacing again in Half-life expression:

[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]

The half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.   

The activation energy of a reaction is related to its rate constant as follows:  

[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex]   (1)

Where:

We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:

[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex]   (2)

Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min

Hence, the rate constant at 652 K is:                            

[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]

Now, from equation (1) we can find the pre-exponential factor (A):

[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]  

With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):

[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]

Finally, the half-life at 629 K is (eq 2):

[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]

Therefore, the half-life at 629 K is 251.1 min.

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Answer:

Ka = 6.15x10⁻⁷

Explanation:

Ka is defined as dissociation constant in the equilibrium of a weak acid with water. The general reaction is:

HA(aq) + H₂O(l) ⇆ H₃O⁺(aq) + A⁻(aq)

And Ka is defined as the ratio between molar concentrations in equilibrium of products over reactants as follows:

Ka = [H₃O⁺] [A⁻] / [HA]

You don't take water in the equilibrium beacuse is a pure liquid

Replacing with the concentrations of the problem:

Ka = [H₃O⁺] [A⁻] / [HA]

Ka = [4.00x10⁻⁴] [4.00x10⁻⁴] / [0.260]

Answer:

D. The Ca[OH]2 solution may have been unsaturated

Explanation:

The solubility product constant Ksp of any given chemical compound is a term used to describe the equilibrium between a solid and the ions it contains solution. The value of the Ksp indicates the extent to which any compound can dissociate into ions in water. A higher the Ksp, implies more greater solubility of the compound in water.

If the Ksp is more than the value in literature, this false value must have arisen from the fact that the solution was unsaturated hence it appears to be more soluble than it should normally be when saturated.

Answer:

d. HF(g)

Explanation:

Hello,

In this case, the standard entropy S° could be predicted by looking at the amount of bonds the compound has, thus, the fewer the number bonds, the lower the standard entropy, it means that d. HF(g) has lowest value as it has one bond only whereas methane has four bonds, water two bonds and ammonia three bonds.

Best regards.

Answer:

Here are four possible voltaic cells.  

Explanation:

1. Standard reduction potentials

                                         E°/V

I₂(s) + 2e⁻ ⟶ 2I⁻(aq);        0.54

Cu²⁺(aq) + 2e⁻ ⟶ Cu(s);   0.34

Fe²⁺(aq) + 2e⁻ ⟶ Fe(s);   -0.41

Zn²⁺(aq) + 2e⁻ ⟶ Zn(s);   -0.76

2. Possible Voltaic cells

(a) Zn/I₂

                                                                       E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                 0.76

Cathode:  I₂(s) + 2e⁻ ⟶ 2I⁻(aq);                     0.54

Cell:         Zn(s) +  I₂(s) ⟶  Zn²⁺(aq) + 2I⁻(aq); 1.30

Zn(s)|Zn²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)

Zn is the anode; graphite is the cathode.

(b) Zn/Cu²⁺

                                                                          E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                    0.76

Cathode:  Cu²⁺(aq) + 2e⁻ ⟶ Cu(s);                  0.34

Cell:          Zn(s) +  Cu²⁺(s) ⟶  Zn²⁺(aq) + Cu(s); 1.10

Zn(s)|Zn²⁺(aq)∥Cu²⁺(aq)|Cu(s)

Zn is the anode; Cu is the cathode.

(c) Zn/Fe²⁺

                                                                            E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                     0.76

Cathode:  Fe²⁺(aq) + 2e⁻ ⟶ Fe(s);                    -0.41

Cell:          Zn(s) +  Fe²⁺(s) ⟶  Zn²⁺(aq) + Fe(s);  0.35

Zn(s)|Zn²⁺(aq)∥Fe²⁺(aq)|Fe(s)

Zn is the anode; Fe is the cathode.

(d) Fe/I₂

                                                                         E°/V

Anode:     Fe(s) ⟶ Fe²⁺(aq) + 2e⁻;                   0.41

Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq);                     0.54

Cell:         Zn(s) +  I₂(s) ⟶  Zn²⁺(aq) + 2I⁻(aq); 0.95

Fe(s)|Fe²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)

Fe is the anode; graphite is the cathode.