Which one of the following combinations would NOT form a precipitate in aqueous solution?
A) Sr(NO3)2 and NaOH.B) AgNO3 and KBr.C) Zn(C2H3O2)2 and Na2S.D) Pb(NO3)2 and Na2SO4.
E) All of the combinations will form precipitates.

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DVCS makes it easy with electronic bookkeeping services. Electronic bookkeeping service includes entering bills, scan and storage of same, deposits, provision of bills to be paid reports, electronic payments, reconciliation of bank and credit card accounts, and much more.

Among the many services offered by electronic bookkeeping are bill entry, bill scanning and storage, deposits, bill-to-be-paid reports, electronic payments, bank and credit card account reconciliation, and much more.

Single-entry bookkeeping and double-entry bookkeeping are the two main types of bookkeeping used in accounting. Depending on their needs, businesses may select either one or both.

Source documents and accounting records reside in digital form rather than on paper in e-accounting. Internationally, this idea is acknowledged. It helps any sort of company cut costs by saving a great deal of time, effort, and paper.

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20 %

Explanation:

becuz 25 divided by 125 multiplied by 100 is 20

When a chemist dissolves solid potassium chloride in a given amount of water until no more will dissolve at that temperature to prepare a saturated solution and concentrated solution.

A saturated solution is defined as a solution that contains the maximum amount of solute that can be dissolved under the condition at which the solution exists. Every solution can reach a status of saturation. A concentrated solution is defined as a solution that has a relatively large amount of dissolved solute. This solution has a large amounts of solute in the given solvent. For example, solid potassium chloride dissolves in a given amount of water until no more will dissolve at that temperature. A saturated solution contains the maximum concentration of a solute dissolved in the solvent while a concentrated solution contains a high amount of a solute dissolved in the solvent.

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Option ,1,3 and 4 are correct.

Nitrogen oxides, sulfur dioxide, and carbon monoxide are the gases that cause respiratory illness when exposed over the long run.

These gases/oxides are harmful because of the following reasons-

An oxide is a chemical compound that contains at least one oxygen atom and one other element.It can be either inorganic or organic. Oxides are compounds with a wide range of physical and chemical properties.

Respiratory illness is a broad term used to describe a variety of respiratory system disorders. The respiratory system is made up of the lungs, airways, and muscles that enable you to breathe.Respiratory disorders can be caused by a variety of factors, including genetic, environmental, and lifestyle factors. Asthma, chronic obstructive pulmonary disease (COPD), bronchitis, and emphysema are some examples of respiratory disorders caused by exposure to environmental pollutants.

Carbon dioxide-Carbon dioxide is a colorless, odorless gas that is a byproduct of respiration. Carbon dioxide is a greenhouse gas that contributes to global warming when released into the atmosphere. However, it is not considered to be harmful to human health when exposed in low concentrations.

Sulfur dioxide-Sulfur dioxide is a colorless gas that is produced by the burning of fossil fuels. Sulfur dioxide is a hazardous air pollutant that causes respiratory illnesses such as asthma and bronchitis

Nitrogen oxides-Nitrogen oxides are produced by combustion processes such as the burning of fossil fuels. Nitrogen oxides can cause respiratory illnesses such as asthma and chronic obstructive pulmonary disease (COPD). When nitrogen oxides combine with water vapor, they form acid rain.

Carbon monoxide-Carbon monoxide is a colorless, odorless gas that is produced by the incomplete combustion of fossil fuels. Carbon monoxide poisoning can cause headaches, nausea, dizziness, and even death when exposed in high concentrations. Carbon monoxide is also a contributor to air pollution.

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The volume of the balloon at the new pressure and temperature will be approximately 0.721 L.

This is a problem involving the ideal gas law, which relates the pressure, volume, number of moles, and temperature of a gas:

PV = nRT

where P is pressure, V is volume, n is number of moles, R is  ideal gas constant, and T is temperature in Kelvin.

We can use this equation to solve for the final volume of the balloon at a different pressure and temperature, assuming that the number of moles of gas remains constant.

Firstly, we need to convert temperature to Kelvin:

T₁ = 273.15 + 15 = 288.15 K

T₂ = 273.15 - 50 = 223.15 K

Next, we can rearrange the ideal gas law equation to solve for the final volume:

V₂ = (nRT₂)/P₂

where P₂ is the new pressure.

To find the new pressure, we can use the ideal gas law equation with the given conditions:

P₁V₁ = nRT₁

Solving for P₁:

P₁ = (nRT₁) / V₁

P₁ = (6.112 mol)(0.0821 L·atm/mol·K)(288.15 K) / (2.29932 L)

P₁ = 16.2 atm

Now we can use the new pressure and temperature to calculate the final volume:

V₂ = (nRT₂)/P₂

V₂ = (6.112 mol)(0.0821 L·atm/mol·K)(223.15 K) / (16.2 atm)

V₂ = 0.721 L

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The type of chemical reactions that may result in the propulsion is chemical reaction that combines two or the more kinds of the chemicals and will makes the different chemical as the product.

Chemical propulsion is the propulsion in which the thrust will be provided by the product of the chemical reaction, usually the burning or the oxidizing the fuel. The chemical reaction that combines two or the more kinds of the chemicals and that makes the different chemical as the product.

The Chemical Propulsion Systems is uses the chemical reactions that will release the energy and it accelerate the gases to the generate thrust.

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Answer:

190%

Explanation:

((BP of glycerin - BP of water) / BP of water) * 100%

((290 - 100) / 100) * 100%

Answer:

11

Explanation:

mass no.= No. of protons+ no.of neutrons

= 5+6

=11

Answer:

(a) At 20 °C the average kinetic energy of the particles is higher than at 60 °C.

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The neutralization reaction is:

O2H 3PO4 + 3Ba(OH) 2 → 6H 2O+ Ba(PO 4)2

A neutralization reaction is a type of chemical reaction that occurs between an acid and a base, resulting in the formation of a salt and water.

In this reaction, the hydrogen ions (H+) from the acid react with the hydroxide ions (OH-) from the base to form water (H2O), which neutralizes the acidic and basic properties of the reactants. The remaining ions then combine to form a salt.

The general equation for a neutralization reaction is:

Acid + Base → Salt + Water

For this question, the chosen answer is the neutralization reaction  because an acid (H3PO4) and a base (Ba(OH)2) react to form water and a salt (Ba(PO4)2).

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A) The conjugate acid of C2H5NH2 is C2H5NH3+, B) The conjugate acid of C6H5NH2 is C6H5NH3+, C) The conjugate acid of HPO42- is H2PO4-, D) The conjugate acid of HCO3- is H2CO3.

According to the balanced chemical equation, the conjugate acid has the formula C6H5NH+3 C 6 H 5 N H 3 +. The gain of a single proton defines the conjugate acid.

H2PO4- is hence capable of acting as both an acid and a base. Hence, HPO42- is the conjugate base of H2PO4-. The conjugate base is HPO42- and the acid is H2PO4-.

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Assuming no change in pressure, the gas' volume at 120.0 °C is 916.7 mL.

To solve this problem, we can use the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas. The ideal gas law is given by the formula:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

To find the volume of the gas at 120.0 °C, we need to convert the temperatures to Kelvin. The temperature in Kelvin is given by:

T (K) = T (°C) + 273.15

Therefore, the temperature at 27.0 °C is 300.15 K, and the temperature at 120.0 °C is 393.15 K.

We can set up the ideal gas law equation for the initial and final states of the gas:

P₁V₁ = nRT₁ (for the initial state at 27.0 °C)

P₂V₂ = nRT₂ (for the final state at 120.0 °C)

Assuming the pressure did not change, we can set P₁ = P₂, which allows us to solve for V₂:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = V₁ * T₂/T₁

Substituting the given values, we get:

V₂ = 700.0 mL * (393.15 K) / (300.15 K) = 916.7 mL

Therefore, the volume of the gas at 120.0 °C assuming pressure did not change is 916.7 mL.

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At 1100 C, the composition of the first solid phase that forms will be closer to the composition of copper than to the composition of silver.

There are a few different steps that need to be taken in order to answer this question properly.

The first step is to identify what the 20 wt% Ag - 80 wt% Cu concentration means. This concentration means that the material is made up of 20% silver and 80% copper by weight.

The second step is to identify the temperature that the material is being cooled from. In this case, the temperature is 1100 C.

The third step is to use phase diagrams to determine what the composition of the first solid phase that forms will be. Phase diagrams are diagrams that show the relationship between temperature, pressure, and the phases of a substance.

In this case, we will be using a binary phase diagram, which shows the relationship between two components. In order to use a phase diagram, we need to know the exact composition of the material that we are working with.

In this case, we know that the material is 20 wt% Ag - 80 wt% Cu. We can use this composition to find the corresponding point on the phase diagram.

Once we have found the point on the phase diagram that corresponds to our material, we can use the diagram to determine the composition of the first solid phase that forms. This composition will depend on the temperature that the material is being cooled from.

At 1100 C, the composition of the first solid phase that forms will be approximately 67% copper and 33% silver. The reason for this composition is that at 1100 C, the material is in the liquid phase.

As the material cools, it will begin to solidify. The first solid phase that forms will be a solid solution of copper and silver. This solid solution will have a composition that is between the composition of pure copper (which is 100% copper) and the composition of pure silver (which is 100% silver).

At 1100 C, the composition of the first solid phase that forms will be closer to the composition of copper than to the composition of silver. This is because copper has a lower melting point than silver, so it will solidify first. As the temperature continues to decrease, the composition of the solid phase will shift towards the composition of silver.

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Over-titrating the impure sample of KHP results in D)a too small calculation of the percent purity of the impure KHP sample.

When a titration is over-titrated, it means that more titrant than needed was added to the titrated solution, indicating the endpoint has passed. It is typical in acid-base titrations when the titration mixture's pH goes beyond the equivalence point's neutral value, indicating that more base has been added than was necessary to react with the acid solution in the flask. If the titration solution is over-titrated, the equivalence point will be passed, and the solution's pH will go beyond the endpoint. This implies that the reaction between the two reagents is no longer complete and that unreacted reagents remain in the solution.

The concentration of the titrant in the burette is calculated, and the moles of the titrant required to neutralize the sample's unknown acid or base are calculated to determine the quantity of the unknown acid or base in the sample. When the titration is over-titrated, the volume of titrant needed to reach the endpoint will be too high. This will cause the calculation of the unknown acid or base's quantity to be too low, implying that the percentage purity of the impure KHP sample is too small.

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The colour of the solution will be blue and the colour of the solid will be white.

When solutions of blue copper sulfate and colorless barium chloride solutions are mixed, an insoluble precipitate is formed. The color of the solution will be blue, while that of the solid will be white.

When two or more aqueous solutions are mixed, they react and form an insoluble solid. This insoluble solid is called a precipitate.

A precipitate is a solid substance that separates from a solution. As a result, when two solutions are mixed, a solid compound is formed that is insoluble in water is referred to as a precipitate.

Precipitation reactions are those in which the reactants, which are usually ions, react to form an insoluble solid called a precipitate. Precipitation reactions are vital in the creation of salt.

In chemistry, a salt is any compound that can be formed by the neutralization reaction between an acid and a base. The cation of a base and the anion of an acid react to form a salt. Salts are ionic compounds that are formed from a neutralization reaction, and they are commonly used as chemical compounds.

Precipitation reactions are common in everyday life, and many useful items are made through these reactions. For example, the precipitation of copper and zinc from their ore minerals can produce bronze, which is a vital metal in the production of various tools and art objects.

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Each element in the periodic table is represented by a one or two-letter symbol, as stated in option A.

These symbols are derived from the name of the element in either English or Latin. For example, the symbol for carbon is "C" and the symbol for gold is "Au" (from the Latin word "aurum"). The elements are not arranged based on alphabetical order, as stated in option B, nor are they listed as an abbreviation of the first letters of their name, as stated in option C. Instead, the elements are arranged in their own boxes based on their atomic number, which is the number of protons in their nucleus.

The periodic table is organized in rows and columns based on the electronic structure and chemical properties of the elements.

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Answer:

To prepare a 0.5M 500cm^3 solution of MgSO4 from a 2.5M stock solution of MgSO4, you can use the following formula:

C1V1 = C2V2

where C1 is the concentration of the stock solution, V1 is the volume of the stock solution needed, C2 is the desired concentration of the final solution, and V2 is the final volume of the solution.

Rearranging the formula, we get:

V1 = (C2 x V2)/C1

Substituting the values we have:

V1 = (0.5 x 0.5)/2.5 = 0.1 L = 100 mL

This means we need 100 mL of the stock solution to prepare the final solution. To prepare the final solution, we can use a volumetric flask and add 100 mL of the stock solution to it. We can then add distilled water to the flask until the volume reaches 500 mL, giving us a 0.5M 500cm^3 solution of MgSO4.

the measure of the intensity of heat in degrees, reflecting the average kinetic energy of the molecules, regardless of volume. It is called temperature.

Temperature is a proportion of the typical kinetic energy of the atoms or molecules in the framework.

The zeroth law of thermodynamics says that no intensity is moved between two items in a warm balance; hence, they are a similar temperature.

Intensity and temperature are two distinct yet firmly related ideas. Note that they have various units: temperature normally has units of degrees Celsius (C degrees, begin a text, C, end text) or Kelvin (K Kstart text, K, end text), and intensity has units of energy, Joules (J Jstart text, J, end text). Temperature is a proportion of the typical kinetic energy of the atoms or molecules in the framework. The water molecules in some hot espresso have a higher typical kinetic energy than the water molecules in some chilled tea, which likewise implies they are moving at a higher speed. Temperature is likewise a serious property, and that implies that the temperature doesn't change regardless of the amount of a substance you have (for however long it is all at a similar temperature!). For this reason, scientific experts can utilize the dissolving point to assist with recognizing an unadulterated substance short the temperature at which it liquefies is a property of the substance with no reliance on the mass of an example.

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the complete question is:

the measure of the intensity of heat in degrees, reflecting the average kinetic energy of the molecules, regardless of volume. It is called_____________.

The ground-state electron configuration of Cr²⁺ is [Ar] 4s¹ 3d⁴.
The ground-state electron configuration of Cu²⁺ is [Ar] 3d⁷.
The ground-state electron configuration of Co²⁺ is [Ar] 3d⁵.

The ground-state electron configurations of the Cr²⁺, Cu²⁺, and Co²⁺ ions using the abbreviated noble gas notation can be found as follows:
1. Chromium ion (Cr²⁺):
First, write the electron configuration of neutral chromium (Cr): [Ar] 4s² 3d⁵.
Now, remove two electrons from the highest energy orbitals to form Cr²⁺. In this case, one electron is removed from the 4s orbital and another from the 3d orbital.
So, the electron configuration of Cr²⁺ is: [Ar] 4s¹ 3d⁴.
2. Copper ion (Cu²⁺):
First, write the electron configuration of neutral copper (Cu): [Ar] 4s² 3d⁹.
To form Cu²⁺, remove two electrons from the highest energy orbitals. In this case, both electrons are removed from the 4s and 3d orbitals.
So, the electron configuration of Cu²⁺ is: [Ar] 3d⁷.
3. Cobalt ion (Co²⁺):
First, write the electron configuration of neutral cobalt (Co): [Ar] 4s² 3d⁷.
To form Co²⁺, remove two electrons from the highest energy orbitals. In this case, both electrons are removed from the 4s and 3d orbitals.
So, the electron configuration of Co²⁺ is: [Ar] 3d⁵.
In summary:
- The ground-state electron configuration of Cr²⁺ is [Ar] 4s¹ 3d⁴.
- The ground-state electron configuration of Cu²⁺ is [Ar] 3d⁷.
- The ground-state electron configuration of Co²⁺ is [Ar] 3d⁵.

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An atom with 6 protons and 2 electrons would have a net charge of +4, indicating that it has lost two electrons. This atom is now referred to as a cation, specifically a helium cation (He₂⁺). The ion notation for this atom would be written as He₂⁺.

The notation is a shorthand way of representing the atomic structure of an ion. It includes the chemical symbol of the element followed by the charge of the ion written as a superscript. In this case, the chemical symbol is He, which represents helium. The superscript of 2+ indicates that the helium atom has lost two electrons, leaving it with a net positive charge.

It is important to note that the number of protons in the nucleus of the helium ion remains the same as in a neutral helium atom. However, the number of electrons in the ion has decreased, resulting in a change in its chemical and physical properties. This ion is now more likely to bond with other ions or atoms to regain its lost electrons and become more stable.

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In IR spectroscopy, vibrational frequency is directly proportional to the energy of the vibrational mode, which in turn is proportional to the strength of the chemical bond and the masses of the atoms involved in the bond.

In infrared (IR) spectroscopy, vibrational frequency is directly proportional to the energy of the vibrational mode. Specifically, the vibrational frequency of a molecule is directly proportional to the strength of the chemical bond and the masses of the atoms involved in the bond. As the vibrational frequency increases, the energy of the bond also increases. This is because the bond becomes stiffer and requires more energy to vibrate. Therefore, IR spectroscopy can be used to identify the types of chemical bonds present in a molecule by measuring the vibrational frequencies of the bonds.

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Dispersion forces increase as the length of the hydrocarbon chain increases for the series of alcohols due to its increase in surface area.

For knowing what is the reason of increases in dispersion force we need to know about -dispersion force

Dispersion forces are weak intermolecular forces that exist between non-polar molecules. These forces are the result of a temporary imbalance of electron distribution around an atom or molecule.

Dispersion forces increase as the length of the hydrocarbon chain increases for the series of alcohols.

The reason why they change is due to the increase in the surface area of the molecule. Since the size of the molecule increases with the addition of carbon atoms, there is more surface area available for electrons to create an imbalance in electron distribution. As a result, the dispersion forces increase.

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1.82 × 10²⁰ ozone molecules in 5.00 L of air under the given stratospheric conditions.

To calculate the number of ozone molecules in 5.00 L of air under the given conditions, we need to use the ideal gas law, which relates the number of gas molecules to the pressure, volume, temperature, and the gas constant (R).

The ideal gas law will be written as:

PV = nRT

where P is the pressure in atmospheres (atm), V is the volume in liters (L), n is the number of moles of gas, R is the gas constant (0.08206 L·atm/mol·K), and T is temperature in Kelvin (K).

First, we need to convert the pressure to units of atmospheres:

1.49 × 10⁻³ atm = 1.46 × 10⁻³ atm (to three significant figures)

Next, we can use the ideal gas law to calculate the number of moles of air in the given volume:

PV = nRT

n = PV / RT

n = (1.46 × 10⁻³ atm) (5.00 L) / [(0.08206 L·atm/mol·K) (235 K)]

n = 0.1008 mol air

Assuming that the concentration of ozone in the stratosphere is 3 ppm (parts per million), which means 3 ozone molecules per million air molecules, we can calculate the number of ozone molecules in 5.00 L of air:

3 ppm = 3 ozone molecules / 1 million air molecules

So the number of ozone molecules in 5.00 L of air is:

n_ozone = (3 ozone molecules / 1 million air molecules) (0.1008 mol air)

n_ozone = 0.0003024 mol ozone

Finally, we can convert the number of moles of ozone to the number of ozone molecules using Avogadro's number:

N_ozone = n_ozone × N_A

where N_A is Avogadro's number (6.022 × 10²³ molecules/mol).

N_ozone = (0.0003024 mol) × (6.022 × 10²³ molecules/mol)

N_ozone = 1.82 × 10²⁰ ozone molecules

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The monoprotic acid's molar mass is 138.1 g/mol as a result.

By dividing the acid sample's mass by the moles of acid identified for each experiment, you may compute the unknown acid's molar mass. Calculate the average acid molar mass and record the result on your data sheet.

The correct answer is:

Acid + KOH → Salt + Water

The amount and concentration of KOH used in the titration may be used to determine the moles of acid:

Moles of KOH = concentration of KOH × volume of KOH

= 0.100 mol/L × 0.0320 L

= 0.00320 mol KOH

We may use the following formula to get the acid's molar mass:

Molar mass = mass of sample / moles of acid

Plugging in the values we know:

Molar mass = 0.442 g / 0.00320 mol

= 138.1 g/mol.

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At this point, pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid. Solving for pKa gives pKa = pH - log([A-]/[HA]).

a) The pH at the equivalence point of this titration is 7.0. At the equivalence point, the number of moles of acid is equal to the number of moles of base added.

b) To make the solution equalized, 1 mole of the strong base must be added per mole of a weak acid. This is because the equivalence point is reached when all of the weak acids have reacted with an equal amount of strong base.

c) The pKa for this acid can be determined by finding the half-equivalence point on the pH curve. The half-equivalence point occurs when half of the weak acid has been neutralized by a strong base. At this point, pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid. Solving for pKa gives pKa = pH - log([A-]/[HA]).

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If you incorrectly measured your starting mass too high, the percent recovery results would be lower than they should be. This is because percent recovery is calculated as the ratio of the actual yield to the theoretical yield, multiplied by 100%.

The actual yield is the amount of product that you actually obtained from the reaction, while the theoretical yield is the amount of product that you would expect to obtain based on stoichiometric calculations.

If you measured your starting mass too high, you would have used more reactant than you should have, resulting in a larger amount of product than you would have obtained if you had used the correct amount of reactant. This means that your actual yield would be higher than it should be, while your theoretical yield would remain the same.

As a result, the ratio of actual to theoretical yield would be lower than it should be, and your percent recovery results would be underestimated.

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The stability of the aromatic ring in phenol makes it a weak acid because it is not as willing to donate a proton as aliphatic alcohols, which lack the resonance stabilization of the negative ion.

Phenol is a weak acid because it has a relatively stable aromatic ring structure that makes it difficult to lose a proton. The hydroxyl group in phenol can donate a proton to form a negatively charged phenoxide ion, but this is not as favorable as it is in aliphatic alcohols due to resonance stabilization of the phenoxide ion. In phenol, the negative charge is delocalized over the entire ring, which reduces the stability of the phenoxide ion.

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the final temperature of the water is 50.75°C when a 100.0 g sample of water at 27.8°C is added to a 100.0 g sample of water at 73.7°C, assuming no heat is lost to the surroundings.

the final temperature of the water, we need to use the principle of heat conservation, which states that the total heat lost by the hot water is equal to the total heat gained by the cold water. This principle is expressed by the following equation:

Q_hot = - Q_cold

To calculate the heat lost or gained, we use the specific heat capacity (c) of water, which is 4.18 J/g°C. The formula for calculating the heat (Q) absorbed or released by a substance is:

Q = m × c × ΔT

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

Using this formula, we can calculate the heat lost by the hot water and the heat gained by the cold water as follows:

Q_hot = m_hot × c × (T_final - T_hot)

Q_cold = m_cold × c × (T_final - T_cold)

heat mass of the two water samples is equal (100 g each), so we can simplify the equation by setting m_hot = m_cold = m = 100 g. Substituting the values, we get:

100 g × 4.18 J/g°C × (T_final - 73.7°C) = - 100 g × 4.18 J/g°C × (T_final - 27.8°C)

Simplifying and solving for T_final, we get:

T_final = (100 g × 4.18 J/g°C × 73.7°C + 100 g × 4.18 J/g°C × 27.8°C) / (2 × 100 g × 4.18 J/g°C)

T_final = 50.75°C

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The two specific pollutants in the stormwater runoff are pathogens and the bacteria.

The common types of the pollutants found in the stormwater runoff will include the sediment, nutrients, the pathogens and the bacteria, heavy metals like the copper, zinc, and the lead, trash and the debris, organic matter, and the microorganisms.

As the storm water flows across streets, sidewalks, lawns and golf courses, it can pick up harmful pollutants and push them into storm drains, rivers and streams. The Stormwater runoff is generated from the rain and the snowmelt that will flows over the land or the impervious surfaces, such as the paved streets, parking lots, and the building rooftops, and it does not soak in the ground.

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