Which value is most likely to be the pH of salt that is formed by the reaction of a week acid and a strong base ?

A. Zinc (Zn)

Like all the other metals, it has a cloud of an electron because of the free electrons in the valence shell. These electrons are easier to move.

When some electric force is applied, the electrons start moving towards the positive end due to the attraction. This causes the flow of electrons, and, ultimately, the generation of electricity. Zinc one has 27%

pOH = 3.08

NX3 + H2O <----> NHX3+ + OH-  

Kb = 4.0 x 10^-6

Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).

c(NH₄⁺) = c(OH⁻) = x.

x² = Kb · c(NH₃)

x² = 4.0 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.

x = c(OH⁻) = √(7.0 × 10⁻⁷)

    = 8.367 × 10⁻⁴

pOH = -log(c(OH⁻))

        =- log ( 8.367 × 10⁻⁴)

        = 3.08

Final answer:

To calculate the pOH of a 0.175 M aqueous solution of NX₃ with Kb of 4.0×10⁻⁶, we simplify the dissociation equation and find the OH- concentration to be approximately 1.67×10⁻³ M, resulting in a pOH of about 2.78.

Explanation:

To find the pOH of a 0.175 M aqueous solution of NX₃, given that Kb for NX3 is 4.0×10⁻⁶, we start by setting up the base dissociation reaction NX₃ + H₂O → NX₃H+ + OH−. The equilibrium expression associated with this dissociation is Kb = [NX3H+][OH−]/[NX3]. We can assume the concentrations of NX₃H+ and OH− to be equal (x) and much smaller than the initial concentration of NX₃, thus [NX₃] approximately equals 0.175 M. With these assumptions, Kb can be rewritten as 4.0×10⁻⁶ = x₂ / (0.175 − x).

Assuming x is much smaller than 0.175, we simplify to 4.0×10⁻⁶ ≈ x2 / 0.175. Solving for x gives us x ≈ √(4.0×10⁻⁶ ∗ 0.175) ≈ 1.67×10⁻³ M, which is the concentration of OH−. The pOH is calculated using the formula pOH = −log[OH−], resulting in pOH ≈ 2.78.

i believe B is the answer

Answer: Alkene

Explanation: Apex

Polar: IF, PCl3, IF5  

Nonpolar: CS2, SO3, SF6

The change in free energy of the system under the given conditions is -40.5 KJ/mol.

N2 (g) + 3H2 (g) → 2NH3 (g)  

We have to use the relation; ΔG = ΔG⁰ + RT ln Q

ΔG = Free energy change under the given conditions

ΔG⁰ = standard free energy change

R = gas constant

Q = reaction quotient

We can obtain the reaction quotient from;

Q = [NH3]^2/[N2] [H2]^3

Q = [0.65]^2/[1.9] [1.6]^3

Q = 0.4225/7.7824

Q= 0.0543

Substituting the values;

ΔG⁰ = -33.3 KJ/mol

R = 8.314 JK-1mol-1

T = 298 K

Q = 0.0543

ΔG = -33.3 KJ/mol + (8.314 JK-1mol-1 × 298 K) ln (0.0543)

ΔG =   -33.3 KJ/mol + (-7.2  KJ/mol)

ΔG = -40.5 KJ/mol

Learn more: https://brainly.com/question/13155407

To find the value of δG for the reaction mixture, use the equation ΔG = ΔG° + RT ln(Q), where ΔG° is the standard Gibbs free energy change, R is the gas constant, T is the temperature in Kelvin, and Q is the reaction quotient.

The value of δG at 298 K for a reaction mixture that consists of 1.9 atm N2, 1.6 atm H2, and 0.65 atm NH₃ can be calculated using the equation ΔG = ΔG° + RT ln(Q), where ΔG° is the standard Gibbs free energy change, R is the gas constant, T is the temperature in Kelvin, and Q is the reaction quotient.

First, calculate the reaction quotient using the given pressures: Q = (NH₃)² / (N₂)(H₂)³. Substitute the given pressures into the equation to find Q. Then, use the equation ΔG = ΔG° + RT ln(Q), where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (298 K), ΔG° is the standard Gibbs free energy change (-33.3 kJ/mol), and Q is the reaction quotient. Calculate the value of ΔG at 298 K for the given reaction mixture using the calculated Q value.

https://brainly.com/question/26667299

#SPJ3

Question 1: Fission

Question 2: Chain reaction

Question 3: Controlled nuclear fission

Question 4: Hard to find safe nuclear disposal of...

Question 5: The ionising ability of radiation to affect cell functions.

Answer:

n = 2 to n = 1

Explanation:

The electrons in the closest orbitals to the atomic nucleus experience the strongest attraction. Therefore, for an electron to jump up from the ground state to n = 2 a lot of energy is required to overcome the attractive force of the nucleus (also means this energy will be dispensed when the electron jumps down). The energy required to jump up to other energy levels decreases as these orbitals are shielded from the attractive force of the nucleus.

Answer:

From n = 2 to n = 1

Explanation:

There are 7 energy levels, numbered from 1 to 7, and in which electrons are distributed, logically in order according to their energy level. Electrons with less energy will be spinning at level 1.  Level 1 is the innermost level or closest to the nucleus and is the one with the lowest energy level. Level 7 is the outermost or furthest from the core and is the level that has the highest energy level.

Then the electrons are spinning around the nucleus forming layers, as previously mentioned. In each of them, the energy that the electron possesses is different. In the layers very close to the nucleus, the force of attraction between it and the electrons is very strong, so they will be strongly bound (joined).  The opposite occurs in remote layers, in which electrons are weakly bound, so it will be easier to make electronic exchanges in the last layers.

So, since electrons are more strongly attracted to the nucleus, for an electron to jump from the fundamental state at level n = 2 to level 1, a lot of energy is required to overcome the aforementioned force of attraction. At the other levels, the attraction of the nucleus is lower, so the energy required to jump to other energy levels will also be lower.

The spoon will get warmer as it takes in the heat of the water

D strong electrolytes

Strong bases are strong electrolytes that completely dissociate in water to form hydroxide ions.

Strong bases are strong electrolytes. They completely dissociate in water, forming hydroxide ions (OH-) that are able to conduct electricity. Examples of strong bases include sodium hydroxide (NaOH) and potassium hydroxide (KOH).

https://brainly.com/question/31824660

#SPJ6

Answer:

[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.

Explanation:

HF + H₂O ⇄ H₃O⁺ + F⁻.

[H₃O⁺] = [F⁻].

∵ [H₃O⁺] = √Ka.C,

Ka = 6.8 x 10⁻⁴, C = 0.710 M.

∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.

∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.2 x 10⁻²) = 4.55 x 10⁻¹³.

The value of concentration of  [H₃O⁺] & [F⁻] is 2.2 x 10⁻² M and [OH⁻] is 4.55 x 10⁻¹³M in 0.710M HF.

How we calculate acid dissociation constant?

Acid dissociation constant for any HA acid will be calculated as:

Ka = [H⁺][A⁻] / [HA].

ICE table for given reaction is written as:

                           HF + H₂O ⇄ H₃O⁺ + F⁻

Initial:                 0.710                 0       0

Change:               -x                    +x      +x

Equilibrium:     0.710-x                +x     +x

Given value of Ka = 6.8 × 10⁻⁴

Putting all values on the above value of Ka, we get

6.8 × 10⁻⁴ = x.x / 0.710-x

We take the value of 0.710-x as 0.710, because value of x is negligible as compare to 0.710.

6.8 × 10⁻⁴ = x.x / 0.710

x² = 6.8 × 10⁻⁴ × 0.710

x = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M

So, [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M

We know that pH + pOH = 14, we write this equation in base 10 form as:

[H₃O⁺][OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 2.2 x 10⁻² = 4.55 x 10⁻¹³

Hence, value of [H₃O⁺] & [F⁻] is 2.2 x 10⁻² M and [OH⁻] is 4.55 x 10⁻¹³M.

To know more about acid dissociation constant, visit the below link:

https://brainly.com/question/26998

The branch of science that relates with the interaction of matter and electromagnetic radiation is spectroscopy. Thus, option C is correct.

Electromagnetic radiations are given as the beam of protons with the presence of the perpendicular magnetic and electrical waves travelling at the speed of light.

The interaction of the electromagnetic radiations with the matter results in the propagation of the valence electrons of matter to the higher energy level and creating transitions.

These interactions between the electromagnetic radiations and matter are studies with spectroscopy. Thus, option C is correct.

Learn more about electromagnetic radiation, here:

https://brainly.com/question/10759891

#SPJ2

The answer is that you have to write this in your own words.

To help you on how to get started, here's what you need to do:

Just say what planet you want to go and why you want to go there. Then, say what you wish to learn on that planet.

Template: My space mission would be going to planet _______. I would like to go there because ___________________. I hope to learn that ________________.

Our space mission involves studying Mars' geological features and potential for life. This will enhance our understanding of the possibility of life beyond our solar system and specifically detail the environment on Mars. The mission might also reveal evidence of past or present life.

Our space mission will aim to explore the unique geological features and potential habitability of Mars. Mars is a planet of particular interest due to evidence suggesting its past was warmer and wetter, hinting at the possibility of life beyond our solar system.

We wish to explore this aspect of the solar system because it could fundamentally alter our understanding of life's existence in the universe. The mission will involve the use of hi-tech equipment to identify potential biomarkers, signs of past or present life.

Through this mission, we hope to learn more about the environment on Mars, detail its capacity for sustaining life, and possibly find evidence of life itself. Our findings will not only influence future Mars missions but also add a significant chapter to our understanding of life in the universe.

https://brainly.com/question/3654134

#SPJ3

Answer : The standard enthalpy of the reaction is 20.2 kJ

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]

The equation used to calculate enthalpy change is of a reaction is:  

[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]

The equilibrium reaction follows:

[tex]CS_2(g)+2H_2O(l)\rightleftharpoons CO_2(g)+2H_2S(g)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=[n_{(CO_2)}\times \Delta H^o_f_{(CO_2)}+n_{(H_2S)}\times \Delta H^o_f_{(H_2S)}]-[n_{(H_2O)}\times \Delta H^o_f_{(H_2O)}+n_{(CS_2)}\times \Delta H^o_f_{(CS_2)}][/tex]

We are given:

[tex]\Delta H^o_f_{(CS_2(g))}=116.7kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2S(g))}=-20.60kJ/mol[/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[(1mol\times -393.5kJ/mol)+(2mol\times -20.60kJ/mol)]-[(1mol\times 116.7kJ/mol)+(2mol\times -285.8kJ/mol)]=20.2kJ[/tex]

Therefore, the standard enthalpy of the reaction is 20.2 kJ

The standard enthalpy of reaction is 20.2  kJ/mol.

We can calculate the standard enthalpy of reaction from the standard heat of formation using the formula;

ΔHreaction = ∑ΔH∘f products - ΔH∘f reactants

The equation of the reaction is; CS2(g)+2H2O(l)→CO2(g)+2H2S(g)

ΔH∘f CS2 = 116.7 KJ/mol

ΔH∘f  H2O(l) = −285.8 KJ/mol

ΔH∘f CO2 = −393.5 kJ/mol

ΔH∘f H2S= −20.60 kJ/mol

Substituting values, we have;

ΔHreaction = ∑[(−393.5) + 2 ×(−20.60) - (116.7) + 2 ×(−285.8)] kJ/mol

ΔHreaction = 20.2  kJ/mol

Learn more: https://brainly.com/question/7827769

No. When water first begins to cool down, it contracts. However, as it gets colder and eventually freezes, it begins to expand.

You can test this by freezing water in a water bottle: when you take it out of the freezer, the cap might have popped off or cracks may have formed in the sides of the bottle.

Answer: Water expands when frozen, not contracts.

Answer:

(a) The amounts of reactants decrease.

(b) The amounts of reactants decrease.

(c) The amounts of reactants decrease.

(d) the amounts of reactants increase.

Explanation:

(a) C₂H₂(g) + H₂O(g) ⇌ CH₃CHO(g) ΔH°rxn = − 151 kJ.

So, the reaction can be represented as:

C₂H₂(g) + H₂O(g) ⇌ CH₃CHO(g) + heat.

As the temperature is decreases, it is like that decreasing the concentration of products side, that shifts the reaction towards the right side (products side) to attain the equilibrium again. So, the amounts of reactants decrease.

So, the right choice is reaction is: The amounts of reactants decrease.

(b) CH₃CH₂OH(l) + O₂ (g) ⇌ CH₃CO₂H(l) + H₂O(g) ΔH°rxn = − 451 kJ.

So, the reaction can be represented as:

CH₃CH₂OH(l) + O₂ (g) ⇌ CH₃CO₂H(l) + H₂O(g) + heat.

As the temperature is decreases, it is like that decreasing the concentration of products side, that shifts the reaction towards the right side (products side) to attain the equilibrium again. So, the amount of reactants decrease.

So, the right choice is reaction is: The amounts of reactants decrease.

(c) 2C₂H₄(g) + O₂(g) ⇌ 2CH₃CHO(g) (exothermic).

So, the reaction can be represented as:

2C₂H₄(g) + O₂(g) ⇌ 2CH₃CHO(g) + heat.

As the temperature is decreases, it is like that decreasing the concentration of products side, that shifts the reaction towards the right side (products side) to attain the equilibrium again. So, the amounts of reactants decrease.

So, the right choice is reaction is: The amounts of reactants decrease.

(d) N₂O₄(g) ⇌ 2NO₂(g) (endothermic).

N₂O₄(g) + heat ⇌ 2NO₂(g)

As the temperature is decreased, it is like that decreasing the concentration of reactants side, that shifts the reaction towards the lift side (reactants side) to attain the equilibrium again. So, the amounts of reactants increase.

So, the right choice is reaction is: the amounts of reactants increase.

Answer:

Explanation:

1) Initial conditions of the helium gas:

2) Ideal gas equation:

3) Solve for n:

4) At STP:

B.  It can damage molecules in skin cells such as DNA. Prolonged exposure can actually cause skin cancer.

Ultraviolet light from the Sun primarily damages molecules in skin cells, particularly the DNA, which can lead to skin aging, skin cancer, and wrinkle formation. However, it also aids in Vitamin D production in the skin that has several health benefits. Apart from this, UV light is also used effectively for disinfection purposes.

Ultraviolet light from the Sun primarily damages molecules in skin cells, particularly the DNA. This happens through the formation of bonds between an adjacent pair of pyrimidine nucleotides, thymine, and cytosine, on the same strand of DNA. The extent of the damage is often believed to be proportional to the amount of ultraviolet radiation received, which is known as the linear hypothesis.

All types of UV radiation can damage collagen fibres, causing an acceleration of skin aging and wrinkle formation. Overexposure to the Sun when young has been linked to the development of skin cancer like melanoma in later life. However, UV-B radiation from sunlight does have some beneficial effects too, such as Vitamin D production in the skin, reducing risks of certain types of cancer and osteoporosis.

Lastly, UV light can sometimes be used effectively for disinfection. It forms thymine dimers in the DNA of microbes, leading to mutations that can kill the microorganisms. These properties of UV rays are used in water purification systems and germicidal lamps.

https://brainly.com/question/8574874

#SPJ3

I think the best way is persuasive. plz forgive me if this is wrong or not the answer that was expected

Answer:

C) 8.9x10-19M

Explanation:

Whe disolving a solid in water, this solid splits in anions and cations, in our example for CuS solid:

CuS (s) -------> Cu+2 (ac)  +   S-2 (ac)

Whe this solution is saturated, and balance is reached, this balance can be represented with Ksp formula (solubility constant), as following:

Ksp= [Cu+2][S-2], and for tgis ecuation, you only consider aqueous ions, but not the solid salt

When balance is reached, you may suppose that "X" moles of CuC gives you "X" moles of Cu+2 and "X" moles of S-2 (according to the stoicheometry of the above reaction):

XCuS (s) ------> XCu+2(ac)  +  XS-2(ac) , for which:

[Cu+2]=X and [S-2]=X so this gives us: Ksp=8x10-37=X.X=X²

X= √8x10-37 and finally X=[Cu+2]=8.9x10-19M (expressed in molar units)

Answer: hydrogen (H⁺) or hydronium (H₃O⁺) and hydroxide (OH⁻)

Explanation:

1) Arrhenius definitions

The chemist Svante Arrhenius, winner of the Noble Prize in chemistry by 1903, defined acids and bases based on the ability of such substances to give ions in water solutons.

2) Acids

Arrhenius defined an acid as a substance that dissociates in water to form hydrogen ions (H⁺), also called protons. This process is called protonation (formation of protons). Due to its small size and high activity, H⁺ does not exist in that form, but it forms H₃O⁺ ions in aqueous solution.

3) Bases

Arrhenius defined a base as a substance that dissociates in water to form hydroxide (OH⁻) ions.

4) Examples

That was the first definitions of acids and bases and is restricted to water solutions and does not include many compounds (like NH₃) which nowadays are identified as acids or bases.

Some examples of Arrhenius acids and bases are:

According to the Arrhenius definition, acids release hydrogen ions (H⁺) and bases release hydroxide ions (OH⁻) in aqueous solutions. These two ions are central to identifying Arrhenius acids and bases.

According to the Arrhenius definition, acids and bases are classified based on the ions they release in an aqueous solution. An Arrhenius acid is a compound that produces hydrogen ions (H⁺), while an Arrhenius base is a compound that produces hydroxide ions (OH⁻). For example, HCl dissociates in water to form H⁺ ions and Cl⁻ ions, categorizing it as an Arrhenius acid. Conversely, NaOH dissociates to form Na⁺ ions and OH- ions, making it an Arrhenius base.

The central ions in these definitions are thus hydrogen ions (H⁺) for acids and hydroxide ions (OH⁻) for bases.

1. 1.35 atm to psi = 19.83953 psi.

2. 100.2 kPa to mm Hg = 751.5617 mm / 29.589043 Hg

3. 10.83 psi to kPa = 74.670221 kPa

Hope this helps,

Davinia.

Answer:

The pH of the solution is 6.35.

Explanation:

Concentration of hydroxide ions = [tex][OH^-]=2.20\times 10^{-8} M[/tex]

pH + pOH = 14

pH = 14 - pOH

[tex]pOH=-\log[OH^-][/tex]

[tex]pOH=-\log[2.20\times 10^{-8} M]=7.65[/tex]

[tex]pH=14-pOH=14-7.65=6.35[/tex]

The pH of the solution is 6.35.

The pH of the solution with a [tex]\( 2.20 \times 10^{-8} \)[/tex] M hydroxide ion concentration is approximately [tex]\( 6.34 \)[/tex].

To find the pH of a solution with a given hydroxide ion concentration [tex](\( [\text{OH}^-] \))[/tex], we can use the relationship between [tex]\( [\text{OH}^-] \)[/tex] and [tex]\( \text{pOH} \)[/tex], and then convert [tex]\( \text{pOH} \)[/tex] to pH.

The relationship between [tex]\( [\text{OH}^-] \)[/tex] and [tex]\( \text{pOH} \)[/tex] is:

[tex]\[ \text{pOH} = -\log([\text{OH}^-]) \][/tex]

Given [tex]\( [\text{OH}^-] = 2.20 \times 10^{-8} \) M[/tex], we can calculate [tex]\( \text{pOH} \)[/tex]:

[tex]\[ \text{pOH} = -\log(2.20 \times 10^{-8}) \][/tex]

[tex]\[ \text{pOH} = -\log(2.20) - \log(10^{-8}) \][/tex]

[tex]\[ \text{pOH} = -(\log(2.20) + (-8 \times \log(10))) \][/tex]

[tex]\[ \text{pOH} = -(0.3424 - 8 \times 1) \][/tex]

[tex]\[ \text{pOH} = -(0.3424 - 8) \][/tex]

[tex]\[ \text{pOH} = -(7.6576) \][/tex]

[tex]\[ \text{pOH} = -(-7.6576) \][/tex]

[tex]\[ \text{pOH} = 7.6576 \][/tex]

Now, we can convert [tex]\( \text{pOH} \)[/tex] to pH using the relationship:

[tex]\[ \text{pH} + \text{pOH} = 14 \][/tex]

[tex]\[ \text{pH} = 14 - \text{pOH} \][/tex]

[tex]\[ \text{pH} = 14 - 7.6576 \][/tex]

[tex]\[ \text{pH} = 6.3424 \][/tex]

HNO2 IS WEAK ACID IF U LOOK AT Ka VALUES U FIND ONE LISTED FOR HNO2

Answer:

1) Add more sulfate ions.

2) 1 M K₂SO₄.

Explanation:

1) Which of the following actions would shift this reaction toward solid barium sulfate?

Add more sulfate ions will increase the concentration of the products side, so the reaction will be shifted towards the reactants side to relieve the stress and attain the equilibrium again.

So, the right choice is: Add more sulfate ions.

2) In which of the following would barium sulfate be least soluble?  

The answer is 1 M K₂SO₄, it has a common ion effect of SO₄⁻ ions that increase the concentration of the products side, so the reaction will be shifted towards the reactants side to relieve the stress and attain the equilibrium again.

So, the solubility will decrease.

1. b) Adding more sulfate ions shifts the reaction towards solid barium sulfate.

2. Barium sulfate is least soluble in a c) 1 M K₂SO₄ solution.

1. Le Châtelier's principle states that if a stress is applied to a system at equilibrium, the system will shift to relieve that stress.

For the reaction:

To shift this reaction towards solid barium sulfate, one must consider the factors that affect the equilibrium. Using Le Châtelier's principle:

Hence, adding more sulfate ions would shift the reaction towards the solid barium sulfate. The correct option is b).

2. Solubility Consideration

Barium sulfate would be least soluble in a 1 M K₂SO₄ solution. This is because K₂SO₄ provides additional SO₄²⁻ ions, which increases the common ion effect, thus decreasing the solubility of BaSO₄. The correct option is c).

Question: Le Châtelier's principle states that if a stress is applied to a system at equilibrium, the system will shift to relieve that stress. This is true for all types of equilibrium, including the dissolution of salts that are only sparingly soluble.

Here is the chemical reaction that represents solid barium sulfate dissolving in water:

BaSO₄(s) ⇌ Ba₂+(aq) + SO₄²⁻(aq), with Ksp = 1.1×10⁻¹⁰

1. Which of the following actions would shift this reaction toward solid barium sulfate?

a) Add more barium sulfate.

b) Add more sulfate ions.

c) Remove sulfate ions.

d) Remove barium ions.

2. In which of the following would barium sulfate be least soluble?

a) pure water

b) 1 M NaNO3

c) 1 M K2SO4

Remark

HA is an acid that produces only 1 Hydrogen when it breaks down.

The NaOH is a base and a good strong one.

When you do the titration you can be guaranteed that ever mole of NaOH that is consumed represents 100% of what is needed.

So your first step is to find out how many moles of NaOH is needed.

Givens

C = 0.1 mol/L

V = 23.64 mL * [ 1 L / 1000 mL] = 0.02364 L

Formula

C = mol / L

Solution

What you are looking for is the number of mols of NaOH used.

0.1 mol/L =  mol / 0.02364 L         Multiply both sides by 0.02364

0.1 mol/L * 0.02364 L = mol

mol = 0.002364

Now the number of mols of HA is going to be exactly the same. That's because the titration formula is

HA + NaOH ==> NaA + HOH

==========================

mols = given mass / molar Mass

Molar Mass = given mass / mols

mols = 0.002364 moles

given mass = 0.5632

Molar Mass = ??

Molar Mass = 0.5632 grams / 0.002364

Molar Mass = 238.24 grams

Answer: b. One atom transferring electrons to another atom

Explanation: An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element and the element which accepts the electrons is known as electronegative element. This bond is formed between a metal and an non-metal.

Covalent bonds are formed by sharing of electrons between non metals

For example, In calcium iodide the one electron from calcium metal gets transferred to iodine atom and thus form an ionic bond to give [tex]CaI_2[/tex]

Electronic configuration of calcium:

[tex][Ca]=1s^22s^22p^63s^23p^64s^2[/tex]

Calcium atom will lose two electron to gain noble gas configuration and form calcium cation with +2 charge.

[tex][Ca^{2+}]=1s^22s^22p^63s^23p^6[/tex]

Electronic configuration of iodine:

[tex][I]=1s^22s^22p^63s^23p^64s^23d^{10}4p^5[/tex]

Iodine atom will gain one electron to gain noble gas configuration and form iodide ion with -1 charge.

[tex][I^-]=1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]

Older people had no money to buy stuff

Answer:

The recent graduate invests taking risks and betting on profitability.

The retired man protects the estate by sacrificing the gain.

Explanation:

PV=nRT  

P = 85.0 x 10^3 Pa  

T = 273 + 20 = 293 K  

n = 2.0 moles  

R = 8.314 m3. pa / mol .K  

V = 2.0 x 293 x 8.314 / 85.0x10^3 =  0.195623529 m^3

Competitive inhibitor

 CO3^2- and O3