48g of Magnesium reacts with excess oxygen to produce 84g of magnesium oxide. Calculate how much magnesium oxide would be produced if 2g, 10, and 4kg were reacted with excess oxygen. 0.4 Moles of iron reacts with 0.3 moles of oxygen gas. Calculate the formula of the iron oxide produced and write a balanced equation for it.

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Calcium carbonate is the active ingredient in agricultural lime.

Calcium carbonate is the calcium salt with formula CCaO3. It has a role as an antacid, a food coloring, a food firming agent and a fertilizer and much more. It is a calcium salt, a carbonate salt, a one-carbon compound and an inorganic calcium salt.

Calcium carbonate is used therapeutically as a phosphate buffer.

488 moles of calcium carbonate (CaCO3) can be calculated by using its molecular weight. Calcium carbonate has a molecular weight of 100.0869 g/mol. To calculate the mass of 488 moles, we can multiply the molecular weight by the number of moles:

mass = molecular weight x number of moles

mass = 100.0869 g/mol x 488 mol

mass = 48,842.4072 g

So, the mass of 488 moles of calcium carbonate is approx. 48,842.4072 grams.

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Element (Si), Element (Oxygen) in molecular form (O2), Compound (H2O2), Compound (Fe2O3), Compound (CH4).

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the concentration of glucose inside a cell is 0.1 mM and the cell is suspended in a glucose solution of 0.01 mM: =  [tex]5.93* 10^{-6} KJ[/tex]

Part A:  Glucose concentration inside cell = 0.1 mm - concentration of glucose outside cell= 0.01 mm

Temperature = 2.73 + 37 = 310k

Gibb's free energy

ΔG = RT in {(Glucose)in/(Glucose)out }

= 8.314 J/molk * 310K * ln (0.1/0.001) = 5934.54 J/mol

Energy transported = (5934.54J/mol [tex]* 10^{-6}[/tex] mol)

                                  = (5.93 [tex]* 10^{-6}[/tex] ) KJ

Part B: Glucose concentration in cell = 1 mm

Glucose concentration outside cell = 10 mm

Temperature = 273+37 = 310 K

Gibb's free energy

ΔG = RT ln {(glucose)in (glucose)out}

= 8.314 J/molK [tex]* 310 *[/tex] ln (0.1 mm/ 10.00 mm)

= -5934.54 J/mol

energy transported = (-5934.34.54 J/mol [tex]*[/tex] [tex]10^{-6}[/tex] ) moles

                                 = (-5.934 [tex]* 10^{-6}[/tex] ) KJ

Part C:

ΔG of ArP hydrolysis = -32.2 KJ/mol

The mumber of moles of ATP formedis calculated =

moles of ATP = (-5.934[tex]* 10^{-6}[/tex] KJ/-32.2 KJ/mol )

                       = (1.84 [tex]* 10^{-7}[/tex] ) moles

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a. 7.623 × 10⁵ nm is expressed as 762300nm

b. 5.34 × 10⁻³ L is expressed as 0.00534

Following numerals are explain below:

A numeral is a name or symbol that represents a certain number. Examples include the numerals three, four, and twelve. So, the numeral is how we express the number, which is a concept. Reason: A numeral is a word that describes a number, whereas a number is conveyed via digits.

c. 5.453 × 10⁶ m is expressed as 5453000

d. 6.791 × 10⁻⁵pm is expressed as 6791m

e. 2.4 × 10⁹pg is expressed as 2400000000

f. 3.65 × 10⁵ is expressed as 365000

g. 3.4287 × 10⁻⁶L is expressed as 0.0000034287

h. 9.2 × 10²mL is expressed as 920L

i. 6.927 × 10⁴  is expressed as 69270m

j. 9.23 × 10⁻³ is expressed as 0.00923

Thus, a. 7.623 × 10⁵ nm is expressed as 762300nm.

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You would need to add approximately 0.2 μL of 6M HCl acid to 800 mL of 100 mM TRIS buffer at.

To calculate how much 6M HCl you need to add to lower the pH of a 800 mL solution of 100 mM TRIS from pH 10.2 to pH 7.0, you need to determine the initial and final concentrations of the TRIS buffer, and use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where;

First, calculate the initial concentration of TRIS- and HTRIS:

TRIS- concentration = 100 mM × 10^(-10.2) = 6.31 × 10^(-6) M

HTRIS concentration = 100 mM - 6.31 × 10^(-6) M = 99.99999369 mM

Next, determine the desired final concentration of TRIS- and HTRIS:

TRIS- concentration = 10^(-7.0) = 1 × 10^(-7) M

HTRIS concentration = 100 mM - 1 × 10^(-7) M = 99.9999999 mM

Now, use the Henderson-Hasselbalch equation to calculate the amount of HCl needed to achieve the desired pH:

pH = pKa + log([A-]/[HA])

7.0 = 8.1 + log([1 × 10^(-7)]/[99.9999999])

log([1 × 10^(-7)]/[99.9999999]) = -1.1

[1 × 10^(-7)]/[99.9999999] = 0.0794

[1 × 10^(-7)] = 0.0794 × [99.9999999]

[1 × 10^(-7)] = 7.93999993 × 10^(-6) M

[HA] = 100 mM - 7.93999993 × 10^(-6) M = 99.99999206 mM

The difference in [H+] concentration between the initial and final buffer conditions is:

[H+] concentration difference = [HA]initial - [HA]final

= 99.99999369 mM - 99.99999206 mM

= 0.00000163 M

To calculate the amount of 6M HCl needed to achieve this change in [H+], use the following formula:

moles of HCl = volume of buffer (L) × concentration difference (M)

where;

moles of HCl = 0.800 L × 0.00000163 M = 1.304 × 10^(-6) moles

Finally, convert moles of HCl to volume of 6M HCl needed:

volume of 6M HCl = moles of HCl ÷ concentration of HCl

The concentration of 6M HCl is 6 mol/L or 6,000 mM.

volume of 6M HCl = (1.304 × 10^(-6) moles) ÷ (6,000 mM)

= 0.219 μL or approximately 0.2 μL (microliters)

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The correct answer is A) NH4+. When a strong base is added to a solution containing NH4+, it reacts with water to produce ammonia gas (NH3) and hydroxide ions (OH-), according to the following equation:

NH4+ + OH- → NH3 + H2O

The ammonia gas produced is responsible for turning the moist red litmus blue, indicating the presence of a base. This reaction is commonly referred to as the ammonium ion hydrolysis reaction.

In contrast, adding a strong base to a solution containing CO32- or Cl- ions will not produce a gas that can cause the litmus paper to turn blue. The carbonate ion (CO32-) would react with a strong acid to release carbon dioxide gas, but it will not release gas upon addition of a strong base. Similarly, the chloride ion (Cl-) is a spectator ion and does not participate in the reaction with the base.

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Zinc: 0.75%, Copper: 99.25%. Zn and Cu are the symbols for zinc and copper, two chemical elements.

Generally, The composition of a penny is 97.5% zinc and 2.5% copper by weight. The zinc core of a penny is covered by a thin copper plating.

The copper plating is what gives the penny its distinct copper color.

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The number of moles of 77.4 grams of C₅H₁₀O₅ is 0.52 mol.

To determine the number of moles of C₅H₁₀O₅ in 77.4 grams, we need to divide the mass by the molar mass of the compound.

The molar mass of C₅H₁₀O₅ is:

5(12.01 g/mol) + 10(1.01 g/mol) + 5(16.00 g/mol) = 150.15 g/mol

So, the number of moles of C₅H₁₀O₅ in 77.4 grams is:

n = m/M = 77.4 g / 150.15 g/mol

n = 0.515 mol (rounded to 2 decimal spaces)

Therefore, 77.4 grams of C₅H₁₀O₅ is equivalent to 0.52 mol of C₅H₁₀O₅.

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the pressure will be increased by a factor of 2.

what is the relationship between the pressure, volume and the temperature ?

The relationship between the pressure, volume, and temperature of a gas is described by the ideal gas law, which is expressed as PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the absolute temperature. According to this law, the pressure of a gas is directly proportional to its temperature, assuming that the volume and number of moles remain constant.

In the given scenario, the gas is heated from 200 K to 400 K, while the container remains sealed and rigid, which means that the volume and number of moles of gas remain constant. As a result, the pressure of the gas must increase by a factor of 2, since the temperature has also increased by a factor of 2.

This can also be explained by the kinetic theory of gases, which states that the pressure of a gas is related to the average kinetic energy of its particles. When the gas is heated, its particles gain kinetic energy and move faster, which causes them to collide with the walls of the container more frequently and with greater force. This increase in collision frequency and force results in an increase in pressure, as described by the ideal gas law.

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Answer:

Solution:

1. Calculate the total volume of the two solutions:

Volume 1 = 100 cm3

Volume 2 = 50 cm3

Total Volume = 150 cm3

2. Calculate the total mass of each solution:

Mass of Solution 1 = (Volume x Density) x Mass Fraction HCl

= (100 cm3 x 1.180 g/cm3) x 0.36

= 43.2 g

Mass of Solution 2 = (Volume x Density) x Mass Fraction HCl

= (50 cm3 x 1.100 g/cm3) x 0.20

= 11 g

3. Calculate the total mass of the two solutions combined:

Total Mass = 43.2 g + 11 g

= 54.2 g

4. Calculate the mass fraction of hydrochloric acid in the new solution:

Mass Fraction HCl = Total Mass of HCl / Total Mass of Solutions

= 54.2 g / (43.2 g + 11 g)

= 0.55

Therefore, the mass fraction of hydrochloric acid obtained by merging 100 cm3 of a solution of hydrochloric acid with a mass fraction of HCl 36%, and 50 cm3 of a solution of hydrochloric acid with a mass fraction of HCl 20%, is 0.55.

The equilibrium expression Kc for the reaction can be written as -

                                                 [tex]Kc = \frac{O_{3}^{2} }{O_{2} ^{3} }[/tex]

Equilibrium constant of a chemical reaction at equilibrium is defined as the ratio of concentration of products to the concentration of reactants, each raised to their respective stoichiometric coefficients.

The expression that links Kc to the equilibrium concentration of reactants and products and stoichiometry of the equation is called equilibrium expression.

Kc is the equilibrium constant where subscript c refers to the fact that concentrations have been used in the calculations. The units of Kc depend on the equilibrium expression.

Therefore, the equilibrium expression Kc for the reaction can be written as -

                                                 [tex]Kc = \frac{O_{3}^{2} }{O_{2} ^{3} }[/tex]

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Answer: [A]t = -kt + [A]₀ or when you rearrange the equation [A]t - [A]₀ = -kt

Explanation:

Below are the integrated rate laws for zero, first and second order reactions.

Zero Order

[A]t = -kt + [A]₀

First Order

ln[A]t = -kt + ln[A]₀

Second Order

1/[A]t = kt + 1/[A]₀

K = Ae^-(Ea/RT) is the Arrhenius equation

The compound is cyclobutane with molecular formula C4H8 and its molecular ion peaks are discussed in the below section.

A hydrocarbon C4H8 with a molecular ion peak at m/z = 66. This is a compound that absorbs IR radiation at 1720 cm-1 and exhibits a molecular ion peak at m/z = 70.  Infrared (IR) spectroscopy is an absorption approach extensively utilized in each qualitative and quantitative analyses. The infrared vicinity of the spectrum consists of electromagnetic radiation that could adjust the vibrational and rotational states of covalent bonds in natural molecules. The IR spectroscopy idea makes use of the idea that molecules generally tend to soak up precise frequencies of mild which can be feature of the corresponding shape of the molecules. The energies are reliant at the form of the molecular surfaces, the related vibronic coupling, and the mass similar to the atoms.

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In a balanced equation a molecule is made up of two or more atoms that have been chemically composed.

The term balanced equation is defined as the mass of reactant is exactly similar to the mass of product in chemical reaction.

In a balanced equation two or more than two atoms are combined to produced molecule tha is also called as product. The right side of atoms known as reactant whereas the left side atoms or molecule are known as product.

Thus, Two or more than two atoms joined top formed molecule.

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The mass could vary within the range of 15.00 - 15.04 g. So, the correct option is B.

When a measurement is made, there is always some level of uncertainty associated with it due to limitations in the precision of the instrument being used or other factors. The level of uncertainty can be expressed using the significant figures of the measurement. In this case, the mass is recorded to two decimal places, so the uncertainty is typically assumed to be ±0.01 g.

Therefore, the range of possible values for the mass would be 15.02 ± 0.01 g, or 15.01 - 15.03 g. Option a is the closest range to this value. It's important to keep in mind that the range of uncertainty may be larger or smaller depending on the instrument used and the level of precision required for a particular application.

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The theoretical yield of lead(II) chloride is 2.992 g.

The percent yield is 79.6%.

(a) To find the theoretical yield of lead(II) chloride, we need to determine the limiting reagent in the two reactions.

First, calculate the moles of PbCO₃:

molar mass of PbCO₃ = 267.2 g/mol

moles of PbCO₃ = 2.871 g / 267.2 g/mol = 0.01075 mol

Next, we use the mole ratio between PbCO₃ and PbCl₂ in the second reaction to determine the moles of PbCl₂ produced:

moles of PbCl₂ = moles of PbCO3 × (1 mol PbCl₂ / 1 mol PbCO3) = 0.01075 mol

Finally, we convert the moles of PbCl₂ to grams using the molar mass of PbCl₂:

molar mass of PbCl₂ = 278.1 g/mol

theoretical yield of PbCl₂ = moles of PbCl₂ × molar mass of PbCl₂ = 0.01075 mol × 278.1 g/mol = 2.992 g

Therefore, the theoretical yield of lead(II) chloride is 2.992 g.

(b) The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100:

percent yield = (actual yield / theoretical yield) × 100%

The actual yield is given as 2.385 g. Substituting this and the theoretical yield calculated in part (a) gives:

percent yield = (2.385 g / 2.992 g) × 100% = 79.6%

Therefore, the percent yield of lead(II) chloride is 79.6%.

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The answer is Structure 1) very broad, centered about 3330 cm-1.; 2) strong, sharp 1717 cm-1 ; 3) broad with spikes at 3367cm-1 and 3392cm-1.; 4) sharp, 2254cm-1; 5) strong, slightly broadened, 1645cm-1.

The structure with their respective characteristics are:

Structure 1: very broad, centered about 3330 cm-1.

Structure 2: strong, sharp 1717 cm-1.

Structure 3: broad with spikes at 3367cm-1 and 3392cm-1.

Structure 4: sharp, 2254cm-1 .

Structure 5: strong, slightly broadened, 1645cm-1.

OH (hydroxide), a chemical compound made up of a single oxygen atom bonded to a single hydrogen atom. It is usually written OH- in aqueous solutions.

O, [tex]NH_{2}[/tex] and CN are chemical symbols. O is oxygen, NH2 is an amine function and CN is a cyanide function. It is functional group consisting of the nitrogen atom bonded to one or two hydrogen atoms. Cyanide is a functional group consisting of a carbon atom bonded to a nitrogen atom. Both functional groups are very reactive and are also found in many organic compounds.

Full Question:

These five structures all have distinguishing absorption in the IR. Match each structure with its characteristic absorption.

(a) sharp, 2254 cm-1

(b) very broad, centered about 3330cm-1

(c) strong, slightly broadened, 1645cm-1

(d) broad with spikes at 3367cm-1 and 3392cm-1

(e) strong, sharp 1717cm-1

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The number of significant figures that should be recorded for the answer will be only one which is 8.

Algebra is the study of abstract symbols, while logic is the manipulation of all those ideas.

The acronym PEMDAS stands for Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction. This approach is used to answer the problem correctly and completely.

The numbers are 43.6 and 21.2. Then the addition of the numbers is given as,

⇒ 43.6 + 21.2

⇒ 64.8

The number of significant figures that should be recorded for the answer will be only one which is 8.

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Answer: 3

Explanation:

Option( A) It maintains the same speed, (B) It continues in the same direction and option ( E) It has zero acceleration  is true about a car with constant velocity.

Constant velocity refers to a state of motion where an object moves in a straight line at a constant speed. In other words, an object that moves at a constant velocity has a consistent speed and direction of motion.

It's important to note that constant velocity does not mean that the object is stationary or moving at a slow speed. It simply means that the object is moving at a constant speed in a straight line without changing direction.

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The balance chemical equation is 2H₃PO₄ + 3Mg(OH)₂ → Mg₃(PO₄)₂ + 6H₂O

This is an example of an acid and base neutralizing one another to create salt and water. The reaction is exothermic, emits heat, and can be balanced by adding balancing numbers in front, or molecules, to make sure that the sum of the atom counts for each element on the left and right sides of the equation is the equal.

The law of conservation of matter, which stipulates that matter cannot be generated or destroyed, must be taken into account in order to balance a chemical equation.

H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O

This law states that whatever is available on the reactant side must also be available on the product side.

We must count the atoms on each side of the equation to balance it, and we might also use the number of moles.

The equation is balanced when

2H₃PO₄ + 3Mg(OH)₂ → Mg₃(PO₄)₂ + 6H₂O

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The term mole concept is used here to determine the number of atoms in 55.2 g of NaOH. The number of atoms in 55.2 g of NaOH is 8.31 × 10²³.

One mole of a substance is defined as that quantity of it which contains as many entities as there are atoms exactly in 12 g of carbon - 12. The formula used to calculate the number of moles is:

Number of moles (n) = Given mass / Molar mass

n = 55.2 / 39.99 = 1.380 moles

a. Number of atoms = n × 6.022 × 10²³

1.380 × 6.022 × 10²³ = 8.31 × 10²³

b. Number of moles = 4.58 × 10²³ / 6.022 × 10²³ = 0.7605 moles

Mass of (NH₄)₂CO₃ = 0.7605 × 96.09 = 73.07 g

c. Number of moles = 1.46 × 10²⁴ / 6.022 × 10²³ = 0.242 moles

Mass of CaCl₂ =  0.242 × 110.98 = 26.85 g

d. Number of moles = 1.230 / 187.56 = 0.0065 moles

Number of atoms = 0.0065 × 6.022 × 10²³ = 3.91 × 10²¹

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The mass of water produced when 9.60 g of butane reacts with excess oxygen is 14.85 gram.

The chemical reaction between substances, usually considering oxygen and usually accompanied by the generation of heat and light in the form of flame is called as combustion reaction.

C4H10(g) + 13 /2 O2(g) → 4CO2(g) + 5H2O(l)

Moles of butane = 9.60⋅g / 58.12⋅g⋅mol−1

= 0.165mol.

We assume complete combustion, and know that each mole of butane gives off 5 mol of water upon combustion.

Given that we know the molar quantity of butane, we simply perform the operation,

= 5 × 0.165 × 18.01

= 14.85 gram

Thus, the mass of water produced when 9.60 g of butane reacts with excess oxygen is 14.85 gram.

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The pressure of the dry oxygen gas alone is 736.2 torr.To determine the pressure of the dry oxygen gas, we need to first account for the contribution of water vapor to the total pressure. At a temperature of 25°C, the vapor pressure of water is 23.8 torr [1].

Since the sample of oxygen was collected over water, the total pressure is the sum of the pressure of the oxygen gas and the vapor pressure of water. Thus:

total pressure = pressure of oxygen gas + vapor pressure of water

760.0 torr = pressure of oxygen gas + 23.8 torr

Solving for the pressure of the oxygen gas alone, we get:

pressure of oxygen gas = 760.0 torr - 23.8 torr

pressure of oxygen gas = 736.2 torr

Therefore, the pressure of the dry oxygen gas alone is 736.2 torr.

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If the hydrate compounds are heated, they lose the water molecules by evaporation from their crystal lattice and they become anhydrous compounds. Thus, copper sulphate penta hydrate will form CuSO₄ solid on heating.

Hydrates are compounds with absorbed water molecules within their crystal lattice. The ability of absorption of moisture from the surrounding and form hydrates is called deliquescence.

Copper sulphate forms its various hydrates such as dihydrates, pentahydrates, hexahydrates etc. The hydrate  Copper Sulfate Pentahydrate,  CuSO₄. 5 H₂O contains 5 water molecules per one formula unit.

When this compound is heated, the water molecules will evaporate from the hydrate and become dry form called anhydrous. Thus, it will form anhydrous  CuSO₄.

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To calculate the R_f values in the student's experiment, you would need to measure the distance traveled by each solute spot and the distance traveled by the solvent front. Then, you would divide the distance traveled by each solute spot by the distance traveled by the solvent front to get the R_f value for each spot.

The R_f values are the ratio of the distance traveled by the solute to the distance traveled by the solvent. It is calculated by dividing the distance traveled by the solute by the distance traveled by the solvent front.

In chromatography, the R_f value is used to identify compounds by comparing the R_f value of a known compound to the R_f value of an unknown compound.


For example, if a solute spot traveled 4 cm and the solvent front traveled 8 cm, the R_f value for that spot would be 4/8 = 0.5. If another solute spot traveled 6 cm and the solvent front traveled 8 cm, the R_f value for that spot would be 6/8 = 0.75.

By calculating the R_f values for each spot, the student can compare the R_f values to known compounds to identify the components of the mixture.

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The following is the correct Newman projection for the following compound as viewed down the indicated bond. The correct wedge and dash conformation for the following Newman projection is:  IV

A Newman projection is a way of visualizing the conformation of a molecule around a single bond. The projection is created by looking directly down the bond axis, with one atom in front and one atom in back. The front atom is typically represented by a dot, and the back atom is usually hidden behind the front atom.

The 3D structure of organic molecules is depicted using the wedge and dash conformation. It is frequently used in organic chemistry to display how substituents are oriented around chiral centers. The atoms in this conformation are dots, and the bonds in the molecule are represented by lines. Dashes and wedges depict bonds entering and exiting the paper's or screen's plane, respectively.

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Compound A's (ethanimine) conjugate acid will be CH3-CH=NH2+. Compound B's (acetonitrile) conjugate acid will be CH3-CNH2+.

It is a colorless, flavorless, and odorless gas that makes up the majority of the atmosphere on Earth and is a component of all living things. Nitrogen is a chemical element with the symbol N and atomic number 7. Nitrogen is a nonmetal and the lightest element in the pnictogens, group 15 of the periodic table. It is a common element in the universe and is thought to be the eighth most abundant element overall in the Milky Way and the Solar System.

The primary source of nitrogen for atmospheric deposition is the combustion on fossil fuels like coal and oil. Nitrogen may be deposited in the atmosphere as dry particulates, gases, or droplets or as a wet form such as rainfall, snow, hail, fog, or freezing rain. Nitrogen primarily comes from the atmosphere. It is made comprised of this harmless, colorless gas in 78 percent of its composition.

Compound B (acetonitrile), which possesses a nitrogen atom with a higher positive charge and more electronegative properties that make it more stable after receiving a proton, is the strongest acid.

Compound A (ethanimine) has the strongest base because its nitrogen atom has a higher electron density and is better equipped to receive a proton.

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The standard Gibbs free energy of reaction is -25.4 kJ/mol. And  the equilibrium constant (K) for the reaction is approximately 4.85. The value of AG for the first reaction is approximately -31.9 kJ/mol. The value of AS for the second reaction is approximately 174 J/mol/K.

We may use the following equation to determine the standard Gibbs free energy of reaction for the first portion of the question:

G=-RT ln K

where T is the temperature in Kelvin, K is the equilibrium constant, and R is the gas constant. Given that K and T are known, we may solve for G° by substituting these values:

G° is equal to -(8.314 J/mol K)/293 K*ln (6.7 10–10).

-25.4 kJ/mol for G°

As a result, the reaction's standard Gibbs free energy is -25.4 kJ/mol.

The following equation can be used to determine the equilibrium constant K for the reaction in the second section of the question:

G=-RT ln K

But, since G and T are provided in this instance, we can modify the equation to account for K:

K = e^(-ΔG°/RT)

K is equal to e(-121000 J/mol / 8.314 J/mol K * 303 K).

K ≈ 4.85

As a result, the reaction's equilibrium constant is roughly 4.85.

We can complete the table as follows to answer the third component of the question:

Response AH (kJ/mol)

J/mol K, AS

(kJ/mol) AG

Graphite spontaneity = CO2 107 397 -31.9

Neither P2O5? -44? nor 2H3PO4?

We can use the following equation to determine AG for the initial reaction:

ΔG = ΔH - TΔS

where H stands for the change in enthalpy, S for the change in entropy, and T for temperature. Given that H and S are known, we can use these values as substitutes to find G:

G is equal to 107000 J/mol - 298 K * 397 J/mol K.

-31.9 kJ/mol for G

As a result, the AG value for the first reaction is roughly -31.9 kJ/mol.

The following response, Given that H and G are provided, we must determine S. We can apply the formula:

ΔG = ΔH - TΔS

then reorder it to account for S:

ΔS = (ΔH - ΔG) / T

"S" is equal to (116000 J/mol - 41000 J/mol) (286 K)

174 J/mol K for S.

As a result, the second reaction's AS value is roughly 174 J/mol K.

We can use the sign of G to ascertain each reaction's spontaneity. If G is less than zero, the reaction moves ahead on its own. The reaction is spontaneously in the opposite direction if G > 0. The system is in equilibrium and neither the forward nor the backward reaction is spontaneous if G = 0.

We discovered that G is negative for the initial reaction, indicating that the reaction is spontaneous in the forward direction.

Because we don't know the values of H and G, we lack the knowledge necessary to determine the spontaneity of the second reaction.

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The partial pressure of the each gas and the total pressure if the temperature of the gas is 18 °C is the partial pressure of O₂ is 3.74 atm  and the partial pressure of He is 19.44 atm . The total pressure of the is 27.59 atm.

The ideal gas equation is as :

P = n R T / V

Moles of O₂ = mass / molar mass

                     = 50.3 / 32

                    = 1.57 mol

Partial pressure of O₂ = ( 1.57 × 0.082 × 291 ) / 10

                                   = 3.74 atm

Moles of He = mass / molar mass

                      = 32.6 / 4

                      = 8.15 mol

Partial pressure of He = ( 8.15 × 0.082 × 291 ) / 10

                                     = 19.44 atm

The total pressure = partial pressure of He + partial pressure of  O₂

                               = 19.44 + 8.15

                               = 27.59 atm

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The metals that would act as the sacrificial anode that is cathodic protection for the iron is the correct option is a. Al and b. Zn.

a) Al is the sacrificial anode as it is higher than the iron in the reactivity series. Therefore, it is the more reactive than the iron.

b) Zn  is the sacrificial anode as it is higher than the iron in the reactivity series. Therefore, it is the more reactive than the iron.

c) Pb cannot acts as the sacrificial anode for the iron as it is lower than the iron in the reactivity series. Therefore, it is less reactive than the iron.

d) Na cannot acts as the sacrificial anode for the iron as it is lower than the iron in the reactivity series. Therefore, it is less reactive than the iron.

e) Ni cannot acts as the sacrificial anode for the iron as it is lower than the iron in the reactivity series. Therefore, it is less reactive than the iron.

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