why is glucose a macromolecule?

To produce 22.4 L of O2 at STP, we need 2 moles of H2O.

Standard temperature and pressure (STP) is a set of reference conditions used in chemistry and physics. It is defined as a temperature of 0 degrees Celsius (273.15 Kelvin) and a pressure of 1 atmosphere (760 mmHg or 101.3 kPa). At STP, 1 mole of any ideal gas occupies a volume of 22.4 liters.

The balanced chemical equation is :

2H2O(l) → 2H2(g) + O2(g)

From the equation, we can see that for every 2 moles of H2O, 1 mole of O2 is produced.

At STP (standard temperature and pressure), 1 mole of any ideal gas occupies 22.4 L.

Therefore, to produce 22.4 L of O2 at STP, we need 2 moles of H2O.

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The car's average speed is 116 kilometers per hour.

The entire distance traveled divided by the total time elapsed is the definition of average speed. It is typically stated in terms of the amount of time or distance per unit, such as meters per second (m/s) or kilometers per hour (km/hr).

Average speed = total distance ÷ total time

In this case, the total distance is 464 km and the total time is 4 hours.

Average speed = 464 km ÷ 4 hours

Average speed = 116 km/hr

Therefore, the car's average speed is 116 kilometers per hour.

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Due to the principle of overlay, in an intact sequence of sedimentary stones, every single layer of rock is elder that the one above and younger that the one below it, with the oldest rocks at the bottom as the latest ones at the top

The multiple vertical layers of sedimentary material demonstrate the original horizontality concept (see Figure below). The law in deposition states that the youngest layers of rocks occupy the highest point whereas the earliest were at the bottom.

The Theory of superposition, that asserts that in an unbroken horizontal succession of rocks, the most ancient layers are on the bottom, and gradually younger ones on top, aids geologists in correlating rock strata all around the world.

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Two alkenes may be produced when sodium ethoxide is used to treat 2-Bromo-3-methylbutane in ethanol. Saytzeff's rule governs the reaction pathway, leading to the formation of 2 methyl 2 butene as the main product, which is a more substituted alkene.

Inorganic synthesis uses sodium ethoxide, 21% w/w in ethanol, as a powerful base. It is used in a variety of chemical processes, including esterification, alkoxylation, condensation, and etherification. It participates actively in the Wolf-Kishner reduction, Stobbe reaction, and Claisen condensation.

Because ethoxide ions are Brnsted-Lowry bases and remove hydrogen ions from water molecules to form hydroxide ions, which raise the pH, the solution is extremely alkaline.

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Answer:

Volume required ≈ 20.8 mL

When a solution is diluted, the total number of moles of solute (n), does not change. Only the total solution volume changes. As n=cV (molarity formula, where n = moles, c = concentration, V = volume), then the value of cV is a constant when diluting solutions. This can be expressed by the ratio:

c₁V₁ = c₂V₂, where subscripts 1 and 2 represent the concentrated and dilute solutions.

To dilute a solution of 6.0 M HCl to 0.25 M with a total volume of 0.500 L, we can use the dilution formula, where initial and final concentrations = 6.0 M and 0.25 M respectively, and final volume = 0.500 L (500 mL).

Hence, 6.0×V₁ = 0.25×0.500 = 0.125

V₁ = 0.125/6.0 = 0.02083 L

Therefore, the volume we require is ≈ 20.8 mL

If 355 mL of 0.325 M Pb(NO)₂ (molar mass 331.2 g/mol) is added to 255 mL of 0.225 M K₃PO₄, (molar mass 212.27 g/m), 2.57 grams of Pb(PO₄)₂ (molar mass 811.54 g/mol) precipitate will be formed.[tex]\frac{355ml * 0.325M}{331.2 g/ml}[/tex]

The word "precipitation reaction" refers to a "chemical reaction occurring in an aqueous solution where two ionic bonds combine, yielding the formation of an insoluble salt." The insoluble salts produced by precipitation reactions are known as precipitates. The precipitate, a solid type of residue, is formed during precipitation reactions, which are frequently double displacement processes. When two or more solutions with different salt concentrations are combined, these reactions lead to the formation of insoluble salts that precipitate out of the solution.

One of the best examples of a precipitation reaction is the chemical interaction between potassium chloride and silver nitrate, which results in the precipitation of solid silver chloride. The insoluble salt is a result of the precipitation reaction.

The number of grams of Pb(PO4)2 precipitate that is formed can be calculated using the following equation:

Moles Pb(NO₃)₂ x Molar Mass Pb(PO₄)₂ / Moles K₃PO₄ = Grams of Pb(PO₄)₂

So the calculation is:

[tex]\frac{355ml * 0.325M}{331.2 g/ml}[/tex] x [tex]\frac{811.54g/mol}{255ml * 212.27g/mol}[/tex]) = 2.57 g Pb(PO₄)₂

Therefore, 2.57 g of Pb(PO₄)₂will be formed when 355 mL of 0.325 M Pb(NO₃)₂ and 255 mL of 0.225 M K₃PO₄ is mixed.

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The expected result of the experiment is that the rate of precipitation of sulfur will be faster with hydrogen chloride than with acetic acid.

What is Precipitation?

Precipitation refers to the process in which a substance in a solution comes out of solution and forms a solid. This occurs when the concentration of the substance exceeds its solubility limit in the solution, resulting in the formation of solid particles that settle at the bottom of the solution.

Procedure:

Prepare a series of solutions with different monobasic acids. For example, you can prepare solutions of sodium thiosulphate (Na2S2O3) with different monobasic acids, such as hydrochloric acid (HCl), sulfuric acid (H2SO4), and nitric acid (HNO3), at a constant concentration and volume.

Place equal volumes of the thiosulphate solution and each monobasic acid solution in separate beakers or test tubes.

Start the timer or stopwatch as soon as the two solutions are mixed. The thiosulphate ions will react with the hydrogen ions from the monobasic acid, resulting in the formation of sulphur precipitate according to the following chemical equation:

S2O3^2- + 2H+ -> H2O + S + SO2

Observe the reaction and record the time taken for the formation of sulphur precipitate in each solution.

Compare the rates of precipitation of sulphur for each monobasic acid solution to study the dependence of the rate on the nature of the monobasic acid. Note any trends or differences in the rates of precipitation among the different acids used.

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The running pressure of the HPLC procedure in torr is approximately 2068.2687 torr.

What is Chromatography?

Chromatography is a technique used to separate, identify, and analyze the components of a mixture based on their differential affinities to a stationary phase and a mobile phase. It is widely used in chemistry, biochemistry, pharmaceuticals, environmental science, forensics, and other fields for purification, separation, and analysis of chemical compounds.

To convert pressure from pascals (Pa) to torr, we can use the following conversion:

1 torr = 101325 Pa

So, to convert the pressure from pascals to torr, we can divide the pressure in pascals by 101325.

Given:

Pressure in pascals (Pa) = 2.09 × [tex]10^{8}[/tex] Pa

Using the conversion factor:

Pressure in torr = (2.09 × [tex]10^{8}[/tex]Pa) / 101325 Pa/torr

Calculating the numerical value in torr:

Pressure in torr = 2068.2687 torr (rounded to four decimal places)

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Taking into account the reaction stoichiometry, 2.14 grams of H₂O are formed when 20.0 grams of NaHCO₃ decomposes.

In first place, the balanced reaction is:

2 NaHCO₃ → H₂O + CO₂ + Na₂CO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The following rule of three can be applied: if by reaction stoichiometry 168 grams of NaHCO₃ form 18 grams of H₂O, 20 grams of NaHCO₃ form how much mass of H₂O?

mass of H₂O= (20 grams of NaHCO₃× 18 grams of H₂O)÷168 grams of NaHCO₃

mass of H₂O= 2.14 grams

Finally, 2.14 grams of H₂O are formed.

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The molarity of the solution is 0.0156 M (or mol/L).

More precisely, a solution of iron(III) nitrate, will react with a solution of sodium hydroxide, to generate iron(III) hydroxide, which precipitates out of solution. The insoluble solid is not shown as ions, as you will see.

The molar mass of Iron(III) nitrate is:

Iron: 1 atom x 55.85 g/mol = 55.85 g/mol

Nitrogen: 3 atoms x 14.01 g/mol = 42.03 g/mol

Oxygen: 9 atoms x 16.00 g/mol = 144.00 g/mol

Total molar mass is equal to 241.88 g/mol (55.85 + 42.03 + 144.00).

So, the number of moles of Iron(III) nitrate in 8.55 g is:

moles = mass/molar mass = 8.55 g/241.88 g/mol = 0.0353 mol

The solution's volume in litres must then be determined. 2.26 L is listed as the volume.

The unit of measurement for molarity is moles of solute per litre of solution.

Molarity = 0.0353 mol / 2.26 L = 0.0156 M

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Answer:

a. SOLUTION:

Step 1: Write the balanced chemical equation.

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

Step 2: Calculate the number of moles of CO₂ formed by each reactant.

• Using C₂H₄

Based on the balanced chemical equation, 1 mole of C₂H₄ is stoichiometrically equivalent to 2 moles of CO₂.

The molar mass of C₂H₄ is 28.054 g/mol.

• Using O₂

Based on the balanced chemical equation, 3 moles of O₂ is stoichiometrically equivalent to 2 mole of CO₂.

The molar mass of O₂ is 31.998 g/mol.

Step 3: Determine the limiting reagent.

Since O₂ produced less amount of CO₂ than C₂H₄, O₂ is the limiting reagent.

Step 4: Determine the mass of CO₂ formed.

Note that the (maximum) mass of a product that can be formed is dictated by the limiting reagent. In this case, we will start at the number of moles of CO₂ formed from the limiting reagent (O₂) which is equal to 0.11022 mol.

The molar mass of CO₂ is 44.009 g.

Hence, 4.85 g of CO₂ can be formed.

------------------------------------------------------------

b. ANSWER:

The LR is O₂ and the ER is C₂H₄.

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c. SOLUTION:

The theoretical yield of the reaction is 4.85 g.

Hence, the percent yield of the reaction is 87.6%.

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Explanation:

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The percent yield for the reaction is 13.1%. it is the weight of the product that was obtained to the theoretical yield as a percentage.

The percentage yield is the figure computed to represent the discrepancy between the theoretical yield and the actual yield of an experiment. Both the desired products and the by-products are produced when experimenting with various solutions or when preparing chemical solutions.

We must convert the amounts of NO and O2 to moles since 1 mol of NO reacts with 1/2 mol of O2:

moles of NO = 33.6 g / 30.01 g/mol = 1.12 mol

moles of O2 = 26.9 g / 32.00 g/mol = 0.84 mol

Since O2 is limiting, Our calculations will be based on the amount of oxygen present.

moles of NO2 = 0.84 mol O2 × (2 mol NO2 / 1 mol O2) × (2 mol NO / 1 mol NO2)

                       = 3.36 mol NO2

The molar mass of NO2 is 46.01 g/mol,

the theoretical yield in grams is:

mass of NO2 = 3.36 mol × 46.01 g/mol = 154.34 g

percent yield = (20.2 g / 154.34 g) × 100% = 13.1%

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Total, 2 moles of iron (II) nitrate would be needed to produce a 4 moles of lithium nitrate.

The balanced chemical equation for the reaction is;

2Li + 1Fe(NO₃)₂ → 1Fe + 2LiNO₃

From this equation, we can see that 2 moles of lithium nitrate are produced for every 1 mole of iron (II) nitrate used.

To determine how many moles of iron (II) nitrate are needed to produce 4 moles of lithium nitrate, we can use the mole ratio from the balanced equation;

moles of LiNO₃ = 4

moles of Fe(NO₃)₂ = (4 mol LiNO₃) x (1 mol Fe(NO₃)₂ / 2 mol LiNO₃)

= 2 mol Fe(NO₃)₂

Therefore, 2 moles of iron (II) nitrate is needed to produce 4 moles of lithium nitrate.

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The  volume of acetone would be required to purify a 5.0g sample of compound is 58.82ml

Let use the solubility data of acetone provided to calculate the volume.

First we will need to calculate the solubility of Y in 25 degree.

   Solubility of Y= 0.12g/10ml= 0.012g/ml.

Solubility of y in boiling acetone = 0.85g/10ml =0.085g/ml.

In order to purify y, we need to dissolve the amount of y in the 5.0g sample.

The amount of acetone needed to dissolve y in the sample is

amount acetone= mass of y/ solubility of y in boiling acetone.

amount acetone= 5.0g/0.085g/ml= 58.82ml.

Therefore, the  volume of acetone would be required to purify a 5.0g sample of compound is 58.82ml

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The pH of the student's stomach acid can affect the egg salad sandwich once inside the student's body in several ways. When the sandwich enters the stomach, the hydrochloric acid (HCl) present in the stomach acid starts to break down the proteins and other components of the sandwich.

The optimal pH range for the activity of the digestive enzyme pepsin, which breaks down protein, is between pH 1.5 and 2.5, so the acidic environment of the stomach acid is necessary for proper digestion of the sandwich. Additionally, the acidic environment of the stomach can denature and break down any harmful bacteria that might be present in the sandwich, reducing the risk of foodborne illness.

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--The complete Question is, How does the pH of the student's stomach acid affect the egg salad sandwich once inside the student's body? --

The balanced equation for the reaction of acetylene (C₂H₂) and oxygen (O₂) is; 2 C₂H₂(g) + 5 O₂(g) → 4 CO₂(g) + 2 H₂O(g) + heat

The coefficients in the balanced equation represent the stoichiometric ratio of the reactants and products in the chemical reaction. In this case, 2 molecules of acetylene (C₂H₂) react with 5 molecules of oxygen (O₂) to produce 4 molecules of carbon dioxide (CO₂) and 2 molecules of water (H₂O), along with the release of heat.

The balanced equation shows that the number of atoms of each element is the same on both the sides of the equation, in accordance with the law of conservation of mass.

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The final temperature with Specific heat = 0.385 J/g °C for copper is 57.7 °C.

Specific heat is a measure of the amount of heat energy required to raise the temperature of a unit mass of a substance. It is a physical property of a material and is usually measured in units of J/(g °C) or J/(g K).

q = m * c * ΔT

where q is the heat absorbed (in joules), m is the mass of the substance (in grams), c is the specific heat (in J/(g °C)), and ΔT is the change in temperature (in °C).

Plugging in the values,

6.30 kJ = 500.0 g * 0.385 J/(g °C) * ΔT

Simplifying the equation, we get:

ΔT = (6.30 kJ) / (500.0 g * 0.385 J/(g °C))

ΔT = 32.7 °C

The final temperature of the copper is as follows:

25 °C + 32.7 °C = 57.7 °C

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Enthalpy of the reaction of the following equation is the same as the enthalpy of formation. 2 Fe (s) + 3 Cla (g) → 2 FeCl (s).

The difference between total reactant and total product molar enthalpies, computed for substances in their standard states, is the standard enthalpy of reaction for a chemical reaction. As long as the enthalpy of mixing is taken into consideration as well, this can then be used to forecast the total chemical bond energy liberated or bound during reaction.

A reaction's standard enthalpy is defined so that it solely depends on the conditions that are prescribed for it, rather than just on the actual circumstances in which reactions take place. The two main circumstances in which thermochemical measurements are actually performed.

The change in enthalpy during the synthesis of 1 mole of a compound from its component elements in their reference state, with all substances in their standard states, is known as the standard enthalpy of formation or standard heat of formation of a compound. The IUPAC recommends using the standard pressure value of 105 Pa (100 kPa = 1 bar), while previous to 1982, the value of 1 atm (101.325 kPa) was used.

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Consequently, 1.57 litres is the volume of a 3.00 M solution prepared using 4.70 moles of LIF.

Probably the most often used unit of measurement for solution concentration is molarity, abbreviated as (M). The ratio of the number of litres of solution to the number of moles of solvent is known as molarity. For instance, we may use the terms "1.5 molar solution" or "1.5 M" to describe a solution.

To calculate the volume of a 3.00 M solution made with 4.70 moles of LIF, we can use the formula:

Molarity is equal to the moles of solute times the litres of solution.

Rearranging the formula to solve for the volume of the solution, we get:

Volume of solution = moles of solute ÷ (Molarity x 1000)

Where 1000 is the conversion factor from milliliters to liters.

Plugging in the values given, we get:

Volume of solution = 4.70 moles ÷ (3.00 mol/L x 1000)

Volume of solution = 1.57 L

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A 225-gram a sample of oxygen contains 84.684375 X 1023 atoms of oxygen.

One oxygen atom, designated O. Two molecules of oxygen combine to form the molecule O2. Our bodies have developed to breathe O2, which is the main component of our atmosphere.

Because each molecule is made up of two atoms of oxygen fused together, its formula is written as O2.

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The pressure of the H2 gas is 739.1 torr.

What is Pressure?

The SI unit of pressure is Pascal (Pa), which is defined as one Newton per square meter (N/m^2). Other common units of pressure include atmospheres (atm), millimeters of mercury (mmHg), torr, and pounds per square inch (psi). Pressure is a fundamental concept in physics, chemistry, and engineering, and it plays a crucial role in various natural and man-made processes, such as fluid mechanics, thermodynamics, and weather patterns.

To find the pressure of the H2 gas, we need to subtract the vapor pressure of water from the total pressure inside the eudiometer.

Given:

Pressure of eudiometer (total pressure) = 760.0 torr

Vapor pressure of water = 20.9 torr

Pressure of H2 gas = Pressure of eudiometer - Vapor pressure of water

Pressure of H2 gas = 760.0 torr - 20.9 torr

Pressure of H2 gas = 739.1 torr

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The theoretical yield of Acetylsalicylic acid is found out to be: 3.01 grams.

The limiting reagent is a substance that prevents a chemical reaction from occurring completely.

When a limiting reagent is used in a chemical reaction, the atoms, molecules, or ions of the other reactant that it (the limiting reagent) reacts with will either stay free or unreacted.

Popular compounds called acids and bases interact with one another to create salt and water. The Latin word "acere," which meaning "sour," is where the term "acid" originates.

According to the problem, we have 2.31 grams of salicylic acid. We need to determine the theoretical yield of acetylsalicylic acid.

The molar mass of salicylic acid is 138.12 g/mol. Therefore, the number of moles of salicylic acid we have is:

n = mass / molar mass

n = 2.31 g / 138.12 g/mol

n = 0.0167 mol

According to the balanced equation, 1 mole of salicylic acid reacts with 1 mole of acetic anhydride to produce 1 mole of acetylsalicylic acid. Therefore, the number of moles of acetylsalicylic acid produced is also 0.0167 mol.

The molar mass of acetylsalicylic acid is 180.16 g/mol. Therefore, the theoretical yield of acetylsalicylic acid is:

mass = n x molar mass

mass = 0.0167 mol x 180.16 g/mol

mass = 3.01 g

Therefore, the theoretical yield of acetylsalicylic acid is 3.01 grams.

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C is the correct option as it contains oxygen along with carbon and hydrogen. Aliphatic hydrocarbons that have had one or more hydrogen atoms .

Aliphatic hydrocarbons that have had one or more hydrogen atoms replaced by a halogen, such as those that have been fluorinated, chlorinated, brominated, or iodized, are known as halogenated aliphatic hydrocarbons.

The group of chemical molecules known as chlorinated aliphatic hydrocarbons is diverse and is characterised by an open-chain structure or a variable number of bonds, which can be single, double, or triple. C is the correct option as it contains oxygen along with carbon and hydrogen.

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The percentage concentration of potassium sulfate in the solution is 20%.

What is Percentage Concentration?

Percentage concentration, also known as percent concentration or mass percent, is a measure of the amount of a solute present in a solution, expressed as a percentage of the total mass or volume of the solution. It is commonly used in chemistry and related fields to express the concentration of a solute in a solution.

To find the percentage concentration of potassium sulfate in the solution, we need to calculate the mass of potassium sulfate in the solution and express it as a percentage of the total mass of the solution.

Given:

Mass of potassium sulfate (K2SO4) = 80.0 g

Mass of water (H2O) = 320 g

Total mass of the solution = Mass of potassium sulfate + Mass of water

= 80.0 g + 320 g

= 400 g

Now, we can calculate the percentage concentration of potassium sulfate:

Percentage concentration = (Mass of potassium sulfate / Total mass of the solution) × 100

Percentage concentration = (80.0 g / 400 g) × 100

= 20%

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a. On introducing 12g of HCl, 0.047 moles of sodium chloride are produced and b. On introducing 2.1 moles of sodium dichromate, 4.2 moles of chromium (III) chloride are produced.

Number of moles = Mass of the molecule (in grams) / Molar mass of the molecule (in grams per mole)

a. To solve this problem, we need to balance the chemical equation first:

Na₂Cr₂O7 + 14HCl → 2CrCl₃ + 2NaCl + 7Cl₂ + 7H₂O

The balanced equation shows that for every 14 moles of HCl, 2 moles of NaCl are produced.

So, to find the number of moles of NaCl produced from 12g of HCl, we first need to convert grams to moles using the molar mass of HCl.

Molar mass of HCl = 1.008 + 35.453 = 36.461 g/mol

Number of moles of HCl = mass / molar mass = 12g / 36.461 g/mol = 0.329 mol

Now, we can use the mole ratio from the balanced equation to find the number of moles of NaCl produced:

2 moles NaCl / 14 moles HCl * 0.329 mol HCl = 0.047 moles NaCl

Therefore, 0.047 moles of sodium chloride are produced.

b. Using the balanced equation from part a, we can see that the mole ratio between sodium dichromate and chromium (III) chloride is 1:2.

So, if 2.1 moles of sodium dichromate are introduced, we can find the number of moles of chromium (III) chloride produced by multiplying the number of moles of sodium dichromate by the mole ratio:

2.1 moles Na₂Cr₂O₇ * 2 moles CrCl₃ / 1 mole Na₂Cr₂O₇ = 4.2 moles CrCl₃

Therefore, 4.2 moles of chromium (III) chloride are produced.

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The empirical formula of the oxygenated hydrocarbon is CH2O.

We need to determine the mole ratios of the elements present in the compound using the given masses and the molar masses of the elements.

First, we need to find the moles of carbon dioxide and water vapor produced in the reaction:

moles of CO2 = 1.76 g / 44.01 g/mol = 0.0400 mol CO2

moles of H2O = 1.08 g / 18.02 g/mol = 0.0599 mol H2O

Next, we need to find the moles of carbon and hydrogen in the compound. We can do this by using the masses of carbon dioxide and water vapor produced, the molar masses of the elements, and the balanced chemical equation for the combustion of hydrocarbons:

CxHyOz + (x + y/4 - z/2) O2 -> x CO2 + (y/2) H2O

From the balanced equation, we can see that the mole ratio of carbon to carbon dioxide is 1:1, and the mole ratio of hydrogen to water vapor is (1/2):(1/2) or 1:1. Therefore:

moles of C = moles of CO2 = 0.0400 mol CO2

moles of H = 2 × moles of H2O = 0.1198 mol H2O

Finally, we can find the mole ratio of carbon to hydrogen in the compound:

mol ratio of C to H = (moles of C) / (moles of H) = 0.0400 mol CO2 / 0.1198 mol H2O ≈ 0.334

To get whole numbers for the mole ratio, we can multiply both sides by a common factor, such as 3:

mol ratio of C to H ≈ 0.334 ≈ 1/3

mol ratio of C to H = 1

mol ratio of O to H = (moles of O) / (moles of H) = (moles of CxHyOz - moles of C - moles of H) / (moles of H) = (0.0400 mol CO2 + 0.0599 mol H2O - 0.0400 mol C - 0.1198 mol H) / 0.1198 mol H = 0.213

To get whole numbers for the mole ratio, we can multiply both sides by a common factor, such as 5:

mol ratio of O to H ≈ 0.213 ≈ 2/9

mol ratio of O to H = 2

Therefore, the empirical formula of the oxygenated hydrocarbon is CH2O.

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The new temperature of a sample of gas initially at 0°C and contains 100L is 546K.

The temperature of a gas can be calculated using the Charles' law equation as follows:

V₁/T₁ = V₂T₂

Where;

According to this question, the volume of a sample of a gas at 0°C is 100 liters. If the volume of the gas is increased to 200 liters at constant pressure, the new temperature is as follows:

100/273 = 200/T₂

200 × 273 = 100T₂

T₂ = 546K

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Unitless calculations and rounding the final answer to the proper number of significant figures : a) 3.323   b)  5.8  c)  42.02  d) 15.7025

Significant figures represent the meaningful and reliable digits in a number.

a) 3.41 - 0.086652 = 3.323348

Since both numbers have four significant figures, the final answer should also have four significant figures. Therefore, rounding the final answer to four significant figures gives: 3.323

b) 17.441 / 3 = 5.813666666666666

The least precise value in this calculation is 3, which has only one significant figure. Therefore, rounding the final answer to one significant figure gives: 5.8

c) 21.01 * 2 = 42.02

Both numbers have four significant figures, so the final answer should also have four significant figures. Therefore, the final answer is: 42.02

d) 18.7644 - 3.472 + 0.4101 = 15.7025

All three numbers have five significant figures, so the final answer should also have five significant figures. Therefore, the final answer is: 15.7025.

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