Which of these describes the phase change that occurs during sublimation?

166 g HCl

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Stoichiometry

Step 1: Define

[RxN - Unbalanced] Mg (s) + HCl (aq) → MgCl₂ (aq) + H₂ (g)

[RxN - Balanced] Mg (s) + 2HCl (aq) → MgCl₂ (aq) + H₂ (g)

[Given] 2.28 mol Mg (s)

Step 2: Identify Conversions

[RxN] 1 mol Mg = 2 mol HCl

Molar Mass of H - 1.01 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of HCl - 1.01 + 35.45 = 36.46 g/mol

Step 3: Stoichiometry

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

166.258 g HCl ≈ 166 g HCl

Answer:

approximately 1 week

Explanation:

Hope this helps, feel free to give feedback

1 to 2 weeks but if they smell a litte more stink then usually you better get rid of them or they my make you sick. i got sick for old boiled egg from the school cafeteria a long time ago.

Explanation:

In the pair of sodium and potassium the second element has a higher first ionization energy but a lower second ionization energy then the first element.

Ionization energy can be defined as that amount of energy which is required in order to pull out an electron from the vicinity of an atom completely.

When we remove out first electron from the atom it is called the first ionization energy and when we pull out the second electron from the atom it is known as the 2nd ionization energy.

As per the question when we talk about Sodium and potassium, potassium has a higher ionization energy but a lower second ionization energy then sodium because of the size difference between the two atoms.

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The number of moles of the cyclopropane that remains after 1.00 s is  6.1 * 10^-4 M.

By the use of the equation of the first order reaction, we can be able to find out the concentration at a given time.

The equation of the first order reaction can be written as;

ln[A] = ln[Ao] - kt

[A] = Concentration at time t

[Ao] = Initial concentration

k = rate constant

t = time

Then;

[A] = e^ln[Ao] - kt

[A] = e^ln6.00 - (9.2 * 1)

[A] = 6.1 * 10^-4 M

The amount of the cyclopropane left is  6.1 * 10^-4 M.

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The snack containing 10 g of carbohydrates, 2 g of protein, and 5 g of fat has a total of 93 calories.

To calculate the total number of calories in the snack, we need to consider the caloric content of each macronutrient (carbohydrates, proteins, and fats).

Carbohydrates: Carbohydrates provide 4 calories per gram. In this case, 10 g of carbohydrates would contribute 10 g * 4 calories/g = 40 calories.

Proteins: Proteins also provide 4 calories per gram. With 2 g of protein, we have 2 g * 4 calories/g = 8 calories.

Fats: Fats provide 9 calories per gram. For the 5 g of fat in the snack, we have 5 g * 9 calories/g = 45 calories.

Now, to find the total calories, we add up the calories from each macronutrient: 40 calories (carbohydrates) + 8 calories (protein) + 45 calories (fat) = 93 calories.

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Answer:

i think its a if not sorry i have it in a test right now

Explanation:

Answer:

3) 333.75 Grams

Explanation:

2Al + 6HCl --- 2AlCl3 + 3H2

Work out and label what we know and workout Ar and Mr

Al = 67.15grams         AlCl3

Al = Ar=27                  Mr = 133.5

Mass/Mr == Moles

Al = 2.5 Moles          AlCl3 = 333.75

Moles * Mr = Mass

2.5 * 133.5 = 333.75 Grams

4) Apply the same concept (numbers in front of elements are ratios, in this case 12 and 8) (when working out MR/AR ignore numbers in front)

1) Workout Mr/Ar (The relative atomic masses added up)

2) Do 36.72/(Mr of O2) = Your Moles

3) Do workout what the moles would be in the ratio 12:8 (Times/divide moles to find out the correct moles for 12:8

4) Times the new moles by the Mr of SO3

5) = Answer

Chemical reactions that break down complex molecules to form simpler structures are known as catabolic reactions or catabolism. These reactions release energy.

What is catabolism?

Catabolism is a set of metabolic processes that break down complex molecules into simpler ones, releasing energy in the process. This energy is typically stored in the chemical bonds of the complex molecules and is released when these bonds are broken. Catabolic reactions are often involved in the breakdown of nutrients such as carbohydrates, lipids, and proteins, as well as in the breakdown of cellular components such as organelles, to provide the energy and building blocks needed for cellular processes.

These reactions release energy, which is stored in the chemical bonds of the complex molecules and is released when the bonds are broken. Therefore, catabolic reactions are also known as exergonic reactions, which means they release energy in the form of heat, light, or electrical energy. This energy can then be used by the cell to perform work, such as powering muscle contractions, synthesizing new molecules, or transporting ions and molecules across cell membranes.

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Complete question is: Chemical reactions that break down complex molecules to form simpler structures are known as catabolic reactions or catabolism. These reactions release energy.

The correct order of molecular conversions for the Calvin Cycle are as follows: The Calvin Cycle is the most significant light-independent reaction for the reduction of carbon dioxide in photosynthesis. It takes place in the stroma of the chloroplasts and is made up of three stages which are carbon fixation, reduction, and regeneration.

The reaction sequence of the Calvin cycle is shown below:Stage 1: Carbon Fixation: RuBP carboxylase/oxygenase, also known as Rubisco, is the enzyme that catalyzes this step, and it occurs in the chloroplast stroma. Carbon dioxide (CO2) diffuses into the stroma and combines with a five-carbon sugar, ribulose-1,5-bisphosphate (RuBP), which results in a six-carbon compound (6C).Stage 2: Reduction: In this step, the 6-carbon compound is converted into two molecules of 3-phosphoglyceric acid (3-PGA) using ATP and NADPH. The carbon compound is reduced as a result of this process and the ATP is transformed to ADP, whereas NADPH is converted to NADP+.Stage 3: Regeneration: The molecule RuBP is regenerated in this final step using ATP. 5 out of 6 of the RuBP molecules are recycled, whereas 1 is retained to proceed to the next round.

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The molarity of the silver nitrate solution is approximately 0.382 M.

To calculate the molarity of the silver nitrate solution, we need to determine the number of moles of AgNO₃ present in the solution and then divide it by the volume of the solution in liters.

Given;

Mass of AgNO₃ = 65.00 grams

Volume of solution = 3.0 L

First, we need to calculate the number of moles of AgNO₃ using the molar mass of AgNO₃, which is the sum of the atomic masses of its constituent elements.

Molar mass of AgNO₃;

Ag = 107.87 g/mol

N = 14.01 g/mol

O = 16.00 g/mol (x3, since there are three oxygen atoms in AgNO₃)

Molar mass of AgNO₃ = 107.87 + 14.01 + (16.00 × 3) = 169.87 g/mol

Next, we calculate the number of moles of AgNO₃;

moles of AgNO₃ = mass of AgNO₃ / molar mass of AgNO₃

moles of AgNO₃ = 65.00 g / 169.87 g/mol

Now, we divide the number of moles of AgNO₃ by the volume of the solution in liters to calculate the molarity;

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

Molarity (M) = moles of AgNO₃ / 3.0 L

Substituting the calculated value for the moles of AgNO₃;

Molarity (M) = (65.00 g / 169.87 g/mol) / 3.0 L

Simplifying and calculating this expression, we find;

Molarity (M) ≈ 0.382 M

Therefore, the molarity of the silver nitrate solution is approximately 0.382 M.

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Answer: Yes.

Explanation:

the number of protons in the nucleus determines an element's identity. Chemical changes do not affect the nucleus, so chemical changes cannot change one type of atom into another. The number of protons in a nucleus does change sometimes, however. The identity of the atom, therefore, changes.

In order to answer the question first we must write the atomic number of each element:

Zirconium (Zr): 40

Cadmium (Cd): 48

Iridium (Ir): 77

Iron (Fe): 26

Then, we have to complete the distribution of electrons in each orbital for each atom:

The first 4 levels have the following distribution:

Level1: 1s

Number of electrones: 2

Level 2: 2s, 2p

Number of electrones 8 (2 in the s orbital and 6 in the p orbitals).

Level3: 3s, 3p, 3d

Number of electrones 18 (2 in the s orbital, 6 in the p orbital and 10 in the d orbitals)

Level 4: 4s, 4p, 4d, 4f

Number of electrones 32 (2 in the s orbital, 6 in the p orbitals, 10 in the d orbitals and 14 in the f orbitals)

The order in which the orbitlas are completed depends on the energy of each level. For example the 4s orbitals will be completed before the 3d orbitals because their energy is lower.

The order is as follows:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p...

Now, knowing the atomic number we can answer the question:

For Zirconium (total 40 electrones):

2 electrones are predicted in the 4d orbital

For Cadmium (total 48 electrones):

10 electrones are predicted in the 4d orbital

For iridium, as it has an atomic number higher than Cadmium we can predict tha it also complets the 4d orbital, then it has also 10 electrones in it.

For iron (total 26 electrones)

Iron has no electrones in the 4d orbitals

Answer:

differential

Explanation:

According to openstax.org, "Gram-staining is a differential staining technique that uses a primary stain and a secondary counterstain to distinguish between gram-positive and gram-negative bacteria."

High-quality scientific research is characterized by several key criteria. Three examples of such criteria include: rigorous experimental design and methodology, reliable data analysis and interpretation, and clear and transparent reporting of results.

These criteria ensure that research is conducted in a systematic and reliable manner, leading to trustworthy and valid findings.

Rigorous Experimental Design and Methodology: High-quality scientific research requires a well-designed experimental approach. This involves careful planning, proper control groups, randomization, and replication. A rigorous methodology ensures that experiments are conducted under controlled conditions, minimizing bias and confounding variables, and allowing for accurate and reliable data collection.

Reliable Data Analysis and Interpretation: After data collection, high-quality research involves thorough and appropriate analysis of the data. This includes using appropriate statistical methods to evaluate the significance of the results and drawing valid conclusions. Proper data analysis helps researchers identify patterns, trends, and relationships, supporting or refuting their hypotheses in an objective and reliable manner.

Clear and Transparent Reporting of Results: High-quality research demands transparent reporting of the methods, procedures, and findings. This includes providing detailed descriptions of the experimental setup, data collection processes, and statistical analyses used. Clear reporting allows other researchers to replicate the study and verify its results. Additionally, complete reporting ensures that readers can understand the research methodology and draw their own conclusions based on the evidence presented.

By adhering to these criteria, high-quality scientific research maintains integrity, credibility, and reproducibility. It fosters trust among the scientific community and facilitates the advancement of knowledge by building upon reliable foundations.

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The pH at the equivalence point is 8.72.

A mixture of a weak acid and its conjugate base is called a buffer solution. The acid-base reaction for the titration of a weak acid using a strong base is shown in the attached picture. The equivalence point is reached when there is enough base to neutralize the acid.

The number of moles of acetic acid can be calculated by multiplying molarity (0.10 M) and volume (50 mL or 0.05L), resulting in 5 x 10 -3 mol. This will also be equal to the moles of conjugate base produced.

For the next reaction, the concentration of the conjugate base must be determined by dividing its number of moles by the total volume. The calculation of the NaOH volume used is shown below.

molarity of NaOH = moles NaOH / volume NaOH

volume NaOH = 5 x 10-3 moles / 0.10 M

volume NaOH = 0.05 L

total volume = volume NaOH + volume acetic acid

total volume = 0.05 L + 0.05 L = 0.10 L

concentration of conjugate base CH3COO- = 5x 10-3 mol / 0.10 L = 0.05 M

Then, we will determine the value of x from the second reaction. The base dissociation constant or Kb is used to solve for x. The value of acid dissociation constant or Ka of acetic acid is 1.8x10-5.

Kw = 1 X 10 -14 = Ka * Kb

Kb = 1x10-14 / Ka = 1x10-14 / 1.8x10-5

Kb= 5.56 x 10-10

Kb = { [CH3COOH]*[OH-]} / [CH3COO-]

Use the equilibrium value from the second reaction to solve for x

\(5.56x10x^{-10}=\frac{[x][x]}{[0.05-x]}\)

\((5.56x10^{-10})(0.05)- (5.56x10^{-10})(x)=x^{2}\)

\(x^{2} +5.56x10^{-10}x-2.78x10^{-11}=0\)

Solving for x through the quadratic equation, x is found to be equal to 5.27x10-6. This is also equal to the concentration of -OH.

pOH = -log (5.27x10-6) = 5.27

pH = 14 - pOH = 8.72

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The energy required to ionize a muonic hydrogen atom compare to that required to ionize a regular hydrogen atom:

Muonic hydrogen is an exotic hydrogen atom, where a muon (instead of an electron) orbits proton. Because muon is 200 times heavier than the electron, the muon's orbit is 200 times closer to the proton in muonic hydrogen than that of the electron in regular hydrogen.

The extremely precise extraction of proton radius obtained by Pohl et al. from measured energy difference between the 2P and 2S states of muonic hydrogen disagrees significantly with that extracted from electronic hydrogen or elastic–proton scattering. This discrepancy is proton radius puzzle.

In muonic hydrogen electron is replaced by muon, μ which is 200 times heavier than electron

ΔE= En - Ei

for hydrogen like atom,

ΔE= 13.6(Z square /n square) - (-13.6Z square/1 square)

For, Z=1 and n=x (for ionization)

I.E.=13.6eV for electron

I.E.∝ mass (as binding energy proportional to mass)

I.E.=13.6eV*200

    =2720eV

    =2.720keV

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Answer:

aswer is A

Explanation:

because the CL,K,AR is correct since it is order right

Propanal will form silver mirror upon heating with Tollen's reagent while propanoic acid will not respond.

The sample will degrade at a rate of 151.74 decays per minute in 1000 years and 10.24 decays per minute in 50000 years, respectively.

To calculate the decay rate of the carbon sample in Part A and Part B, we need to consider the half-life of carbon-14. The half-life of carbon-14 is approximately 5730 years.

Part A:

To find the decay rate of the sample in 1000 years, we need to determine the number of half-lives that have passed in 1000 years. We can do this by dividing the time elapsed (1000 years) by the half-life of carbon-14 (5730 years):

Number of half-lives = 1000 years / 5730 years ≈ 0.1748

Since each half-life halves the initial quantity, we can calculate the remaining fraction of the sample after 0.1748 half-lives:

Remaining fraction = (1/2)^(0.1748) ≈ 0.9391

The decay rate is given as 161.5 decays/minute, so we can calculate the decay rate of the sample in 1000 years:

Decay rate in 1000 years = Remaining fraction * Initial decay rate

= 0.9391 * 161.5 decays/minute

≈ 151.74 decays/minute

Therefore, the decay rate of the sample in 1000 years is approximately 151.74 decays/minute.

Part B:

To find the decay rate of the sample in 50000 years, we need to determine the number of half-lives that have passed in 50000 years:

Number of half-lives = 50000 years / 5730 years ≈ 8.7257

Using the same logic as in Part A, the remaining fraction after 8.7257 half-lives is:

Remaining fraction = (1/2)^(8.7257) ≈ 0.0632

Now we can calculate the decay rate in 50000 years:

Decay rate in 50000 years = Remaining fraction * Initial decay rate

= 0.0632 * 161.5 decays/minute

≈ 10.24 decays/minute

Therefore, the decay rate of the sample in 50000 years is approximately 10.24 decays/minute.

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The age of the shell can be determined by using the half-life of Carbon-14, which is 5730 years, and the fact that the shell has 12.5% of its original 14C left.
The shell is approximately 17,190 years old.

Carbon-14 is a radioactive isotope that decays over time. When an organism dies, it stops taking in carbon-14, and the carbon-14 in its body begins to decay into nitrogen-14. The half-life of carbon-14 is the time it takes for half of the carbon-14 in a sample to decay.

In this case, the shell has 12.5% of its original 14C left. This means that 87.5% of the original 14C has decayed into nitrogen-14. We can use the half-life of carbon-14 to determine how long it would take for 87.5% of the 14C to decay.

One half-life of carbon-14 is 5730 years. Therefore, two half-lives would be 2 x 5730 years = 11,460 years. Three half-lives would be 3 x 5730 years = 17,190 years, and so on.

To find the age of the shell, we can use the formula:

age = number of half-lives x half-life of carbon-14

If 87.5% of the 14C has decayed, then 12.5% is left. This means that 1/8th of the original 14C is left. One half-life of carbon-14 is the time it takes for half of the 14C to decay, so if 1/8th is left, then three half-lives have passed.

Therefore, the age of the shell can be calculated as:

age = 3 x 5730 years = 17,190 years

Long Answer: The age of the shell can be determined by using the concept of radioactive decay and the half-life of Carbon-14.

Carbon-14 is a radioactive isotope of carbon that decays over time. When an organism dies, it stops taking in carbon-14, and the carbon-14 in its body begins to decay into nitrogen-14. The rate of decay of carbon-14 is constant, and it is measured by its half-life.

The half-life of carbon-14 is the time it takes for half of the carbon-14 in a sample to decay. This means that if we know the original amount of carbon-14 in a sample and the amount of carbon-14 that is left, we can calculate how many half-lives have passed and use this information to determine the age of the sample.

In this case, the shell has 12.5% of its original 14C left. This means that 87.5% of the original 14C has decayed into nitrogen-14. We can use the half-life of carbon-14 to determine how long it would take for 87.5% of the 14C to decay.

One half-life of carbon-14 is 5730 years. Therefore, two half-lives would be 2 x 5730 years = 11,460 years. Three half-lives would be 3 x 5730 years = 17,190 years, and so on.

To find the age of the shell, we can use the formula:

age = number of half-lives x half-life of carbon-14

If 87.5% of the 14C has decayed, then 12.5% is left. This means that 1/8th of the original 14C is left. One half-life of carbon-14 is the time it takes for half of the 14C to decay, so if 1/8th is left, then three half-lives have passed.

Therefore, the age of the shell can be calculated as:

age = 3 x 5730 years = 17,190 years

In conclusion, the shell is approximately 17,190 years old.

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The model that correctly shows how light waves interact with red jackets is model 3.

When light waves interact with a red jacket, they are absorbed, reflected and/or transmitted by the material of the jacket.

In the case of a red jacket, it appears red because it selectively reflects and absorbs different wavelengths of visible light. Red is the color of light that is reflected by the material of the jacket and reaches our eyes, while all the other colors of light are absorbed or transmitted through the material. This is because the material of the jacket contains pigments or dyes that selectively absorb certain colors of light, and reflect other colors.

Red light has a longer wavelength and lower frequency than blue light, which is why it is able to penetrate deeper into the material of the jacket before being absorbed or reflected. This also means that the material of the jacket is less likely to heat up when exposed to red light compared to blue light, which has a higher frequency and shorter wavelength.

Overall, the interaction between light waves and a red jacket is determined by the properties of the material and the wavelength/frequency of the light.

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Answer:

the 3rd one

Explanation i took it and it was right

Answer:

its a

Explanation:

Depending on the toxin, yes. Many things, including proteins, will surely be denatured or degraded and finally turn harmless if we adopt the colloquial definition of the word and include toxins and poisons.

For instance, ricin, botulinum, tetrodotoxin, etc. They are rather delicate, so I'm hoping this toxin has the lowest shelf life.

For a poison, "not up to quality" might indicate anything. It might be more harmful or less toxic. It might be more toxic than before, which would make it less effective against the intended target but extremely harmful to other species. It only indicates that a single chemical may have begun to breakdown, and the final result of that deterioration may be something entirely else.

Small organic molecules make up a lot of other toxins. These include nerve agents like nicotine, sarin, and VX, all of which are capable of being broken down through air oxidation, UV exposure, hydrolysis, and other processes. Numerous nerve agents have a shelf life of several years, and studies have been done to extend their use in weapons.

Lead, mercury, and cadmium are just a few examples of metals that are hazardous because they are toxic not only in their elemental forms but also as inorganic and organic compounds. Although the toxicity of the various forms can varies substantially (see, for example, methylmercury and elemental mercury), the majority of them are nevertheless at least harmful.

These can persist for a very long period since the reactions that can happen under usual circumstances might not be safe, such as a sizable piece of cinnabar left on a desk even throughout the course of a geological time scale. any substantial modification to ensure safety.

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We first verify that the equation is balanced. We have 5 carbons (C), 2 sulfurs (S), and 4 oxygens (O) on each side of the reaction. So the reaction is balanced.

a) Now if we look at the reaction we can see that when 2 moles of SO2 react, 1 mole of CS2 is produced. That is, the ratio is 2 to 1. For each mole of SO2 half as many moles of CS2 will be produced.

So if we have 9.5 moles of SO2 we will have 9.5/2 moles, that is 4.75 moles of CS2.

Answer a) By reacting 9.50 moles of SO2 with an excess of it would be produced 4.75 moles of CS2.

Now, for the following parts of the question, we can apply the ideal gas law. This is because the reaction is in the gas phase and the law applies only to gases.

Where,

P= Pressure at STP = 1 atm

T= Temperature at STP = 273.15K

R= Ideal law constant = 0.08206 (atm L)/(mol K)

V= Volume of the gas

n= Numer of moles

b)We clear n and we replace the known values of SO2 to find the number of moles of SO2 that react.

Now, for each mole of SO2 that reacts we need 5/2 moles of C, that is 0.24x5/2=0.61 moles of C.

We use mass molar of C to calculate the grams.

Mass molar of C=12.01g/mol

Mass of C= Moles of C x Mass Molar

Mass of C= 0.61 mol x 12.01 g/mol = 7.37 g

So, To fully react 5.5 L of SO2 at STP we will need 7.37 g of C.

c)We apply the gas law again but this time we clear the volume.

We also take into account that for each mole of C, 4 moles of CO are produced, so if we have 20 moles of C we will produce 20x4=80 moles of CO.

So, from 20.0 moles of C at STP can be produced 1793.18 liters of CO

The statement is FALSE. Non-ferrous metals can be hardened by heat treatment, although the mechanisms and processes involved may differ from ferrous metals.

Heat treatment techniques such as precipitation hardening can be used to increase the strength of non-ferrous metals. Non-ferrous metals are metals or alloys that do not include iron (or iron allotropes, such as ferrite, etc.) in significant quantities. Non-ferrous metals are employed because they have desired qualities like reduced weight (for example, aluminium), greater conductivity (for example, copper), non-magnetic characteristics, or corrosion resistance (for example, zinc), even though they are often more expensive than ferrous metals. In the iron and steel sectors, several non-ferrous materials are also employed. Bauxite, for instance, is used as a flux in blast furnaces, whereas wolframite, pyrolusite, and chromite are utilised to create ferrous alloys.

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36.2L of N2 can be produced.

1st) According to the stoichiometry of the reaction, 2 moles of NaN3 produce 3 moles of N2. Using the molar mass of NaN3 (65.0g/mol ) and N2 (28.0g/mol) we can convert the moles to mass, and we can see that with 130.0g of NaN3 we can produce 84.0g of N2.

Now, we can use a mathematical rule of three to calculate the grams of N2 that can be produced from 71.0g of NaN3:

So, 45.9g of N2 are produced from 71.0g of NaN3.

2nd) It is necessary to convert the grams of N2 produced to moles, so we can use it in the Ideal gas equation:

Now we know that 1.6mol of N2 are produced.

3rd) To calculate the volume of N2, it is necessary to use the Ideal gas equation and replace the values of Pressure (P), Temperature (T, in Kelvin) and Number of moles (n):

Finally, 36.2L of N2 can be produced from 71.0g of NaN3.

The fair test that he could do to answer the question is C. Try to scratch the marble with each of the minerals, and group the minerals that do scratch the marble together.

When he try to scratch the marble with each of the minerals in her group he can observe the results for a fair test. however One that scratches the other is harder than one that has been scratched.

Hence, Given that marble is a well-known mineral, any mineral that scratches it is harder, while those that do not are less so. and the hardness of minerals can be determined using  with the Moh's scale, with diamond being the hardest mineral and talc being the least hard.

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complete question;

Lewis has the collection of minerals shown in the picture below.He knows that a harder mineral will scratch a softer mineral. He wants to design an experiment that will answer the following question:

Which of the minerals in the collection have a greater hardness than a rock made of marble?

Which of these is a fair test that he could do to answer the question?

A.

Separate the minerals into light and dark colors, and then try to scratch the marble with the light colored minerals.

B.

Separate the minerals into ones that feel heavier and lighter, and then try to scratch the marble with the heavy minerals.

C.

Try to scratch the marble with each of the minerals, and group the minerals that do scratch the marble together.

D.

Try to scratch the pink quartz with each of the minerals, and group the minerals that do not scratch the pink quartz together.