2. Radical chlorination of 2-methylpentane yields a mixture of monochlorinated products. (i) Draw all monochlorinated products resulting from reaction of 2 methylpentane. (ii) Identify the major product. Iii) Show step-wise mechanism for the radical halogenation of the major product. (8) i. mixture of isomers 2-methylPentane

The weight, in grammes, of 2.00 x 1023 F2 molecules. As a result, 12. 62 g equals 21023 moles of F2.

The precise formula for determining a substance's Gram Molecular Mass is: Formula in grammes Mass is equal to the product of the solute's mass and its formula. It is consistently expressed in terms of grammes per mole (g/mol).

Number-wise, the mass with one mole (or formula unit) of atomic mass units is equal to the mass with one mole (or formulas unit) in grammes. One mole of O2 molecules, for instance, weighs 32.00 g and a single O2 molecule, 32.00 u.

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Answer:

To determine the empirical formula of a compound, we need to know the relative amounts of each element in the compound. Given the masses of iron and sulfur, we can calculate the number of moles of each element present:

moles of iron = 39.27 g / 55.85 g/mol = 0.703 mol

moles of sulfur = 33.77 g / 32.06 g/mol = 1.053 mol

We can then find the ratio of these moles by dividing both by the smallest number of moles (0.703):

0.703 mol Fe : 1.053 mol S

0.667 Fe : 1.000 S

This ratio indicates that there are approximately 0.667 atoms of iron for every 1 atom of sulfur in the compound.

To convert this ratio to a whole-number ratio of atoms, we need to multiply by a factor that will give us whole numbers. We can do this by dividing both sides of the ratio by the smallest number of atoms:

0.667 Fe : 1.000 S

0.667/0.667 Fe : 1.000/0.667 S

1.000 Fe : 1.498 S

This gives us a ratio of approximately 1 atom of iron for every 1.498 atoms of sulfur. To get a whole-number ratio, we can multiply both sides by 2:

2.000 Fe : 2.996 S

Rounding to the nearest whole number, we get a ratio of:

2 Fe : 3 S

Therefore, the empirical formula of the compound is Fe2S3.

Answer:

mass = 25.0 g

density = 1.48 g / cm^3

volume = ?

volume = mass / density

volume = 25 / 1.48

volume = 16.892 cm^3

There are 0.03675 moles of NaI in 175.0 mL of 0.210 M NaI solution.

To calculate the number of moles of NaI in 175.0 mL of 0.210 M NaI solution, we can use the formula:

moles of solute = concentration (in M) x volume (in liters)

First, we need to convert the volume from milliliters to liters:

175.0 mL = 175.0/1000 L = 0.175 L

Now we can plug in the values:

moles of NaI = 0.210 M x 0.175 L = 0.03675 moles of NaI

Therefore, there are 0.03675 moles of NaI in 175.0 mL of 0.210 M NaI solution.

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There are approximately 2.526 × 10^26 atoms of O in 46.6 moles of Al2(CO3)3.

To find the number of atoms of O in 46.6 moles of Al2(CO3)3, we need to first calculate the total number of moles of O in the compound and then multiply it by Avogadro's number.

The molecular formula of Al2(CO3)3 shows that there are three O atoms per one CO3 group, and there are three CO3 groups per one Al2(CO3)3 molecule. Therefore, the total number of O atoms in one Al2(CO3)3 molecule is:

3 O atoms/CO3 group × 3 CO3 groups/Al2(CO3)3 molecule = 9 O atoms/Al2(CO3)3 molecule

Next, we need to calculate the total number of moles of O atoms in 46.6 moles of Al2(CO3)3:

46.6 moles Al2(CO3)3 × 9 O atoms/Al2(CO3)3 molecule = 419.4 moles O atoms

Finally, we can calculate the total number of O atoms by multiplying the number of moles of O by Avogadro's number:

419.4 moles O atoms × 6.022 × 10^23 atoms/mole = 2.526 × 10^26 O atoms

Therefore, there are approximately 2.526 × 10^26 atoms of O in 46.6 moles of Al2(CO3)3.

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No, two samples of NaCl and CoCl2 do not have the same mass, but they have the same number of particles.

Mole is the amount of material containing 6.02214 × 10²³ particles.

Molar mass of NaCl (sodium chloride) is 58.44 g/mol, while molar mass of CoCl2 (cobalt chloride) is 129.84 g/mol. Since both samples have same number of moles, it means that they contain same number of particles of their respective compounds. However, mass of each sample will be different due to the difference in molar mass.

For example, if we assume that both samples contain 1 mole of their respective compounds, then mass of NaCl sample will be 58.44 g, while  mass of CoCl2 sample will be 129.84 g. So, two samples have different masses but the same number of particles.

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In 0.05μg, there are  [tex]6.9 * 10^{15} molecules[/tex]  of batrachotoxin.

To calculate the number of molecules of batrachotoxin, we first need to calculate the molar mass of the molecule. This can be done by adding up the atomic masses of the elements in the compound. The elements in batrachotoxin are carbon (C), hydrogen (H), nitrogen (N), and oxygen (O). The atomic masses of these elements are 12 g/mol for C, 1 g/mol for H, 14 g/mol for N, and 16 g/mol for O. Therefore, the molar mass of batrachotoxin is:

Molar mass = 12 g/mol C + 1 g/mol H + 14 g/mol N + 16 g/mol O

Molar mass = 43 g/mol

We then need to calculate the mass of 0.05 μg of batrachotoxin. This can be done by converting 0.05 μg to grams. To do this, we divide 0.05 μg by 1,000,000. This gives us:

[tex]\frac{0.05\mu g }{ 1,000,000 }= 0.00000005 g[/tex]

Now we can calculate the number of molecules of batrachotoxin in 0.05 μg by dividing the mass in grams by the molar mass:

Number of molecules =[tex]\frac{ 0.00000005 g }{ 43 g/mol}[/tex]

Number of molecules = [tex]1.16 * 10^{-8} mol[/tex]

Finally, we can calculate the number of molecules by multiplying the number of moles by Avogadro's number:

Number of molecules =[tex]\frac{1.16 * 10^{-8 }mol * 6.022 * 10^{23 }molecules}{mol}[/tex]

Number of molecules = [tex]6.9 * 10^{15} molecules[/tex]

Therefore, there are  [tex]6.9 * 10^{15} molecules[/tex]  of batrachotoxin in 0.05 μg.

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complete question:Batrachotoxin, C31H42N2O6 an active component of South American arrow poison, is so toxic that 0.05μg can kill a person.

How many molecules is this? Express your answer as an integer

The cation of zinc sulfate is Zn²⁺, which is a positively charged ion of zinc.

The anion of zinc sulfate is SO₄²⁻, which is a negatively charged ion of sulfate.

A cation is a positively charged ion that is formed when an atom loses one or more electrons. The loss of electrons causes the atom to have a net positive charge, and it is then referred to as a cation. Cations are formed from metals, which tend to lose electrons to form positively charged ions.

An anion, on the other hand, is a negatively charged ion that is formed when an atom gains one or more electrons. The gain of electrons causes the atom to have a net negative charge, and it is then referred to as an anion. Anions are formed from non-metals, which tend to gain electrons to form negatively charged ions.

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Based on the given information, it can be inferred that a chemical reaction occurred in test tube #4, resulting in a change of color.

Silver nitrate (AgNO3) is commonly used as a reagent in chemical reactions. It is known to react with halide ions, such as chloride (Cl-), bromide (Br-), and iodide (I-), forming insoluble silver halide compounds. For example, the reaction between silver nitrate and chloride ions can be represented by the following equation:

AgNO3(aq) + Cl-(aq) -> AgCl(s) + NO3-(aq)

In this reaction, silver nitrate reacts with chloride ions to form silver chloride, which is insoluble in water and precipitates out of solution. This reaction can cause a color change, as the precipitate has a different color than the original solution.

The effect of the reaction on the equilibrium position will depend on the specific equilibrium involved in the experiment. If the reaction involves the formation of a product that is also a reactant in the equilibrium, the reaction will shift the equilibrium to favor the reactants. Conversely, if the reaction involves the formation of a product that is not a reactant in the equilibrium, the equilibrium will shift to favor the products.

In conclusion, the role of silver nitrate in the reaction depends on the specific experiment, but it is known to react with halide ions to form insoluble silver halide compounds. The effect on the equilibrium position will depend on the specific equilibrium involved in the reaction.

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Answer:

At STP (standard temperature and pressure), the conditions are:

Temperature (T) = 273.15 K Pressure (P) = 1 atm = 101.3 kPa Volume (V) = 22.4 L (for one mole of gas)

So, for a gas at STP with a volume of 5 L, we can use the following formula to calculate the number of moles present:

n = V / Vm

where: n = number of moles V = volume of gas (in liters) Vm = molar volume of gas at STP (22.4 L/mol)

Plugging in the values, we get:

n = 5 L / 22.4 L/mol n = 0.2232 mol (rounded to four significant figures)

Therefore, there are approximately 0.2232 moles of the gas present.

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The measurement of 0.03412 kg has 5 significant figures. Significant figures are used to provide an indication of the precision of a measurement and should be used when performing calculations.

Measurements are the numerical values used to quantify the size, amount, or extent of something. Measurements are used to describe physical attributes and characteristics, such as length, width, weight, height, area, volume, speed, temperature, pressure, force, energy, power, and time. Measurements are also used to determine the amount of a certain material that is needed for a given purpose, such as a recipe. Measurements are essential for scientific investigation, engineering, and other activities that require precision and accuracy.

The measurement of 0.03412 kg has 5 significant figures. Significant figures are the meaningful digits of a number that are important in providing information about the measurement. In this measurement, the 4 digits after the decimal point (3412) provide the precision of the measurement, and therefore are significant figures. The leading zero does not provide any additional precision and is not a significant figure. Therefore, the measurement of 0.03412 kg has 5 significant figures.

Significant figures are used to provide an indication of the precision of a measurement. For example, a measurement of 0.03412 kg has a precision of 0.00001 kg, which is greater than a measurement of 0.3412 kg, which has a precision of 0.0001 kg. The more significant figures a measurement has, the more precise it is.

Significant figures are also important in calculations. To ensure accuracy, all calculations should be done using the same number of significant figures as the measurements being used. For example, if two measurements, 0.03412 kg and 0.02384 kg, are used in a calculation, the result should be reported to 5 significant figures.

In conclusion, the measurement of 0.03412 kg has 5 significant figures. Significant figures are used to provide an indication of the precision of a measurement and should be used when performing calculations.

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The measurement of 0.03412 kg has 5 significant figures. Significant figures are used to provide an indication of the precision of a measurement and should be used when performing calculations.

Measurements are the numerical values used to quantify the size, amount, or extent of something. Measurements are used to describe physical attributes and characteristics, such as length, width, weight, height, area, volume, speed, temperature, pressure, force, energy, power, and time. Measurements are also used to determine the amount of a certain material that is needed for a given purpose, such as a recipe. Measurements are essential for scientific investigation, engineering, and other activities that require precision and accuracy.

The measurement of 0.03412 kg has 5 significant figures. Significant figures are the meaningful digits of a number that are important in providing information about the measurement. In this measurement, the 4 digits after the decimal point (3412) provide the precision of the measurement, and therefore are significant figures. The leading zero does not provide any additional precision and is not a significant figure. Therefore, the measurement of 0.03412 kg has 5 significant figures.

Significant figures are also important in calculations. To ensure accuracy, all calculations should be done using the same number of significant figures as the measurements being used. For example, if two measurements, 0.03412 kg and 0.02384 kg, are used in a calculation, the result should be reported to 5 significant figures.

In conclusion, the measurement of 0.03412 kg has 5 significant figures. Significant figures are used to provide an indication of the precision of a measurement and should be used when performing calculations.

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The complete question is as follows:

How many significant figure are in 0.03412​ kg?

Answer:

To calculate vapor pressure lowering:

Use ΔP = X2 * P2° equation, where X2 is the mole fraction of the solute and P2° is the vapor pressure of the pure solvent at the same temperature.

Calculate mole fraction of glycerol by using moles of glycerol and water.

Calculate vapor pressure of pure water at 50°C using the Antoine equation.

Substitute the values and calculate the vapor pressure lowering, which is 0.459 torr for this problem.

The vapor pressure lowering is 0.213 torr when 10.0 mL of glycerol is added to 500 mL of water at 50°C.

Pressure exerted by vapor in thermodynamic equilibrium with its condensed phases at a certain given temperature in closed system is called vapor pressure.

ΔP = X2 * P0 * (1 - (ρ1 / ρ2))

ΔP is vapor pressure lowering, X2 is mole fraction of the solute (glycerol), P0 is vapor pressure of solvent (water), and ρ1 and ρ2 are the densities of solvent and solution, respectively.

As, moles of glycerol = mass of glycerol / molar mass of glycerol

moles of glycerol = (10.0 mL)(1.26 g/mL) / (92.09 g/mol)

moles of glycerol = 0.136 mol

and moles of water = mass of water / molar mass of water

= (500 mL)(0.988 g/mL) / (18.02 g/mol)

moles of water = 27.5 mol

So, total moles = moles of glycerol + moles of water

total moles = 0.136 mol + 27.5 mol

total moles = 27.6 mol

X2 (mole fraction of glycerol) = moles of glycerol / total moles

= 0.136 mol / 27.6 mol

X2 = 0.00493

ΔP = X2 * P0 * (1 - (ρ1 / ρ2))

= (0.00493)(92.5 torr) * (1 - (0.988 g/mL / 1.250 g/mL))

ΔP = 0.213 torr

Therefore, the vapor pressure lowering is 0.213 torr when 10.0 mL of glycerol is added to 500 mL of water at 50°C.

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Therefore, 0.160 moles of aluminum chloride could be produced from 17.0g of chlorine gas, Chlorine.

Aluminum chloride is created when aluminum and chlorine gas combine, as shown by the equation below. 3Chlorine + 2aluminum = 2Aluminum chloride if the reaction between 20 aluminum ions and 45 chlorine gas molecules.

First, let's determine how many molecules of each reactant we have:

moles of aluminum = 12.0 g / 26.98 g/mol = 0.445 mol

moles of Chlorine = 17.0 g / 70.90 g/mol = 0.240 mol

To determine the limiting reagent, we need to compare the actual mole ratio of aluminum to Chlorine with the stoichiometric ratio. The actual ratio is:

0.445 mol aluminum / 0.240 mol Chlorine = 1.854

From the balanced equation, 3 moles of Chlorine react with 2 moles of Aluminum chloride to form 2 moles of Aluminum chloride. Therefore, the number of moles of Aluminum chloride that can be produced is:

0.240 mol Chlorine × (2 mol Aluminum chloride / 3 mol Chlorine) = 0.160 mol Aluminum chloride

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The empirical formula of Aniline is C9H10N and its molecular formula is C9H9.9N.

To determine the empirical formula of Aniline, we need to calculate the number of moles of each element present in the given mass of the compound and then find the smallest whole-number ratio between them.

Given:

Mass of Aniline = 9.71 mg

Mass of water produced = 6.63 mg

Mass of N2 produced = 1.46 mg

The molecular weight of Aniline = 93 amu

First, let's calculate the number of moles of water and nitrogen produced:

moles of H2O = 6.63 mg / 18.015 g/mol = 0.3680 mmol

moles of N2 = 1.46 mg / 28.014 g/mol = 0.0521 mmol

Next, we need to calculate the number of moles of carbon and nitrogen present in Aniline:

moles of C = (9.71 mg - (0.3680 mmol x 12.011 g/mol)) / 12.011 g/mol = 0.4811 mmol

moles of N = 0.0521 mmol

Now, we need to find the smallest whole-number ratio between these elements. We can divide the number of moles of each element by the smallest value, which is 0.0521 mmol:

moles of C = 0.4811 mmol / 0.0521 mmol = 9.231 ≈ 9

moles of N = 1

moles of H = not calculated, but we can find it using the difference in mass between Aniline and the products formed.

Now, let's calculate the mass of carbon and nitrogen in Aniline:

mass of C = 9 x 12.011 g/mol = 108.099 g/mol

mass of N = 1 x 14.007 g/mol = 14.007 g/mol

Finally, we can calculate the mass of hydrogen by taking the difference between the mass of Aniline and the sum of the masses of carbon, nitrogen, water, and nitrogen:

mass of H = 9.71 mg - (108.099 g/mol + 14.007 g/mol + 6.63 mg + 1.46 mg) = 2.47 mg

Now, we can calculate the number of moles of hydrogen:

moles of H = 2.47 mg / 1.008 g/mol = 2.449 mmol

Finally, we can express the empirical formula of Aniline as: C9H10N

To find the molecular formula, we need to calculate the molecular weight of the empirical formula:

Empirical formula weight = (9 x 12.011 g/mol) + (10 x 1.008 g/mol) + (1 x 14.007 g/mol) = 93.126 g/mol

Now, we can find the molecular formula by dividing the molecular weight of Aniline by the empirical formula weight and multiplying each subscript in the empirical formula by the result:

The molecular weight of Aniline = 93 amu

The molecular weight of the empirical formula = 93.126 g/mol

Molecular formula = empirical formula x (Molecular weight of Aniline / Empirical formula weight)

= C9H10N x (93 amu / 93.126 g/mol)

= C9H9.9N

Therefore, the empirical formula of Aniline is C9H10N and its molecular formula is C9H9.9N.

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Answer : Option (c) is correct.

During electrolysis, a process in which an electric current is passed through a solution containing ions, two electrodes are immersed in the electrolyte solution. The cathode is the negatively charged electrode, while the anode is the positively charged electrode. The electrolyte solution contains both positively charged ions (cations) and negatively charged ions (anions).

At the anode, oxidation occurs as the positively charged ions (cations) in the electrolyte are attracted to the negatively charged anode. The cations lose electrons and become neutral atoms, and these electrons are transferred to the anode. This loss of electrons by the cations results in their oxidation.

The anode, being the electrode where oxidation occurs, has an excess of electrons, which are attracted by the positively charged cations in the electrolyte. The excess of electrons is due to the fact that the anode is connected to the positive terminal of the power source, which supplies electrons to the electrode.

Answer:

it is "no"

Explanation:

do u understand

Arrhenius base -  Releases OH ions when dissolved in water

Arrhenius acid - Releases H+ ions when dissolved in water

Bronsted-Lowry base  - Accepts a proton

Bronsted-Lowry acid - donates a proton

An Arrhenius base is a molecule that decomposes into an OH- or hydroxide in solution when dissolved in water. Look for a molecule ending in OH that does not follow CHx, which denotes an alcohol, to identify the Arrhenius base. Examples of Arrhenius bases include sodium hydroxide, or NaOH.

A substance that raises the concentration of H+ ions in an aqueous solution is known as an Arrhenius acid. Traditional Arrhenius acids are highly polarized covalent substances that dissociate in water to form an anion (A-) and the cation H+. Often, the H+ is referred to as a proton.

Count the hydrogens on each component before and after the reaction to determine if it is an acid or a basic. If there are fewer hydrogens, then the substance is acid (donates hydrogen ions). The material is the base if the hydrogen count has increased (accepts hydrogen ions).

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When two aqueous solutions are combined, a precipitation process takes place in which an insoluble substance (precipitate) develops. Reactions 1, 2, 3, and 5 in the list are precipitation reactions.

When an impermeable material called a precipitate separates from the solution, a reaction known as a precipitation reaction takes place. As an example, a white precipitate of barium sulphate and sodium chloride solution is created when sodium sulphate solution and barium chloride solution are mixed.

An insoluble material is called a precipitate. For instance, barium sulphate and sodium chloride solution is created when sodium sulphate solution and barium chloride solution are combined.

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Answer:

there are approximately 0.004118 moles in 0.1 g of magnesium.

Explanation:

The molar mass of magnesium is approximately 24.31 g/mol. To calculate the number of moles in 0.1 g of magnesium, we can use the following formula:

Number of moles = Mass / Molar mass

Number of moles = 0.1 g / 24.31 g/mol

Number of moles = 0.004118 mol (rounded to 3 significant figures)

Therefore, there are approximately 0.004118 moles in 0.1 g of magnesium.

Answer:

Explanation:

To calculate the number of moles of magnesium in 0.1 g of magnesium, we first need to determine the molar mass of magnesium. The molar mass of magnesium is 24.31 g/mol.

Using this information, we can use the following formula to calculate the number of moles of magnesium:

moles of magnesium = mass of magnesium / molar mass of magnesium

moles of magnesium = 0.1 g / 24.31 g/mol

moles of magnesium ≈ 0.00412 mol

Therefore, there are approximately 0.00412 moles of magnesium in 0.1 g of magnesium.

Potassium (K) reacts with water (H2O) to form potassium hydroxide (KOH) and hydrogen gas (H2).

What is Balanced Chemical Equation?

A balanced chemical equation is a representation of a chemical reaction that shows the relative numbers of reactant and product molecules that participate in the reaction. It obeys the law of conservation of mass, which states that matter cannot be created or destroyed, only transformed from one form to another.

The given chemical equation represents a chemical reaction between potassium (K) and water (H2O), which results in the formation of potassium hydroxide (KOH) and hydrogen gas (H2).

When potassium is added to water, it reacts vigorously, releasing hydrogen gas and forming an alkaline solution of potassium hydroxide. The balanced chemical equation shows that for every 2 moles of potassium (K) and 2 moles of water (H2O), 2 moles of potassium hydroxide (KOH) and 1 mole of hydrogen gas (H2) are produced.

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Answer:

To calculate the moles of Zn(NO3)2 in 173.50 g of the substance, we first need to know the molar mass of Zn(NO3)2.

The molar mass of Zn(NO3)2 can be calculated by adding the atomic masses of its constituent elements:

Zn: 1 x 65.38 g/mol = 65.38 g/mol

N: 2 x 14.01 g/mol = 28.02 g/mol

O: 6 x 16.00 g/mol = 96.00 g/mol

Molar mass of Zn(NO3)2 = 65.38 + 28.02 + 96.00 = 189.40 g/mol

Now we can use this molar mass to calculate the moles of Zn(NO3)2 in 173.50 g of the substance:

moles = mass/molar mass

moles = 173.50 g/189.40 g/mol

moles = 0.9164 mol

Therefore, there are 0.9164 moles of Zn(NO3)2 in 173.50 g of the substance.

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For Au-196.96656, the values of the atomic number and mass number are given below:

The atomic number is the number of protons in the nucleus of an atom. It determines the element to which the atom belongs.

The mass number, on the other hand, is the sum of the number of protons and neutrons in the nucleus of an atom. It is represented by the letter A.

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The mass proportion of sugar in the initial combination, assuming complete dissociation of the salt, is 55.0%.

There are 58.44 g/mol of NaCl in the mixture, which equals the following:

Moles of NaCl are equal to the mass of NaCl divided by the molar mass of NaCl. For example, 10.40 g - x g = 58.44 g/mol

NaCl dissolves into two particles, hence there are actually two particles in the solution.

effective particles equal 2 moles of sodium chloride plus 1 mole of sugar.

2.24 °C = 1.86 °C mol/kg x (27.63 - 0.605 x) mol/kg

x = 5.72 g

(Mass of sugar / Total Mass of Mixture) x 100%, where % Sugar

% sugar = 55.0%

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4. To determine how many moles of NaOH are needed to make a 6 liters of a 3.0 M NaOH solution, we can use the formula:

moles = concentration (in M) x volume (in L)

Substituting the values given, we have:

moles = 3.0 M x 6 L = 18 moles of NaOH

Therefore, 18 moles of NaOH are needed to make a 6 liters of a 3.0 M NaOH solution.

5. To determine how many grams of NaOH are needed to make 4 liters of a 6.0 M NaOH solution, we can use the formula:

moles = concentration (in M) x volume (in L)

Then, we can use the molar mass of NaOH to convert moles to grams:

mass (in g) = moles x molar mass

The molar mass of NaOH is 40 g/mol.

Substituting the values given, we have:

moles = 6.0 M x 4 L = 24 moles of NaOH

mass = 24 moles x 40 g/mol = 960 grams of NaOH

Therefore, 960 grams of NaOH are needed to make 4 liters of a 6.0 M NaOH solution.

Answer:

A

Explanation:

They're easy to dislodge a difficult to move through it the crystal

Wear personal defence tools, follow the guidelines and be careful with chemicals. Measure 14.72g of [tex]H_2SO_4[/tex], dissolve in distilled water, and make up to 400mL in a volumetric flask.

1. Always wear appropriate personal protective equipment such as gloves, goggles, and lab coat.

2. Read and follow the instructions carefully before handling any chemical.

3. Handle the chemicals in a well-ventilated area to prevent inhalation of harmful fumes.

To prepare a 0.2M solution of tetraoxosulphate (VI) acid in a [tex]400cm^3[/tex]  volumetric flask:

Calculate the amount of tetraoxosulphate (VI) acid required using the formula:

Mass = (Molarity x Volume x Molecular weight) / 1000

Where:

Molarity = 0.2M

Volume =[tex]400cm^3[/tex]

Molecular weight = (4x16) + 32 + (6x16) = 98g/mol

Mass = (0.2 x 400 x 98) / 1000 = 7.84g

Weigh out 7.84g of tetraoxosulphate (VI) acid using a balance.

Transfer the weighed tetraoxosulphate (VI) acid into the [tex]400cm^3[/tex] volumetric flask using a funnel.

Add distilled water to the flask until the volume reaches the [tex]400cm^3[/tex] mark on the neck of the flask.

Stopper the flask and mix the solution thoroughly by inverting the flask several times.

It is important to specify the density of the tetraoxosulphate (VI) acid, as this will affect the mass required for the solution. In this case, the percentage purity of the acid is also given, which can be used to calculate the actual mass of the tetraoxosulphate (VI) acid needed.

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The IUPAC name of the amine shown is N-ethylbutan-1-amine.

The IUPAC name of an organic compound is a systematic way of naming organic chemical compounds as recommended by the International Union of Pure and Applied Chemistry (IUPAC).

It is a standard method to ensure that every organic compound has a unique name, which reflects its structure and the functional groups present in the molecule.

The IUPAC name usually consists of several parts, including the parent chain, substituent groups, and functional groups, arranged in a specific order according to a set of rules.

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We need 8.8 moles of O2

We need 1.266 moles of K

We will produce 0.96 moles of water

We know that;

2H2O + O2 → 2H2O2

If 1 mole of O2 makes 2 moles of H2O2

x moles of O2 makes 17.6 moles of H2O2

x = 8.8 moles

Again;

2K + HgCl2 → 2KCl + Hg

If 2 moles of K reacts with 1 mole of HgCl2

x moles of K reacts with 0.633 moles of HgCl2

x = 1.266 moles

Again;

C3H8 + 5O2 → 3CO2 + 4H2O

5 moles of O2 produces 4 moles of H2O

1.2 moles of O2 would produce x moles of water

x = 0.96 moles

Again;

2NH3 ---->N2 + 3H2

If 1 moles of N2 is formed when 3 moles of H2 are formed

3 moles of N2 will lead to the formation of 3 * 3/1

= 9 moles of H2

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