Calculate the actual yield for the production of ammonia gas (nh3) from hydrogen and nitrogen gases if the percent yield is 68.2% and you begin with 2.00 kg of nitrogen gas

An alloy, such as brass or bronze, is formed when two pieces of different metals are stuck together lengthwise. These materials combine the properties of the constituent metals, often resulting in superior characteristics.

When two pieces of different metals are stuck together lengthwise, the resulting material is known as an alloy. An alloy is a mixture composed of two or more elements, with at least one being a metal. Some common examples of alloys include brass (an alloy of copper and zinc) and bronze (an alloy of copper and tin). These combinations of metals result in a material with properties that are superior to the pure metals themselves. For example, bronze, first used around 2400 B.C., is harder and more durable than either of its constituent metals, copper and tin.

The material formed from sticking two different metals together lengthwise is called an alloy. Alloys like bronze and brass are mixtures of two or more elements that possess improved properties compared to the pure component metals.

When two pieces of different metals are stuck together lengthwise, the material formed is known as an alloy. An alloy is a mixture composed of two or more elements, at least one of which is a metal. Alloys are designed to have properties superior to those of the pure metals from which they are made. For example, bronze is a well-known alloy consisting mainly of copper and tin. The creation of alloys like bronze and brass, which is an alloy of copper and zinc, exemplifies how metals can be combined to achieve desired physical and chemical properties for various applications, from tools and weapons to musical instruments.

Sample Response: The flat sheet will completely rust before the iron cube. Since they both have the same volume, the flat sheet has more surface area than the small cube. This means more particles are exposed on the flat sheet that can react in a chemical reaction.

The upper bound estimate of the composite's elastic modulus is approximately 650 GPa, and the lower bound estimate is approximately 560 GPa, using the Reuss model.

To estimate the elastic modulus of the composite, we use the rule of mixtures. The Voight model and the Reuss model are the two estimates.

Upper Bound (Voigt Model): The Voigt model assumes that the strain in both materials is the same. The formula is:

Lower Bound (Reuss Model): The Reuss model assumes that the stress in both materials is the same. The formula is:

Result: The upper bound estimate of the elastic modulus of the 10 vol% Co – 90 vol% WC composite is 650 GPa, and the lower bound estimate is 560 GPa.

The solubility of lead hydroxide in pure water is approximately 1.5 × [tex]10^{-7}[/tex] M.

Solubility of Lead Hydroxide in Pure Water

To find the solubility of lead hydroxide, Pb(OH)2, in pure water, we start with the dissociation equation:

Pb(OH)2 (s) → Pb2+ (aq) + 2 OH- (aq)

The solubility product constant (Ksp) for this reaction is given as 1.4 × [tex]10^{-20}[/tex]. Let s be the molar solubility of Pb(OH)2, which means s moles of Pb(OH)2 dissolve to form s moles of Pb2+ and 2s moles of OH-:

Ksp = [Pb2+][tex][OH-]^2[/tex]

Substitute the concentrations: Ksp = (s)(2s)2

Ksp = 4[tex]s^3[/tex]

Given Ksp = 1.4 × [tex]10^{-20}[/tex], we solve for s:

1.4 ×[tex]10^{-20}[/tex] = 4[tex]s^3[/tex]

[tex]s^3[/tex] = (1.4 × [tex]10^{-20}[/tex]) / 4

[tex]s^3[/tex] = 3.5 × [tex]10^{-21}[/tex]

s = (3.5 × [tex]10^{-21}[/tex])[tex]^{1/3}[/tex]

s ≈ 1.5 × [tex]10^{-7}[/tex] M

Thus, the solubility of lead hydroxide in pure water is approximately 1.5 × [tex]10^{-7}[/tex] M.

The solubility of lead hydroxide in pure water, neglecting the contribution of OH- from water, is approximately [tex]\( 1.52 \times 10^{-7} \)[/tex] M.

The solubility of lead hydroxide, [tex]Pb(OH)_2[/tex], in pure water can be determined from its solubility product constant (Ksp). The Ksp expression for [tex]Pb(OH)_2[/tex] is given by:

[tex]\[ K_{sp} = [Pb^{2+}][OH^-]^2 \][/tex]

Let [tex]\( s \)[/tex] represent the molar solubility of [tex]Pb(OH)_2[/tex], which is the concentration of [tex]Pb^{2+}[/tex] ions in solution at equilibrium. Since for every mole of [tex]Pb(OH)_2[/tex] that dissolves, 1 mole of [tex]Pb^{2+}[/tex] and 2 moles of OH- are produced, the concentration of OH- ions will be [tex]\( 2s \)[/tex].

Therefore, the Ksp expression can be rewritten as:

[tex]\[ K_{sp} = s(2s)^2 \][/tex]

[tex]\[ K_{sp} = 4s^3 \][/tex]

Given that [tex]\( K_{sp} = 1.4 \times 10^{-20} \)[/tex], we can solve for [tex]\( s \)[/tex]:

[tex]\[ 1.4 \times 10^{-20} = 4s^3 \][/tex]

[tex]\[ s^3 = \frac{1.4 \times 10^{-20}}{4} \][/tex]

[tex]\[ s^3 = 3.5 \times 10^{-21} \][/tex]

[tex]\[ s = \sqrt[3]{3.5 \times 10^{-21}} \][/tex]

Taking the cube root of both sides gives us the solubility [tex]\( s \)[/tex]:

[tex]\[ s = (3.5 \times 10^{-21})^{1/3} \][/tex]

[tex]\[ s \approx 1.52 \times 10^{-7} \text{ M} \][/tex]

Hydrolysis of [tex]NH_{4}ClO_{4}[/tex] is given as:

[tex]NH_{4}ClO_{4}+H_{2}O\rightleftharpoons NH_{4}OH + HClO_{4}[/tex]

Here, [tex]NH_{4}OH[/tex] is a   weak base and [tex]HClO_{4}[/tex] is a strong acid. Thus, solution is more acidic

Hydrolysis of [tex]KBr[/tex] is given as:

[tex]KBr+H_{2}O\rightleftharpoons KOH  +   HBr[/tex]

Here, [tex]KOH[/tex] is a strong base and [tex]HBr[/tex] is a strong acid.Thus, solution is neutral.

Hydrolysis of [tex]NaCl[/tex] is given as:

[tex]NaCl+H_{2}O\rightleftharpoons NaOH  + HCl[/tex]

Here, [tex]NaOH[/tex] is a strong base and [tex]HCl[/tex] is a strong acid.Thus, solution is neutral.

[tex]Na_{2}CO_{3}+2H_{2}O\rightleftharpoons  2NaOH + H_{2}CO_{3}[/tex]

Here, [tex]NaOH[/tex] is a strong base and [tex]H_{2}CO_{3}[/tex] is a weak acid.Thus, solution is basic.      

[tex]NaHCO_{3}+H_{2}O\rightleftharpoons  NaOH + H_{2}CO_{3}[/tex]

Here, [tex]NaOH[/tex] is a strong base and [tex]H_{2}CO_{3}[/tex] is a weak acid.Thus, solution is basic.        

Hence, an aqueous solution of  [tex]NaHCO_{3}[/tex] and [tex]Na_{2}CO_{3}[/tex] will produce a basic solution.

In the formation of sodium chloride, two steps result in a release of energy: attraction of Na+ and Cl- to form NaCl, and the conversion of Na(s) to Na(g) as the sodium atom donates an electron to chlorine atoms.

The formation of an ionic bond involves the transfer of electrons from one atom to another. In the case of sodium chloride (NaCl), two steps result in a release of energy. The first step is the attraction of Na+ and Cl- ions to form NaCl. The second step is the conversion of Na(s) to Na(g) as the sodium atom donates an electron to chlorine atoms.

To find the amount of argon to fill a 0.475 L light bulb at STP, one must calculate the number of moles using the ideal gas law and then multiply by the molar mass of argon to convert moles to grams.

To determine how many grams of argon it would take to fill a light bulb with a volume of 0.475 L at STP (standard temperature and pressure), we can use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. At STP, P is 1 atm and T is 273.15 K. The volume of the light bulb is already given as 0.475 L.

First, we can find the number of moles of argon that would occupy the 0.475 L volume at STP. The gas constant R in units that match our pressure in atm, volume in liters, and temperature in Kelvin is 0.0821 L·atm/K·mol. Therefore, n = PV / RT = (1 atm × 0.475 L) / (0.0821 L·atm/K·mol × 273.15 K). Calculating this value gives us the number of moles of argon.

Now, to find the mass in grams, we need to use the molar mass of argon which is 39.95 g/mol. The mass m can be calculated by multiplying the number of moles n by the molar mass of argon. The mass m = n × molar mass of argon will give us the final answer in grams.

Answer:

Π2py and Π2pz

Explanation:

In the molecular orbital electron configuration of the boron molecule B2, the electrons are arranged as follows:

KK,σ2s^2,σ*2s^2, Π2py , Π2pz.

The two highest level orbitals are degenerate and are both singly occupied in the boron molecule as shown in the molecular orbital electron configuration above.

Note KK refers to the core electrons.

Answer:

b. an attraction between the positive end of one molecule and the negative end of another

Explanation:

An attraction between the positive end of one molecule and the negative end of another. Hence, option B is correct.

Dipole-Dipole forces are the interaction between molecules of the permanent dipole. It occurs between the partially charged positive molecules and partially charged negative molecules.

Dipole-dipole forces are attractive forces between the positive end of one polar molecule and the negative end of another polar molecule.

Hence, an attraction between the positive end of one molecule and the negative end of another is a cause of dipole-dipole force.

Hence, option B is correct.

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Redox reaction is the chemical reaction in which both oxidation (loss of electrons) and reduction (gain of electrons) takes place or transfer of electrons between two species occurs.

The given chemical equation is :

[tex]Zn(s)+2HCl(aq)\rightarrow ZnCl_{2}(aq)+H_{2}(g)[/tex]

Here, in reactant side, the oxidation state of zinc is zero (0), after reacting with hydrogen chloride, the oxidation sate of zinc becomes two (+2).  In this reaction, electrons are transferred from zinc atoms to the hydrogen atoms. Thus, zinc is oxidized by losing electrons.

Hence, oxidation state of zinc increases due to loses electrons i.e. Option (A) is correct.

Answer:

A - 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s), ∆H = -3,926kj

Explanation:

Thermochemical equations:

The correct thermochemical equation is :

4Fe(s)  +  3O₂(g)  --->  Fe₂O₃(s)     ΔH = -3,926 kJ. The correct option is first.

The thermochemical equation is the equation which is balanced chemical equation that shows magnitude of the enthalpy value The enthalpy value with the sign that is the positive sign means it is endothermic process, and the negative sign is an exothermic process.

4Fe(s)  +  3O₂(g)  --->  Fe₂O₃(s)     ΔH = -3,926 kJ

This is the thermochemical equation.

NH₄Cl -->  NH₄⁺  +Cl⁻

This is not the thermochemical as it does not the enthalpy value.

C₃H₈(g)   +  O₂(g)  ---> CO₂(g)  +  H₂O(l)    ΔH = -2,220 kJ/mol

This is also not the thermochemical equation.

2C₈H₁₈  +  25O₂  ---> 6CO₂ +  18H₂O      ΔH = - 5,471 kJ/mol

This is not the thermochemical equation it is not balanced equation. The first option is correct.

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Final answer:

To synthesize 2-methyl-2-butene using the Wittig reaction, ethyltriphenylphosphonium bromide reacts with acetaldehyde. The resulting oxaphosphetane intermediate yields 2-methyl-2-butene and triphenylphosphine oxide upon decomposition.

Explanation:

To synthesize 2-methyl-2-butene using the Wittig reaction, the chemist should choose appropriate reactants that will couple to form the desired alkene. The necessary phosphonium ylide would need to possess a suitable leaving group that, after the reaction with the carbonyl-containing molecule, results in 2-methyl-2-butene.

For this specific synthesis, the chemist would need ethyltriphenylphosphonium bromide as the ylide precursor. This compound, when deprotonated, forms a ylide that can react with acetaldehyde (ethanal). Upon reacting, the phosphonium ylide and acetaldehyde undergo nucleophilic addition to form the oxaphosphetane intermediate, which then yields 2-methyl-2-butene and triphenylphosphine oxide upon reductive elimination.

The Wittig reaction is a reliable method for synthesizing alkenes, and it has the advantage that the location of the double bond in the product is fixed, yielding a specific geometric isomer rather than a mixture of isomers.

Based on the values of δH°rxn, reactions can be classified as thermodynamically favorable or unfavorable. Reaction 1 and reaction 3 are thermodynamically favorable, while reaction 2 is thermodynamically unfavorable.

Based on the values of δH°rxn, we can determine whether a reaction is thermodynamically favorable or unfavorable. A negative δH°rxn indicates a thermodynamically favorable reaction, while a positive δH°rxn indicates a thermodynamically unfavorable reaction. Let's analyze the given reactions:

Therefore, reaction 1 and reaction 3 are thermodynamically favorable, while reaction 2 is thermodynamically unfavorable.

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Final answer:

The nitrogen atom in HNO3 and the resulting nitrate ion after dissociation in water both exhibit sp² hybridization. Thus, there is no change in hybridization of the nitrogen atom during this dissociation process.

Explanation:

When nitric acid, HNO3, dissociates in water, the nitrogen atom undergoes a change in hybridization. In nitric acid, the nitrogen is sp² hybridized as it is bonded to three oxygen atoms and has one lone pair of electrons. Upon dissociation into nitrate ions (NO3⁻) and hydronium ions (H3O⁺), the nitrogen atom in the nitrate ion becomes sp² hybridized as the lone pair used to create the coordinate bond with hydrogen in HNO3 is lost and replaced by a bond with an oxygen atom to form the nitrate ion. Therefore, the hybridization of nitrogen does not change; it remains sp² before and after dissociation in water.

Note: hydrogen ion concentration = 3.16 x 10⁻¹² M

          hydroxide ion concentration = 3.16 x 10⁻³ M

          pH NH₃ = 11.50

Asked: concentrations of hydrogen ions and hydroxide ions in household ammonia?

Answer: pH (NH₃) = 11.50.

              pH = -log [H⁺]

              [H⁺] = 10∧ (-pH)

              [H⁺] = 10∧ (-11.5)

               [H⁺] = 3.16 · 10⁻¹² M

              [H⁺] x [OH⁻] = 10⁻¹⁴ M²

              [OH⁻] = 10⁻¹⁴ M² ÷ 3.16 x 10⁻¹² M

               [OH⁻] = 0.00316 M = 3.16 x 10⁻³ M

Thus, the concentration of hydrogen ions and hydroxide ions in household ammonia is 3.16 · 10⁻³M

In chemistry, concentration is a measure that describes the amount of substance in a mixture divided by the total volume of the mixture. There are four kinds of quantitative descriptions of concentration, namely mass concentration, molar concentration, total concentration, and volume concentration. The term concentration can be applied to all types of mixtures, but it is most often used to describe the amount of solute in the solution. Molar concentrations have variations such as normal concentration and osmotic concentration.

Solution Concentration

Concentration is a way to express the quantitative relationship between solute and solvent. Expressing the concentration of the solution there are several types, including:

1. MOL FRACTION

The mole fraction is the ratio between the number of moles of a component with the number of moles of all components contained in a solution.

The mole fraction is denoted by X.

2. PERCENT WEIGHT

Percent weight states the gram weight of the solute in 100 grams of solution.

3. MOLALITY (m)

Molality states the mole of solute in 1000 grams of solvent.

4. MOLARITY (M)

Molarity states the number of moles of solute in 1 liter of solution.

5. NORMALITY (N)

Normality represents the number of moles equivalent of solute in 1 liter of solution. For acids, 1 mole is equivalent to 1 mole of H + ions. For bases, 1 mole is equivalent to 1 mole of OH- ion.

Between Normality and Molarity there is a relationship:

N = M x valence

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Subject: Chemistry

keywords: concentration

A racemic mixture of d-2-butanol and l-2-butanol has no net optical rotation, resulting in a rotation of 0 degrees.

A racemic mixture is composed of equal amounts of enantiomers, which in this case are d-2-butanol and l-2-butanol.

When the d and l isomers are present in equal amounts, as they are in a racemic mixture, their optical activities cancel each other out. This is because the rotation caused by one enantiomer is exactly counteracted by the rotation caused by the other enantiomer. Therefore, the overall rotation of polarized light by the racemic mixture is the sum of the individual rotations:

Rotation of d-2-butanol: +13.5 degrees

Rotation of l-2-butanol: -13.5 degrees

Sum of rotations: +13.5 degrees + (-13.5 degrees) = 0 degrees

Thus, the racemic mixture does not rotate polarized light, and the rotation of the polarization of light of the mixture is 0 degrees.

To find the mass of solid sea salt needed to produce the given gas volume of hydrochloric acid (HCl), we need to use a balanced chemical equation and the molar ratio between the reactants. The volume of HCl(g) can be converted to moles using the ideal gas law, and then the moles of the other reactant can be calculated using the molar ratio. The mass of the solid sea salt can be found by converting the moles to grams using the molar mass.

To find the mass of solid sea salt needed to produce the given gas volume of hydrochloric acid (HCl), we need to use the balanced chemical equation. The equation you provided is not directly related to the reaction between hot lava and seawater, so it cannot be used to determine the mass of sea salt. However, we can use the equation HCl(aq) + Mg(OH)2(aq) → H2O(l) + MgCl2(aq) as an example.

In this equation, we can see that 1 mole of HCl reacts with 1 mole of Mg(OH)2 and produces 1 mole of water. From the equation, we can calculate the molar ratio between HCl and Mg(OH)2.

If we know the volume of HCl(g) produced, we can use the ideal gas law to convert it to moles. Then, using the molar ratio, we can calculate the moles of Mg(OH)2 needed. Finally, we can convert the moles of Mg(OH)2 to grams using its molar mass to find the mass of solid sea salt needed.

Considering the organization of the periodic table, (option C) Bromine (Br) and Iodine (I) are the most similar in their chemical properties because they are both halogens in group 17.

Elements in the same group (vertical column) have similar chemical properties because they have the same number of valence electrons.

Other pairs listed belong to different groups and therefore have different chemical properties.

In conclusion, the most similar pair of elements in terms of their chemical properties is Bromine (Br) and Iodine (I).

The complete question is as follows:

Which pair of elements do you expect to be most similar in their chemical properties?

A) Ne and N

B) O and Si

C) Br and I

D) K and Fe

Answer:

[tex]21.994KJ[/tex]

Explanation:

The acetone has the following molecular formula :

C3H6O

The molar mass of the acetone is

μ(C3H6O) = [tex]58.08 \frac{g}{mole}[/tex]

You can find this value in any table.

The molar mass means the mass that has 1 mole of C3H6O

Now, the heat follows this equation :

[tex]Q=(n).(LHV)[/tex]

Where Q is the heat, n is the number of moles of the substance and LHV is the latent heat of vaporization of the substance.

Let's calculate n

If 1 mole of C3H6O has a mass of 58.08 g ⇒

x moles of C3H6O will have a mass of 43.9 g ⇒

[tex]x=\frac{(43.9)}{(58.08)}=0.7558mole[/tex]

In 43.9 g of C3H6O there is 0.7558 moles of C3H6O

The LHV for C3H6O is [tex]29.1\frac{KJ}{mole}[/tex]

You can find this value in any table

To find the amount of heat :

[tex]Q=(n).(LHV)=(0.7558mole)(29.1\frac{KJ}{mole})=21.994KJ[/tex]

We find out that 21.994 KJ are required to vaporize 43.9 g of acetone (C3H6O).

Using Dalton's Law to find the partial pressure of hydrogen gas, we can subsequently use the Ideal Gas Law to calculate the number of moles of hydrogen gas in the mixture.

This question revolves around the concepts of gas law chemistry, specifically Dalton's Law and the Ideal Gas Law. According to Dalton's Law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of its components. Hence, the partial pressure of hydrogen gas can be obtained by subtracting the partial pressure of the water vapor from the total pressure (755 torr - 24 torr = 731 torr). Then, we can apply the Ideal Gas Law, PV = nRT, to calculate the amount of hydrogen gas in moles. We substitute the known values: P=731 torr, V=1.55 L, R=62.36 L Torr K−1mol−1, and T=298 K. Solving for 'n' (number of moles), we get n = PV/RT, which gives us the required quantity of hydrogen gas in the mixture.

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Answer: Option (c) is the correct answer.

Explanation:

Entropy means the degree of randomness present in a substance or within the reactants in a chemical reaction.

Change in entropy is represented by [tex]\Delta S[/tex]. More is the degree of randomness present more positive will be the value of [tex]\Delta S[/tex]. Similarly, less is the degree of randomness present within a substance lesser will be the value of [tex]\Delta S[/tex].

(a)    [tex]2NO(g) + O_{2}(g) \rightarrow 2NO_{2}(g)[/tex]

Here, 3 moles of reactants are giving 2 moles of product. Hence, entropy is decreasing so, the value of [tex]\Delta S[/tex] is negative.

(b)   [tex]2N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)[/tex]

Here, 5 moles of reactants are giving 2 moles of product.  Hence, entropy is decreasing so, the value of [tex]\Delta S[/tex] is negative.

(c)   [tex]C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)[/tex]

Here, 6 moles of reactants are giving 7 moles of product.  Hence, entropy is increasing so, the value of [tex]\Delta S[/tex] is positive.

(d) [tex]Mg(s) + Cl_{2}(g) \rightarrow MgCl_{2}(s)[/tex]

Here, 2 moles of reactants are giving 1 mole of product.  Hence, entropy is decreasing so, the value of [tex]\Delta S[/tex] is negative.

(e)  [tex]C_{2}H_{4}(g) + H_{2}(g) \rightarrow C_{2}H_{6}(g)[/tex]

Here, 2 moles of reactants are giving 1 mole of product.  Hence, entropy is decreasing so, the value of [tex]\Delta S[/tex] is negative.

Thus, we can conclude that [tex]\Delta S[/tex] is positive for the reaction [tex]C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)[/tex].

The volume of NaOH dispensed is 23.86 mL. The number of moles of NaOH dispensed is 0.00370 moles. The balanced molecular equation for the neutralization reaction is HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq).

The volume of NaOH dispensed can be calculated by subtracting the initial buret reading from the final buret reading. In this case, the initial reading is 0.33 mL and the final reading is 24.19 mL. So the volume of NaOH dispensed is 24.19 mL - 0.33 mL = 23.86 mL.

To calculate the number of moles of NaOH dispensed, we can use the formula:

moles = concentration × volume (L)

Since the concentration of NaOH is 0.1550 M and the volume is 23.86 mL (or 0.02386 L), the number of moles of NaOH dispensed is 0.1550 M × 0.02386 L = 0.00370 moles.

The balanced molecular equation for the neutralization reaction between HCl and NaOH is:

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

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Final answer:

After 4 half-lives, a radioactive substance initially weighing 64 g would be reduced to 4.0 g. The mass lost during a fusion reaction that yields 4.50 x 10^9 kJ is 5.00 x 10^-5 kg. The energy yield most likely from a fission or fusion reaction is 2.0 x 10^11 kJ/mol.

Explanation:

To solve each problem, we need to understand the concept of half-life, which is the time required for half the atoms in a sample of a radioactive substance to decay. We also use Einstein's mass-energy equivalence principle for converting between mass lost and energy produced.

Question 1:

If you have 4.0 g of a radioactive substance left after 4 half-lives have passed, the original amount can be found by doubling the remaining mass for each half-life. So, after 1 half-life, we would have 8 g; after 2 half-lives, 16 g; after 3 half-lives, 32 g; and after 4 half-lives, 64 g.

Answer: b. 64 g

Question 2:

Using the formula E=mc2, where E is energy, m is mass, and c is the speed of light, we can rearrange to find m=E/c2. Plugging in the given energy and the value for the speed of light, we find that 4.50 x 109 kJ of energy corresponds to a mass loss of 5.00 x 10-5 kg.

Answer: a. 5.00 x 10

-5 kg

Question 5:

Fission and fusion reactions typically release a vast amount of energy compared to chemical reactions. Option d, which represents an energy yield of 2.0 x 1011 kJ/mol, is far greater than the others and is thus most likely to be from a fission or fusion reaction.

Answer: d. 2.0 x 10

11 kJ/mol

The partial pressures of nitrogen and hydrogen, after decomposing ammonia are calculated using the reaction stoichiometry and Avogadro's law. The nitrogen's partial pressure is 2.1325 x 102 mmHg and the hydrogen's partial pressure is 6.3975 x 102 mmHg.

To calculate the partial pressures of N2 and H2, we will depend on the reaction stoichiometry given by the equation N₂(g) + 3H₂(g) = 2NH3(g). In this chemical reaction, one molecule of nitrogen gas reacts with three molecules of hydrogen gas to produce two molecules of ammonia gas. An important thing to remember here is that gases react in definite and simple proportions by volume, which is derived from Avogadro's law.

The decomposition of ammonia produces 1 volume of nitrogen for every 3 volumes of hydrogen - so, in a decomposed sample, the hydrogen will always be three times as concentrated as the nitrogen. So if we let nitrogen's partial pressure be x, the hydrogen's will be 3x. As the total pressure is given as 853 mmHg, it will be the sum of the nitrogen and hydrogen pressures and we can write:

x + 3x = 853,

Solving for x gives x = 213.25 mmHg, so the nitrogen's partial pressure is 213.25 mmHg and hydrogen's pressure is 3 * 213.25 = 639.75 mmHg. Therefore, the partial pressures of nitrogen and hydrogen are 2.1325 x 102 mmHg and 6.3975 x 102 mmHg respectively in scientific notation.

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