A + B = AB is an example of a ________reaction

Answer:

21% percent

Explanation:

Air has small amounts of other gases, such as carbon dioxide,neon, and hydrogen. Oxygen shares some parts of air basically. The highest percentage of oxygen we can breathe is approximately 60℅

Answer: calm water.

Explanation:

Based οn the wοrds prοvided, a pοssible title fοr this sectiοn οf Sheila's nοtes cοuld be Mass.

In chemistry, prοperties οf matter refer tο the characteristics οr attributes that can be used tο describe and identify a substance. These prοperties can be divided intο twο categοries: physical prοperties and chemical prοperties.

Physical attributes are thοse that can be examined οr measured withοut changing the substance's makeup. Mass, vοlume, density, cοlοr, melting pοint, bοiling temperature, and sοlubility are examples οf physical qualities.

Chemical prοperties, οn the οther hand, describe hοw a substance interacts with οther substances tο prοduce new substances.

Understanding the prοperties οf matter is impοrtant in chemistry because it allοws scientists tο identify and classify different substances based οn their unique characteristics. This knοwledge can alsο be used tο predict hοw substances will behave under different cοnditiοns and tο design new materials with specific prοperties fοr variοus applicatiοns.

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Complete question:

Sheila spilled tea on her notes and is now unable to read some words.

What is the correct title for this section of Sheila’s notes?

If the volume is increased to 1125 in El with a new pressure of 99.7 K PA, the new temperature will be around 808 K.

Once the piston's pressure has doubled, it goes downward until the gas's pressure and the piston pressure are equal. The gas has now lost half of its original volume. The volume of gas falls to one-fourth of its initial volume if the pressure on the piston is once more increased by a factor of two.

This issue can be resolved using the coupled gas law:

(P1V1) / T1 = (P2V2) / T2

Using the following conversions, we can first change the starting volume to litres and the original pressure to atmospheres (atm):

1 mL = 0.001 L

1 kPa = 0.00987 atm

V1 = 850 mL = 0.85 L

P1 = 11.4 kPa = 0.1126 atm

T1 = 467 K

The new volume and pressure can also be converted to litres and atmospheres:

V2 = 1125 mL = 1.125 L

P2 = 99.7 kPa = 0.984 atm

Now we can plug in the values and solve for T2:

(P1V1) / T1 = (P2V2) / T2

(0.1126 atm * 0.85 L) / 467 K = (0.984 atm * 1.125 L) / T2

T2 = (0.984 atm * 1.125 L * 467 K) / (0.1126 atm * 0.85 L)

T2 = 808 K

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Answer:

548 K

I hope this helps! Cheers ^^

It will take about 10990 milliliters to fill 106 vials of loratadine.

Based on the information in the statement, to obtain the number of milliliters needed to fill 106 vials of loratadine, just construct the following simple rule of three:

14 liters of loratadine - 135 vials

x liters of loratadine - 106 vials

14 * 106 = 135x

1484 = 135x

x ≅ 10.99 liters of loratadine (value rounded to two decimal places).

Note, however, that the statement asks for the volume of loratadine in milliliters. To convert from liters to milliliters, just multiply the value obtained by above 1000:

10.99 * 1000 = 10990 milliliters.

The number or amount of moles of magnesium sulfide in 100g of MgS is 1.774 moles.

The number of moles in a substance can be calculated by dividing the mass of the substance by its molar mass as follows:

Mole is the base unit of amount of substance i.e. the amount of substance of a system which contains exactly 6.02214076 × 10²³ elementary entities.

no of moles = mass ÷ molar mass

According to this question, 100g of magnesium sulfide is given. The molar mass of magnesium sulfide is 56.38 g/mol.

moles = 100g ÷ 56.38g/mol

moles = 1.774 moles

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Answer:

Ca²+O²

Explanation:

The Ca gives out 2 electrons and oxygen gains 2 elctrons

The percentage by mass of hydrogen can be calculated from the problem as 2.016

To calculate the mass percent of an atom in a compound, you first need to determine the molar mass of the compound and the molar mass of the atom of interest.

Determine the molar mass of the compound by adding up the atomic masses of all the atoms in the compound.

Determine the number of moles of the atom of interest in one mole of the compound. This is done by dividing the atomic mass of the atom by the molar mass of the compound.

We know that the relative molecular mas of the compound is; 300 g/mol

Then;

Percent by mass of hydrogen is; 6/300 * 100/1

= 2.016%

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Answer:

sulphur has the most valence electron

Answer:

[tex]valence \: electrons \: are \: the \\ electrons \: in \: the \: outermost \\ shell \: or \: energy \: level \: of \: an \: \\ atom. [/tex]

Cyclohexene and sulfuric acid can react to form a cyclic carbonium ion intermediate, which can then undergo various reactions.

One common reaction is the addition of a nucleophile, such as water or an alcohol, to form an alcohol product. This reaction is often referred to as acid-catalyzed hydration.

The general reaction mechanism can be summarized as follows:

Protonation of the alkene: The sulfuric acid protonates the alkene double bond, creating a more electrophilic carbocation intermediate.

Formation of a cyclic carbonium ion: The positive charge is then delocalized throughout the cyclohexene ring, forming a cyclic carbonium ion intermediate.

Addition of a nucleophile: A nucleophile, such as water, can then attack the carbon atom of the carbonium ion, forming a new bond and creating an alcohol product.

The overall reaction can be written as follows:

Cyclohexene + H2SO4 + H2O → Cyclohexanol + HSO4-

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Enthalpy change, H, for this reaction per mole of X is thus equal to 0 J/mol.

The amount of heat that the calorimeter, q cal, gains may be calculated using the formula qcal = Ccalt, where t is the change in temperature that the mixture experiences.

The equation: can be used to compute the enthalpy change, H.             ΔH = q / n

The heat absorbed by the water can be calculated using the equation:

q1 = m1 x c1 x ΔT1

m1 = 21.0 g = 0.0210 kg (since the density of water is 1.00 g/mL)

c1 = 4.18 J/(g⋅∘C)

ΔT1 = 25.0 ∘C - 13.5 ∘C = 11.5 ∘C

q1 = (0.0210 kg) x (4.18 J/(g⋅∘C)) x (11.5 ∘C) = 1.09 J

The heat absorbed by X can be calculated using the equation:

q2 = m2 x ΔHfus

where m2 is the mass of X and ΔHfus is the enthalpy of fusion of X.

m2 = 1.70 g = 0.00170 kg

ΔHfus = ΔH / n = ΔH / (m2/M)

where M is the molar mass of X.

We can rearrange this equation to solve for ΔH:  ΔH = q2 x (m2/M)

With this assumption, we can calculate ΔH as follows:

ΔH = q / n = (q1 + q2) / n

ΔH  = (1.09 J + q2) / (0.00170 kg / 77.0 g/mol)

ΔH  = (1.09 J + q2) / 0.0000221 mol

The fact that the heat absorbed by X is equal to the heat emitted by the solution can be used to solve for q2:  q2 = -q1

Therefore,  ΔH = (1.09 J - q1) / 0.0000221 mol

Substituting q1, we get:

ΔH = (1.09 J - 1.09 J) / 0.0000221 mol

ΔH = 0 J/mol

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The basic units of the human body are called cells, and they carry out specialized tasks to keep the body alive. Cells in a city might be compared to people who perform particular duties and contribute to the general efficiency of the city.

Organizational hierarchy is built up from lower levels. Consequently, molecules combine to create molecules, molecules create cells, cells create tissues, tissues create organs, organs create organ systems, and organ systems create beings. The body's next level of organisation. A tissue is made up of related cells with a common purpose. Human tissues can be divided into four categories: epithelial, muscular, nerve, and connective.

The major levels of organisation in the body, from the most basic to the most complex, are atoms, molecules, organelles, cells, tissues, organs, organ systems, and the human body.

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Explanation:

answer: 3057.44 m^3

Explanation:

This is Charles Law which is V1/T1=V2/T2  and temp must be in kelvin

So v2=V1 x T2/T1

v2= 3100 X 359.15 / 364.15  = 3057.44 m^3

t1= 91 +273.15 = 364.15

t2=86+273.15 =359.15

Answer:

The density of an object is constant.

Density is a derived unit of measure.

The density of an object determines whether it will sink or float.

Answer:

option a

Velocity is plotted on the x-axis of a graph and speed is plotted on the y-axis.

Answer:b

Explanation:i just did it

Answer: Sucrose is a glycosyl glycoside formed by glucose and fructose units joined by an acetal oxygen bridge from hemiacetal of glucose to the hemiketal of the fructose.

Explanation:

Answer: 5.5 atm

Explanation:

Answer: 3.96 moles of CO2 are required in the photosynthesis reaction.

Explanation: The balanced equation for photosynthesis tells us that for every mole of glucose produced, 6 moles of CO2 are required. Therefore, to determine how many moles of CO2 are required for 0.66 moles of glucose, we can set up a proportion:

6 moles CO2 / 1 mole glucose = x moles CO2 / 0.66 moles glucose

Solving for x, we get:

x = (6 moles CO2 / 1 mole glucose) x (0.66 moles glucose / 1) = 3.96 moles CO2

Therefore, to produce 0.66 moles of glucose, 3.96 moles of CO2 are required in the photosynthesis reaction.

Answer: 204.63 K

Explanation:

Ideal gas law

Pv=nRT

T=PV/nR

T= 5.6 X 12/(4 x0.0821)= 204.63 K

R= ideal gas constant = 0.0821

Answer:

3.8%

Explanation:

Ca × 1 = 50

C ×4 = 48

H × 6 = 6

O × 4 = 64

158

6÷158×100% = 3.8%

Substance that gives out ions when dissolved in water, which are able to conduct electricity describes an electrolyte.

The original mass of Tritium (hydrogen-3) was 13.4mg if its current mass is 0.42 micrograms.

The formula for radioactive decay is given by:

N = N0 x (1/2)^(t/T)

where,

N = final number of radioactive atoms

N0 = initial number of radioactive atoms

t = time elapsed

T = half-life of the radioactive substance

Let's substitute the given values into the formula:

0.42 μg = N0 x (1/2)^(60/12)

0.42 μg = N0 x (1/2)^5

0.42 μg = N0 x 1/32

N0 = 0.42 μg x 32

N0 = 13.44 μg

Therefore, the original mass of the tritium sample was 13.44 micrograms.

What is radioactive decay?

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting ionizing radiation, such as alpha particles, beta particles, and gamma rays. This process can result in a change in the number of protons and/or neutrons in the nucleus, leading to the transformation of one element into another. The rate of decay is typically characterized by a half-life, which is the time required for half of the atoms in a sample to decay.

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As the concentration of a solution increases, the vapor pressure decreases.

Raoult's law states that the vapor pressure of a solution is proportional to the mole fraction of the solvent present in the solution.

When a solute is added to a solvent, it reduces the mole fraction of the solvent, and therefore the vapor pressure of the solution decreases. This is because the solute molecules occupy space on the surface of the solution, which reduces the number of solvent molecules that can escape into the gas phase, and thus reduces the vapor pressure of the solution.

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Complete question:

As the concentration of a solution increases, the vapor pressure _____

Answer:

1. The balanced equation is 2KCIO3 → 2KCI + 3O2. According to the law of conservation of mass, the mass of the reactants must equal the mass of the products. Therefore, the mass of oxygen produced is:

Mass of oxygen = Mass of KCIO3 - Mass of KCI

Mass of oxygen = 500 g - 303 g

Mass of oxygen = 197 g

Therefore, 197 g of O2 are produced.

2. The balanced equation is N2 + 3H2 → 2NH3. We need to find out how much H2 is needed to react with 100 g of N2 to produce 121 g of NH3. First, we need to calculate the number of moles of N2 and NH3:

Moles of N2 = Mass of N2 / Molar mass of N2

Moles of N2 = 100 g / 28 g/mol

Moles of N2 = 3.57 mol

Moles of NH3 = Mass of NH3 / Molar mass of NH3

Moles of NH3 = 121 g / 17 g/mol

Moles of NH3 = 7.12 mol

According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the number of moles of H2 needed is:

Moles of H2 = Moles of N2 x (3/1)

Moles of H2 = 3.57 mol x 3

Moles of H2 = 10.71 mol

Finally, we can calculate the mass of H2 needed:

Mass of H2 = Moles of H2 x Molar mass of H2

Mass of H2 = 10.71 mol x 2 g/mol

Mass of H2 = 21.42 g

Therefore, 21.42 g of H2 are needed.

3. The balanced equation is 4Fe + 3O2 → 2Fe2O3. We need to find out how much oxygen is needed to react with 350 g of Fe to produce 500 g of Fe2O3. First, we need to calculate the number of moles of Fe and Fe2O3:

Moles of Fe = Mass of Fe / Molar mass of Fe

Moles of Fe = 350 g / 55.85 g/mol

Moles of Fe = 6.26 mol

Moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3

Moles of Fe2O3 = 500 g / 159.69 g/mol

Moles of Fe2O3 = 3.13 mol

According to the balanced equation, 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3. Therefore, the number of moles of O2 needed is:

Moles of O2 = Moles of Fe x (3/4)

Moles of O2 = 6.26 mol x (3/4)

Moles of O2 = 4.69 mol

Finally, we can calculate the mass of O2 needed:

Mass of O2 = Moles of O2 x Molar mass of O2

Mass of O2 = 4.69 mol x 32 g/mol

Mass of O2 = 150.08 g

Therefore, 150.08 g of O2 are needed.

4. The balanced equation is CH2 + 2O2 → CO2 + 2H2O. We know that 16 g of CH2 reacts with 64 g of O2 to produce 44 g of CO2. We need to find out how much water is produced. First, we need to calculate the number of moles of CH2 and CO2:

Moles of CH2 = Mass of CH2 / Molar mass of CH2

Moles of CH2 = 16 g / 14 g/mol

Moles of CH2 = 1.14 mol

Moles of CO2 = Mass of CO2 / Molar mass of CO2

Moles of CO2 = 44 g / 44 g/mol

Moles of CO2 = 1 mol

According to the balanced equation, 1 mole of CH2 reacts with 2 moles of O2 to produce 2 moles of H2O. Therefore, the number of moles of H2O produced is:

Moles of H2O = Moles of CH2 x (2/1)

Moles of H2O = 1.14 mol x 2

Moles of H2O = 2.28 mol

Finally, we can calculate the mass of H2O produced:

Mass of H2O = Moles of H2O x Molar mass of H2O

Mass of H2O = 2.28 mol x 18 g/mol

Mass of H2O = 41.04 g

Therefore, 41.04 g of H2O are produced.

5. The balanced equation is CaCO3 → CaO + CO2. We need to find out how much CO2 is produced from the decomposition of 200 g of CaCO3 if 112 g of CaO are produced. First, we need to calculate the number of moles of CaCO3 and CaO:

Moles of CaCO3 = Mass of CaCO3 / Molar mass of CaCO3

Moles of CaCO3 = 200 g / 100.09 g/mol

Moles of CaCO3 = 1.999 mol

Moles of CaO = Mass of CaO / Molar mass of CaO

Moles of CaO = 112 g / 56.08 g/mol

Moles of CaO = 1.999 mol

According to the balanced equation, 1 mole of CaCO3 produces 1 mole of CaO and 1 mole of CO2. Therefore, the number of moles of CO2 produced is:

Moles of CO2 = Moles of CaCO3 x (1/1)

Moles of CO2 = 1.999 mol

Finally, we can calculate the mass of CO2 produced:

Mass of CO2 = Moles of CO2 x Molar mass of CO2

Mass of CO2 = 1.999 mol x 44 g/mol

Mass of CO2 = 87.96 g

Therefore, 87.96 g of CO2 are produced.

80 g of 15% solution and 120 g of 20% solution are needed to prepare 200 g of 17% solution.

What is Mass Percentage?

Mass percentage is the percentage of the mass of the solute in a solution to the total mass of the solution. It is calculated by dividing the mass of the solute by the total mass of the solution and multiplying by 100%. The mass percentage is often used in chemistry to express the concentration of a solution.

Let x be the mass of the 15% solution needed and y be the mass of the 20% solution needed to prepare 200 g of 17% solution. Then we have the system of equations:

x + y = 200 (total mass of the solution)

0.15x + 0.20y = 0.17(200) (total amount of solute in the solution)

Solving for x and y, we get:

x = 80 g of 15% solution

y = 120 g of 20% solution

Therefore, 80 g of 15% solution and 120 g of 20% solution are needed to prepare 200 g of 17% solution.

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The experimental percentage of water by mass, given that the hydrate is heated and it produces 9.87 g of anhydrous, is 22.6%

First, we shall determine the mass of the water in the hydrate. Details below:

Mass of water = Mass of hydrate - Mass of anhydrous

Mass of water = 12.75 - 9.87

Mass of water = 2.88

Finally, we shall determine the experimental percentage of water by mass. Details below:

Experimental percentage of water = (mass of of water / mass of hydrate) × 100

Experimental percentage of water = (2.88 / 12.75) × 100

Experimental percentage of water = 22.6%

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Answer: The percentage of fluorine as F- in a HF solution at pH 2.33 is 9.09%.