The synthesis of methanol from carbon monoxide and hydrogen gas is described by the following chemical equation:

CO(g)+2H2(g)?CH3OH(g)

The equilibrium constant for this reaction at 25 ?Cis Kc=2.3×104. In this tutorial, you will use the equilibrium-constant expression to find the concentration of methanol at equilibrium, given the concentration of the reactants.

The equilibrium-constant expression is a mathematical equation that can be rearranged to solve for any of the variables in it. Rearrange the equilibrium-constant expression to solve for [CH3OH].

Kc[CO][H2]^2

Suppose that the molar concentrations for CO and H2 at equilibrium are [CO] = 0.02 M and [H2] = 0.06 M.

Use the formula you found in Part B to calculate the concentration of CH3OH.

Express your answer to one decimal place and include the appropriate units.

Common cations that can form insoluble sulfate salts that can form insoluble salts with the anion include calcium, barium, strontium, and lead.

A precipitate forms only when [tex]Na_2SO_4[/tex] is added to the solution of the unknown soluble ionic compound. This indicates that the anions in [tex]Na_2SO_4 (SO_4^2-)[/tex] react with certain cations in the unknown compound to form an insoluble compound.

To form an insoluble compound with sulfate ions [tex](SO_4^2-)[/tex], the cations need to form an insoluble sulfate salt. Common cations that can form insoluble sulfate salts include [tex]Ca^{2+}, Ba^{2+}, Sr^{2+}, Pb^{2+}[/tex].

A precipitate forms when [tex]Na_2SO_4[/tex] is added to the unknown solution, it suggests that the unknown compound could contain one or more of these cations. Other cations that do not form insoluble sulfate salts would not result in a precipitate with [tex]Na_2SO_4[/tex].

For the determination of the cation present in the compound, a confirmatory test is almost necessary.

Therefore, the common cations that can form insoluble sulfate salts that can form insoluble salts with the anion include calcium, barium, strontium and lead.

To know more about cations, visit:

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The cation that can be present in the unknown soluble ionic compound can be Ba2+ since Barium sulfate(BaSO4) is known to be an insoluble sulfate salt which forms a precipitate.

The unknown soluble ionic compound can have cations such as Ba2+ (barium). Here, the solubility is determined based on the interaction of ions in the solution. When a precipitate forms only with the addition of Na2SO4, it suggests that the sulfates form an insoluble compound with the cation present in the unknown solution. However, not with KCl or NaOH. According to solubility guidelines, sulfate salts (SO42-) are typically soluble, with exceptions for salts containing Pb2+, Sr2+, Ca2+, and Ba2+. So, one possible cation in the unknown soluble ionic compound is Ba2+.

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Answer:

outer electron moved to a higher energy state

Explanation:

To know this, let's write again the electron arrangements:

[A] = 1s² 2s² 2p^6 3s^1

[B] = 1s² 2s² 2p^6 5s^1

Now, let's discart the options.

First option cannot be, because if you count the inner electrons, which are the numbers uppering (besides the s and p), in both cases, you have the same number of electrons, which is 11 in both cases.

Second option cannot be either, because electron configuration, always go from lower to higher state. In the case, that one electron move to a lower state, it should move fro 3s to the 2p level. In this case, it was not, mainly because the lower level of 2, has already maxed out it's capacity to hold electrons.

Therefore, option 3 it's the more accurate option, you can see that from 2p6, it jump to the energy level 5, skipping 3 and 4th level. Therefore this is the correct option.

Answer:

The outer electron of atom B has moved to a higher energy state.

Explanation: (Graded correct)

Answer:

Molarity = 1.53 M

Explanation:

At STP,  

Pressure = 1 atm  

Temperature = 273.15 K

Given, Volume = 27.0 L

Using ideal gas equation as:

[tex]PV=nRT[/tex]

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × 27.0 L = n × 0.0821 L.atm/K.mol × 273.15 K  

⇒n = 1.20 moles

Given that volume = 785 mL

Also,  

[tex]1\ mL=10^{-3}\ L[/tex]

So, Volume = 785 / 1000 L = 0.785 L

Considering:

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

[tex]Molarity=\frac{1.20}{0.785}[/tex]

Molarity = 1.53 M

Answer:

Final temperature is 71·31 °C

Explanation:

where

m is the mass of the substance

s is the heat capacity of the substance

ΔT is the temperature difference

where

m is the mass of the substance

L is the latent heat of the substance

Let the final temperature be T

As the density of the water is 1 g/ml

∴ The mass of 90ml water will be 90 g

Heat gained by the ice when it is attained at 0°C and converted to water is

(10×2·08×11) +  10×6 = 288·8

But the heat loss by the water when water attains 0°C will be 90×4·18×80 = 30,096

As the heat loss is more therefore there will be further rise in temperature

Now the heat gain = 288·88 + 10×4·18×T

heat loss = 90×4·18×(80-T)  

Here in the case of heat loss as we are already mentioning that the heat is lost so we are taking the magnitude of the heat change otherwise it would be 90×4·18×(T-80)

288·88 + 41·8×T = 9×41·8×(80-T)

41·8×9×80 - 288·8 = 10×41·8×T

∴T=71·31 °C

To determine the spontaneity of the reaction between N2 and O2 to form 2NO and the equilibrium constant K, one would use the Gibbs Free Energy (ΔG) formula and the equilibrium constant (K) formula. While exact values can't be computed without data for the standard enthalpy and entropy changes, it can be inferred that this reaction is typically not spontaneous at room temperature (298 K), but becomes more spontaneous as temperature increases toward 2000 K.

The given chemical reaction is between nitrogen and oxygen to form nitrogen monoxide: N2(g) + O2(g) = 2 NO(g). To answer your question, we'll need to use the Gibbs Free Energy (ΔG) formula and the equilibrium constant (K) formula which are linked by the equation ΔG = -RTlnK. These formulas are what we use to determine the spontaneity of a reaction and the equilibrium state at a given temperature.

Without the specific values of the standard enthalpy (ΔH) and entropy (ΔS) changes for this reaction, it's impossible to calculate ΔG and K exactly. However, at room temperature (298 K), the reaction is typically not spontaneous because the formation of NO requires high-energy conditions - usually above 2000 K. This is inferred by the presence of NO only in extreme conditions such as lightning strikes or in high-temperature combustion processes.

At 2000 K, although I can't give an exact numerical estimate without the ΔH and ΔS values, we can say that the reaction becomes more spontaneous as temperature increases since high energy and high temperature favor the formation of NO. It's always important to remember that whether a chemical reaction is spontaneous or not depends not only on the change in enthalpy (ΔH), but also on the change in entropy (ΔS) and the absolute temperature (T).

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The reaction N2(g) + O2(g) = 2NO(g) is not spontaneous at 298 K, with a very small equilibrium constant (K). However, at 2000 K, the reaction becomes more spontaneous as the free energy change (ΔG) is negative.

The reaction N2(g) + O2(g) = 2NO(g) has a standard free energy change (ΔG°) of 173.4 kJ/mol at 298 K. To find the equilibrium constant (K) at this temperature, we use the equation ΔG° = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin. Rearranging the equation, we get K = e-ΔG°/(RT).

Substituting the values into the equation and solving, we find K to be 6.95 x 10-31, which is a very small value. Since K is less than 1, the reaction is not spontaneous at 298 K.

To estimate ΔG° at 2000 K, we can use the equation ΔG = ΔH - TΔS, where ΔH is the enthalpy change and ΔS is the entropy change. Assuming ΔH and ΔS do not change significantly with temperature, we can use the same values as at 298 K.

Substituting the values into the equation, we find ΔG to be -545.4 kJ/mol at 2000 K. Since ΔG is negative, the reaction becomes more spontaneous as temperature increases.

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The thermosphere is a layer of Earth's atmosphere located 80-700km above sea level. It supports long-distance communication by reflecting radio waves back to Earth and is also where auroras form due to the collision of charged particles and gas molecules.

The thermosphere is the fourth layer of Earth's atmosphere, located above the mesosphere and extend from about 80 to 700 kilometers above sea level. This layer holds a unique property due to the presence of electrically charged particles, or ions which allows it to support long-distance communication by reflecting radio waves back to Earth, bypassing the need for satellites. This is also the layer in which auroras form. Auroras occur when these charged particles from the sun collide with gas molecules in the Earth's atmosphere, releasing energy that manifests as visible, colorful light.

The Earth's atmosphere is divided into five main layers: The troposphere, the stratosphere, the mesosphere, the thermosphere and the exosphere. The Troposphere being the one closest to the Earth's surface and is where weather is generally observed. This layer expands to a height of roughly 12 km above the sea level and makes up around 80% of the atmosphere's mass. The Thermosphere is not the layer of atmosphere closest to the Earth’s surface where weather typically occurs, this property represents the troposphere.

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The third option is correct.

3. The thermosphere is the layer where auroras form when electrically charged particles from the sun collide with gas molecules releasing energy visible as light of different colors.

Answer:

A) Work (positive)

B) Heat (negative) and work (positive)

C) Work (negative) and heat (negative)

Explanation:

A) This energy change is work given that the first ball is using this energy to move the other one. The system (fist ball) is doing the work so it has + sign.

B) The book loses part of its energy by friction (heat) and part by applying a force to the floor (work). The heat is taken from the system so it has - sign and the work is done by the system and it has + sign.  

C) The daughter is receiving the work done by the father. This work done to the system has - sign. Also, you can say that the system (daughter and swing) loses energy by friction (heat) and because of that it slows down.

Note: the sign of work and heat is defined by convention.

Answer:

(C) substance B is not involved in the rate-determining step of the mechanism but is involved in subsequent steps.

Explanation:

A. Is incorrect because if the order of the reaction with respect B was one then the rate would increase by the same multiple that B is increased by.

B. If B is reactant then it must be involved in the mechanism of the reaction and in the formation of the product

D. If B was a catalyst then increasing it's amount would affect the rate

E. That is just factually untrue. Effect of reactants on rates can only be found experimentally, not stoichiometrically.

The most probable explanation for this observation is (C) substance B is not involved in the rate-determining step of the mechanism but is involved in subsequent steps.

Answer:

a. If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g).

Explanation:

The answer is a) because the value for Kp is really close to zero (having x10⁻¹⁵), this means that at equilibrium O₂ and F₂ will be significantly more present than OF₂.

Final answer:

For the equilibrium O2(g) + 2F2(g) ↔ 2OF2(g) with Kp = 2.3×10^-15, the reaction favors the reactants, making the correct answer that a mixture initially containing only OF2(g) will consist of essentially only O2(g) and F2(g) at equilibrium.

Explanation:

The question considers the equilibrium O2(g) + 2F2(g) ↔ 2OF2(g); Kp = 2.3×10-15 and asks which statement is true. Given the extremely low value of Kp, this indicates a strong preference for reactants at equilibrium. Therefore, the correct answer is: If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g). This is because a very small Kp value means the equilibrium lies heavily on the side of the reactants, making statement (a) true.

Answer:

(a)The molar mass of the gene fragment is 18220.071g/mol = 18.22 kg/mol

(b)The freezing point for the aqueous solution is [tex]-3.413\times10^{-5}[/tex] C

Explanation:

The osmotic pressure (π) is given by the following equation:

[tex]\pi =cRT[/tex]

[tex]c[/tex]= Concentration of solution

R = universal gas constant = 62.364 [tex]\frac{L\times torr}{mol\times K}[/tex]

T = temperature

Weight of solute = w = 10.0 mg

Let the molecular weight of the solute be m g/mol.

Concentration = [tex]c=\frac{n}{V}\\ n=\frac{w}{m}\\ n=\frac{10\times10^{-3}}{m}\\c=\frac{10\times10^{-3}}{m\times30\times10^{-3}}M[/tex]

[tex]m=\frac{RT}{3\pi}[/tex]

m = 18220.071g/mol

Therefore, the molar mass of the gene fragment is 18220.071g/mol = 18.22 kg/mol

[tex]\Delta T_{f}=K_{f}m[/tex]

m is the molality of the solution.

m = [tex]1.835\times10^{-5}[/tex] mol/kg

[tex]\Delta T_{f}=1.86\times m[/tex]

[tex]\Delta T_{f}[/tex] = [tex]3.413\times10^{-5}[/tex] C

The freezing point for the aqueous solution is [tex]-3.413\times10^{-5}[/tex] C

Answer:

-1.71 J/K

Explanation:

To solve this problem we use the formula

ΔS = n*ΔH/T

Where n is mol, ΔH is enthalpy and T is temperature.

ΔH and T are already given by the problem, so now we calculate n:

Molar Mass C₂H₅OH = 46 g/mol

2.71 g C₂H₅OH ÷ 46g/mol = 0.0589 mol

Now we calculate ΔS:

ΔS = 0.0589 mol * −4600 J/mol / 158.7 K

ΔS = -1.71 J/K

The vapor pressure of the solution can be calculated using Raoult's law. First, we need to calculate the mole fraction of glycerol and water in the solution. Then, we can use the mole fraction to calculate the vapor pressure of the solution.

In order to find the vapor pressure of the solution, we can use Raoult's law. Raoult's law states that the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.

First, we need to calculate the mole fraction of glycerol in the solution. To do this, we need to convert the mass percent of glycerol to moles. Since the molar mass of glycerol is 92.1 g/mol and the solution contains 34.4% glycerol by mass, we can calculate the moles of glycerol:

Moles of glycerol = (34.4 g / 92.1 g/mol)

Next, we need to calculate the mole fraction of water in the solution. Since the solution is 34.4% glycerol, the mass percent of water is 100% - 34.4% = 65.6%. Using the molar mass of water (18.0 g/mol), we can calculate the moles of water:

Moles of water = (65.6 g / 18.0 g/mol)

Now we can calculate the mole fraction of glycerol and water:

Mole fraction of glycerol = (moles of glycerol) / ((moles of glycerol) + (moles of water))

Mole fraction of water = (moles of water) / ((moles of glycerol) + (moles of water))

Finally, we can calculate the vapor pressure of the solution using Raoult's law:

Vapor pressure of solution = (mole fraction of water) * (vapor pressure of water)

The jar contains A Mixture

Explanation:

A mixture consists of different atoms that are not chemically bonded. It is of two types of the heterogeneous mixture and homogeneous mixture. The chemical substances can be differentiated visually in the heterogeneous mixture.

In a mixture, the proportion of atoms can be slightly changed without changing  or modifying any substance. This is because the individual substances in mixture keep their properties when mixed together. Further, mixtures have variable compositions and substances. these mixtures can be are separated using physical methods.  

The jar contains A Mixture

Explanation:

A mixture consists of different atoms that are not chemically bonded. It is of two types of the heterogeneous mixture and homogeneous mixture. The chemical substances can be differentiated visually in the heterogeneous mixture.

In a mixture, the proportion of atoms can be slightly changed without changing  or modifying any substance. This is because the individual substances in mixture keep their properties when mixed together. Further, mixtures have variable compositions and substances. these mixtures can be are separated using physical methods.

Answer:

The correct answer is: Oxidation number

Explanation:

According to the IUPAC nomenclature for the inorganic compounds, roman numerals are used to denote the oxidation number of a positively charged ion, called cation.

This rule is only applicable to elements that can form more than one cation with different oxidation state or charge.

The oxidation state or charge of each cation of such an element is denoted or represented by a Roman numeral in the parentheses followed by the name of the element.

Therefore, the oxidation number of a cation is denoted by a roman numeral which is written after the name of the element.

Answer:

The answer to your question is letter A, Argon

Explanation:

Isotope               Atomic mass                      Percent (%)

    1                       35.9675                              0.337

    2                      37.9627                              0.063

    3                      39.9624                            99.6

Average atomic mass = (Mass isotope 1)(percent 1) + (Mass isotope 2)(percent  

                                        2) + (Mass isotope 3)(percent 3)

Average atomic mass = (35.9675)(0.00337) + (37.9627)(0.00063) +

                                       (39.9624)(0.996)

Average atomic mass = 0.1212 + 0.0239 + 39.8025

Average atomic mass = 39.9476

                   Theoretical  Atomic mass

a) Ar                         39.95

b) K                          39.10

c) Cl                         35.45

d) Ca                       40.08

Answer: [tex]156.4^0C[/tex]

Explanation:

[tex]2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542kJ[/tex] [tex]\Delta H=-6542kJ[/tex]

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.700g}{78.11g/mol}=0.9858moles[/tex]

According to stoichiometry :

2 moles of [tex]C_6H_6[/tex] releases = 6542 kJ of heat

0.9858 moles of [tex]C_6H_6[/tex]  release =[tex]\frac{6542}{2}\times 0.9858=3224kJ[/tex] of heat

Thus heat given off by burning 7.700 g of [tex]C_6H_6[/tex]  will be absorbed by water.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat absorbed= 3224kJ = 3224000J   (1kJ=1000J)

m= mass of water = 5691 g

c = specific heat capacity = [tex]4.184J/g^0C[/tex]

Initial temperature of the water = [tex]T_i[/tex] = 21.0°C

Final temperature of the water = [tex]T_f[/tex]  = ?

Putting in the values, we get:

[tex]3224000J =5691g\times 4.184J/g^0C\times (T_f-21)[/tex]

[tex]T_f=156.4^0C[/tex]

The final temperature of the water is [tex]156.4^0C[/tex]

Answer: 181 kJ

Explanation:

The balanced chemical reaction is;

[tex]C_2H_5OH(l)+3O_2(g)\rightarrow 2CO_2(g)+3H_2Ol)[/tex] [tex]\Delta H=-555jJ[/tex]

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{15.0g}{46.0g/mol}=0.326moles[/tex]

According to stoichiometry:

1 mole of [tex]C_2H_5OH[/tex] on complete combustion give= 555 kJ

Thus 0.326 moles of [tex]C_2H_5OH[/tex] on complete combustion give=[tex]\frac{555}{1}\times 0.326=181kJ[/tex]

Thus the enthalpy change for combustion of 15.0 g of ethanol is 181 kJ

Answer:

Personal Referral.

Explanation:

This technique of finding employees is called Personal Referral. In personal referral the existing employee of the company uses hi/her own connections to find employees for the company. He/she may ask his/her asks current customers for names of friends, customers and other businesses that might be interested in job provided by the company.

Final answer:

Mary is employing a 'Word of Mouth' marketing strategy by asking current customers for referrals, tapping into the informal network to reach potential customers with the trust built through personal recommendations.

Explanation:

Mary, a sales representative, is utilizing a marketing strategy known as Word of Mouth. By asking current customers for referrals, she is tapping into her business's informal network to identify potential customers who might be interested in the cleaning services her company provides.

This approach takes advantage of the trust and reliability found within personal networks, as individuals are more likely to believe and try services recommended by people they know and trust. Additionally, Mary's strategy could involve engaging with information hubs or influencers within her network who are more likely to share information about the cleaning services with a broader audience.

In the context of increasing sales, connecting with customers on a more personal level and leveraging their networks can be an effective method to gain new leads. This is particularly effective as it has been shown that a vast majority of jobs, similar to customer opportunities, are found through social networks, highlighting the power of personal contacts and referrals. Mary's approach can be seen as a grassroots method to increase customer outreach and foster organic growth for the business.

Answer:

Covalent network solids typically have high melting points and high boiling points

Explanation:

Covalent solids are the solids which have covalent bonds as intermolecular or interatomic interactions.

As covalent bonds are very strong, these solids are generally characterized by high melting points and high boiling points.

The examples of such solids are: Diamond and Graphite.

The other classes of solids are

i) ionic

ii) molecular

iii) metallic.

Answer:

The answer is NO₂-NO₂, N₂O₄

Explanation:

If 23.0 g of this gas were found to occupy 5.6 L at STP we can apply the Ideal gas Law to determinate the quantity of mols of the oxide.

As I have the mass, I can find out the molar weight.

STP are 1 atm and 273, 1 K so:

P . V = n . R . T

1 atm .  5.6L = n . 0.082 L.atm/mol.K . 273.1K

(1 atm .  5.6L)/ 0.082 mol.K/L.atm . 273.1K = n

0.250 mol = n

Molar weight = mass/mols

23 g /0.250 mol = 92 g/m

The percentages are:  30.4% N and 69.6%O

100% ____ 92 g

30.4% _____ (30.4 .92)/100 = 28 g → 2 mols of N

100% _____ 92 g

69.6% ____ (69.6 . 92)/ 100 = 64 g → 4 mols of O

Molecular formula N₂O₄

Empirical formula: NO₂-NO₂

Answer: a. Synthesis

Explanation:

a. Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

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

Thus magnesium in its elemental form is combining with oxygen to form magnesium oxide.

b. Double displacement reaction is one in which exchange of ions take place.  

Example: [tex]2NaOH(aq)+(NH_4)_2SO_4(aq)\rightarrow 2NH_4OH(aq)+Na_2SO_4(aq)[/tex]

c. Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

Example: [tex]Li_2CO_3\rightarrow Li_2O+CO_2[/tex]

Final answer:

Nonmetals generally gain electrons to form anions in reactions with metals, creating ionic compounds, or they share electrons in reactions with each other to form molecular compounds.

Explanation:

When nonmetals react with other elements, they typically undergo a process where they gain electrons to form anions, or negative ions, as they combine with metals to produce ionic compounds. In reactions with each other, nonmetals often form molecular compounds with covalent bonding, where electrons are shared rather than transferred. The most reactive nonmetals, like fluorine, have a particularly high electron affinity and can take part in vigorous redox reactions, changing their oxidation states. Considering the diverse group of nonmetals, these elements can form a variety of different compounds, both ionic and covalent, depending on the elements they react with and the environmental conditions.

Answer:

Metalloids have properties of metals and non-metals.

Explanation:

A metalloid is a type of chemical element which has properties in between, or that are a mixture of, those of metals and nonmetals.

The six commonly recognized metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium.

Metalloids have a metallic appearance, but they are brittle (nonmetal property) and only fair conductors of electricity. Chemically, they behave mostly as nonmetals. They can form alloys with metal.

Most of their other physical properties and chemical properties are intermediate in nature.

That means Metalloids  have properties of both metal and nonmetal.

Answer:

2Mg  +  O2 →  2MgO

Explanation:

In all conbustion you should know, that reactans are an specific compound and O2, so the products must be CO2 and H2O, or in this case, the corresponding oxide.

The balanced equation for the combustion of Magnesium is 2Mg + O2 -> 2MgO. It's a redox reaction where Magnesium is oxidised and Oxygen is reduced.

The combustion of Magnesium in Oxygen is a well-known reaction in Chemistry. The balanced equation for this reaction is 2Mg + O2 -> 2MgO. In this reaction, two moles of Magnesium react with one mole of Oxygen gas to produce two moles of Magnesium Oxide. This is a redox reaction where Magnesium gets oxidised and Oxygen gets reduced.

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Answer:can change as new researches and experiments are done

Explanation:

Answer:

55.7%

Explanation:

The reaction that takes place is:

With the volume and concentration of NO₃⁻ solution, we can calculate the moles of Sn²⁺ that reacted:

Now we convert moles of Sn to mass, using its atomic weight:

Finally we calculate the percent by mass of Sn:

Answer:

mass water = 32.4 g

Explanation:

specific heat iron = 0.450 J/g°C

specific heat water = 4.18 J/g°C

32.8 x 0.450 ( 59.1 - 22.4) + mass water x 4.18 ( 59.1- 63.1)=0

541.7 - mass water x 16.7 = 0

mass water = 32.4 g

Answer:

d. n H2(g) = 0.034 mol

Explanation:

                              2 - Na - 2

                              4 - H - 4

                               2 - O - 2

∴ n Na(s) = 0.068 mol

⇒ n H2(g) = ( 0.068 mol Na(s) )( mol H2(g) / 2 mol Na(s) )

⇒ n H2(g) = 0.034 mol

Based on the stoichiometric relationship between sodium and hydrogen in the given balanced chemical equation, 0.068 mol of sodium would yield 0.034 mol of hydrogen gas.

The correct answer is (b) 0.034 mol H2. The balanced chemical equation given states that 2 moles of sodium metal (Na) react with 2 moles of water molecules (H2O) to produce 2 moles of sodium hydroxide (NaOH) and 1 mole of hydrogen gas (H2). This implies a stoichiometric ratio of 2:1 between sodium and hydrogen. Therefore, if we have 0.068 moles of sodium, we would obtain half of this number in moles of hydrogen because of this stoichiometric relationship. 0.068 mol ÷ 2 = 0.034 mol H2.

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The density of a metal that crystallizes in a face-centered cubic unit cell can be calculated by determining the number of atoms per unit cell, calculating the volume of a unit cell, and finally calculating the mass of the atoms in the unit cell using the given molar mass. The density is then found by dividing the mass by the volume.

To answer this question, we need to know a couple of key pieces of information. First, a face-centered cubic unit cell consists of four atoms: one-eighth of an atom at each of the eight corners and half of an atom on each of the six faces (1/8*8+1/2*6=4 atoms).

Secondly, we need to find the volume of the atom that is engulfed in the cubic unit cell. Given that the radius of the metal is 234 picometers (or 234*10^-12 meters), the volume of the unit cell can be found by applying the formula for the volume of a cube (side^3) where the side equals 2*sqrt(2)*r.

We can then calculate the number of moles of atoms in the unit cell, convert them to grams using the molar mass, and finally calculate the density by dividing the calculated mass by the calculated volume.

We first convert the volume of the unit cell from cubic meters to cubic centimeters (1m^3 = 10^6 cm^3).

Then, we calculate the mass of the atoms in the unit cell using the molar mass (195.08 g/mol).

Finally, we calculate the density. Density = Mass/Volume.

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To calculate the density of a metal in a face-centered cubic structure, you need to determine the volume of the unit cell and the molar mass of the metal.

The density of a metal can be calculated using the formula:

Density = (Molar Mass ×Avogadro's Number) / (Volume of Unit Cell)

First, let's calculate the volume of the face-centered cubic (FCC) unit cell. The FCC unit cell consists of 4 atoms at the corners and 8 atoms at the face centers. The radius of the atom can be used to calculate the edge length of the unit cell using the formula:

Edge Length = 4×Radius / √2

Once we have the edge length, we can calculate the volume of the unit cell using the formula:

Volume of Unit Cell = Edge Length³

Finally, we can plug the values into the density formula and calculate the density of the metal.

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Answer:

2Fe + 3Br2 = 2FeBr3

Explanation: