which substances are a weak electrolytes? select all which apply. question 13 options: nh3 caso4 agcl ch3oh

Gay-Lussac's Law-

[tex] \:\:\:\:\:\:\star\longrightarrow \underline{\sf \boxed{\sf \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}}}[/tex]

[tex] \:\:\:\:\:\:\star\longrightarrow \sf \underline{T_2=\dfrac{T_1 \:P_2}{P_1}}[/tex]

Where-

As per question, we are given -

Now that we are given all the required values, so we can put them into the formula and solve for T₂:-

[tex] \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\star\longrightarrow \sf \underline{T_2=\dfrac{T_1 \:P_2}{P_1}}[/tex]

[tex]\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf T_2=\dfrac{298 \times 175}{75.5}[/tex]

[tex]\:\:\:\:\: \:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf T_2=\dfrac{52150}{75.5}[/tex]

[tex]\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf T_2=690.728476......[/tex]

[tex] \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf T_2=690.73 \:K[/tex]

[tex]\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf T_2=(690.73-273)°C [/tex]

[tex] \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf \underline{T_2=417.73\:°C} [/tex]

Therefore, the temperature of the gas at 175 kPa will become 690.73 K or, 417.73°C.

The following are categorized into isometric and isotonic:
Isometric: Pushing constantly against a concrete wall, Holding a bench-press bar in the same position, Doing a plank exercise (holding a push-up position), and Balancing on tiptoes.

Isotonic: Running up a hill, Swimming freestyle, Pedaling a bicycle on a flat surface, and Doing bicep curls.

Isometric exercises are exercises where a muscle or group of muscles contracts without any change in length. These exercises involve the muscle or group of muscles contracting against an immovable object or against an opposing force. Examples of isometric exercises are pushing against a wall, holding a plank, and doing a wall sit.

Isotonic exercises are exercises that involve the muscle or group of muscles contracting against resistance, and causing a change in the muscle’s length. These exercises involve the muscle or group of muscles contracting and then extending against a fixed weight or external force.

Examples of isotonic exercises are running up a hill, swimming freestyle, pedaling a bicycle on a flat surface, doing bicep curls, and doing leg extensions.

1. Pushing constantly against a concrete wall - Isometric Exercise
2. Running up a hill - Isotonic Exercise
3. Swimming freestyle - Isotonic Exercise
4. Pedaling a bicycle on a flat surface - Isotonic Exercise
5. Holding a bench-press bar in the same position - Isometric Exercise
6. Doing a plank exercise (holding a push-up position) - Isometric Exercise
7. Balancing on tiptoes - Isometric Exercise
8. Doing bicep curls - Isotonic Exercise

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The statement "the instantaneous rate of a reaction can be read directly from the graph of molarity versus time at any point on the graph." is False

Instantaneous rate of reaction is the rate at which a chemical reaction is occurring at a specific moment in time.

It is the slope of the tangent line at that particular point in time. The instantaneous rate of a reaction can be determined from a graph of concentration vs time.

We must draw a tangent line at the point in the graph that we are interested in, and the slope of that tangent line is the instantaneous rate of the reaction.

The rate can also be determined by finding the slope of a secant line over a very small time interval.

So, the given statement, “the instantaneous rate of a reaction can be read directly from the graph of molarity versus time at any point on the graph” is not true, and the correct answer is False.

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Atoms are made up of three subatomic particles: protons, neutrons, and electrons. Protons are positively charged particles found in the nucleus of an atom.

They have a mass of approximately 1 atomic mass unit (AMU). Neutrons, also found in the nucleus, have no charge and have a mass of approximately 1 amu. Electrons are negatively charged particles that orbit around the nucleus in shells or energy levels. They have a much smaller mass than protons and neutrons and are responsible for chemical reactions and bonding between atoms. Understanding the properties and behaviors of these subatomic particles is crucial in understanding the nature of matter and the physical world around us.

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The overall balanced nuclear equation for the process is:

234[tex]Th_{90}[/tex] → 230[tex]Th^{90}[/tex] + 2α + 2β-1 + 2νe

What is nuclear equation?

The overall nuclear reaction for the decay of thorium-234 to thorium-230 via protactinium-234 and uranium-232 can be written as follows:

234[tex]Th_{90}[/tex] → 234[tex]Pa_{91}[/tex] + 0β-1 + νe

234[tex]Pa_{91}[/tex] → 232[tex]U_{92}[/tex] + 2β-1 + 2νe

232[tex]U_{92}[/tex] → 230[tex]Th_{90}[/tex] + 2α

Combining these three equations, we can write the overall balanced nuclear equation for the process:

234Th90 → 230Th90 + 2α + 2β-1 + 2νe

This equation represents the complete decay of thorium-234 to thorium-230 via the intermediate steps of protactinium-234 and uranium-232. It shows that the decay process involves the emission of a beta particle and two alpha particles.

Note that the conservation of mass and atomic numbers are satisfied in this overall reaction, since the mass number and atomic number are the same on both sides of the equation.

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There are 3.6312 x 10²⁴ atoms in 6 moles of KCIO3 because there are 6 moles of KCIO3 and 6.022 x 10²³ atoms per mole.

Potassium One atom of potassium, three atoms of oxygen, and one atom of chlorine make up the inorganic substance chlorate.

The molar mass of KCIO3 is:

K = 39.10 g/mol

Cl = 35.45 g/mol

I = 126.90 g/mol

O3 = (16.00 x 3) g/mol = 48.00 g/mol

Molar mass of KCIO3 = K + Cl + I + 3O3

= 39.10 + 35.45 + 126.90 + 3(48.00)

= 307.35 g/mol

6 KCIO3 = 6 moles KCIO3

Number of atoms = 6 moles [tex]KCIO3 x 6.022 x 10^23[/tex] atoms/mol

[tex]= 3.6312 x 10^24 atoms[/tex]

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the correct option is D. between 7 and 14 best describes the pH at the equivalence point of a titration of a weak base with a strong acid.

The pH at the equivalence point of a titration of a weak base with a strong acid is somewhere in the range of 7 and 14. At the equivalence point, the quantity of moles of the base is equivalent to the number of moles of acid added. The pH at the equivalence point depends on the strength of the weak base. On the off chance that the weak base is exceptionally weak, the pH at the equivalence point will be closer to 7.0. If the weak base is stronger, the pH at the equivalence point will be closer to 14.0.

A weak base is a substance that, when dissolved in water, does not break totally into its constituent ions. Weak bases have some other properties, such as their solutions being unfortunate conductors of power. They are also categorized as weak electrolytes. Examples of weak bases incorporate smelling salts (NH3), pyridine (C5H5N), and trimethylamine (N(CH3)3).

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Answer: In the Picture all the question has been done

Explanation:

The oxidation number of manganese (Mn) in potassium permanganate (KMnO4) is B.  +7.

Potassium permanganate is a strong oxidizing agent that is commonly used in organic chemistry to oxidize alcohols, aldehydes, and ketones to carboxylic acids. Its formula is KMnO4.The oxidation state of manganese in KMnO4 is determined by subtracting the sum of the oxidation states of potassium, oxygen, and hydrogen from the overall charge of the molecule. The oxidation state of potassium is +1, and the oxidation state of oxygen is -2. Hydrogen's oxidation state is +1, and it is usually removed from the organic compound in question, so it is not included in the final calculation. For KMnO4, the overall charge is -1. Using this information,

we can calculate the oxidation state of manganese as follows: Oxidation state of Mn + Oxidation state of K + 4(Oxidation state of O) = -1Oxidation state of Mn + 1 + 4(-2) = -1Oxidation state of Mn - 7 = -1 Oxidation state of Mn = +7Therefore, the oxidation state of manganese in potassium permanganate is +7. Therefore the correct option is B

The complete question is :

What is the oxidation number of manganese (MN) in potassium permanganate (KMnO4)?

A. +5,

B. +7,

C. +8,

D. +9

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In the solid phase, the atoms of a substance are closely packed together and held in a fixed position by intermolecular forces. In the case of ice, which is a solid form of water, the atoms are arranged in a regular pattern, forming a crystal lattice structure.

The motion of atoms in the solid phase is limited to small vibrations around their fixed positions. These vibrations are caused by thermal energy, but they are not significant enough to overcome the intermolecular forces that hold the atoms in place. As a result, the shape of the solid, including ice, is definite and maintains a fixed shape.

The volume or amount of space taken up by a solid, including ice, is also definite. The intermolecular forces between the atoms prevent them from moving around freely, and as a result, the solid has a fixed volume.

Overall, in the solid phase, such as ice, the arrangement of atoms is regular and fixed, and their motion is limited to small vibrations around their fixed positions. The volume or amount of space taken up is definite, and the shape is also definite and maintains a fixed shape.

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The ligand field splitting energy in the complex in units of kilojoules per mole when the solutions of the [V(OH2)6]2 ion are lilac and absorb light of wavelength 806 nm is 331kJ/mol. To calculate the ligand field splitting energy in the complex.

The ligand field splitting energy in the complex can be calculated by the following formula:Δoct = hc/λ Where,Δoct = Ligand field splitting energy hc = Planck's constant (6.626 x 10^-34 J s)λ = wavelength of light absorbed In the given problem, the absorbed wavelength of light is 806 nm. So we need to convert it into meters as we have to use the value in the formula. c = speed of light = 3.0 x 10^8 m/sλ = 806 nm = 806 x 10^-9 m Substituting the values in the formula:Δoct = hc/λ = (6.626 x 10^-34 J s) x (3.0 x 10^8 m/s) / (806 x 10^-9 m) = 2.477 x 10^-19 J= 2.477 x 10^-19 J x 1 kJ/1000J x NA Where NA is Avogadro's number (6.022 x 10^23)Thus, Δoct = 331 kJ/mol (approx.)Therefore, the ligand field splitting energy in the complex in units of kilojoules per mole is 331 kJ/mol.

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It would take approximately 4858.35 years for 250 kg of radium to decay down to less than 10 kg.

The amount of time it takes for a given amount of radium to decay to a certain level can be determined using the radium's half-life. This can be done using the following formula:

Amount of time = Half-life x ln (Initial amount/Final amount)

Given that radium has a half-life of 1500 years, we can use this formula to determine the amount of time it takes for 250 kg of radium to decay down to less than 10 kg.

Initial amount of radium = 250 kg

Final amount of radium = 10 kg

Half-life of radium = 1500 years

Amount of time = Half-life x ln (Initial amount/Final amount)

Amount of time = 1500 x ln (250/10)

Amount of time = 1500 x ln (25)

Amount of time = 1500 x 3.2189

Amount of time ≈ 4858.35 years

Therefore, it would take 4858.35 years.

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The initial concentration of the HCl solution was 0.1251 M.

The balanced chemical equation for the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is:

HCl + NaOH → NaCl + H2O

We can use the balanced equation and the volume and concentration of the NaOH solution to calculate the number of moles of NaOH used in the titration:

moles of NaOH = volume of NaOH x concentration of NaOH

moles of NaOH = 0.02746 L x 0.1138 mol/L

moles of NaOH = 0.003127 mols

Since HCl and NaOH react in a 1:1 ratio, the number of moles of HCl used in the titration is also 0.003127 moles.

Now, we can use the number of moles of HCl and the volume of HCl to calculate the initial concentration of HCl:

moles of HCl = concentration of HCl x volume of HCl

0.003127 mol = concentration of HCl x 0.02500 L

concentration of HCl = 0.1251 M

Therefore, the initial concentration of the HCl solution was 0.1251 M.

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Answer:

15. Flow

16. Volume

Explanation:

That must be the right answer

According to the VSEPR theory, the shape of a molecule is determined by the repulsion between its electron pairs, both bonding and non-bonding.

In sulfuryl fluoride (SO2F2), the sulfur atom is surrounded by four regions of electron density: two single bonds with fluorine atoms and two double bonds with oxygen atoms. These four regions of electron density repel each other and arrange themselves as far apart as possible to minimize repulsion.

Since the molecule has no lone pairs of electrons, the electron density is evenly distributed among the four regions around sulfur. This results in a tetrahedral shape, where the fluorine atoms and oxygen atoms are arranged symmetrically around the central sulfur atom, with bond angles of approximately 109.5 degrees.

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Aldohexoses have four chiral focuses and, in this way, 24 = 16 isomers. There will be eight d-isomers and eight l-isomers. Thusly, the quantity of enantiomer matches is 8 (2n−1).

An aldohexose has four chiral focuses.

So there are

24=16

optical isomers.

Their perfect representations are the L-aldohexoses, the other 8 of the 16.

Their names are L-allose, L-altrose, L-glucose, and so on.

Consequently, every one of the 16 aldohexoses has its own name.

The four chiral focuses in glucose show there might be upwards of sixteen (24) stereoisomers having this constitution. These would exist as eight diastereomeric sets of enantiomers, and the underlying test was to figure out which of the eight related to glucose. This challenge was acknowledged and met in 1891 by German physicist Emil Fischer. His fruitful exchange of the stereochemical labyrinth introduced by the aldohexoses was a sensible masterpiece, and it is fitting that he got the 1902 Nobel Prize for science for this achievement. At the time Fischer embraced the glucose project laying out the outright setup of an enantiomer was unrealistic. Thusly, Fischer pursued an erratic decision for (D)- glucose and laid out an organization of related aldose setups that he called the D-family. The identical representations of these arrangements were then assigned the L-group of aldoses. To represent utilizing present-day information, Fischer projection formulas and names for the D-aldose family (three to six-carbon atoms) are displayed underneath, with the uneven carbon atoms (chiral focuses) hued red.

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Understanding matter and its properties helped humans move from the stone age to the iron age by enabling them to develop new materials and technologies. In the stone age, humans used materials such as stone, bone, and wood to create tools and weapons. However, these materials had limitations in terms of their strength, durability, and ability to hold a sharp edge.

As humans began to understand the properties of matter, they were able to experiment with new materials and processes. For example, they learned that by heating certain rocks, they could extract metals such as copper and tin, which could be used to create stronger tools and weapons. By combining copper with tin, they discovered that they could create bronze, which was even stronger and more durable.

Eventually, humans discovered how to extract iron from iron ore and use it to create even stronger tools and weapons. This knowledge of matter and metallurgy enabled humans to advance technologically and culturally, leading to the development of civilizations and the rise of empires.

In summary, understanding the properties of matter helped humans move from the stone age to the iron age by enabling them to experiment with new materials and develop new technologies. This process of discovery and innovation continues to this day, as scientists and engineers continue to push the boundaries of our knowledge and understanding of the natural world.

The presence of CaCl2 affects the molar solubility of Ca(OH)2 by decreasing it.

When CaCl2 is added to a solution containing Ca(OH)2, it ionizes to produce Ca2+ ions and Cl- ions. This addition of Ca2+ ions to the solution containing Ca(OH)2 creates a common ion effect, which decreases the solubility of Ca(OH)2 by shifting the equilibrium to the left. Experimentally, this effect can be observed by adding different concentrations of CaCl2 to a solution containing Ca(OH)2 and measuring the resulting molar solubility of Ca(OH)2. As the concentration of CaCl2 is increased, the molar solubility of Ca(OH)2 decreases.

The presence of CaCl2 in a solution containing Ca(OH)2 has an effect on the molar solubility of Ca(OH)2 by decreasing it. This is due to the common ion effect, where the addition of Ca2+ ions to the solution containing Ca(OH)2 decreases its solubility by shifting the equilibrium to the left. Experimentally, this can be observed by adding different concentrations of CaCl2 to a solution of Ca(OH)2 and measuring the molar solubility of Ca(OH)2. As the concentration of CaCl2 increases, the molar solubility of Ca(OH)2 decreases.

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The formula for one neutral uncharged atom and the four ions will be: Cl (neutral) + Cl⁻ + Cl⁺ + Cl₃⁺ + Cl₅⁺.

The electron configuration of an atom or ion refers to the arrangement of electrons in its shells or subshells. For example, the electron configuration of a neutral oxygen atom is 1s₂ 2s₂ 2p₄, which means that it has two electrons in its first shell, two electrons in its second shell, and four electrons in its third shell.

To write a formula for one neutral uncharged atom and any three ions which have the same electron configuration, we need to first identify an element that has four ions with the same electron configuration. Let's take chlorine (Cl) as an example.

The electron configuration of the neutral chlorine atom will be 1s₂ 2s₂ 2p₆ 3s₂ 3p₅. Chlorine can form four different ions by either gaining or losing electrons:

Chlorine ion with a -1 charge (Cl⁻) has the same electron configuration as a neutral argon atom: 1s₂ 2s₂ 2p₆ 3s₂ 3p₆

Chlorine ion with a +1 charge (Cl⁺) has the same electron configuration as a neutral neon atom: 1s₂ 2s₂ 2p₆

Chlorine ion with a +3 charge (Cl₃⁺) has the same electron configuration as a neutral magnesium ion (Mg₂⁺): 1s₂ 2s₂ 2p₆

Chlorine ion with a +5 charge (Cl₅⁺) has the same electron configuration as a neutral neon ion (Ne2+): 1s₂ 2s₂ 2p₅

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To solve this problem, we can use the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas:

PV = nRT

where P is the pressure in Pa, V is the volume in m³, n is the number of moles, R is the ideal gas constant (8.31 J/(molₓK)), and T is the temperature in Kelvin.

First, we need to convert the pressure from kPa to Pa and the volume from L to m³:

P = 1.70x10² kPa x 1000 Pa/kPa = 1.70x10⁵ Pa

V = 9.300x10¹ L x 0.001 m³/L = 0.093 m³

Next, we need to calculate the number of moles of ammonia using its molar mass:

molar mass of ammonia (NH3) = 14.01 g/mol + 3(1.01 g/mol) = 17.03 g/mol

moles of NH3 = 5.9000x10¹ g ÷ 17.03 g/mol = 3.462 mol

Now we can rearrange the ideal gas law to solve for the temperature:

T = PV ÷ nR

T = (1.70x10⁵ Pa)(0.093 m³) ÷ (3.462 mol)(8.31 J/(molₓK))

T = 686.3 K

Therefore, at a temperature of 686.3 K (413.1 °C or 775.6 °F), 5.9000x10¹ g of ammonia gas would exert a pressure of 1.70x10² kPa in a 9.300x10¹ L container.

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Answer:

A Single replacement/displacement reaction will occur, making the equation:

Calcium + Lead (II) nitrate → Calcium Nitrate + Lead (II)

Ca + Pb(NO3)2 → Ca(NO3)2 + Pb

Explanation:

Calcium is more reactive than Lead in the reactivity series so theoretically will replace in Lead in Lead Nitrate, forming Calcium Nitrate.

Hopefully this helps!!!

The ion with the smallest number of unpaired electrons is Ni2+.

The number of unpaired electrons in an ion depends on its electronic configuration, specifically the number of electrons in its partially filled d-orbitals.

Out of the given ions, V2+ and Cr2+ have the electronic configurations [Ar] 3d3 and [Ar] 3d4 respectively, and both have three unpaired electrons.

Ni2+ has the electronic configuration [Ar] 3d8, and it has two unpaired electrons.

Co2+ has the electronic configuration [Ar] 3d7, and it has three unpaired electrons.

Fe3+ has the electronic configuration [Ar] 3d5, and it has five unpaired electrons.

Therefore, Ni2+ has the smallest number of unpaired electrons among the given ions with two unpaired electrons.

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The correct option is c. 75.36g.  The given oxide is named Potassium Superoxide, as it contains O2-anion which is known as superoxide.

As per the mole concept,

1 mol of the compound holds a similar measure of the compound as demonstrated by its molar mass.

In this way, here,

Molar mass is= 94.2 g/mol

Implies, 1 mol of compound holds 94.2 g of it

In this way, 0.8 mol of compound holds (94.2×0.8) g of it = 75.36 g

Thus, the right response is-75.36 g.

The mole concept is a helpful technique for communicating how much a substance is. Any estimation can be separated into two sections - the mathematical extent and the units that the greatness communicated. For instance, when the mass of a ball is estimated to be 2 kilograms, the size is '2' and the unit is 'kilogram'.

While managing particles at a nuclear (or molecular) level, even one gram of an unadulterated component is known to contain an enormous number of iotas. This is where the mole concept is generally utilized. It basically centers around the unit known as a 'mole', which is a count of an extremely enormous number of particles.

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Enzyme feedback inhibition is a process that regulates enzyme movement in cells. It works by impeding the development of products by changing the configuration of enzymes.

This will keep the cells from becoming harmful. Feedback inhibition is significant in enzyme and hereditary regulation because it prevents cells from wasting energy and substrates on chemical reactions that are not necessary at the time. For instance, a cell does not have to separate glucose on the off chance that there is sufficient energy accessible for the cell to use.

Enzyme regulation is a process that controls the movement of enzymes in cells. There are three types of enzyme regulation: allosteric regulation, hereditary and covalent modification, and enzyme inhibition.

Allosteric regulation is a natural illustration of control loops, such as feedback from downstream products or feedforward from upstream substrates. Long-range allostery is especially significant in cell signaling. Examples of allosteric enzymes incorporate Aspartate Transcarbamoylase, Glucokinase, and Acetyl-CoA Carboxylase.

The hereditary and covalent modification involves changes to the enzyme's structure that influence its movement. For instance, phosphorylation can activate or deactivate an enzyme.

Enzyme inhibition is a process that reduces or stops enzyme movement. There are two types of enzyme inhibition: cutthroat inhibition and non-serious inhibition.

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We can use the solubility product constant (Ksp) for lead(II) chloride, which is 1.17 x 10-5 to determine the concentration of the lead ion (Pb2+) that must be exceeded to precipitate PbCl2 from a 1.00 x 10-2 M solution of chloride ions (Cl-).

The equilibrium constant for a solid material dissolving in an aqueous solution is denoted by the solubility product constant, which is abbreviated as Ksp. It is a measurement that indicates how much of a solute may be dissolved in a solution. The Ksp value of a material is greater when it has a higher degree of solubility.

The dissociation reaction for PbCl2 in water is:
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)

The Ksp expression for this reaction is:
Ksp = [Pb2+][Cl-]^2

We are given the concentration of Cl- as 1.00 x 10^-2 M. Let [Pb2+] = x, so we can plug in the values into the Ksp expression:
1.17 x 10^-5 = x(1.00 x 10^-2)^2

Now, solve for x:
x = (1.17 x 10^-5) / (1.00 x 10^-2)^2
x ≈ 1.17 x 10^-1

Therefore, the concentration of lead ion (Pb2+) that must be exceeded to precipitate PbCl2 from the solution is approximately 1.17 x 10^-1 M.

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The forces that connect two hydrogen atoms to an oxygen atom in a water molecule are intramolecular, but the forces that hold water molecules close together in an ice cube are intermolecular.

Intramolecular forces are those that bind atoms to other atoms within the same molecule. These are typically very strong forces such as covalent bonds and hydrogen bonds. In the case of a water molecule, the two hydrogen atoms and the oxygen atom are held together by covalent bonds and hydrogen bonds, respectively.

On the other hand, intermolecular forces are those that bind different molecules together. In the case of water molecules in an ice cube, the hydrogen bonds between molecules create a weak attraction between them, allowing them to remain close together in an ordered structure.

This creates a lattice of water molecules, which is what makes ice so solid. Thus, while the intramolecular forces of the water molecules hold them together, it is the intermolecular forces that give the structure of the ice cube its stability.

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The molar mass of the compound is 128.4 g/mol.

The empirical formula is an empirical formula that represents the lowest whole-number ratio of the atoms present in a compound. The empirical formula for the molecular compound is calculated using the percentage composition of the elements present in the compound. The steps used to find the empirical formula are as follows:

Find the mass of each element present in the compound.Convert each mass to moles.Divide each mole value by the smallest number of moles.Round to the nearest whole number and write the subscripts.The molar mass is the mass of a substance that contains 6.02 × 10²³ atoms or molecules. To calculate the molar mass of a compound, add the masses of all the atoms present in the compound.

C=0.707g,12.01 g/mol=0.0588 molCnH=0.2372 =1.01g/m=0.235 mol H

nH=4nC

The empirical formula of the compound is CH4. The molar mass of the compound can be calculated using the empirical formula.

M=12.01 g/mol+4(1.01 g/mol)=16.05 g/mol

The molar mass of the compound is 8 times the empirical mass, so the actual molar mass is;

M=8(16.05{g/mol})=128.4g/mol. The molar mass of the compound is 128.4 g/mol.

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The partial pressure of nitrogen is 0.116 atm and the partial pressure of helium is 0.583 atm.

To find the partial pressures of nitrogen and helium, we can use the fact that the total pressure of the mixture is equal to the sum of the partial pressures of each component:

P_total = P_O2 + P_He + P_N2

We are given the partial pressure of oxygen as P_O2 = 0.300 atm. We can find the total number of moles of gas in the mixture as:

n_total = n_O2 + n_He + n_N2 = 1.50 mol + 2.50 mol + 0.500 mol = 4.50 mol

We can use the mole fraction of each component to find the partial pressure of nitrogen and helium.

X_N2 = n_N2 / n_total = 0.500 mol / 4.50 mol = 0.111

X_He = n_He / n_total = 2.50 mol / 4.50 mol = 0.556

The mole fraction of oxygen is then:

X_O2 = n_O2 / n_total = 1.50 mol / 4.50 mol = 0.333

We can use these mole fractions to find the partial pressures of nitrogen and helium:

P_N2 = X_N2 * P_total = 0.111 * P_total

P_He = X_He * P_total = 0.556 * P_total

Substituting the given value for P_O2 and adding the expressions for P_N2 and P_He, we get:

P_total = P_O2 + P_N2 + P_He

P_N2 + P_He = P_total - P_O2 = (1.000 atm - 0.300 atm) = 0.700 atm

Substituting the expressions for P_N2 and P_He in terms of mole fractions, we get:

0.111 * P_total + 0.556 * P_total = 0.700 atm

0.667 * P_total = 0.700 atm

P_total = 1.049 atm

Now we can use the expressions for P_N2 and P_He in terms of mole fractions to find their partial pressures:

P_N2 = 0.111 * P_total = 0.116 atm

P_He = 0.556 * P_total = 0.583 atm]

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The balanced equation for the reaction of N2 and H2 to produce

NH3 is:N2(g) + 3H2(g) → 2NH3(g)

From this equation, we can infer that for every one mole of N2, we need 3 moles of H2 to produce 2 moles of NH3. Therefore, to determine the moles of H2 needed to produce 59.2 grams of NH3, we must first determine the number of moles of NH3 produced by 59.2 grams. The molar mass of NH3 is 17.03 g/mol. Therefore, the number of moles of NH3 produced is:59.2 g / 17.03 g/mol = 3.47 mol NH3Now we can use the mole ratio from the balanced equation to determine the number of moles of H2 required. For every 2 moles of NH3, we need 3 moles of H2. Therefore:3.47 mol NH3 x (3 mol H2 / 2 mol NH3) = 5.21 mol H2

Therefore, 5.21 moles of H2 are needed to produce 59.2 grams of NH3 if sufficient N2 is present.

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