Which set of reactants will be the most efficient (LEAST wasteful of materials) for the reaction?

Answer:

if a rabbit hears or detects a sound then it will raise or twitch its ears.

The gas with the greatest volume at STP is 80 g of HF (2nd option)

To know the gas which have the greatest volume at STP, we shall determine the volume of each gas. This is illustrated below:

For HCl

1 mole of HCl = 22.4 L at STP

Thus,

2.19 moles of HCl = 2.19 × 22.4

2.19 moles of HCl = 49.06 L

Thus, the volume of HCl is 49.06 L

For HF

1 mole of HF = 22.4 L at STP

Thus,

4 moles of HF = 4 × 22.4

4 moles of HF = 89.6 L

Thus, the volume of HF is 89.6 L

For HBr

1 mole of HBr = 22.4 L at STP

Thus,

0.99 mole of HBr = 0.99× 22.4

0.994 mole of HBr = 22.2 L

Thus, the volume of HBr is 22.2 L

From the above calculation, we can conclude that the gas with the greatest volume is 80 g of HF (2nd option)

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The atomic radius is the distance between the nucleus of an atom and the outside border of the electron cloud. Atomic radius falls during a period and grows down a group in general. As electron shielding remains constant over time, effective nuclear charge grows.

Atomic radius reduces with time because valence electrons are added to the same energy level while the nucleus increases in protons. The increased nuclear charge attracts electrons more strongly, causing them to move closer to the nucleus.

The electron enters the same shell as the atomic number grows throughout a period.

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According to the problem  the theoretical yield of Ba(NO3)2 for this reaction is 4.85 g.

Reaction is an action or response to a stimulus or event. It is an immediate response to what is happening around us. It is a physical or emotional response that occurs when something touches a nerve, triggers a thought, or stimulates an emotion. Reactions can be positive or negative and can vary from person to person depending on their individual experiences, beliefs, values, and perspectives. Reactions can range from a smile and a nod of agreement to a full-blown emotional outburst. In most cases, reactions are instinctive, although we can also be taught to respond in a certain way.

The theoretical yield of Ba(NO3)2 for this reaction can be calculated using the following formula:

Mass of reactant = Mass of product

9.7 g AgNO3 = x g Ba(NO3)2

Rearranging the equation, we can solve for the theoretical yield of Ba(NO3)2:

x = (9.7 g AgNO3) / (2)

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Answer:sry

Explanation:i just needed points

Answer:

Eletrons

Explanation:

Electrons are a type of subatomic particle with a negative charge.

Answer:

hope this helps

Explanation:

A redox equation can be balanced using the following stepwise procedure: (1) Divide the equation into two half-reactions. (2) Balance each half-reaction for mass and charge. (3) Equalize the number of electrons transferred in each half-reaction. (4) Add the half-reactions together. Dec 2, 2020

Answer:

Sodium bicarbonate, also known as baking soda, has a wide range of uses in various industries and applications. Here are some of the most common uses of sodium bicarbonate:

Baking: Sodium bicarbonate is widely used in baking as a leavening agent. It reacts with acids, such as cream of tartar, buttermilk, or vinegar, to produce carbon dioxide gas, which causes the dough or batter to rise and become lighter.

Cleaning: Sodium bicarbonate is also used as a natural and effective cleaning agent. It can be used to clean and deodorize carpets, upholstery, kitchen appliances, and bathroom fixtures.

Personal care: Sodium bicarbonate is used in various personal care products, such as toothpaste, deodorants, and bath salts. It is effective in neutralizing odors and as an exfoliant.

Medical: Sodium bicarbonate is used in some medical treatments, such as as an antacid for heartburn or to treat metabolic acidosis, a condition where the body produces too much acid.

Fire extinguisher: Sodium bicarbonate is also used in some fire extinguishers because it can release carbon dioxide gas, which can displace oxygen and extinguish small fires.

Overall, sodium bicarbonate is a versatile substance with many uses in various industries and applications.

1.) New pressure is 341 kPa ; 2.) New volume of O2 is approx. 15 L ; 3.)  New pressure is 794 kPa ; 4.) New initial pressure of the O2 sample is 164.2 kPa ; 5.) New pressure is slightly higher than initial pressure, but it doesn't double ; 6.) Pressure of gas will increase to 1.366 atm at 100 degrees Celsius.

Boyle's law states that : pressure and volume of gas have an inverse relationship.

1.)Boyle's law equation: P1V1 = P2V2

P1 and V1 are initial pressure and volume, respectively, and P2 and V2 are final pressure and volume, respectively.

Given; P1 = 155 kPa   V1 = 22 L   V2 = 10 L

P2 = (P1V1) / V2

= (155 kPa x 22 L) / 10 L

P2 = 341 kPa

Therefore, new pressure is 341 kPa.

2.)  Gay-Lussac's law equation: (V1 / T1) = (V2 / T2)

Given, V1 = 10 L

T1 = -25°C + 273.15 = 248.15 K

T2 = 100°C + 273.15 = 373.15 K

V2 = (V1T2) / T1

= (10 L x 373.15 K) / 248.15 K

V2 = 15 L (rounded to two significant figures)

Therefore, new volume of O2 is approximately 15 L.

3.)  combined gas law equation: (P1V1) / T1 = (P2V2) / T2

Given, P1 = 501 kPa

V1 = 5.2 L

T1 = 25°C + 273.15 = 298.15 K

V2 = 7 L

T2 = 100°C + 273.15 = 373.15 K

P2 = (P1V1T2) / (V2T1)

= (501 kPa x 5.2 L x 373.15 K) / (7 L x 298.15 K)

P2 = 794 kPa

Therefore, new pressure is 794 kPa.

4.) Combined gas law equation again: (P1V1) / T1 = (P2V2) / T2

Given, V1 = 105 L   T1 = 50°C + 273.15 = 323.15 K   V2 = 55 L

T2 = -25°C + 273.15 = 248.15 K

P2 = 105.4 kPa

P1 = (P2V2T1) / (V1T2)

= (105.4 kPa x 55 L x 323.15 K) / (105 L x 248.15 K)

P1 = 164.2 kPa

Therefore, new initial pressure of the O2 sample is 164.2 kPa.

5. As (P1/T1) = (P2/T2)

(P1/283) = (P2/293)

P2 = (P1 × T2) / T1 = (1 atm × 293 K) / 283 K = 1.034 atm

So new pressure is slightly higher than the initial pressure, but it doesn't double.

6.) Gay-Lussac's law: P1/T1 = P2/T2

At 100 degrees Celsius (373 K), pressure would be:

P2 = (P1 × T2) / T1 = (1 atm × 373 K) / 273 K = 1.366 atm

So pressure of the gas will increase to 1.366 atm at 100 degrees Celsius.

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Answer: The pH of a 0.036 M solution of NH4Cl at 25°C is approximately 3.2.

Explanation:

The dissociation of NH4Cl in water can be represented as follows:

NH4Cl(s) → NH4+(aq) + Cl-(aq)

The NH4+ ion can act as a weak acid by donating a proton (H+) to water to form the hydronium ion (H3O+):

NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)

The equilibrium constant expression for this reaction is:

Ka = [NH3][H3O+] / [NH4+]

Since NH3 is a weak base, we can assume that its concentration is negligible compared to that of NH4+. Therefore, we can simplify the expression to:

Ka ≈ [H3O+]^2 / [NH4+]

Substituting the values given in the problem, we get:

1.8 × 10^-5 = [H3O+]^2 / 0.036

Solving for [H3O+], we get:

[H3O+] = √(1.8 × 10^-5 × 0.036) ≈ 6.1 × 10^-4 M

The pH of the solution can be calculated as:

pH = -log[H3O+] ≈ 3.2

The complete forms of the chemical reactions are as follows:

A chemical equation is a symbolic representation of a chemical reaction where reactants are represented on the left, and products on the right.

In a chemical reaction, the atoms of substances called reactants are rearranged to form new substances called products.

According to this question, two chemical reactions are given incompletely. The complete forms of these reactions showing the missing reactants and products are as illustrated above.

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D. Consult with an orthopedist about how to best accommodate potential consumers in terms of physical comfort and pain alleviation.

Researching the foot condition involves gaining knowledge about the condition, its causes, and its impact on individuals. Consulting with an orthopedist, who is a medical professional specializing in foot and ankle conditions, would provide valuable insights into how the shoe can best address the condition and alleviate pain. The other options, such as finding a rubber manufacturer and determining consumer willingness to pay, are important steps in the design and marketing process but should come after the research stage. Testing the shoe in a trial is also important but would come later in the product development process.

Ice cores can be used to study carbon dioxide levels in the atmosphere from long ago because air bubbles get trapped in the ice as it forms.

An ice core is a cylindrical sample of ice that is drilled from a glacier or ice sheet. The air trapped in ice core is a record of the atmosphere's composition at the time the ice was formed, which can provide information on the past levels of atmospheric carbon dioxide.

Scientists can extract air from these bubbles in ice cores and measure the carbon dioxide levels using a variety of techniques, including gas chromatography and laser absorption spectroscopy.

By studying the carbon dioxide levels in ice cores from different periods in history, scientists can reconstruct a timeline of past atmospheric carbon dioxide concentrations and compare them to current levels. This helps to inform our understanding of the Earth's climate and how it has changed over time.

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The following ionic compounds are mentioned below.

What is compound ?

A compound combines one or more additional substances to form a new product. The combination of two substances or elements is equal to the mass ratio for making that product. It is attached firmly, and they have equal in all ways to form a combination. The elements both are bonded together. Each compound has its chemical structure; if it breaks, it can either split into two atoms or molecules or in a single. Different types of compounds are present.

What is ionic compound ?

An ion with a positive charge is a cation, whereas one with a negative charge is an anion.

a) KCN = Potassium cyanide

b) NaBrO2 = Sodium bromite

c) Sr(OH)2 = Strontium hydroxide

d) CoTe = Cobalt telluride

e) Fe2(CO3)3 = Iron (III) Carbonate

f) Cr(NO3)3 = Chromium(III) nitrate

g) (NH4)2SO3 = Ammonium Sulfite

h) NaH2PO4 = Sodium dihydrogen phosphate

i) KMnO4 = Potassium permanganate

j) Ag2Cr2O7 = Silver dichromate

Therefore, The following ionic compounds are mentioned above.

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Therefore, the final answer, with only positive exponents, is:[tex]27m^12 n^6.[/tex]

First, we can simplify the expression inside the parentheses by canceling out the common factor :

Starting with the expression inside the parentheses:

[tex]3m^4 n^4/6n^2[/tex]

We can simplify by dividing the numerator and denominator by 2:

[tex]3m^4 n^4/6n^2[/tex]

= [tex](3/2) m^4 n^(4-2)[/tex]

= [tex](3/2) m^4 n^2[/tex]

Now we can raise this simplified expression to the power of 3 using the power of a product rule:

[tex][(3/2) m^4 n^2]^3 \\= (3/2)^3 m^(43) n^(23) \\= 27m^12 n^6[/tex]

Therefore, the final answer is [tex]27m^12 n^6.[/tex]

An exponent is a mathematical operation that represents repeated multiplication of a number by itself. It is written as a small number above and to the right of the base number. The exponent tells you how many times the base number is multiplied by itself. For example, in the expression 3^4, the base number is 3 and the exponent is 4, which means that 3 is multiplied by itself 4 times. The result of this operation is 81. Exponents are commonly used in algebra and calculus, as well as in many scientific and engineering fields to represent large and small numbers.

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Answer:

Explanation:

Lead (II) chloride and lead (IV) chloride have different charges on the lead and chloride ions in each compound.

In lead (II) chloride, also known as plumbous chloride, the lead ion has a charge of +2 and the chloride ion has a charge of -1. The chemical formula for lead (II) chloride is PbCl2.

In lead (IV) chloride, also known as plumbic chloride, the lead ion has a charge of +4 and the chloride ion has a charge of -1. The chemical formula for lead (IV) chloride is PbCl4.

So the key difference between these two compounds is the oxidation state of the lead ion. In lead (II) chloride, the lead ion has a +2 oxidation state, while in lead (IV) chloride, the lead ion has a +4 oxidation state.

the percent yield of the reaction is 7.25%. in order to obtain 5.95 g of ethyl nitrate.

How to calculate  the yield percentage ?

To calculate the percent yield, we need to know the theoretical yield and the actual yield of the reaction. The theoretical yield is the maximum amount of product that can be produced based on the amount of limiting reactant used in the reaction. The actual yield is the amount of product obtained from the reaction in the lab.

Assuming the reaction is:

Ethanol + Butyric acid → Ethyl butyrate + Water

The balanced equation is:

2 C₄H₈O₂ + 2 C₂H₅OH → 2 C₆H₁₂O₂ + 2 H₂O

The molar mass of ethyl butyrate (C₆H₁₂O₂) is 144.17 g/mol.

First, we need to calculate the theoretical yield of ethyl butyrate:

1 mol of butyric acid produces 1 mol of ethyl butyrate.

The molar mass of butyric acid (C₄H₈O₂) is 88.11 g/mol.

Therefore, the theoretical yield of ethyl butyrate can be calculated as follows:

Theoretical yield = (mass of butyric acid used / molar mass of butyric acid) x (molar mass of ethyl butyrate / 2)

Let's assume the chemist used 10 g of butyric acid:

Theoretical yield = (10 g / 88.11 g/mol) x (144.17 g/mol / 2) = 81.9 g

Next, we can calculate the percent yield using the actual yield and theoretical yield:

Percent yield = (actual yield / theoretical yield) x 100%

Given that the chemist obtained 5.95 g of ethyl butyrate, the percent yield can be calculated as follows:

Percent yield = (5.95 g / 81.9 g) x 100% = 7.25%

Therefore, the percent yield of the reaction is 7.25%.

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The correct standard cell notation for a galvanic cell with aluminum and gold electrodes is option C: Al(s) | Al³+ (aq) || Au+ (aq) | Au(s).

The cell notation consists of two half-cells, separated by a double vertical line to indicate a salt bridge or porous barrier. The left side of the notation represents the anode, where oxidation occurs, and the right side represents the cathode, where reduction occurs.

In this case, the anode is aluminum (Al) and its ion Al³+ in aqueous solution (aq), while the cathode is gold (Au) and its ion Au+ in aqueous solution (aq). The solid state of each metal is represented in parentheses after its chemical symbol.

Therefore, the correct standard cell notation is Al(s) | Al³+ (aq) || Au+ (aq) | Au(s)

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Using an excess of barium ions in the precipitation reaction is desirable to ensure complete precipitation of the sulfate ions and accurate determination of the sulfate content of the original sample.

When precipitating a sulfate compound using barium chloride, it is desirable to use an excess of barium ions to ensure complete precipitation of the sulfate ions. This is because the precipitation reaction is a equilibrium reaction and the precipitation may not be complete if the concentration of barium ions is not high enough.

In the precipitation reaction, barium ions (Ba2+) react with sulfate ions (SO42-) to form barium sulfate (BaSO4):

Ba2+ + SO42- → BaSO4

If there is not enough barium ions in the solution, some of the sulfate ions may not react with the available barium ions and remain in solution. This would result in an inaccurate determination of the sulfate content of the original sample. Using an excess of barium ions ensures that all the sulfate ions in the sample react with the barium ions, resulting in complete precipitation of the barium sulfate. Any unreacted barium ions will remain in solution and will not interfere with the analysis of the sulfate content of the sample.

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If there are 4 moles of O2, 0.6667 moles of the C6H12O6 are present.

If there are 4 moles of oxygen present. In that case, we can use the balanced chemical equation for glucose combustion to calculate the number of moles of glucose required to react with it.

According to the equation, one mole of glucose reacts with six moles of oxygen to produce six moles of carbon dioxide and six moles of water. The number of moles of glucose required to react with it is calculated by dividing the number of moles of oxygen by six.

So, if 4 moles of oxygen are present, 0.6667 moles of glucose must react with it.

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About 940 kg (to three significant figures) of chalcopyrite must be mined to obtain 325 g of pure Cu.

What is Molar Mass?

It is calculated by adding up the atomic masses of all the atoms in a molecule or formula unit. The atomic masses are taken from the periodic table and are usually given in atomic mass units (amu) or unified atomic mass units (u).

We can start by setting up a proportion, where x is the mass of chalcopyrite in kilograms that must be mined:

(mass of Cu in grams obtained from x kg of CuFeS2) / (mass of CuFeS2 in kg) = (mass of pure Cu obtained) / (mass of Cu in the sample)

From the chemical formula of chalcopyrite (CuFeS2), we know that 1 mole of CuFeS2 contains 1 mole of Cu. The molar mass of Cu is 63.55 g/mol, and the molar mass of CuFeS2 is 183.52 g/mol.

So, the mass of Cu in x kg of CuFeS2 is:

(1 mol Cu / 1 mol CuFeS2) × (63.55 g/mol Cu) × (x kg CuFeS2 / 183.52 g/mol CuFeS2) = 0.346 x kg Cu

The mass of CuFeS2 in x kg is simply x kg. Substituting the given values into the proportion, we get:

0.346 x kg / x kg = 0.325 kg / 0.001 kg

Solving for x, we get:

x = (0.325 kg / 0.001 kg) / 0.346 = 940.18 kg

Therefore, about 940 kg (to three significant figures) of chalcopyrite must be mined to obtain 325 g of pure Cu.

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The molar mass of the compound, given that it is between 250.0 and 350.0 g/mol, is 330 g/mol

We'll begin by obtaining the mass of carbon, hydrogen and oxygen. Details below:

For Carbon

Mass of C = (12 / 44) × 21.33

Mass of C = 5.817 g

For Hydrogen

Mass of H = (2 / 18) × 2.728

Mass of H = 0.303 g

For Oxygen

Mass of O = (mass of compound) - (mass of C + mass of H)

Mass of O = 10 - (5.817 + 0.303)

Mass of O = 3.88 g

Next, we shall determine the formula. details below:

Divide by their molar mass

C = 5.817 / 12 = 0.485

H = 0.303 / 1 = 0.303

O = 3.88 / 16 = 0.2425

Divide by the smallest

C = 0.485 / 0.2425 = 2

H = 0.303 / 0.2425 = 1.25

O = 0.2425 / 0.2425 = 1

Multiply by 8 to express in whole number

C = 2 × 8 = 16

H = 1.25  8 = 10

O = 1 × 8 = 8

×

Thus, the formula of the compound is C₁₆H₁₀O₈

Finally, we shall determine the molar mass. Details below:

Molar mass of C₁₆H₁₀O₈ = (16 × 12) + (10 × 1) + (8 × 16)

Molar mass of C₁₆H₁₀O₈ = 192 + 10 + 128

Molar mass of C₁₆H₁₀O₈ = 330 g/mol

Thus, we can conclude that the molar mass is 330 g/mol

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Answer:

Oxidation reaction; propanone and potassium manganate (VII) are formed.

Oxidation reaction; butan-2-one and potassium manganate (VII) are formed.

Reduction reaction; pentan-3-ol is formed.

Esterification reaction; ethyl ethanoate and water are formed.

The list sorted from smallest to largest:

Mole is a unit used to measure the amount of a substance. One mole of a substance is defined as the amount of that substance that contains the same number of entities (such as atoms, molecules, or ions) as there are atoms in 12 grams of carbon-12.

The number of entities in one mole is called Avogadro's number, which is approximately 6.022 x 10²³. This means that one mole of any substance contains approximately 6.022 x 10²³ atoms or molecules of that substance.

Moles are commonly used in chemistry to perform calculations involving the amount of a substance in a reaction, such as determining the limiting reagent, calculating the theoretical yield, and determining the concentration of a solution.

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Answer:

First, put them all into “units”

1/2 mole = 1/2 x 6.02x1023 = 3.01x1023

5 = 5

2 pair = 2 x 12 = 24

1 ream = 500

4 dozen = 4 x12 = 48

5.5x10-4 = 0.00055

So from smallest to largest  

0.00055

5

24

48

500

3.01x1023

Explanation:

The capacity of a substance to go through a particular chemical transition is referred to as its chemical property. A process that takes place is a chemical reaction.

The definition of a physical reaction is a response in which molecules rearrange one another but do not undergo chemical change. Without such a change in composition, physical reactions can involve changes in state, temperature, texture, and shape.

The paper ought to contain a synopsis of the source material, but the main emphasis should be on your opinions and justifications regarding the topics included in the actual text. This necessitates your consideration of the provided work, followed by a targeted, well-thought-out response supported by outside sources, when appropriate.

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Answer:

To find the number of Na2SO4 formula units in 1.75 moles of Na2SO4, we first need to determine the Avogadro's number (6.022 x 10^23), which is the number of particles (atoms, molecules, or formula units) in one mole of a substance.

Then we can use the following formula:

Number of formula units = moles of substance x Avogadro's number

Substituting the given values, we get:

Number of formula units = 1.75 moles x 6.022 x 10^23 formula units/mole

Number of formula units = 1.0545 x 10^24 formula units

Therefore, there are approximately 1.0545 x 10^24 Na2SO4 formula units in 1.75 moles of Na2SO4.

Answer: (B) High pressure and heat

Explanation: The precursor to coal is peat. Peat is a soft, organic material consisting of partly decayed plant and mineral matter. When peat is placed under high pressure and heat, it undergoes physical and chemical changes (coalification) to become coal.

The amount of heat that would be given out by 5.30 moles of A would be obtained as  305.81 kJ

The enthalpy of reaction (ΔHrxn) is the amount of heat energy that is either absorbed or released during a chemical reaction at a constant pressure. It is a thermodynamic quantity that measures the change in enthalpy of a chemical reaction.

Given that;

1 mole of A would accept  57.7 kJ of heat

5.30 moles of A will accept 5.30 moles *   57.7 kJ /1 mole

= 305.81 kJ

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