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Making Esters
1. Name the following esters and give the name of the alcohol + carboxylic reacted to make each one.
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The energy required to increase 1.000 kg of water ice between 0 °C °C through 79.8 °C °C is equivalent to the energy required to melt one pound of ice (334 kJ).

No surface temp change happens from heat exchange if ice caps melt and has become water ice (i.e., that during phase transition). Imagine about water melting from stalactites that were already dripping on a sun-warmed roof. Alternatively, water boils in an ice bucket cooled by relatively low surroundings.

To melt, the iceberg will take in all the heat that is available. The water stays at zero degrees Fahrenheit until all of the ice has dissolved as when the ice has melted, it changes becomes saltwater at that temperature.

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Answer:

Explanation:

It is a binary molecular compound ( and when the first element is to the right of the zigzag line on the periodic table you will use prefixes (di)

Answer:

c

Explanation:

energy neither created nor distroyed

Answer:

The answer is C

Explanation:

We learned so many times about law conservation energy,states that"Energy neither created or destroyed but can change its form to other."

E.g If we rub our palm,it changes from kinetic energy to heat energy.

Answer:

Explanation:

not sure your elements to choose from

But it would be one with a +2 charge

Examples

Be, Mg, Ca, Sr, Ba, Cu, Fe these are just a few

send your list and I can finish this for you

The response is C. 400mL of 5% glucose solution requires 20g of glucose to make.

To prepare 400mL of 5% glucose solution, we need to determine the amount of glucose required.

5% glucose solution means that 5g of glucose is present in 100mL of the solution.

Therefore, for 400mL of the solution, the amount of glucose required can be calculated as:

5g glucose/100mL solution = x g glucose / 400mL solution

x = (5g glucose/100mL solution) x (400mL solution) = 20g glucose

Therefore, the answer is D. 20g of glucose is needed to prepare 400mL of 5% glucose solution.

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20 mL of 0.20 M H2SO4 is required to completely liquidate 40 mL of 0.10 M Ca (OH)2.

We can use the balanced chemical equation for the neutralization reaction between sulfuric acid (H2SO4) and calcium hydroxide (Ca (OH)2) to determine the stoichiometry of the reaction:

H2SO4 + Ca (OH)2 -> CaSO4 + 2H2O

moles of Ca (OH)2 = Molarity x Volume

= 0.10 mol/L x 0.040 L

= 0.004 mol

moles of solute = Molarity x Volume (in liters)

alter this equation gives:

Volume (in liters) = moles of solute / Molarity

Therefore, the volume of 0.20 M H2SO4 required to supply 0.004 moles of H2SO4 is:

Volume (in liters) = 0.004 mol / 0.20 mol/L

= 0.02 L

Multiplying by 1000 mL/L, we get:

Volume (in mL) = 0.02 L x 1000 mL/L

= 20 mL

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The mole ratio of NaNO₃ to Pb(NO₃)2 is 2:1. Therefore, 6.75 g of Pb(NO₃)₂ will produce 2 x 6.75 g = 13.5 g of NaNO₃.

Mole is a unit of measurement used in chemistry to measure the amount of a substance, usually expressed in terms of the number of atoms, molecules, or other units in a given mass of that substance. The mole, also known as Avogadro's number, is a very large number equal to 6.022 x 10^23. This means that a mole of a substance contains 6.022 x 10^23 atoms or molecules, depending on the substance. The mole is used to measure the amount of a substance in a given sample, and is essential to many chemical calculations.

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The empirical formula of phosphorus sulfide is [tex]P^{1}S^3.[/tex].

To determine the empirical formula of phosphorus sulfide, we need to find the mole ratio between phosphorus and sulfur in the compound. We can use the given masses of phosphorus and phosphorus sulfide to calculate the number of moles of each element.

First, we convert the masses of phosphorus and phosphorus sulfide from milligrams to grams:

Mass of P = 44.0 mg = 0.0440 g

Mass of P4S3 = 158 mg = 0.158 g

Next, we calculate the number of moles of each element:

Moles of P = mass of P / molar mass of P = 0.0440 g / 30.97 g/mol = 0.00142 mol

Moles of S = (mass of P4S3 - mass of P) / molar mass of S = (0.158 g - 0.0440 g) / 32.07 g/mol = 0.00393 mol

We then divide the number of moles of each element by the smallest value to obtain the simplest whole-number ratio of atoms:

Moles of P / 0.00142 mol ≈ 1

Moles of S / 0.00142 mol ≈ 2.77

We can then round the above ratio to the nearest whole number to get the empirical formula:

[tex]P^{1}S^3.[/tex]

Phosphorus sulfide is a chemical compound with the formula P4S3. It is a yellowish-white solid with a garlic-like odor, and is insoluble in water but soluble in carbon disulfide.

It is formed by the reaction of phosphorus with sulfur, and can also be produced by heating a mixture of red phosphorus and sulfur.

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Answer:

n = 0.012 moles

Explanation:

Recall that;

n = V note: V in dm³

22.4

V = 26.5 mL = 0.265 dm³

n = 0.265

22.4

n = 0.012 moles

12.5 mol of the substance contains 7.53 x 10²⁴ formula units of calcium carbonate.

Calcium carbonate (CaCO₃) is a compound that consists of one calcium atom, one carbon atom, and three oxygen atoms. The formula unit of calcium carbonate contains one Ca atom, one C atom, and three O atoms. To calculate the number of formula units in 12.5 mol of calcium carbonate, we need to use Avogadro's number, which relates the number of particles (atoms, molecules, or formula units) in a given amount of substance.

Avogadro's number (Nₐ) is 6.022 x 10²³ particles/mol. Therefore, the number of formula units of calcium carbonate in 12.5 mol can be calculated as:

n = Nₐ * 12.5 mol

n = 6.022 x 10²³ particles/mol * 12.5 mol

n = 7.53 x 10²⁴ particles

Therefore, there are 7.53 x 10⁴ formula units of calcium carbonate in 12.5 mol of the compound.

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Answer:

the formula for sulfide ion is S²⁻, and the formula for calcium ion is Ca²⁺. When combined, they form calcium sulfide with the chemical formula CaS.

Explanation:

Answer: theres a 1 in 4 chance of that happening

Explanation:

Aqueous solutions of aluminum chloride and sodium carbonate are combined to form carbon dioxide and sodium chloride.

When chemicals are combined, a chemical change known as a balanced chemical reaction takes place. The reactants, or the chemicals that are combined, are on the left side of the equation. The reaction's byproducts are on the right side.

Aluminum oxide, sodium chloride, and carbon dioxide gas bubbles are created when aluminum chloride and sodium carbonate combine.

When dissolved substances react, one (or more) solid products are produced, which is known as a precipitation reaction. These kinds of reactions, which are also occasionally known as double displacement, double replacement, or metathesis reactions, frequently involve the exchange of ions between ionic compounds in aqueous solutions.

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The complete question is

Write a balanced chemical equation: aqueous solutions of aluminium chloride and sodium carbonate are combined, producing aluminium carbonate.

Whether the time taken tο displace all οf the water in a measuring cylinder will be greater fοr lumps οf calcium carbοnate οr pοwder are mentiοned belοw.

The transfοrmatiοn οf οne οr mοre reactants intο οne οr mοre new prοducts is referred tο as a chemical reactiοn. Substances are made οf chemical cοnstituents οr cοmpοunds. The transfοrmatiοn οf οne οr mοre reactants intο οne οr mοre new prοducts is referred tο as a chemical reactiοn. Substances are made οf chemical cοnstituents οr cοmpοunds.

One part carbοn and twο parts οxygen make up the gas called carbοn diοxide. Its usage by plants tο create carbοhydrates during a prοcess knοwn as phοtοsynthesis makes it οne οf the mοst significant gases οn the planet.

The cοncentratiοn οf calcium carbοnate cannοt fluctuate because it is a sοlid. Hence, the οnly reactant in this reactiοn that can fluctuate in cοncentratiοn and impact reactiοn rate is hydrοchlοric acid. We need tο graph the cοncentratiοn οf hydrοchlοric acid against time, as yοu shοuld knοw frοm yοur reactiοn rate theοry.

Therefοre, whether the time taken tο displace all οf the water in a measuring cylinder will be greater fοr lumps οf calcium carbοnate οr pοwder are mentiοned abοve.

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The amount of the original sample remaining would be between 1.50 and 3.00 grams. Option I.

To solve this problem, we can use the half-life formula:

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

where:

We are given N0 = 12.0 grams, T = 3.6 days, and t = 10 days. We can plug these values into the formula and solve for N:

N = 12.0 grams x (1/2)^(10/3.6)

N ≈ 1.50 grams

Therefore, about 1.50 grams of the original sample remains after 10 days. The answer is between 1.50 and 3.00 grams.

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Answer: sp3

Explanation:

The correct answer is that Friedel-Crafts reactions can not be done on moderately or strongly deactivated ring systems.

This is a limitation of Friedel-Crafts reactions because they require an activated ring system in order to proceed.

Deactivated ring systems do not have the necessary electron density to facilitate the reaction, and therefore the reaction will not occur.

Friedel-Crafts alkylation and acylation reactions are both types of electrophilic aromatic substitution reactions that involve the formation of a carbon-carbon bond between an aromatic ring and an alkyl or acyl group.

These reactions are commonly used in organic synthesis to introduce new functional groups onto an aromatic ring.

However, they do have other limitations including the fact that Friedel-Crafts alkylation substrates can undergo rearrangement,

Friedel-Crafts acylation often leads to polyacylated products,

and Friedel-Crafts alkylation often leads to polyalkylated products.

These limitations must be taken into consideration when designing a synthetic route that involves a Friedel-Crafts reaction.

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At constant pressure, the systems that work on the surroundings are: A(s)+B(g) → 2C(g), A(g)+B(g) → 3C(g). The correct options are B and C.

In thermodynamics, if the reaction proceeds at constant pressure, it implies that the pressure remains constant during the course of the reaction, whereas the volume might change. The systems that work on the surroundings are those in which work is done on the surroundings by the system, resulting in a decrease in the internal energy of the system.

The reaction's stoichiometric coefficients of the reactants and products are balanced, implying that the number of moles of reactants and products is the same. It means there is no change in the number of moles of gas, which means there is no expansion of gases, resulting in no work done by the system. Thus, this option does not satisfy the criteria for a system working on the surroundings.

The reaction has one gas molecule in the reactants and two gas molecules in the products, implying that there is an increase in the number of moles of gas. As a result, there is an increase in the volume, resulting in work being done by the system. This option satisfies the requirements for a system working on the surroundings.

The reaction has two gas molecules in the reactants and three gas molecules in the products, which means there is an increase in the number of moles of gas. As a result, there is an increase in the volume, resulting in work being done by the system. This option satisfies the requirements for a system working on the surroundings.

The reaction has two gas molecules in the reactants and one gas molecule in the product, implying that there is a decrease in the number of moles of gas. It implies that the volume of the system will decrease, resulting in no work done by the system. Thus, this option does not satisfy the criteria for a system working on the surroundings.

Therefore, the options that work on the surroundings at constant pressure are options b and c.

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The condition where only four chromosomes are present in some cells of a fruit fly is known as haploidy.

Haploidy can occur due to a variety of reasons, including errors during cell division, such as nondisjunction, which is the failure of chromosomes to separate properly during meiosis. In nondisjunction, the chromosomes fail to divide equally, resulting in daughter cells with abnormal numbers of chromosomes. In the case of the fruit fly, this can result in cells with only four chromosomes instead of the normal eight.

Haploidy can also occur naturally in certain stages of development, such as during gamete formation, where cells undergo meiosis to produce haploid gametes with half the number of chromosomes as the parent cell.

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The concentration of O₂(g) in the solution is 8 ppm. Option B is correct.

Parts per million (ppm) is a unit of concentration used to express the amount of a substance in a solution. It represents the number of parts of the substance per one million parts of a solution.

To calculate the concentration of O₂(g) in parts per million (ppm), we need to first determine the amount of O₂ dissolved in the solution per unit mass of the solution.

We are given that the solution contains 0.008 grams of O₂(g) dissolved in 1000 grams of H₂O(₂):

Concentration of O₂(g) = (mass of O₂(g) / mass of solution) x 10⁶

Concentration of O₂(g) = (0.008 g / 1000 g) x 10⁶

= 8 ppm

Hence, B. 8ppm is the correct option.

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--The given question is incomplete, the complete question is

"What is the concentration of O2(g), in parts per million, in a solution that contains 0. 008 gram of O2(g) dissolved in 1000. Grams of H2O(2)?A) 10. 8 ppm B) 8 ppm C) 80 ppm D) 800 ppm"--

Answer:

gasoline

Explanation:

The right response is gasoline

Because the ingredients are evenly distributed and do not settle over time, homogeneous combinations include tomato soup, sugar-water solution, and salt-water solution. Contrarily, gasoline is a heterogeneous mixture since it contains a variety of hydrocarbons and additives that may separate or settle over time.

The output voltage of the device is 12 volts.

The transformer law states that the ratio of the number of turns in the primary coil to the number of turns in the secondary coil is equal to the ratio of the output voltage to the input voltage. Mathematically, we can express this as:

Vp/Vs = Np/Ns

where Vp is the input voltage (120 volts in this case), Vs is the output voltage (what we're trying to find), Np is the number of turns in the primary coil (200 turns in this case), and Ns is the number of turns in the secondary coil (20 turns in this case).

We can rearrange this equation to solve for Vs:

Vs = Vp * Ns/Np

Plugging in the given values, we get:

Vs = 120 volts * 20/200

Vs = 12 volts.

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The pH of the mixture of acids is 0.62, the pH of the mixture of acids is 0.63, the pH of the mixture of acids is 0.39, the pH of the mixture of acids is 1.00.

To find the pH of a mixture of acids, we need to first calculate the total concentration of H⁺ ions in the solution, and then use that value to calculate the pH using the formula pH = -log[H⁺].

The chemical formulas for the two acids are:

HBr - hydrobromic acid

HCHO₂ - formic acid

The dissociation reactions for these acids are:

HBr → H⁺ + Br⁻

HCHO₂ → H⁺ + CHO₂⁻

The concentrations of the two acids are given as:

[HBr] = 0.115 M

[HCHO₂] = 0.125 M

To find the total concentration of H⁺ ions in the solution, we need to consider the dissociation reactions of both acids. The H⁺ ions from each acid will contribute to the total H⁺ ion concentration:

[H⁺] = [H⁺ from HBr] + [H⁺ from HCHO₂]

[H⁺] = [HBr] + [HCHO₂]

[H⁺] = 0.115 M + 0.125 M

[H⁺] = 0.24 M

Now that we know the total concentration of H⁺ ions in the solution, we can calculate the pH:

pH = -log[H⁺]

pH = -log(0.24)

pH ≈ 0.62

The chemical formulas for the two acids are:

HNO₂ - nitrous acid

HNO₃ - nitric acid

The dissociation reactions for these acids are:

HNO₂ → H⁺ + NO₂⁻

HNO₃ → H⁺ + NO₃⁻

The concentrations of the two acids are given as:

[HNO₂] = 0.150 M

[HNO₃] = 0.085 M

To find the total concentration of H⁺ ions in the solution, we need to consider the dissociation reactions of both acids. The H⁺ ions from each acid will contribute to the total H⁺ ion concentration:

[H⁺] = [H⁺ from HNO₂] + [H⁺ from HNO₃]

[H⁺] = [HNO₂] + [HNO₃]

[H⁺] = 0.150 M + 0.085 M

[H⁺] = 0.235 M

Now that we know the total concentration of H⁺ ions in the solution, we can calculate the pH:

pH = -log[H⁺]

pH = -log(0.235)

pH ≈ 0.63

The chemical formulas for the two acids are:

HCHO₂ - formic acid

HC₂H₃O₂ - acetic acid

The dissociation reactions for these acids are:

HCHO₂ → H⁺ + CHO₂⁻

HC₂H₃O₂ → H⁺ + C₂H₃O₂⁻

The concentrations of the two acids are given as:

[HCHO₂] = 0.185 M

[HC₂H₃O₂] = 0.225 M

To find the total concentration of H⁺ ions in the solution, we need to consider the dissociation reactions of both acids. The H⁺ ions from each acid will contribute to the total H⁺ ion concentration:

[H⁺] = [H⁺ from HCHO₂] + [H⁺ from HC₂H₃O₂]

[H⁺] = [HCHO₂] + [HC₂H₃O₂]

[H⁺] = 0.185 M + 0.225 M

[H⁺] = 0.410 M

Now that we know the total concentration of H⁺ ions in the solution, we can calculate the pH:

pH = -log[H⁺]

pH = -log(0.410)

pH ≈ 0.39

The chemical formulas for the two acids are:

CH₃COOH - acetic acid

HCN - hydrocyanic acid

The dissociation reactions for these acids are:

CH₃COOH → H⁺ + CH₃COO⁻

HCN → H⁺ + CN⁻

The concentrations of the two acids are given as:

[CH₃COOH] = 0.050 M

[HCN] = 0.050 M

To find the total concentration of H⁺ ions in the solution, we need to consider the dissociation reactions of both acids. The H⁺ ions from each acid will contribute to the total H⁺ ion concentration:

[H⁺] = [H⁺ from CH₃COOH] + [H⁺ from HCN]

[H⁺] = [CH₃COOH] + [HCN]

[H⁺] = 0.050 M + 0.050 M

[H⁺] = 0.100 M

Now that we know the total concentration of H⁺ ions in the solution, we can calculate the pH:

pH = -log[H⁺]

pH = -log(0.100)

pH = 1.00

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When the concentration (absorbance) of reactants is doubled, the reaction rate and slope of the reaction will increase.

This is due to the increased availability of reactants, leading to more successful collisions between molecules. This in turn results in a faster reaction rate, and therefore a steeper slope. The reaction rate and slope of a reaction depend on the concentration of reactants. When the concentration (absorbance) of the reactants is doubled, the reaction rate increases, and the slope of the graph increases accordingly.

The rate of a reaction is the speed at which the reactants change into products.  If the reaction rate is high, the reaction will occur quickly, whereas if the reaction rate is low, the reaction will take longer to occur. When the concentration (absorbance) of the reactants is doubled, the reaction rate and slope of the graph increase. It occurs because the number of reactant particles available for collision increases.

Hence  increase in the number of reactant particles available for collision leads to more collisions per unit of time, and this increases the reaction rate.

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Based on the appearance of the reaction of aluminum with copper (II) sulfate, the reagent that was consumed was aluminum, and the reagent that had some left over was copper (II) sulfate.

Aluminum is a chemical element that has the symbol Al and the atomic number 13. It is a silvery-white, soft, nonmagnetic, ductile metal in the boron group. Aluminum is the third most prevalent element and the most abundant metal in the Earth's crust after oxygen and silicon.

Copper (II) sulfate is a blue solid that is soluble in water. Its appearance is due to the presence of water of crystallization, which occurs in the crystal structure. It is a compound that is commonly used as a fungicide and algaecide because it is toxic to many fungi and algae. Copper sulfate has also been used to treat various diseases.

Aluminum replaces the copper ions in copper (II) sulfate and creates aluminum sulfate and copper metal when aluminum reacts with copper (II) sulfate. The chemical reaction between aluminum and copper (II) sulfate is as follows:

Al(s) + CuSO4(aq) ⟶ Al2(SO4)3(aq) + Cu(s)

When aluminum is placed in copper (II) sulfate solution, copper ions from copper (II) sulfate move to aluminum, displacing the aluminum ions in the process. As a result, the copper ions from copper (II) sulfate solution are lowered to metallic copper, and aluminum ions combine with sulfate ions from copper (II) sulfate solution to form aluminum sulfate, which is soluble in water. When copper metal is produced, it forms a brown layer on top of the solution and sinks to the bottom.

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the half-life of the reaction is 1.29 s (to three significant figures).

The half-life (t1/2) of a first-order reaction is given by the formula:

t1/2 = ln(2) / k

where k is the rate constant of the reaction.

In this case, the rate law for the reaction is:

rate = k[A]3[B]2[C]

where [A], [B], and [C] are the concentrations of A, B, and C, respectively.

Since the reaction is first order with respect to A, the concentration of A at any time t is given by:

[A]t = [A]0 x

[tex] {e}^{ - kt} [/tex]

where [A]0 is the initial concentration of A.

Given that the initial concentration of A is 0.867 mol/L and the rate constant is 0.538 , we can use the formula for the half-life to calculate the time required for the concentration of A to decrease by half:

t1/2 = ln(2) / k = ln(2) / 0.538 = 1.29 s

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Answer:

The rate of the forward reaction must be equal to the rate of the reverse reaction. In other words, the speed at which products are formed from reactants must be equal to the speed at which reactants are formed from products.

The concentrations (or partial pressures) of the reactants and products must be constant. At equilibrium, the system has reached a state of balance where the amounts of reactants and products are no longer changing with time, and their concentrations (or partial pressures) have stabilized.

Thus, at equilibrium, the reaction is said to be "dynamic," meaning that the forward and reverse reactions are still occurring but at an equal rate, resulting in no net change in the concentrations (or partial pressures) of the reactants and products.

Explanation:

The system is losing heat to the surroundings. The approximate value for the heat exchange is 36770 J.

What is Equilibrium?

In general, equilibrium refers to a state of balance or stability in a system. In the context of physics, specifically thermodynamics, equilibrium is a state in which there is no net flow of heat or matter between a system and its surroundings. In other words, the system and surroundings have reached the same temperature and pressure, and there is no longer any exchange of heat or matter between them.

To solve this problem, we can use the equation for heat exchange:

Q = mcΔT

where Q is the heat exchange, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

For aluminum, the specific heat capacity is 0.903 J/g°C, and for ethyl alcohol, it is 2.44 J/g°C. We can also assume that the heat exchange between the aluminum and the ethyl alcohol is equal and opposite.

A) To find the equilibrium temperature, we can use the equation:

mcΔT = mcΔT

(100.0 g)(0.903 J/g°C)(T - 450.0°C) = (100.0 g)(2.44 J/g°C)(T - 80.0°C)

Solving for T, we get:

T = 198.15°C

Therefore, the equilibrium temperature of the mixture is approximately 198°C.

B) To find the heat exchange, we can use the same equation:

Q = mcΔT

The heat exchange for aluminum is:

Q₁ = (100.0 g)(0.903 J/g°C)(450.0°C - 198.15°C) ≈ 21260 J

The heat exchange for ethyl alcohol is:

Q₂ = (100.0 g)(2.44 J/g°C)(198.15°C - 80.0°C) ≈ 15510 J

The total heat exchange for the system is:

Q = -(Q₁ + Q₂) ≈ -36770 J

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The temperature in the atmosphere will rise as humidity rises. Only in the case of an exothermic chemical reaction will an increase in temperature slow down the reaction and thus reduce the percentage yield.

Energy is involved in all chemical reactions. When reactants break bonds, energy is released, and when products form new bonds, energy is released. Endothermic reactions are those in which less energy is released when new bonds form in the products than is required to break bonds in the reactants. Exothermic reactions are the polar opposite. In an exothermic reaction, breaking bonds in the reactants requires less energy than the energy released when new bonds form in the products.

The amount of water vapor in the air is referred to as humidity. The humidity will be high if there is a lot of water vapor in the air. The higher the humidity, the wetter the weather feels outside. Humidity is usually described as relative humidity in weather reports.

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