I'm making a AD for my special ed class room and I am interviewing people. Make 10 unique questions I can ask my fellow classmates about the things they have learned in this room.

The molarity of the 750 ml solution of BaI₂ was calculated to be 0.787 M.

413 grams of BaI₂corresponds to 1.05 moles and 750 ml of water corresponds to 0.75 liters of water. So the molarity of the solution is calculated as

1.05* 0.75= 0.787 moles.

24) Thus the molarity of the solution is 0.787 M.

25) P₂O₇ is a covalent compound. Both phosphorous and oxygen have similar electronegativity.

SnBr₂ is ionic as the electronegativity difference between the two is less.

Fe(OH)₂ is an ionic compound.

Cl₃O₈ is a covalent compound.

26) (NH₄)₂CO₃ is highly soluble in water while Fe(OH)₂ is insoluble in water. CaOH is poorly soluble in water while PbCl₂is only sparingly soluble in water.

27) In the given reaction FeS is formed as the precipitate and it is highly insoluble in water while the KCl is dissolved in the aqueous solution.

In the second reaction, ZnCl₂ is soluble as a part of the aqueous solution while strontium sulfate forms the precipitate.

28) In salt water salt is the solute and water is the solvent.

29) Air pressure is lower in a higher atmosphere. The pressure is 0.65 atm and the temperature is -15 degrees at the altitude where the balloon has risen. As the balloon rises, the external pressure decreases and the balloon volume increases. However, the internal pressure or ballon volume remains the same.

30) With an increase in the temperature of a substance, the kinetic energy of the substance increases too.

31) With an increase in the pressure, volume decreases while with a pressure decreases volume increases.

32) If the temperature of a gas increases the pressure also increases.

33) When the plunger is pushed in, the air pressure increases. This pushes the bubbles out and reduces the size of the marshmallow. When the plunger is pushed out, the air pressure decreases, causing the marshmallow to expand.

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Answer:

Phosphorus(P) and Oxygen(O)=Covalent bond

Chlorine(Cl) and Sodium(Na) = Ionic bond

Silver (Ag) and Silver (Ag)= Metallic bond

The mass of CaCl₂ required  to make a 1L solution of 3M CaCl₂ is equal to 332.94 g, hence option C is correct.

To find the mass of CaCl₂ required to make a 3M solution, it considers the molar mass of CaCl2 and the desired concentration.

The molar mass of CaCl₂ can be observed as follows:

Molar mass (CaCl₂) = (molar mass of Ca) + 2 × (molar mass of Cl)

= (40.08 g/mol) + 2 × (35.45 g/mol)

= 40.08 g/mol + 2 × 35.45 g/mol

= 40.08 g/mol + 70.90 g/mol

= 110.98 g/mol

Now, by using the formula for molarity to find the mass of CaCl₂ required:

Molarity (M) = (moles of solute) / (volume of solution in liters)

Arrange the formula to solve for moles of solute:

(moles of solute) = (Molarity) × (volume of solution in liters)

It is required to make a 1L solution of 3M CaCl₂:

(moles of CaCl2) = (3 mol/L) × (1 L)

= 3 mol

Finally, find the mass of CaCl₂ using the moles and molar mass:

(mass of CaCl2) = (moles of CaCl₂ × (molar mass of CaCl₂)

= 3 mol × 110.98 g/mol

= 332.94 g

Thus, the mass of CaCl2 required to make a 1L solution of 3M CaCl₂ is  332.94 g.

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We can use the following formula to determine how many grams of AgNO3 are needed to make a 0.30 M solution with a volume of 750 ml:

moles = volume (L) x concentration (M)

The volume provided must first be converted from milliliters to liters:

Volume = 750 ml ÷ 1000 ml/L = 0.75 L

Now we can find the molarity of AgNO3:

moles = 0.30 M × 0.75 L = 0.225 moles

To find the grams of AgNO3, we need to use the molar mass of AgNO3, which is calculated as follows:

Ag: 1 atom × 107.87 g/mol = 107.87 g/mol

N: 1 atom × 14.01 g/mol = 14.01 g/mol

O: 3 atoms × 16.00 g/mol = 48.00 g/mol

Total molar mass of AgNO3:

107.87 g/mol + 14.01 g/mol + 48.00 g/mol = 169.88 g/mol

Now, we can calculate the grams of AgNO3 needed:

grams = moles × molar mass

grams = 0.225 moles × 169.88 g/mol = 38.22 grams

Therefore, approximately 38.22 grams of AgNO3 are needed to prepare 750 ml of a 0.30 M solution.

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The 5 moles of water is equal to 90.075 grams of water and 220 J of energy is equal to 52.636 calories.

To change moles of water to grams, it is required to find the molar mass of the substance. The molar mass of water (H2O) is equal to 18.015 grams/mol.

To change 5 moles of water to grams, by using the following calculation:

5 moles × 18.015 grams/mol = 90.075 grams of water

Thus, 5 moles of water is equal to 90.075 grams of water.

To change joules to calories,  by using the conversion factor:

1 cal = 4.184 J.

To change 220 J of energy to calories, by using the following calculation:

220 J ×  (1 cal / 4.184 J) = 52.636 cal

Thus, 220 J of energy is equal to 52.636 calories.

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Answer:

C. The sun is closer to Earth than other stars.

Explanation:

The sun appears larger and brighter than other stars because it is much closer to Earth. The sun is the closest star to Earth, at a distance of about 93 million miles. Other stars are much farther away, so they appear smaller and less bright in the sky.

If it travells at 330m/s, and it has to travel 5100m just

5100 ÷ 330

because time = distance ÷ speed

= 15.45 s

Answer:

15.4545455 seconds

Explanation:

(5100 m) / (330 (m / s))

5100/330 = 15.4545455 seconds

Answer:  505 grams K3PO4 x (3 x 222 grams HCI)/ (3 x K3PO4) = 555.5 grams KCl

Explanation:

The molar mass of the gas is approximately 20.72 g/mol.

To determine the molar mass of the gas, we can use the ideal gas law equation:

PV = nRT

where:

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = ideal gas constant (0.0821 L·atm/(mol·K))

T = temperature (in Kelvin)

First, let's convert the given values to the appropriate units:

Pressure = 432 torr = 432/760 atm

Volume = 333 mL = 333/1000 L

Temperature = 23°C = 23 + 273.15 K

Substituting the values into the ideal gas law equation:

(432/760) atm * (333/1000) L = n * 0.0821 L·atm/(mol·K) * (23 + 273.15) K

Simplifying the equation:

0.191 atm * 0.333 L = n * 0.0821 L·atm/(mol·K) * 296.15 K

Solving for the number of moles (n):

n = (0.191 atm * 0.333 L) / (0.0821 L·atm/(mol·K) * 296.15 K)

n ≈ 0.01012 moles

Finally, we can calculate the molar mass using the formula:

Molar mass = mass of the gas sample / moles of gas

Molar mass = 0.210 g / 0.01012 moles

Molar mass ≈ 20.72 g/mol

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Answer:

An atom contains three basic particles namely protons, neutrons and electrons. The nucleus of the atom contains protons and neutrons where protons are positively charged and neutrons are neutral. The electrons are located at the outermost regions called the electron shell.

The limiting reagent is chlorine and the correct option is option 2.

In a chemical reaction, the limiting reagent is the reactant that determines the quantity of the products that are produced. Limiting reagents are defined as the substances which are entirely consumed in the completion of a chemical reaction and so a limiting reagent limits the formation of products and determines the amount of products obtained in the reaction.

The limiting reagent can be identified from the number of moles in the reaction, the one that is having the lesser number of moles acts as a limiting reagent in the reaction.

Given,

Moles of hydrogen = 5.3 moles

Moles of chlorine = 4.8 moles

Limiting reagent is the one that has lesser number of moles and thus chlorine is the limiting reagent in this reaction.

Thus, the ideal selection is option 2.

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Answer:

Question 1:

1: a metal and a nonmetal

3: a metal and a polyatomic anion

4: a polyatomic cation and a metal

Question 2:

Cations give away electrons and anions accept those electrons

The extraction of iron from haematite ore involves a process called reduction. Reduction is the chemical reaction in which oxygen is removed from a compound, resulting in the formation of a new substance.

In the case of haematite, the reduction process involves removing the oxygen from the iron oxide (Fe2O3) to obtain elemental iron (Fe). This is typically achieved through a process called smelting, which is carried out in a blast furnace. Before the haematite ore is reduced in a blast furnace, it needs to undergo a series of steps to separate impurities and prepare it for the reduction process. The first step is crushing and grinding the ore into smaller particles. This is done to increase the surface area of the ore, allowing for better contact with the reducing agent. After crushing and grinding, the ore is then subjected to a process called beneficiation, where it is separated from gangue materials and other impurities.

Beneficiation techniques vary, but commonly involve processes such as gravity separation, magnetic separation, and flotation. These methods exploit the differences in physical and chemical properties between the haematite ore and the impurities, allowing for their separation. Once the ore is purified and separated, it is ready to be reduced in a blast furnace, where the smelting process takes place.

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The limiting reactant will be the one that produces fewer moles of BaSO4. The excess reactant will be the one that has moles left over after the reaction.

To determine the grams of BaSO4 produced and the limiting reactant, we need to compare the stoichiometry of the balanced chemical equation for the reaction between Ba(NO3)2 and Na2SO4, which is:

Ba(NO3)2 + Na2SO4 → BaSO4 + 2NaNO3

First, calculate the number of moles for each reactant:

Moles of Ba(NO3)2 = 200.0 g / molar mass of Ba(NO3)2

Moles of Na2SO4 = 100.0 g / molar mass of Na2SO4

Then, calculate the moles of BaSO4 formed by comparing the stoichiometric coefficients:

Moles of BaSO4 formed = Moles of Ba(NO3)2 (according to the stoichiometry ratio)

Next, calculate the grams of BaSO4 formed:

Grams of BaSO4 formed = Moles of BaSO4 formed × molar mass of BaSO4

To identify the limiting reactant, compare the moles of BaSO4 formed from each reactant. The reactant that produces fewer moles of BaSO4 is the limiting reactant.

To determine the excess reactant remaining, calculate the moles of excess reactant and then convert it to grams.

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To calculate the pH of a solution of NaOH (sodium hydroxide), we need to consider that NaOH is a strong base that dissociates completely in water, producing hydroxide ions (OH⁻).

Given:

Concentration of NaOH = 0.005 M

Since NaOH dissociates into one hydroxide ion (OH⁻) per molecule, we can determine the concentration of hydroxide ions in the solution, which will allow us to calculate the pOH. Then, we can convert the pOH to pH using the relationship: pH + pOH = 14.

1. Calculate the concentration of hydroxide ions (OH⁻):

The concentration of OH⁻ ions will be the same as the concentration of NaOH since NaOH dissociates completely.

Concentration of OH⁻ = 0.005 M

2. Calculate the pOH:

pOH = -log[OH⁻]

pOH = -log(0.005)

Using logarithm properties, we can determine the pOH value:

pOH = -log(0.005)

pOH = -(-2.301)

pOH = 2.301

3. Calculate the pH:

pH = 14 - pOH

pH = 14 - 2.301

pH ≈ 11.699

Therefore, the pH of a 0.005 M NaOH solution is approximately 11.699.

The pH of a 0.005 M concentration of NaOH ( sodium hydroxide ) solution is approximately 11.70.

The pH of a solution is defined as the logarithm of the reciprocal of the hydrogen ion concentration  [H+] of the given solution.

From the formula;

pH = -log[ H⁺ ]

pOH = -log[ OH⁻ ]

pH + pOH = 14

Given that; the concentration of solution (molarity) ( OH⁻ ) is 0.005 M.

First, we determine the pOH.

pOH = -log[ OH⁻ ]

Plug in ( OH⁻ ) = 0.005

pOH = -log[ 0.005 ]

pOH = 2.30

Now, plug pOH = 2.30 into the above formula and solve for the pH:

pH + pOH = 14

pH + 2.30 = 14

Subtract 2.30 from both sides:

pH + 2.30 - 2.30 = 14 - 2.30

pH = 14 - 2.30

pH = 11.7

Therefore, the pH of the solution is 11.7.

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The volume of the gas at standard temperature and pressure conditions is approximately 12.77 liters.

To find the volume of the gas at STP, we can use the combined gas law:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

Note that: at STP (Standard Temperature and Pressure) is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm.

Given that:

P₁ = initial pressure = 4.71 atm

V₁ = initial volume = 2.99 L

T₁ = initial temperature = 28.10 °C = ( 28.10 + 273.15 ) = 301.25 K

P₂ = final pressure (STP pressure ) = 1 atm

T₂ = final temperature (STP temperature)  = 0°C = 273.15 K

V₂ = final volume = ?

Substituting the given values into the formula:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\\\frac{4.71\ *\ 2.99 }{301.25} = \frac{1\ *\ V_2}{273.15 }\\\\V_2 = 12.77\ L[/tex]

Therefore, the final volume is 12.77 litres.

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Answer:

Sodium (Na) is a highly reactive metal, while sodium chloride (NaCl) is a compound formed by the combination of sodium and chlorine (Cl). Sodium exists as a pure element, whereas sodium chloride is a stable, crystalline compound.

Sodium is a soft, silvery-white metal that is highly reactive and can easily react with water or air. In contrast, sodium chloride is a white crystalline solid that is highly stable and does not react readily with water or air. Sodium chloride is commonly known as table salt and is widely used as a seasoning and food preservative.

To calculate the pH of the solution obtained by mixing HCl and NaOH, we need to consider the neutralization reaction between the two compounds. The reaction between HCl (hydrochloric acid) and NaOH (sodium hydroxide) produces water (H₂O) and forms a salt (NaCl).

Given:

Volume of HCl solution (V₁) = 40 cm³

Concentration of HCl solution (C₁) = 0.2 M

Volume of NaOH solution (V₂) = 30 cm³

Concentration of NaOH solution (C₂) = 0.1 M

1. Determine the moles of HCl and NaOH used:

Moles of HCl = Concentration (C₁) × Volume (V₁)

Moles of HCl = 0.2 M × 0.04 L (converting cm³ to L)

Moles of HCl = 0.008 mol

Moles of NaOH = Concentration (C₂) × Volume (V₂)

Moles of NaOH = 0.1 M × 0.03 L (converting cm³ to L)

Moles of NaOH = 0.003 mol

2. Determine the limiting reagent:

The stoichiometry of the reaction between HCl and NaOH is 1:1, meaning that they react in a 1:1 ratio. Whichever reactant is present in a smaller amount will be the limiting reagent.

In this case, NaOH is present in a smaller amount (0.003 mol), which means it will be fully consumed during the reaction.

3. Determine the excess reagent and its remaining moles:

Since NaOH is the limiting reagent, we need to find the remaining moles of HCl.

Moles of HCl remaining = Moles of HCl initially - Moles of NaOH reacted

Moles of HCl remaining = 0.008 mol - 0.003 mol

Moles of HCl remaining = 0.005 mol

4. Calculate the concentration of HCl in the resulting solution:

Volume of resulting solution = Volume of HCl solution + Volume of NaOH solution

Volume of resulting solution = 0.04 L + 0.03 L

Volume of resulting solution = 0.07 L

Concentration of HCl in the resulting solution = Moles of HCl remaining / Volume of resulting solution

Concentration of HCl in the resulting solution = 0.005 mol / 0.07 L

Concentration of HCl in the resulting solution ≈ 0.071 M

5. Calculate the pH of the resulting solution:

pH = -log[H⁺]

pH = -log(0.071)

Using logarithm properties, we can determine the pH value:

pH ≈ -log(0.071)

pH ≈ -(-1.147)

pH ≈ 1.147

Therefore, the pH of the solution obtained by mixing 40 cm³ of 0.2 M HCl and 30 cm³ of 0.1 M NaOH is approximately 1.147.

An atom has 17 protons and 17 electrons. The atom's charge is neutral. The positive charge of the 17 protons in this atom is balanced by the negative charge of the 17 electrons.

The ratio of an atom's protons, which have a positive charge, to its electrons, which have a negative charge, determines the charge of the atom. The quantity of protons in an electrically neutral atom is equal to the quantity of electrons.

The positive charge of the 17 protons in this atom is balanced by the negative charge of the 17 electrons, since there are 17 protons and 17 electrons in it. Consequently, the atom is electrically neutral or has a net charge of zero.

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1.To calculate the concentration of ammonium ions in water, we need to determine the number of moles of ammonium ions and then convert it to milligrams per liter (mg/L).

Given:

Volume of water = 30 ml

Volume of ammonia released = 200 ml

First, we need to convert the volume of ammonia released to the volume of water. Since the ammonia was released from the reaction with sodium hydroxide, the volume of ammonia released is equivalent to the volume of water used. Therefore, the volume of water used is 200 ml.

Next, we'll calculate the number of moles of ammonium ions:

Molar volume of water = 18.015 g/mol

Volume of water used = 200 ml = 0.2 L

The molar ratio between sodium hydroxide and ammonium ions is 1:1. Therefore, the number of moles of ammonium ions is equal to the number of moles of sodium hydroxide used.

Now, let's calculate the number of moles of sodium hydroxide used:

Molar mass of sodium hydroxide (NaOH) = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.99 g/mol

The concentration of sodium hydroxide in water is not provided. If you have the concentration of sodium hydroxide, we can use it to determine the number of moles of sodium hydroxide used. Without that information, we cannot calculate the number of moles of ammonium ions and, subsequently, the concentration of ammonium ions in water.

2. To determine the mass of barium in 1 ml of drinking water, we'll use the information given:

Volume of drinking water = 20 ml

Mass of barium carbonate precipitate formed = 20 mg

We need to calculate the mass of barium in the precipitate and then convert it to milligrams per milliliter (mg/ml).

The molar mass of barium carbonate (BaCO₃) is:

Molar mass of barium (Ba) = 137.33 g/mol

Molar mass of carbonate (CO₃) = 12.01 g/mol + (3 × 16.00 g/mol) = 60.01 g/mol

Molar mass of barium carbonate (BaCO₃) = 137.33 g/mol + 60.01 g/mol = 197.34 g/mol

The molar ratio between barium carbonate and barium is 1:1. Therefore, the number of moles of barium in the precipitate is equal to the number of moles of barium carbonate formed.

Now, let's calculate the number of moles of barium carbonate:

Mass of barium carbonate precipitate formed = 20 mg = 0.020 g

Number of moles of barium carbonate = Mass of barium carbonate / Molar mass of barium carbonate

= 0.020 g / 197.34 g/mol

Finally, we'll calculate the mass of barium in 1 ml of drinking water:

Volume of drinking water = 20 ml

Mass of barium in 1 ml of drinking water = (Number of moles of barium carbonate / Volume of drinking water) × Molar mass of barium

= (0.020 g / 197.34 g/mol) / 20 ml × 137.33 g/mol

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The sample containing 7.1 × 10^19 molecules of H2O corresponds to approximately 1.18 × 10^(-4) moles of H2O.

To determine the number of moles of H2O in a sample containing 7.1 × 10^19 molecules, we need to use Avogadro's number, which states that 1 mole of any substance contains 6.022 × 10^23 molecules.

Given that there are 7.1 × 10^19 molecules of H2O in the sample, we can calculate the number of moles using the following formula:

Moles = Number of molecules / Avogadro's number

Moles = 7.1 × 10^19 / 6.022 × 10^23

Moles ≈ 1.18 × 10^(-4) moles

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Based on Table G, the mass of KCl that must be dissolved in 200 grams of H₂O at 10 °C to make a saturated solution is 60 g.

Based on Table G, the solubility of KCl at 10°C is given as 30 g/100 g water.

To calculate the mass of KCl that can be dissolved in 200 grams of water at 10°C, we can set up a proportion:

(30 g KCl / 100 g water) = (x g KCl / 200 g water)

Cross-multiplying and solving for x, we get:

x g KCl = (30 g KCl / 100 g water) * (200 g water)

x g KCl = 60 g KCl

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The amount of time it will take for the reaction to consume 6.02 x 10²³ molecules of reactant is 3.67 × 10² seconds.

The amount of time it will take for a molecule to react can be calculated by dividing the number of molecules in the substance by the rate of time as follows;

Time taken = no of molecules ÷ no of molecules/seconds

According to this question, if a chemical reaction consumes reactants at a steady rate of 1.64 x 10²¹ molecules per second, the amount of time it will take for the reaction to consume 6.02 x 10²³ molecules of reactant is as follows!

Time = 6.02 x 10²³ molecules ÷ 1.64 x 10²¹ molecules per second

Time = 3.67 × 10² seconds

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Dividing the class into groups and assigning them the task of measuring the mass of the same object multiple times promotes scientific inquiry, encourages critical thinking.

It also provides an opportunity to discuss the concepts of precision, accuracy, and the role of statistical analysis in scientific investigations.

When the teacher divides her class into groups and assigns each group the task of measuring the mass of the same object three times, it allows for multiple measurements to be taken in order to obtain more accurate and reliable results. This approach is a common practice in scientific experiments and data collection.

By having multiple groups perform the measurements, several factors come into play:

1. Precision: Each group's measurements may have some inherent variability due to factors such as the sensitivity of the measuring instrument, human error, or slight variations in the experimental conditions. Taking multiple measurements allows for better assessment of the precision of the measurements by evaluating the spread or range of values obtained.

2. Accuracy: While the teacher already knows the mass of the object is 25 g, the purpose of the exercise is to assess the accuracy of the measurements performed by the students. By comparing the measured values from each group to the known value, the teacher can evaluate the accuracy of the measurements and identify any systematic errors or biases.

3. Averaging: Taking multiple measurements allows for the calculation of an average value, which tends to be a more reliable representation of the true value. By averaging the measurements from all the groups, the teacher can obtain a more accurate estimate of the mass of the object.

4. Statistical Analysis: With multiple measurements, the teacher can perform statistical analysis on the data, such as calculating measures of central tendency (mean, median) and measures of dispersion (standard deviation), to further assess the quality and reliability of the measurements.

Overall, dividing the class into groups and assigning them the task of measuring the mass of the same object multiple times promotes scientific inquiry, encourages critical thinking, and helps students understand the importance of repeated measurements in obtaining accurate and reliable data. It also provides an opportunity to discuss the concepts of precision, accuracy, and the role of statistical analysis in scientific investigations.

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Answer:

3rf or 5d cd Yu been successful of a new future

The name of the branched alkane shown below is 2- methylheptane that is in option D as alkane is a type of hydrocarbon, which is a compound consisting of hydrogen and carbon atoms only. 

Alkanes are characterized by having single bonds between carbon atoms and being saturated hydrocarbons, meaning they have the maximum number of hydrogen atoms bonded to each carbon atom. Alkanes are often referred to as "paraffins" and serve as the simplest and most basic form of hydrocarbons. They are relatively unreactive and are commonly found in petroleum and natural gas. The systematic names for alkanes are derived from the prefix "n-" or "normal-," followed by the Greek numerical prefix indicating the number of carbon atoms. For example, "n-pentane" refers to the straight-chain alkane with five carbon atoms.

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When 30.0 g of iron reacts with 258 g lead (Il) nitrate and 67.8 grams of lead form, then the percentage yield is 40.62%.

Given information,

Mass of iron = 30g

Mass of Lead (III) nitrate = 258g

Mass of lead = 67.8g

The balanced equation for the reaction is:

2 Fe + 3 Pb(NO₃)₂ → 3 Pb + 2 Fe(NO₃)₃

The stoichiometric ratio between iron (Fe) and lead (Pb) is 2:3.

The moles of Fe:

Moles of Fe = mass of Fe / molar mass of Fe

Moles of Fe = 30.0/ 55.845

Moles of Pb = (3/2) × moles of Fe

The theoretical yield of Pb:

Mass of Pb (theoretical) = moles of Pb × molar mass of Pb

Mass of Pb (theoretical) = (3/2) × moles of Fe × molar mass of Pb

The percent yield:

Percent yield = (actual yield / theoretical yield) × 100

Actual yield = 67.8 g

Theoretical yield = (3/2) × (30/55.845) × 207.2 = 166.95

Percent yield = 67.8/166.9  × 100 = 40.62%

Thus, the percentage yield is 40.62%.

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7.54 *[tex]10^8[/tex] meters is  75.4 nanometers.

To convert 7.54 *  [tex]10^8[/tex] meters to nanometers, you can multiply the value by [tex]10^9[/tex]

as,  [tex]10^9[/tex]nanometers = 1  meter.

7.54 * [tex]10^8[/tex] m * [tex]10^9[/tex] =  7.54 x [tex]10^1[/tex] nm

Therefore, 7.54 *[tex]10^8[/tex] meters is equal to 75.4 nanometers.

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To convert 7.54 x 10^-8 meters to nanometers, you multiply 7.54 x 10^-8 by 1 x 10^9 to get 75.4 nanometers.

To convert meters to nanometers, you need to know that 1 meter is equivalent to 1 x 109 nanometers. Therefore, if you were to convert 7.54 x 10-8 m to nanometers, you would multiply 7.54 x 10-8 by 1 x 109.

Here's how you'd do it: 7.54 x 10-8 m * 1 x 109 nm/m = 75.4 nm. So, 7.54 x 10-8 meters is equivalent to 75.4 nanometers.

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