Be sure to answer all parts. The sulfate ion can be represented with four S―O bonds or with two S―O and two S═O bonds. (a) Which representation is better from the standpoint of formal charges? four S―O bonds two S―O bonds and two S═O bonds (b) What shape is the sulfate ion, and what hybrid orbitals of S are postulated for the σ bonding? bent tetrahedral trigonal planar trigonal pyramidal spmdn where m = and n = (c) In view of the answer to part (b), what orbitals of S must be used for the π bonds? What orbitals of O? sulfur: 3d 3p 2p 3s oxygen: 3p 2p 2s 3s

Answer:

[tex]Br_2[/tex]

Explanation:

The compound [tex]X_2[/tex] is [tex]Br_2[/tex].

The reaction included are:

[tex]Br_2+NaCl--> No\ reaction[/tex]  (because  bromine is less reactive than chlorine)

But , because of hexane the solution get dilute and its color changes to orange.

Now, NaI is added and we know Br is more reactive than I.

Therefore it replace it.

Reaction: [tex]Br_2+NaI-->NaBr+I_2[/tex]

Purple color develop due to formation of Iodine.

Answer:

1. The theoretical yield is 0.2mols

2. No, aldol condensation reaction does not proceeds via an E1 mechanism.

The radioactive uranium decays to produce thorium and it emits an alpha particle or helium atom. Thus, option A is correct.

Unstable heavy isotopes of elements undergo nuclear decay to produce stable atoms by the emission of charged particle such as alpha or beta particles.

Based on the emitted particle, there are two types of decay process namely alpha decay and beta decay. In alpha decay atoms emits alpha particles which are helium nuclei and the atom losses its mass number by 4 units and atomic number by two units,

In beta decay, electrons are emitted by the atom, where no change occurs in mass number and atomic number increases by one unit. Uranium undergo alpha decay by emitting alpha particle or helium nuclei.

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Answer:

Kp = 3.62 x 10 ^ -2  

Explanation:

The relationship between Kc and Kp is given as:

Kp =  Kc (RT) ^ Δn -------- (1)

Where:

Kp = Equilibrium Constant in terms of Molar concentrations = 61

Kc = Equilibrium Constant in terms of Partial Pressures

R = Gas Constant = 0.0821 litre.atm/mole.K

Δn = (Total No. of moles of products) - (Total No. of moles of reactants)

Δn = (2) - (3+1) = -2

T = Temperature (K) = 500 K

Substituting all values in equation (1), we get,

Kp = 61 x [ 0.0821 x 500 ] ^ (-2)

Kp = 3.62 x 10 ^ -2  

The study of chemicals and bonds is called chemistry. There are two types of elements are there are these are metals and nonmetals.

The correct answer is Kp = [tex]3.62 * 10^{-2[/tex]

The relationship between Kc and Kp is given as:

Kp =  Kc (RT)^Δn -------- (1)

Where:

Δn = (2) - (3+1) = -2

Substituting all values in equation (1), we get,

[tex]Kp = 61 * [ 0.0821 * 500 ]^{-2Kp = 3.62*10^{-2[/tex]

Hence, the correct answer is [tex]Kp = 3.62*10^{-2[/tex].

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Answer:

Mole fraction: A =  8.70%, B = 37.00%, C =  19.60%, D = 34.80%

K = 6.86

Standard reaction free energy change: -4.77 kJ/mol

Explanation:

Let's do an equilibrium chart for the reaction:

2A + B ⇄ 3C + 2D

1.00  2.00  0     1.00    Initial

-2x     -x     +3x   +2x    Reacts (stoichiometry is 2:1:3:2)

1-2x    2-x    3x    1+2x  Equilibrium

3x = 0.9

x = 0.3 mol

Thus, the number of moles of each one at the equilibrium is:

A = 1 - 2*0.3 = 0.4 mol

B = 2 - 0.3 = 1.7 mol

C = 0.9 mol

D = 1 + 2*0.3 = 1.6 mol

The molar fraction is the mol of the component divided by the total number of moles (0.4 + 1.7 + 0.9 + 1.6 = 4.6 mol):

A = 0.4/4.6 = 0.087 = 8.70%

B = 1.7/4.6 = 0.37 = 37.00%

C = 0.9/4.6 = 0.196 = 19.60%

D = 1.6/4.6 = 0.348 = 34.80%

The equilibrium constant is the multiplication of the concentration of the products elevated by their coefficients, divided by the multiplication of the concentration of the reactants elevated by their coefficients. Because the volume remains constant, we can use the number of moles:

K = (nC³*nD²)/(nA²*nB)

K = (0.9³ * 1.6²)/(0.4² * 1.7)

K = 6.86

The standard reaction free energy change can be calculated by:

ΔG° = -RTlnK

Where R is the gas constant (8.314 J/mol.K), and T is the temperature (25°C = 298 K)

ΔG° = -8.314*298*ln6.86

ΔG° = -4772.8 J/mol

ΔG° = -4.77 kJ/mol

The Mole fraction: A is = 8.70%, B = 37.00%, C = 19.60%, D = 34.80%

K is = 6.86

Then, The Standard reaction free energy change: -4.77 kJ/mol

Now, Let's do an equilibrium chart for the reaction:

Then, 2A + B ⇄ 3C + 2D

1.00  2.00  0     1.00    Initial

After that, -2x     -x     +3x   +2x    Reacts (stoichiometry is 2:1:3:2)

1-2x    2-x    3x    1+2x  Equilibrium

Then, 3x = 0.9

x = 0.3 mol

Thus, When the number of moles of each one at the equilibrium is:

A is = 1 - 2*0.3 = 0.4 mol

B is = 2 - 0.3 = 1.7 mol

C is = 0.9 mol

D is = 1 + 2*0.3 = 1.6 mol

When The molar fraction is the mol of the component divided by the total number of moles (0.4 + 1.7 + 0.9 + 1.6 = 4.6 mol):

A is = 0.4/4.6 = 0.087 = 8.70%

B is = 1.7/4.6 = 0.37 = 37.00%

C is = 0.9/4.6 = 0.196 = 19.60%

D is = 1.6/4.6 = 0.348 = 34.80%

When The equilibrium constant is the multiplication of the concentration of the products elevated by their coefficients, Also divided by the multiplication of the concentration of the reactants elevated by their coefficients.

Because When the volume remains constant, we can use the number of moles:

K is = (nC³*nD²)/(nA²*nB)

K is = (0.9³ * 1.6²)/(0.4² * 1.7)

K is = 6.86

When The standard reaction free energy change can be calculated by:

Then, ΔG° = -RTlnK

Where R is the gas constant (8.314 J/mol.K), and T is the temperature (25°C = 298 K)

After that, ΔG° = -8.314*298*ln6.86

Then, ΔG° = -4772.8 J/mol

Therefore, ΔG° = -4.77 kJ/mol

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Answer:

The average diameter of a helium atom is smaller.

Explanation:

Step 1: Data given

An atom gets larger as the number of electronic shells increase; therefore the radius of atoms increases as you go down a certain group in the periodic table of elements.

In general, the size of an atom will decrease as you move from left to the right of a certain period.

Step 2:

Since hydrogen is more left on the periodic table than helium, the diameter of the atom will decrease moving from hydrogen to helium.

Hydrogen has a bigger diameter than helium.

The average diameter of a helium atom is smaller.

Answer:

Ka1 = 4.27x10⁻⁷, Ka2 = 2.78x10⁻¹⁶

Explanation:

The number of moles that had reacted when the pH was 6.37 was:

nH₂A = 0.046 L * 0.15 mol/L = 6.9x10⁻³ mol

nKOH = 0.023 L * 0.15 mol/L = 3.45x10⁻³ mol

Thus, H₂A is in excess, and the base will not react with HA⁻ to form A⁻², so the first reaction is absolute, and after the reaction:

nH₂A = 6.9x10⁻³ - 3.45x10⁻³ = 3.45x10⁻³

nHA⁻ = 3.45x10⁻³

Because the volume will be the same, we can use the number of moles instead of the concentrations. So:

6.37 = pKa1 + log(3.45x10⁻³/3.45x10⁻³)

pKa1 = 6.37

Ka1 = [tex] 10^{-6.37}[/tex]

Ka1 = 4.27x10⁻⁷

After 46 mL of KOH was added:

nKOH = 0.046 * 0.15 = 6.9x10⁻³

So, all the H₂A reacts to form HA⁻, and the absolute equilibrium is:

HA⁻ ⇄ H⁺ + A⁻²

6.9x10⁻³ 0 0 Initial

-x +x x Reacts

6.9x10⁻³-x x x Equilibrium

pH = -log[H⁺]

8.34 = -log[H⁺]

[H⁺] = [tex]10^{-8.34}[/tex]

[H⁺] = 4.57x10⁻⁹ M

The total volume is 46 mL + 46 mL = 92 mL = 0.092 L

x = 0.092 L * 4.57x10⁻⁹ = 4.20x10⁻¹⁰ mol

Thus, nHA⁻ = 6.9x10⁻³ - 4.20x10⁻¹⁰ ≅ 6.9x10⁻³

Then,

8.34 = pKa2 + log(4.20x10⁻¹⁰/6.9x10⁻³)

8.34 = pKa2 - 7.216

pKa2 = 15.556

Ka2 = [tex] 10^{-15.556}[/tex]

Ka2 = 2.78x10⁻¹⁶

Final answer:

The values of Ka1 and Ka2 for the diprotic acid can be determined using the titration data provided. At the halfway point of the titration, the pH of the solution is equal to the pKa1 value. Using the given pH value, we can calculate the pKa1 and Ka1 values. At the endpoint of the titration, the pH of the solution is equal to the pKa2 value, allowing us to calculate the pKa2 and Ka2 values.

Explanation:

To determine the values of Ka1 and Ka2 for the diprotic acid, we can use the titration data provided. When 23.0 mL of 0.15 M KOH was added, the pH of the solution was 6.37. At this point, half of the acid has reacted with the base, indicating that we are at the halfway point between Ka1 and Ka2. Therefore, the pH is equal to the pKa at this point.

Using the pH value of 6.37, we can calculate the pKa by taking the negative logarithm of the [H+] concentration. Since pH = -log[H+], we can rearrange the equation to find [H+]:
[H+] = 10^(-pH).
Therefore, [H+] = 10^(-6.37) = 2.17 x 10^(-7) M.

Since this is the concentration at the halfway point, we can assume that [H2A] = [HA-]. Using the Henderson-Hasselbalch equation, we can write:
pKa1 = pH + log([HA-]/[H2A]) = 6.37 + log(2.17 x 10^(-7)/0.15).
pKa1 = 6.37 + (-3.66) = 2.71.

Therefore, the value of Ka1 for the diprotic acid is 10^(-pKa1) = 10^(-2.71).

Using similar calculations, we can determine the value of Ka2. When 46 mL of 0.15 M KOH was added, the pH of the solution was 8.34. At this point, all of the acid has reacted with the base, indicating that we are at the endpoint of the titration. Therefore, the pH is equal to the pKa2 at this point.

Using the pH value of 8.34, we can calculate the pKa2 by taking the negative logarithm of the [H+] concentration. Since pH = -log[H+], we can rearrange the equation to find [H+]:
[H+] = 10^(-pH).
Therefore, [H+] = 10^(-8.34) = 4.28 x 10^(-9) M.

Since all of the acid has reacted, we can assume that [A-] = 0.15 M. Using the Henderson-Hasselbalch equation, we can write:
pKa2 = pH + log([A-]/[HA]).
pKa2 = 8.34 + log(0.15/0.15).
pKa2 = 8.34 + 0 = 8.34.

Therefore, the value of Ka2 for the diprotic acid is 10^(-pKa2) = 10^(-8.34).

Answer:

B) ) –1615.1 kJ mol^–1

Explanation:

since

SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l) ∆Hºrxn = 4.6 kJ mol–1

the enhalpy of reaction will be

∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr

where ∆Hºrxn= enthalpy of reaction , ∆Hºfp= standard enthalpy of formation of products , ∆Hºfr = standard enthalpy of formation of reactants , νp=stoichiometric coffficient of products, νr=stoichiometric coffficient of reactants

therefore

∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr

4.6 kJ/mol = [1*∆HºfX + 2*(–285.8 kJ/mol)] - [1*(–910.9kJ/mol) + 4*(–320.1 kJ/mol)]

4.6 kJ/mol =∆HºfX -571.6 kJ/mol + 2191.3 kJ/mol

∆HºfX = 4.6 kJ/mol + 571.6 kJ/mol - 2191.3 kJ/mol = -1615.1 kJ/mol

therefore ∆HºfX (unknown standard enthalpy of formation = standard enthalpy of formation of SiF4(g) ) = -1615.1 kJ/mol

Answer:

Empirical C5H3

Molecular

C20H12

Explanation:

Like all other hydrocarbons, benzopyrene would burn in oxygen or air to form carbon iv oxide and water only.

From the mass of carbon iv oxide produced, we can get the mass of carbon and hence the mass of hydrogen.

It can be seen from the question that 12.73mg of carbon iv oxide was produced. Let us convert this to grammes, we simply divide by 1000 = 0.01273g

Now we need to calculate the number of moles of carbon iv oxide produced. To do this, we simply divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles is thus 0.01273/44 = 0.00028392moles

As we can see that there is only one atom of carbon in a molecule of carbon iv oxide. Hence, the number of moles of carbon iv oxide present is equal to 0.00028392moles

From here we can get the mass of carbon which is equal to the number of moles of carbon multiplied by the atomic mass of carbon. The atomic mass of carbon is 12 a.m.u

The mass of carbon in the benzopyrene is thus 0.00028392 * 12 = 0.003472g

The mass of hydrogen in the compound is 0.003649g - 0.003472g = 0.000177g

We can now deduce the empirical formula by dividing the masses by the atomic masses.

H = 0.000177/1 = 0.000177

C = 0.003572/12 = 0.000289333333

We now divide by the smallest which is that of hydrogen.

H = 0.000177/0.000177= 1

C = 0.000289333333/0.000177 = apprx 1.63

Multiply both by 3 to yield C5H3

The molecular formula is as follows:

[(12* 5) + ( 3 * 1)]n = 252.3

63n = 252.3

n = 4

The molecular formula is thus C20H12

Answer:

The heat of combustion of liquid butane is: -2636 kJ/mol

Explanation:

This problem is solved by using Hess law of heats summation:

We have the heat of combustion for gaseous butane:

C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)     ΔHºcomb = -2658 kJ/mol

and we want the heat of combustion for liquid butane.

What we need to do is add the heat of vaporization for liquid butane to this equation:

C₄H₁₀ (l)   ⇒  C₄H₁₀ (g)   ΔHºvap =  22.44 kJ/mol

C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)  ΔHºcomb = -2658 kJ/mol

______________________________________________

C₄H₁₀ (l)   + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)  

So,

ΔH comb liq butane = -2658 kJ/mol + 22.44 kJ/mol = -2636 kJ/mol

Answer:

Option B

Explanation:

Energy required to remove outermost electron from an neutral atom is called ionization energy or first ionization energy ([tex]IE_1[/tex])

Energy required to remove an electron from M+ ion is called second ionization energy ([tex]IE_2[/tex]).

Energy required to remove an electron from M2+ ion is called third ionization energy [tex]IE_2[/tex]).

Therefore, among the given options:

[tex]Al+e^- \rightarrow Al^{+}(g),\ IE_1[/tex]

[tex]Al^++e^- \rightarrow Al^{2+}(g),\ IE_2[/tex]

[tex]Al^{2+}+e^- \rightarrow Al^{3+}(g),\ IE_3[/tex]

Therefore, the correct option is b

The third ionization of aluminum refers to the removal of the third electron from an aluminum atom. The correct representation is: B) Al2+ (g) --------> Al3+(g) + e-. This represents the transition from a double positive ion to a triple positive ion of aluminum.

In Chemistry, the ionization energy of an element refers to the energy required to remove an electron from an atom or an ion. The process of ionization occurs in successive stages for each electron removed. When considering the third ionization of aluminum, it implies that two electrons have already been removed, and we're focusing on the removal of a third electron from the resulting ion.

The correct option that represents the third ionization of aluminum is:

B) Al

2+

(g) --------> Al

3+

(g) + e

-

This is because option B represents the transition of aluminum from a double positive ion (Al2+) to a triple positive ion (Al3+) with the removal of an electron.

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In a chemical reaction, the total mass of reactants is equals to the total mass of the products. Here, A and B react to produce C and D. Given that 21g of A and 22g of B (43g in total) are converted into 17g of C and x grams of D, the mass of D is therefore 43g - 17g = 26g.

The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. In your hypothetical reaction, 21g of A and 22g of B react to create C and D. Combined, the reactants have a total mass of 43g. In the product, we have 17g of C and the mass of D which we're trying to find. Now, given that mass of reactants = mass of products, we can say that 43g (total mass of A+B) = 17g (mass of C) + mass of D. By rearranging the equation, we find that mass of D = 43g - 17g = 26g. So, 26g of D will be produced according to the law of conservation of mass.

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Answer:

-1.37 kJ/mol

Explanation:

The expression for the calculation of the enthalpy of dissolution of [tex[NH_4NO_3[/tex] is shown below as:-

[tex]\Delta H=m\times C\times \Delta T[/tex]

Where,  

[tex]\Delta H[/tex]  is the enthalpy of dissolution of [tex[NH_4NO_3[/tex]

m is the mass

C is the specific heat capacity

[tex]\Delta T[/tex]  is the temperature change

Thus, given that:-

Mass of ammonium nitrate = 5.60 g

Specific heat = 4.18 J/g°C

[tex]\Delta T=17.9-22.0\ ^0C=-4.1\ ^0C[/tex]

So,  

[tex]\Delta H=-1.25\times 4.18\times 3.9\ J=-95.9728\ J[/tex]

Negative sign signifies loss of heat.  

Also, 1 J = 0.001 kJ

So,  

[tex]\Delta H=-0.096\ kJ[/tex]

Also,

Molar mass of [tex[NH_4NO_3[/tex] = 80.043 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{5.60\ g}{80.043 \ g/mol}[/tex]

[tex]Moles= 0.06996\ mol[/tex]

Thus, [tex]\Delta H=-\frac{0.096}{0.06996}\ kJ/mol=-1.37\ kJ/mol[/tex]

Answer:

1. The difference between a chemical change and a physical change is as follows;

In a physical change no new substance is formed while in a chemical change new substance are formed

Chemical change involves great amount of heat changes while physical change does not involves great amount of heat change except the latent heat changes which occur during changes of state

Physical change is easily reversed while a chemical change is not easily reversed

There is no change in the mass of a substance involve in physical change while in chemical change there is a change in the mass of a substance that undergoes such a change

2 . examples of physical and chemical changes

Physical change.

1. The magnetization and demagnetization of iron rods( the iron can be removed and no new substance will be formed)

11. The separation of mixtures by evaporation, distillation, fractional distillation, sublimation.( the separation techniques mentioned are physical change because they can be easily revised to get the original quantity back, like when water evaporates to vapour it can be retrieved back by cooling)

111. Changes in the states of matter such as melting of solids to liquids the freezing of liquids to solids, the vaporization of liquids to gases.( also changes of states are physical change because no new substance will be formed and it is easily reversible)

Chemical change

1. Fermentation and decay of substances .( new substance is formed which is the decayed substance)

11. The rusting of iron( rusting of iron is a chemical change because a new substance is formed. Which is the rusted iron)

111. The addition of water to quick lime I.e the slacking of lime. ( is a chemical change because a new substance is formed)

3. Boiling of water in a pot is a physical change because the water will evaporate into vapour at 100° Celsius and can be regained by cooling

Food digesting is a chemical change because a new substance will be formed , which is faeces.

Cooking an egg for breakfast is a chemical change because a new substance will be formed which is boiled egg and it can not be reversed.

Souring of milk is a chemical change because a new substance will be formed which is the soured taste

1. Chemical changes alter a substance's composition (e.g., burning wood), while physical changes affect its physical properties (e.g., melting ice).

2. Examples: Physical (melting ice), Chemical (burning wood).

3. Boiling water is a physical change; food digestion, cooking an egg, and milk souring are chemical changes due to altered composition.

1. **Difference between Chemical and Physical Changes**:

  - **Chemical Change**: A chemical change involves the alteration of a substance's chemical composition. New substances with different properties are formed. This change is typically irreversible and accompanied by a release or absorption of energy.  

  - **Physical Change**: A physical change, on the other hand, does not alter the substance's chemical identity. It involves changes in physical properties, such as size, shape, phase (solid, liquid, gas), or state (e.g., melting, freezing), but the chemical composition remains the same. Physical changes are generally reversible, and no new substances are created.

2. **Examples of Physical and Chemical Changes**:

  **Physical Changes**:

  - **Ice Melting**: When ice (solid water) melts to form liquid water, it undergoes a physical change. The chemical composition of H₂O remains the same, but the phase changes from solid to liquid.

  - **Cutting Paper**: When you cut a piece of paper into smaller pieces, it's a physical change. The chemical composition of cellulose in the paper remains unchanged.

  - **Boiling Water**: Boiling water represents a physical change. Water changes from a liquid to a gas (steam) without any alteration in its chemical composition.

  **Chemical Changes**:

  - **Burning Wood**: When wood is burned, it undergoes a chemical change. The cellulose and other compounds in wood react with oxygen to produce carbon dioxide and water vapor. New substances are formed, and energy is released.

  - **Digesting Food**: The process of digesting food in your body involves chemical changes. Enzymes break down complex molecules into simpler ones, allowing your body to absorb nutrients.

  - **Milk Souring**: When milk sours, it undergoes a chemical change due to the action of bacteria. Lactose in milk is converted into lactic acid, changing the taste and composition of the milk.

3. **Boiling Water - Physical Change**:

  Boiling water is a physical change because it involves a change in the physical state of water from liquid to gas (steam), but the chemical composition remains H₂O.

  **Chemical Changes**:

  1. **Your Food Digesting**: Digestion involves chemical changes as enzymes break down complex food molecules into simpler ones, which can then be absorbed by your body.

  2. **Cooking an Egg for Breakfast**: Cooking an egg is a chemical change because heat causes proteins in the egg white to denature and coagulate, forming a solid structure with different properties than the raw egg white.

  3. **The Milk Souring in Your Fridge**: Milk souring is a chemical change. Bacteria in the milk metabolize lactose into lactic acid, changing the taste and composition of the milk.

In summary, chemical changes involve a change in chemical composition, while physical changes involve changes in physical properties without altering the chemical identity of a substance. Boiling water is a physical change, while food digestion, cooking an egg, and milk souring are chemical changes due to alterations in chemical composition and properties.

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The balanced net ionic equations for the reactions are:

a. Cr2(SO4)3(aq) + 3(NH4)2CO3(aq) → 2Cr(CO3)3(s) + 6NH4NO3(aq),

b. Ba(NO3)2(aq) + K2SO4(aq) → BaSO4(s) + 2KNO3(aq),

c. Fe(NO3)2(aq) + 2KOH(aq) → Fe(OH)2(s) + 2KNO3(aq)

Net ionic equations represent the actual chemical change occurring in a reaction, excluding spectator ions. Examples include the exchange of ions to form water in a neutralization reaction and the formation of precipitates such as calcium phosphate or silver chloride.

Thus, the net balanced ionic equations of given reactions are:

a. Cr2(SO4)3(aq) + 3(NH4)2CO3(aq) → 2Cr(CO3)3(s) + 6NH4NO3(aq)

b. Ba(NO3)2(aq) + K2SO4(aq) → BaSO4(s) + 2KNO3(aq)

c. Fe(NO3)2(aq) + 2KOH(aq) → Fe(OH)2(s) + 2KNO3(aq)

Answer: 0.100 m [tex]K_2SO_4[/tex]

Explanation:

Elevation in boiling point is given by:

[tex]\Delta T_b=i\times K_b\times m[/tex]

[tex]\Delta T_b=T_b-T_b^0[/tex] = Elevation in boiling point

i= vant hoff factor

[tex]K_b[/tex] = boiling point constant

m= molality

1. For 0.100 m [tex]K_2SO_{4}[/tex]

[tex]K_2SO_4\rightarrow 2K^{+}+SO_4^{2-}[/tex]  

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be [tex]3\times 0.100=0.300[/tex]

2. For 0.100 m [tex]LiNO_3[/tex]

[tex]LiNO_3\rightarrow Li^{+}+NO_3^{-}[/tex]  

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be [tex]2\times 0.100=0.200[/tex]

3.  For 0.200 m [tex]C_2H_8O_3[/tex]

, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be [tex]1\times 0.200=0.200[/tex]

4. For 0.060 m [tex]Na_3PO_4[/tex]

[tex]Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}[/tex]  

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be [tex]4\times 0.060=0.24m[/tex]

Thus as concentration of solute is highest for [tex]K_2SO_4[/tex] , the elevation in boiling point is highest and thus has the highest boiling point.

The aqueous solution of K2SO4 has the highest boiling point among the given options.

The boiling point of a solution depends on the concentration and nature of the solute. In this case, we need to compare the boiling points of different aqueous solutions. Higher concentration and the presence of ionic solutes will generally result in a higher boiling point.

Comparing the given options, K2SO4 and Na3PO4 are both ionic compounds, and their solutions will have higher boiling points due to their strong ionic interactions. LiNO3 is also an ionic compound, but its concentration is lower than K2SO4 and Na3PO4. C3H8O3 is a molecular compound and does not dissociate into ions in solution, so its solution will have a lower boiling point than the ionic compounds. Therefore, the aqueous solution of K2SO4 (option A) will have the highest boiling point among the given options.

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Answer:

b) C = 0.50 J/(g°C)

Explanation:

∴ Q = 50 J

∴ m = 10.0 g

∴ ΔT = 35 - 25 = 10 °C

specific heat (C) :

⇒ C = Q / mΔT

⇒ C = 50 J / (10.0 g)(10 °C)

⇒ C = 0.50 J/(g°C)

Answer:

K remains the same;

Q < K;

The reaction must run in the forward direction to reestablish the equilibrium;

The concentration of [tex]PCl_3[/tex] will decrease.

Explanation:

In this problem, we're adding an excess of a reactant, chlorine gas, to a system that is already at equilibrium. According to the principle of Le Chatelier, when a system at equilibrium is disturbed, the equilibrium shifts toward the side of the equilibrium that minimizes the disturbance.

Since we'll have an excess of chlorine, the system will try to reduce that excess by shifting the equilibrium to the right. Therefore, the reaction must run in the forward direction to reestablish the equilibrium.

The value of K remains the same, as it's only temperature-dependent, while the value of Q will be lower than K, that is, Q < K, as Q < K is the case when reaction proceeds to the right.

As a result, since [tex]PCl_3[/tex] is also a reactant, its concentration will decrease.

Final answer:

Upon decomposition, lithium nitride (Li₃N) breaks down into lithium (Li) and nitrogen (N₂) gas. The balanced chemical equation for this decomposition is 2 Li₃N(s) → 6 Li(s) + N₂(g). This is an example of chemical decomposition in chemistry.

Explanation:

Predicting the products of the decomposition of lithium nitride, Li₃N, involves understanding the reactions of ionic compounds. Lithium nitride is comprised of lithium (Li) ions and nitride (N³⁻) ions. Upon decomposition, lithium nitride would likely break down into its constituent elements, lithium (Li) and nitrogen (N₂). The balanced equation for this decomposition would be: 2 Li₃N(s) → 6 Li(s) + N₂(g)

Here, solid lithium nitride (Li₃N) decomposes into solid lithium (Li) and nitrogen gas (N₂) when subjected to suitable conditions such as heating. This type of reaction demonstrates basic principles of chemical decomposition and stoichiometry in chemistry.

Answer:

The ice will not melt, and the temperature will remain at 0°C.

Explanation:

The reaction of the sodium in water is exothermic because heat is being released. In an isolated system, the change in heat must be 0, so the released heat must be absorbed by the ice.

The molar mass of Na is 23 g/mol, so the number of moles that reacted was:

n = 0.250 g/ 23g/mol

n = 0.011 mol

By the reaction:

2 moles ------- -368 kJ

0.011 mol ----- x

By a simple direct three rule:

2x = -4.048

x = -2.024 kJ/mol

So the ice will absorbs 2.024 kJ/mol, which is less than the necessary to melt it (6.02 kJ/mol). Then, the ice will not melt.

The temperature of a pure substance didn't change until all of it has changed of phase, so the temperature must remain at 0°C.

The question is incomplete, the complete question is attached below.

Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

Explanation : Given,

Mass of NaOH = 60 g

Volume of stock solution = 300 mL

Molar mass of NaOH = 40 g/mol

First we have to calculate the molarity of stock solution.

[tex]\text{Molarity}=\frac{\text{Mass of }NaOH\times 1000}{\text{Molar mass of }NaOH\times \text{Volume of solution (in mL)}}[/tex]

Now put all the given values in this formula, we get:

[tex]\text{Molarity}=\frac{60g\times 1000}{40g/mole\times 300mL}=5mole/L=5M[/tex]

Now we have to determine the volume of stock solution and distilled water mixed.

Formula used :

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of stock solution.

[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted solution.

From data (A) :

[tex]M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]5M\times 20mL=1M\times V_2\\\\V_2=100mL[/tex]

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (B) :

[tex]M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]5M\times 20mL=1M\times V_2\\\\V_2=100mL[/tex]

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (C) :

[tex]M_1=5M\\V_1=60mL\\M_2=1M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]5M\times 60mL=1M\times V_2\\\\V_2=300mL[/tex]

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From data (D) :

[tex]M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]5M\times 60mL=1M\times V_2\\\\V_2=300mL[/tex]

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.

Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

To make a 1 M NaOH solution, you would mix 2 L of the stock solution with 2 L of distilled water.

To prepare a 1 M NaOH solution using a stock solution of 60 g NaOH in 300 mL, we need to calculate the amount of stock solution and distilled water needed.

We can start by calculating the number of moles of NaOH in the stock solution:

moles = mass / molar mass => moles = 60 g / (22.99 g/mol + 16.00 g/mol + 1.01 g/mol) = 2 mol NaOH

Since we want a 1 M NaOH solution, we can use the formula: moles = concentration x volume

2 mol = 1 M x volume => volume = 2 mol / 1 M = 2 L

Therefore, we need to mix 2 L of the stock solution with 2 L of distilled water to obtain a 1 M NaOH solution.

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Answer:see attached image

Explanation:

In organic chemistry, a product may be kinetically or thermodynamically favoured. A kinetically favours product forms faster, it may not necessarily be the more stable product. The thermodynamically favoured product forms at a slower rate but is more stable. Often times, the kinetically favoured product rearranges itself to form the thermodynamically favoured product at equilibrium. The endo product of the Diels Alder reaction mentioned in the question is first formed (kinetically favoured) but rearranges to form the exo product (thermodynamically favoured) at equilibrium. This is clear shown in the reaction mechanism attached below.

Answer:

2 mole H2O = 18.02 g will not work.

Explanation:

Let us check the options one by one:

1 mole of O2 weighs 32 g.

⇒Hence option A is correct.

1 mole of H2O weighs 18 g.

⇒Hence option B is wrong.

From the balanced equation we can say :

1 mole of CH4 reacts with 2 moles of O2 to give 1 mole of CO2 and 2 moles of H2O.

⇒Hence option C and D are correct.

Answer:

Option B: 2 mole H2O = 18.02 g is not correct.

Explanation:

Step 1: The balanced equation

CH4 + 2O2 → CO2 + 2H2O

Step 2: Data given

Molar mass of O2 = 32 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass of CH4 = 16.04 g/mol

Step 3: Calculate number of mol

For 1 mol of CH4 we need 2 moles of O2 to produce 1 mol of CO2 and 2 moles of H2O

This means option C is correct: For 2 moles O2 consumed, we'll get 1 mol of CO2 produced.

Also Option D is corect: For 1 mole of CH4 we get 2 moles of H2O

Mass = moles * molar mass

Mass of 1 mole O2 = 1 mol * 32.00 g/mol = 32.00 grams

Mass of 2 moles H2O = 2 mol * 18.02 g/mol = 36.04 grams  (This means option B is not correct.)

Option B: 2 mole H2O = 18.02 g is not correct.

Answer:

contains the maximum amount of solute that will dissolve in that solvent at that temperature.

Explanation:

Solution= solute+ solvent

The solubility of a solute depends on temperature. A solution containing just the right (maximum) amount of solute that can normally dissolve at a given temperature is said to be saturated.

A saturated solution contains the maximum amount of solute that dissolves in a solvent at a given temperature. Exceeding the solubility leads to a supersaturated solution and solute precipitation. Solubility varies with temperature.

A saturated solution is one that contains the maximum amount of solute that will dissolve in that solvent at that particular temperature. The solubility of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium. In a saturated solution, the solute concentration equals its solubility. If a solute's concentration exceeds its solubility, it forms a supersaturated solution which is an unstable condition and results in solute precipitation when perturbed. The extent to which a substance may be dissolved in water, or any solvent, is quantitatively expressed as its solubility. Solubilities for gaseous solutes decrease with increasing temperature, while those for most, but not all, solid solutes increase with temperature.

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Answer:

Nitrogen is an example of an element (option B)

Explanation:

Water, glucose and salt are sort of compound.

They are compounds that have certain elements linked by specific bonds.

H₂O - Water

C₆H₁₂O₆ - Glucose

NaCl - Salt

Nitrogen is an example of an element. Unlike water, glucose, and salt, which are compounds, nitrogen exists as a diatomic molecule with the chemical formula N₂.

An example of an element is B) Nitrogen. Nitrogen is a chemical substance that is represented by the symbol N and is found in the air we breathe as a diatomic molecule (N₂).

In the context of other options presented in such questions, water (H₂O) is a compound consisting of hydrogen and oxygen, glucose (C₆HO₆) is a carbohydrate compound, and salt commonly refers to sodium chloride (NaCl), which is also a compound.

The concentration of free Ag+ ions at equilibrium is calculated by assuming full complexation of Ag+ with CN- due to a large formation constant, and then considering any small dissociation of the complex back to Ag+ and CN-. The exact concentration can be determined by applying the formation constant for the complex, which is not provided here.

To calculate the concentration of free Ag+ ions at equilibrium in a silver plating operation, we need to consider the following reaction:

Ag+(aq) + 2 CN-(aq) \<--> Ag(CN)2-(aq)

First, we determine the initial concentrations of Ag+ and CN-. AgNO3 completely dissociates in water, so the initial concentration of Ag+ is 0.21 M in 90.0 L. NaCN also dissociates completely, providing 5.0 M of CN- ion in 9.0 L.

Mixing the solutions, we have a total volume of 99.0 L (90.0 L + 9.0 L). The concentration of CN- is calculated as follows:

(5.0 M)(9.0 L) / 99.0 L = 0.4545 M

Since the formation constant (Kf) of the Ag(CN)2- complex is large, we can assume that all the Ag+ will react to form the complex, leaving a negligible concentration of free Ag+ ions. Any remaining CN- ions will affect the equilibrium concentration of free Ag+ ions due to the shift in reaction to the left, as described by Le Chatelier's principle.

We assume that essentially all 0.21 M of the Ag+ ion forms the complex. To find the equilibrium concentration of Ag+, we use the dissociation of the complex, simplified by assuming that the dissociation is so small that it's negligible compared to the remaining CN- concentration.

Since we do not have the exact Kf value provided here, we would normally express the equilibrium concentration of Ag+ in terms of Kf and the concentration of remaining CN-, but to give an exact numerical answer, the specific Kf value for the Ag(CN)2- complex is required from a reliable source like a textbook or scientific database.

Part A:

Therefore the correct statements are as follows.

With a few exceptions, the solubility of most solid solutes in water decreases as the solution temperature increases.(True)

The solubility of gases in water increLases with increasing temperature. (False).

The solubility of a gas in any solvent is increased as the partial pressure of the gas above the solvent increases. (True)

Substances with similar intermolecular attractive forces tend to be insoluble in one another. (False)

solubility.

Part B:

Therefore, hexane (C6H14) is the best solvent for nonpolar solutes.

Part A:

The first statement is true. Generally, the solubility of solid solutes in water tends to decrease with increasing temperature, although there are exceptions.

The second statement is false. The solubility of gases in water typically decreases with increasing temperature. This is described by Henry's law.

The third statement is true. This statement is consistent with Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid

The fourth statement is false. Substances with similar intermolecular attractive forces tend to be more soluble in one another. Like dissolves like is a general principle in solubility.

Part B:

The best solvent for nonpolar solutes is the one with similar characteristics, meaning it should be nonpolar as well

Acetone is a polar solvent. It is not the best solvent for nonpolar solutes.

Ethanol is a polar solvent. It is not the best solvent for nonpolar solutes.

Hexane is a nonpolar solvent and is the best choice for nonpolar solutes.

Water is a polar solvent. It is not the best solvent for nonpolar solutes.

Therefore, hexane (C6H14) is the best solvent for nonpolar solutes.

Answer:

a) 39/19 K : stable nuclide, 40/19 K  : radioactive nuclide.

b) 209B: stable nuclide, 208Bi : radioactive nuclide

c) nickel-58 : stable nuclide, nickel-65 : radioactive nuclide.

Explanation:

As per the rule, nuclides having odd number of neutrons are generally not stable and therefore, are radioactive.

Mass number (A) = Atomic number (Z) + No. of neutrons (N)

Or, N = A - Z

a)

39/19 K and 40/19 K

Calculate no. of neutrons in 39/19 K as follows:

atomic no. = 19, mass no. 39

N = 39 - 19

   = 20 (even no.)

Calculate no. of neutrons in 40/19 K as follows:

atomic no. = 19, mass no. 40

N = 40 - 19

   = 21 (odd no.)

Therefore, 39/19 K is a stable nuclide and 40/19 K is a radioactive

nuclide.

b)

209Bi and 208Bi

Calculate no. of neutrons in 209Bi as follows:

atomic no. = 83, mass no. 209

N = 209 - 83

   = 126 (even no.)

Calculate no. of neutrons in 208Bi as follows:

atomic no. = 83, mass no. 208

N = 208 - 83

   = 125 (odd no.)

Therefore, 209Bi is a stable nuclide and 208Bi is a radioactive nuclide.

c)

nickel-58 and nickel-65

Calculate no. of neutrons in nickel-58 as follows:

atomic no. = 28, mass no. 58

N = 58 - 28

   = 30 (even no.)

Calculate no. of neutrons in nickel as follows:

atomic no. = 28, mass no. 65

N = 65 - 28

   = 37 (odd no.)

Therefore,nickel-58 is a stable nuclide and nickel-65 is a radioactive nuclide.

To determine which nuclide is radioactive and which is stable, we need to analyze the composition of their nuclei. The number of neutrons in a nuclide plays a crucial role in determining its stability. Based on this, we can predict whether a nuclide is radioactive or stable.

In order to predict which nuclide is radioactive and which is stable, we need to determine the composition of their nuclei. Radioactive nuclides have an unstable nucleus that undergoes radioactive decay, while stable nuclides have a more balanced composition.

Therefore, the pairs are:
a. 39/19K (stable) and 40/19K (radioactive)
b. 209Bi (stable) and 208Bi (radioactive)
c. 58Ni (stable) and 65Ni (radioactive)

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Answer:

10 g %  w/v

Explanation:

Weight/volume percent concentration means the grams of solute, in 100 mL of solution.

If we have 1 g of B3 in 10mL of solution, the rule of three for the weight volume percent will be:

10mL ___ 1 g

100 mL ____ 10 g

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

The limiting reagent in a chemical reaction is the reactant that is completely consumed first. To identify the limiting reagent, we need to compare the mole ratios of the reactants to the mole ratio of the products in the balanced chemical equation.

* Reaction mixture a: neither reactant is limiting.

* Reaction mixture b: Mg is the limiting reagent.

* Reaction mixture c: I2 is the limiting reagent.

* Reaction mixture d: Mg is the limiting reagent.

* Reaction mixture e: neither reactant is limiting.

* Reaction mixture f: I2 is the limiting reagent.

* Reaction mixture g: neither reactant is limiting.

* Reaction mixture h: I2 is the limiting reagent.

* Reaction mixture i: Mg is the limiting reagent.

To identify the limiting reagent in a reaction mixture, we need to compare the mole ratios of the reactants to the mole ratio of the products in the balanced chemical equation. The limiting reagent is the reactant that will be completely consumed before the other reactants are consumed.

**Balanced chemical equation:**

Mg(s) + I2 (s) → MgI2 (s)

**Mole ratios:**

Mg : I2 : MgI2 = 1 : 1 : 1

**a. 100 atoms of Mg and 100 molecules of I2**

Since we have the same number of atoms of Mg and molecules of I2, neither reactant is limiting. The reaction will proceed until both reactants are completely consumed.

**b. 150 atoms of Mg and 100 molecules of I2**

Since we have more atoms of Mg than molecules of I2, Mg is the limiting reagent. I2 will be completely consumed before Mg is consumed.

**c. 200 atoms of Mg and 300 molecules of I2**

Since we have more molecules of I2 than atoms of Mg, I2 is the limiting reagent. Mg will be completely consumed before I2 is consumed.

**d. 0.16 mol Mg and 0.25 mol I2**

Since we have less moles of Mg than moles of I2, Mg is the limiting reagent. I2 will not be completely consumed.

**e. 0.14 mol Mg and 0.14 mol I2**

Since we have the same number of moles of Mg and moles of I2, neither reactant is limiting. The reaction will proceed until both reactants are completely consumed.

**f. 0.12 mol Mg and 0.08 mol I2**

Since we have less moles of I2 than moles of Mg, I2 is the limiting reagent. Mg will not be completely consumed.

**g. 6.078 g Mg and 63.455 g I2**

First, we need to convert the masses of Mg and I2 to moles:

Moles of Mg = 6.078 g / 24.31 g/mol = 0.250 mol

Moles of I2 = 63.455 g / 253.808 g/mol = 0.250 mol

Since we have the same number of moles of Mg and moles of I2, neither reactant is limiting. The reaction will proceed until both reactants are completely consumed.

**h. 1.00 g Mg and 2.00 g I2**

First, we need to convert the masses of Mg and I2 to moles:

Moles of Mg = 1.00 g / 24.31 g/mol = 0.0411 mol

Moles of I2 = 2.00 g / 253.808 g/mol = 0.00791 mol

Since we have less moles of I2 than moles of Mg, I2 is the limiting reagent. Mg will not be completely consumed.

**i. 1.00 g Mg and 20.00 g I2**

First, we need to convert the masses of Mg and I2 to moles:

Moles of Mg = 1.00 g / 24.31 g/mol = 0.0411 mol

Moles of I2 = 20.00 g / 253.808 g/mol = 0.0788 mol

Since we have more moles of I2 than moles of Mg, Mg is the limiting reagent. I2 will not be completely consumed.

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