put a small portion in water
excite a small portion of it and identify the spectral lines
smell it. taste a small portion

At equilibrium, the number of moles of \(Cl_2\) (g) will be 0.2025 mol.

1: Write the balanced chemical equation:

\(C_O\)(g) + \(Cl_2\)(g) ⟶ \(C_OCl_2\)(g)

2: Set up an ICE table to track the changes in moles of the substances involved in the reaction.

Initial:

\(C_O\)(g) = 0.3500 mol

\(Cl_2\)(g) = 0.05500 mol

\(C_OCl_2\)(g) = 0 mol

Change:

\(C_O\)(g) = -x

\(Cl_2\)(g) = -x

\(C_OCl_2\)(g) = +x

Equilibrium:

\(C_O\)(g) = 0.3500 - x mol

\(Cl_2\)(g) = 0.05500 - x mol

\(C_OCl_2\)(g) = x mol

3: Write the expression for the equilibrium constant (Kc) using the concentrations of the species involved:

Kc = [\(C_OCl_2\)(g)] / [\(C_O\)(g)] * [\(Cl_2\)(g)]

4: Substitute the given equilibrium constant (Kc) value into the expression:

1.2 x \(10^3\) = x / (0.3500 - x) * (0.05500 - x)

5: Solve the equation for x. Rearrange the equation to obtain a quadratic equation:

1.2 x \(10^3\) * (0.3500 - x) * (0.05500 - x) = x

6: Simplify and solve the quadratic equation. This can be done by multiplying out the terms, rearranging the equation to standard quadratic form, and then using the quadratic formula.

7: After solving the quadratic equation, you will find two possible values for x. However, since the number of moles cannot be negative, we discard the negative solution.

8: The positive value of x represents the number of moles of \(Cl_2\)(g) at equilibrium. Substitute the value of x into the expression for \(Cl_2\)(g):

\(Cl_2\)(g) = 0.05500 - x

9: Calculate the value of \(Cl_2\)(g) at equilibrium:

\(Cl_2\)(g) = 0.05500 - x

\(Cl_2\)(g) = 0.05500 - (positive value of x)

10: Calculate the final value of \(Cl_2\) (g) at equilibrium to get the answer.

Therefore, at equilibrium, the number of moles of \(Cl_2\) (g) will be 0.2025 mol.

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Answer:

0.24 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation,

The mole ratio of the acid, H₂SO₄ (nₐ) = 1

The mole ratio of the base, NaOH (n₆) = 2

Finally, we shall determine the molarity of the acid. This can be obtained as follow:

Volume of acid, H₂SO₄ (Vₐ) = 17.4 mL

Molarity of base, NaOH (M₆) = 0.20 M

Volume of base, NaOH (V₆) = 41.51 mL

Molarity of acid, H₂SO₄ (Mₐ) =?

MₐVₐ / M₆V₆ = nₐ / n₆

Mₐ × 17.4 / 0.2 × 41.51 = 1 / 2

Mₐ × 17.4 / 8.302 = 1 / 2

Cross multiply

Mₐ × 17.4 × 2 = 8.302

Mₐ × 34.8 = 8.302

Divide both side by 34.8

Mₐ = 8.302 / 34.8

Mₐ = 0.24 M

Therefore, the molarity of the acid is 0.24 M

For this question, we have the following reaction:

Fe2O3 + 3 CO -> 2 Fe + 3 CO2

For letter D, we have:

10 grams of Fe2O3, molar mass = 159.7g/mol

6 grams of CO, molar mass = 28g/mol

In this case, we need to find which compound is the limiting and which is the excess reactant, and it is important to have in mind the molar ratio concept, which we use the coefficients in front of the compound to determine the ratios, like for Fe2O3 and CO, the molar ratio is 1:3. Now let's check if Fe2O3 is in excess, finding the number of moles first:

159.7g = 1 mol

10g = x moles

x = 0.063 moles of Fe2O3

According to the molar ratio, we need 3 times this value for CO, therefore, 0.063 * 3 = 0.189 moles of CO, now we need to check if this is the amount that we have of CO, or if we have more than that:

28g = 1 mol

6g = x moles

x = 0.214 moles of CO, this means that we have more CO than we actually need, making it the excess reactant and Fe2O3 the limiting reactant

Now to find the mass of Fe, we will be using the number of moles of the reactant, which is Fe2O3, 0.063 moles, and according to the molar ratio, we have twice this value for Fe, therefore, 0.063 * 2 = 0.126 moles of Fe, now using the molar mass of Fe, 55.84g/mol, we can find the final mass:

55.84g = 1 mol

x grams = 0.126 moles of Fe

x = 7.03 grams of Fe are produced, or if it is possible to round up, 7 grams of Fe, this is letter D

E. In this case, we need to see how much CO remains in the reaction, and since we need only 0.189 moles of CO and we actually have 0.214 moles, we have a 0.025 moles leftover, which is, in grams:

28g = 1 mol

x grams = 0.025 moles

x = 0.7 grams of CO remaining, only 5.3 grams will react

F. The percent yield is calculated having the actual yield, given in the question and the theoretical yield, which is the value we found in letter D, the formula for it is:

%yield = (actual yield/theoretical yield)*100

%yield = (6.75g/7.03g)*100

%yield = 0.964*100

The percent yield will be 96.4% for this reaction

Answer:

14

Explanation:

Step 1: Calculate the moles corresponding to 25 g of Ba(OH)₂

The molar mass of Ba(OH)₂ is 171.34 g/mol.

25 g × 1 mol/171.34 g = 0.15 mol

Step 2: Calculate the molar concentration of Ba(OH)₂

Molarity is equal to the moles of solute divided by the liters of solution.

[Ba(OH)₂] = 0.15 mol/0.250 L = 0.60 M

Step 3: Calculate the molar concentration of OH⁻

Ba(OH)₂ is a strong base according to the following equation.

Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻

The molar ratio of Ba(OH)₂ to OH⁻ is 1:2. The molar concentration of OH⁻ is 2/1 × 0.60 M = 1.2 M.

Step 4: Calculate the concentration of H⁺

We will use the ionic product of water expression.

Kw = 1.0 × 10⁻¹⁴ = [H⁺] × [OH⁻]

[H⁺] = 1.0 × 10⁻¹⁴/[OH⁻] = 1.0 × 10⁻¹⁴/1.2 = 8.3 × 10⁻¹⁵ M

Step 5: Calculate the pH of the solution

We will use the definition of pH.

pH = -log [H⁺] = -log 8.3 × 10⁻¹⁵ = 14

The percent composition of CuBr₂ is approximately 28.46% of Cu and 71.54% of Br.

To determine the percent composition of CuBr₂ (copper(II) bromide), we need to calculate the mass of each element in the compound and then divide it by the molar mass of the entire compound.

The molar mass of CuBr₂ can be calculated by adding up the atomic masses of copper (Cu) and bromine (Br) in the compound. The atomic masses of Cu and Br are approximately 63.55 g/mol and 79.90 g/mol, respectively.

Molar mass of CuBr₂ = (63.55 g/mol) + 2(79.90 g/mol) = 223.35 g/mol

Now, let's calculate the percent composition of each element in CuBr₂:

Percent composition of copper (Cu):

Mass of Cu = (63.55 g/mol) / 223.35 g/mol × 100% ≈ 28.46%

Percent composition of bromine (Br):

Mass of Br = 2(79.90 g/mol) / 223.35 g/mol × 100% ≈ 71.54%

Therefore, the percent composition of CuBr₂ is approximately:

- Copper (Cu): 28.46%

- Bromine (Br): 71.54%

These values represent the relative mass percentages of copper and bromine in the compound CuBr₂.

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NH4Cl+ KOH --> KCl + NH3 + H2O

Zn + Cu (CH3COO)2 --> Zn(CH3COO)2 + Cu . This is not double replacement reactions.

What is meant by H2O ?

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Answer:

B. Single replacement.

Explanation:

The A in compound AB was replaced by C to make CB. Since there was only one instance of replacement, it is a single replacement reaction.

When you see  or visit grasslands you know that :

Grasslands are a type of vegetation found mostly in areas with favourable weather conditions such as mild summers and right amount of precipitation which provides the grasslands with enough water for proper growth.

Other types of vegetation include :

Desert vegetation expreciences very hot summers and little precipitation which leads to the scarcity of green vegetation, while forest vegetation expreciences a very high level of precipitation.

Hence we can conclude that When you see  or visit grasslands you know that :They have mild to hot summers ,They have a lot of precipitation  

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Answer:

They have little precipitation.

They have cold winters.

They have mild to hot summers.

The standard reduction potential of the Cr³+/Cr electrode is -0.46V. None of the option is correct.

To determine the standard reduction potential of the Cr³+/Cr electrode, we can use the Nernst equation, which relates the standard reduction potential to the cell potential under non-standard conditions. The Nernst equation is given by:

E = E° - (0.0592/n) * log(Q)

where E is the cell potential, E° is the standard reduction potential, n is the number of electrons transferred in the half-reaction, and Q is the reaction quotient.

In this case, we have the standard potential of the cell as −0.60V. We know that the standard reduction potential of the Sn/Sn²+ electrode is -0.14V. Therefore, the reduction potential of the Cr³+/Cr electrode can be calculated as:

E = -0.60V - (-0.14V)

E = -0.60V + 0.14V

E = -0.46V

Therefore, the standard reduction potential of the Cr³+/Cr electrode is -0.46V.

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Answer:

b

Explanation:

Answer:

b

Explanation:

The half-reaction is; H₂ → 2H⁺ + 2e⁻, the oxidizing agent is chlorine , and the reducing agent is hydrogen. The oxidation half-reaction is where a species loses electrons, the reduction half-reaction is where a species gains electrons, the oxidizing agent is the species that causes another species to be oxidized by accepting electrons itself, and the reducing agent is the species that causes another species to be reduced by donating electrons itself.

The oxidation half-reaction is the half-reaction where a species loses electrons. In this reaction, hydrogen (H₂) is oxidized to form hydrogen ions (H⁺). The half-reaction can be written as;

H₂ → 2H⁺ + 2e⁻

The reduction half-reaction is the half-reaction where a species gains electrons. In this reaction, chlorine (Cl₂) is reduced to form chloride ions (Cl⁻). The half-reaction can be written as follows;

Cl₂ + 2e⁻ → 2Cl⁻

The oxidizing agent is the species that causes another species to be oxidized by accepting electrons itself. In this reaction, chlorine (Cl₂) is the oxidizing agent because it accepts electrons from hydrogen (H₂), causing it to be oxidized.

The reducing agent is the species that causes another species to be reduced by donating electrons itself. In this reaction, hydrogen (H₂) is the reducing agent because it donates electrons to chlorine (Cl₂), causing it to be reduced.

Therefore, chlorine is the oxidizing agent, and hydrogen is the reducing agent.

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To find the density of the unknown piece of metal, we can use the formula:

Density = mass / volume.

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The number of C atoms in 0.247 mol of C is  1.48 X 10²³ .

The number of C atoms in 1 mol of C is 6.02 X 10²³, which is also called Avogadro’s number .

so, The number of C atoms in 0.247 mol of C is _

0.247x6.02x10²³= 1.48 X 10²³.

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Cargnello and coworkers developed a novel catalyst that advances this objective by boosting the formation of long-chain hydrocarbons during chemical processes.

Thus, The same amounts of carbon dioxide, hydrogen, catalyst, pressure, heat, and time as the standard catalyst, it created 1,000 times more butane—the longest hydrocarbon it could produce at its maximum pressure—than the standard catalyst.

The novel catalyst is made of ruthenium, a platinum group rare transition metal that is coated in a thin coating of plastic. This idea accelerates chemical processes without being consumed in the process, much like any catalyst.

Another benefit of ruthenium is that it is less expensive than other platinum- and palladium-based high-quality catalysts.

Thus, Cargnello and coworkers developed a novel catalyst that advances this objective by boosting the formation of long-chain hydrocarbons during chemical processes.

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The control illustrative or the control group is a name given to a group of samples that are not served with anything. In this experiment.

The control group is the group that is not set for the treatment. For example: “The control group is the plants that were given pure water with no fertilizer.” The experimental group is the group that is slated for the treatment. The plants that do not collect the fertilizer are the control group. An experiment must have a control group. In the control group, nil changed. It is not subjected to the bold variable control group is the group in a study that does not cover the thing being tested and is used as a benchmark to measure the results of the other group and is one of the two groups in any valid experiment.

So we can conclude that  A control group in a scientific experiment is a group part from the rest of the experiment.

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The reactions that result in the emission of light involve the ruthenium label and tripropylamine (TPA), two electrochemically active molecules.

Thus, The electrode surface inside the measurement cell is where the reactions take place.

The ruthenium label is oxidized at the electrode surface as an electrical potential is applied, and TPA is oxidized into a radical cation that spontaneously loses a proton.

When the resultant TPA radical interacts with oxidized ruthenium, the ruthenium label enters an excited state and emits a photon (620 nm) before decaying. The ruthenium label is renewed and ready to carry out numerous light-generating cycles as it goes back to its ground state.

Thus, The reactions that result in the emission of light involve the ruthenium label and tripropylamine (TPA), two electrochemically active molecules.

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According to the question,4.3 grams  would this oxygen (0₂) weigh in its liquid form.

What is the oxygen ?

Oxygen is an essential element for life. It is a colorless, odorless gas that makes up about 21% of the Earth's atmosphere. Oxygen is a vital component for the production of energy in all living things, as it is necessary for the process of respiration. Without oxygen, all living organisms would suffer, as their cells would not be able to produce the energy they need to survive. Oxygen is also necessary for the combustion of fuels, as it helps to create the conditions required for burning to occur. In addition, oxygen is used in many different industrial and scientific processes, ranging from welding to creating rocket fuel. Oxygen is incredibly important to life on Earth and is essential for the survival of all organisms.

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Greater than 0 is the change in entropy (AS) when a solid substance decomposes and produces gaseous products. The correct option is B.

The change in entropy (ΔS) is a measure of the degree of randomness or disorder in a system. When a solid substance decomposes and produces gaseous products, the number of particles (molecules or atoms) in the system increases, and the arrangement of the particles becomes more disordered. This results in an increase in entropy, which is greater than 0.

When a solid substance decomposes and produces gaseous products, the change in entropy (ΔS) is greater than 0. This is because the number of particles in the system increases, leading to an increase in disorder or randomness.

The entropy of a system is a measure of the degree of disorder, and it tends to increase in processes that lead to a greater dispersion of energy or matter. In the case of the decomposition of a solid substance into gaseous products, the transition from a more ordered solid state to a more disordered gas state leads to an increase in entropy.

This phenomenon is a manifestation of the second law of thermodynamics, which states that the entropy of a closed system tends to increase over time.

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The concept stoichiometry is used here to determine the moles of NaNO₃ produced. Chemical stoichiometry refers to the quantitative study of the reactants and products involved in a chemical reaction. Here the moles of NaNO₃ is

Stoichiometry is an important concept in chemistry which helps us to use the balanced chemical equation to find out the amounts of reactants and products. Here we make use of ratios in the balanced equation.

The balanced equation is:

Co(NO₃)₂ + Na₂SO₄ → 2 NaNO₃ + CoSO₄

1 mole of Na₂SO₄ gives 2 moles of NaNO₃.

So 5 moles of Na₂SO₄ gives: 2 NaNO₃ × 5 mole Na₂SO₄ / 1 mole Na₂SO₄ = 10 mole NaNO₃

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The amount of heat energy produced when 1 g of hydrogen and 2 g of nitrogen are reacted, is -6.61 KJ

First, we shall obtain the limiting reactant. Details below:

3H₂ + N₂ -> 2NH₃

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂

Therefore,

2 g of N₂ will react with = (2 × 6) / 28 = 0.43 g of H₂

We can see that only 0.43 g of H₂ is needed in the reaction.

Thus, the limiting reactant is N₂

Finally, we the amount of heat energy produced. Details below:

3H₂ + N₂ -> 2NH₃ ΔH = -92.6 KJ

From the balanced equation above,

When 28 grams of N₂ reacted, -92.6 KJ of heat energy were produced.

Therefore,

When 2 grams of N₂ will react to produce = (2 × -92.6) / 28 = -6.61 KJ

Thus the heat energy produced from the reaction is -6.61 KJ

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The molecules move faster and farther apart when the state changes from the liquid state to the gaseous state.

The matter exists in different forms in nature. Some substances have a fixed shape like wood and stone and can flow, take the shape of their container such as water, while there are forms of matter that do not have certain shapes or sizes.

In solids, the molecules are closely packed and thus solids have a rigid structure. The compact packing of solids means that the molecules are very close to each other and the molecules do not show movement. The kinetic energy of molecules in the solid is the least.

In liquids, the molecules are some distance away from each other and they show movements. Since the molecules in liquid show significant movement, they have some kinetic energy. The kinetic energy of molecules in the liquid is greater than that of solids.

In gases, the molecules are farther from each other and thus their molecules show large amounts of movement. The kinetic energy in gases is the greatest in comparison to solids and liquids.

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The concentration of the hydrogen carbonate ion ([HCO₃⁻]) is approximately 3.15 x 10⁷ and the concentration of the carbonic acid ([H₂CO₃]) is approximately 2.52 x 10⁷.

The carbonate buffer system plays a crucial role in maintaining the pH balance of the extracellular environment. In this system, carbonic acid (H₂CO₃) and hydrogen carbonate (HCO₃⁻) act as a conjugate acid-base pair. Given that the ratio of carbonic acid to hydrogen carbonate is 1.25:1, we can assume that the initial concentration of carbonic acid is higher. Let's denote the initial concentration of carbonic acid as [H₂CO₃] and the concentration of hydrogen carbonate as [HCO₃⁻]. The dissociation of carbonic acid can be represented by the equation: H₂CO₃ ⇌ H⁺ + HCO₃⁻. The equilibrium constant (Ka) for this reaction is given as 4.5 x 10⁻⁷. At physiological pH (7.35), the concentration of H⁺ is determined by the dissociation of carbonic acid and is tightly regulated. To calculate the concentration of hydrogen carbonate ion ([HCO₃⁻]), we need to make use of the Henderson-Hasselbalch equation:

pH = pKa + log([HCO₃⁻]/[H₂CO₃])

Substituting the given values, we have:

7.35 = -log(4.5 x 10⁻⁷) + log([HCO₃⁻]/[H₂CO₃])

Rearranging the equation, we find:

log([HCO₃⁻]/[H₂CO₃]) = 7.35 + log(4.5 x 10⁻⁷)

Taking antilog of both sides, we get:

[HCO₃⁻]/[H₂CO₃] = 10^(7.35 + log(4.5 x 10⁻⁷))

Simplifying the right-hand side, we have:

[HCO₃⁻]/[H₂CO₃] ≈ 3.15 x 10⁷

Since the initial ratio of H₂CO₃ to HCO₃⁻ is 1.25:1, we can set up the equation:

[HCO₃⁻] = 1.25 x [H₂CO₃]

Substituting the value of [HCO₃⁻]/[H₂CO₃] from above, we find:

1.25 x [H₂CO₃] = 3.15 x 10⁷

Solving for [H₂CO₃], we get:

[H₂CO₃] ≈ 2.52 x 10⁷

Therefore, the concentration of the hydrogen carbonate ion ([HCO₃⁻]) is approximately 3.15 x 10⁷ and the concentration of the carbonic acid ([H₂CO₃]) is approximately 2.52 x 10⁷.

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19.42g of fluorine is produced upon decomposition of sodium fluoride.

The mass of a given substance is converted to moles using the molar mass of this substance in the periodic table. Moles of a given substance are then converted to moles of an unknown substance using the molar ratios from the balanced chemical formulas.

Mass ratio is defined as the percentage composition of the masses of elements in a molecule or compound. A compound always has a defined mass fraction of the corresponding element.

Mass ratio of sodium to fluorine = 1.21:1

If the mass of sodium fluoride produced is 23.5 g

Using dimensional Analysis,

(23.5g of sodium/sodium fluoride)×(1 g of Fluorine/1.21 g of sodium)

= 19.42g(g of fluorine/g of sodium fluoride)

Mass of fluorine produced = 19.42g

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A change in the entropy of a system indicate that the disorder of the system has changed. Option A

A change in entropy indicates is a measure of the disorder, or randomness, in a system.

An increased value of entropy means more disorder, and a decreased value of entropy means less disorder.

These changes in entropy occur spontaneously

Thus, a change in the entropy of a system indicate that the disorder of the system has changed. Option A

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Answer:these remove the nitrate from. The water - plants

These produce waste that contains ammonia- fish

These convert ammonia into nitrate-bacteria

Explanation:

Plants remove the nitrate from the water, fishes produce waste that contains ammonia, and bacteria convert ammonia into nitrate.

An aquaponics system is a sustainable method of agriculture that combines aquaculture with hydroponics. In this system, fish or other aquatic animals are kept in a tank or pond. The fish produce waste in the form of ammonia-rich water.

This wastewater is then circulated to the hydroponic component of the system, where plants are grown in a soil-less medium or directly in water. Beneficial bacteria in the system convert ammonia into nitrites and then nitrates, which are essential nutrients for plants.

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The median reaction at the second end point in the HCl and NaOH titration is: HCl + NaOH → NaCl + H2O

In a titration between hydrochloric acid (HCl) and sodium hydroxide (NaOH), the reaction involved is the neutralization reaction between an acid and a base. The balanced equation for this reaction is:

HCl + NaOH → NaCl + H2O

In this reaction, one mole of HCl reacts with one mole of NaOH to form one mole of NaCl (sodium chloride) and one mole of water.

During the titration process, the reaction occurs gradually as the base is added to the acid solution.

The first end point of the titration is reached when the moles of HCl and NaOH are stoichiometrically equivalent, meaning they react in a 1:1 ratio. At this point, all the HCl has been neutralized by the NaOH, and no excess of either reagent remains.

However, if the titration is continued beyond the first end point, the reaction between HCl and NaOH can still occur, albeit in a different ratio.

The second end point refers to the point where the moles of NaOH added exceed the stoichiometrically required amount to neutralize the HCl completely. As a result, any excess NaOH added after the second end point reacts with the excess HCl in a 1:1 ratio.

Therefore, the median reaction at the second end point in the HCl and NaOH titration is:

HCl + NaOH → NaCl + H2O

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Answer:

See explanation

Explanation:

The molecular geometry of an atom is connected to the number of electron pairs that surround it(whether lone pairs or bonding pairs) as well as its hybridization state. We shall now examine the N, P, or S atoms in each of the following compounds.

a)

In H3PO4, P has a tetrahedral molecular geometry and is sp3 hybridized.

b) In NH4NO3

N is sp3 hybridized in NH4^+ and sp2 hybridized in NO3^-. Also, N is tetrahedral in NH4^+ but trigonal planar in NO3^-.

c) In S2Cl2, we expect a tetrahedral geometry but as a result of the presence of two lone pairs on each sulphur atom, the molecular geometry is bent. The sulphur is sp3 hybridized.

d) In K4[O3POPO3], each phosphorus atom is in a tetrahedral molecular geometry and is sp3 hybridized.

151.3 grams of carbon are required to produce 12.6 moles of methane.

The balanced chemical equation for the reaction is:

C + 2H₂ → CH₄

According to the stoichiometry of the equation, 1 mole of carbon reacts with 2 moles of hydrogen gas to produce 1 mole of methane gas. Therefore, the number of moles of carbon required to produce 12.6 moles of methane is:

12.6 moles CH₄ × 1 mole C / 1 mole CH₄ = 12.6 moles C

To convert moles to grams, we need to use the molar mass of carbon, which is 12.01 g/mol.

Therefore, the mass of carbon required to produce 12.6 moles of methane is:

12.6 moles C × 12.01 g/mol = 151.3 g

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