of the following, which are not polyprotic acids? (select all that apply) A. HI b) hno3 c) hcl d) h2so4

In 2CaCO3 there are 3 elements

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To measure and determine the flow in an arroyo and the Rio Grande during a summer rainy season in New Mexico, one can use flow measurement techniques such as velocity measurements and cross-sectional area calculations.

For determining the pollutant concentration in the arroyo, water samples collected before, during, and after the rain can be analyzed in a laboratory to compare pollutant concentrations and assess the impact of rainfall on water quality. Adhering to proper sampling protocols and employing suitable equipment and methods are crucial for accurate measurements and analysis.

Flow Measurement in the Arroyo:

a. Select a suitable location in the arroyo for flow measurement.

b. Use a flowmeter, such as a current meter or an acoustic Doppler flowmeter, to measure the velocity of water in the arroyo.

c. Take multiple velocity measurements across the cross-section of the arroyo to obtain an average velocity.

d. Measure the cross-sectional area of the arroyo using methods like the salt dilution or the velocity-area method.

e. Calculate the flow rate by multiplying the average velocity by the cross-sectional area.

Flow Measurement in the Rio Grande:

a. Choose a representative location in the Rio Grande for flow measurement.

b. Use a flowmeter or a current meter to measure the velocity of water in the river.

c. Take multiple velocity measurements across the river's cross-section to obtain an average velocity.

d. Measure the cross-sectional area of the river using methods like the salt dilution or the velocity-area method.

e. Calculate the flow rate by multiplying the average velocity by the cross-sectional area.

Determining Pollutant Concentration in the Arroyo:

a. Collect water samples from the arroyo at different times: before, during, and after the rain event.

b. Analyze the water samples in a laboratory to determine the pollutant concentration.

c. Compare the pollutant concentrations recorded before, during, and after the rain event to assess the impact of rainfall on pollutant levels.

d. Calculate the change in pollutant concentration by subtracting the pre-rain concentration from the concentrations measured during and after the rain.

e. Analyze the data to identify any trends or patterns in pollutant concentrations and assess the overall impact of the rain event on water quality in the arroyo.

It is important to note that specific techniques and equipment may vary depending on available resources and the desired level of accuracy for the measurements and analysis. Additionally, adherence to proper sampling protocols and laboratory methods is essential for reliable results.

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Answer

Form the sentences as you read along the blanks

Explanation:

1. Classical Era

2. Homophonic

3. Dynamics

4. Symphony

5. Sonata

6. Concerto

7. Composers

8. Ludwig

9. Wolfgang

10. Franz

The major products formed when benzene reacts with the following reagents are :

(a) tert-butylbenzene

(b) bromobenzene

(c) mixture (ortho-nitrobenzene, meta-nitrobenzene, para-nitrobenzene)

(d) benzaldehyde

(e) nitrobenzene

(a) The major product formed when benzene reacts with tert-butyl bromide and \(AlCl_3\) is tert-butylbenzene.

(b) The major product formed when benzene reacts with bromine and a nail (iron) is bromobenzene.

(c) The major product formed when benzene reacts with iodine and \(HNO_3\) is a mixture of ortho-nitrobenzene, meta-nitrobenzene, and para-nitrobenzene.

(d) The major product formed when benzene reacts with carbon monoxide, HCl, and \(AlCl_3\)/CuCl is benzaldehyde.

(e) The major product formed when benzene reacts with nitric acid and sulfuric acid is nitrobenzene.

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Answer:

105.98844 g/mol

Explanation:

Answer:

Temperature is the average kinetic energy of particles of an object. Warmer objects have faster particles and higher temperatures. If two objects have the same mass, the object with the higher temperature has greater thermal energy. Temperature is measured with a thermometer.

Answer:

the sun is directly behind your head

Explanation:

The reason why we can see a rainbow is caused by sunlight and atmospheric conditions

Answer:

K₅Mn₅O₁₆

Explanation:

From the question given above, the following data were obtained:

Potassium (K) = 27.5 %

Manganese (Mn) = 37.5 %

Oxygen (O) = 35 %

Empirical formula =?

The empirical formula for the compound can be obtained as follow:

K = 27.5 %

Mn = 37.5 %

O = 35 %

Divide by their molar mass

K = 27.5 / 39 = 0.705

Mn = 37.5 / 55 = 0.682

O = 35 / 16 = 2.188

Divide by the smallest

K = 0.705 / 0.682 = 1

Mn = 0.682 / 0.682 = 1

O = 2.188 / 0.682 = 3.2 = 32/10 = 16/5

K = 1

Mn = 1

O = 16/5

Multiply by 5 to express in whole number

K = 1 × 5 = 5

Mn = 1 × 5 = 5

O = 16/5 × 5 =16

Thus, the empirical formula for the compound is K₅Mn₅O₁₆

To determine the ratio of the concentrations of acetate ion and undissociated acetic acid at pH 5.22, we can use the Henderson-Hasselbalch equation, which relates the pH, pKa, and the concentrations of the acid and its conjugate base.

Given to us is

pH = 5.22

pKa = 4.76

\(pH = pKa + log\frac{[A-]}{[HA]}\)

In this equation, [A-] represents the concentration of acetate ion, and [HA] represents the concentration of undissociated acetic acid.

We can rearrange the Henderson-Hasselbalch equation to solve for the ratio \(\frac{A-}{HA}\):

\(\frac{A-}{HA} = 10^(pH - pKa)\)

Substituting the given values:

\(\frac{A-}{HA} = 10^(5.22 - 4.76)\)

\(\frac{A-}{HA} = 10^{0.46}\)

Using logarithmic properties, we can calculate:

\(\frac{A-}{HA} = 2.82\)

Therefore, at pH 5.22, the ratio of the concentrations of acetate ion to undissociated acetic acid is approximately 2.82.

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Respuesta:

Los modelo atómicos han permitido representar el modo de funcionamiento de los átomos. A lo largo de la historia han surgido un numero de modelos atómicos diferentes incluyendo los modelos de Bohr, Thomson, Rutherford, Sommerfeld, Dalton y Schrödinger.

Explicación:

El modelo atómico propuesto por John Dalton (1808) demostró que las sustancias químicas reaccionan en proporciones fijas y cómo mediante su combinación se producen elementos diferentes. Dalton fue el primero en postular la existencia de elementos indivisibles llamados átomos. A continuación, Thomson (1904) desarrolló un modelo en el cual el átomo estaba compuesto por protones con carga positiva y electrones con carga negativa los cuales se incrustaban uniformemente dentro de este átomo, asemejándose a las pasas de uva de un budín. En 1911, Ernest Rutherford desarrolló un nuevo modelo donde la masa principal del átomo tenía carga positiva y se localizaban en el núcleo, mientras que los electrones con carga negativa se posicionaban en la región externa del átomo. Subsecuentemente, Niels Bohr (1913) represento el funcionamiento del átomo de hidrógeno mediante un protón inmóvil en el núcleo atómico y un electrón girando a su alrededor. El modelo atómico de Sommerfeld permitió generalizar el diagrama de Bohr a otros tipos de átomos mas allá del Hidrógeno, incluyendo diferentes niveles energéticos para cada átomo particular. El modelo de Schrödinger​ (1926) permitió corregir aquellas discordancias surgidas del modelo atómico de Bohr. Schrödinger incluyó diferentes niveles y subniveles de energía a los electrones e incorporó órbitas elípticas a su movimiento, con lo cual permitiendo predecir los efectos relativos de los campos magnético y eléctrico sobre el movimiento de los electrones.

the molar concentration of \(AsF_5\) (g) at equilibrium is 0.0226.

We  consider the percent decomposition and the initial molar concentration of  \(ASF_5\)(g).

The percent decomposition of 27.7% means that 27.7% of the original moles of \(ASF_5\)(g) have decomposed. Therefore, the remaining moles of \(ASF_5\)(g) at equilibrium would be 100% - 27.7% = 72.3% of the original moles.

[ASF5] equilibrium = (72.3/100) * [ASF5]₀

= 0.723 × 0.0313 M = 0.0226 M

This equation gives us the molar concentration of \(ASF_5\)(g) at equilibrium.

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The amount of heat absorbed in turning 3.00 grams of ethanol from a liquid to a vapor is approximately 7.86 kJ. This can be calculated using the molar heat of vaporization of ethanol, which is 38.6 kJ/mol.

To calculate the heat absorbed, we first need to convert grams of ethanol to moles. The molar mass of ethanol (C2H5OH) is approximately 46.07 g/mol, so 3.00 grams is approximately 0.065 moles.

Next, we use the molar heat of vaporization (ΔHvap) of ethanol, which is 38.6 kJ/mol. Multiplying ΔHvap by the number of moles gives us the heat absorbed:

Heat absorbed = ΔHvap * moles

= 38.6 kJ/mol * 0.065 moles

≈ 2.51 kJ

Therefore, approximately 7.86 kJ of heat is absorbed when 3.00 grams of ethanol is converted from a liquid to a vapor.

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The pressure of the gas in the flask in kilopascals is given by the term 100.3 kPa, option E.

The pressure of any gas is a crucial characteristic. In contrast to qualities like viscosity and compressibility, we have some experience with gas pressure. Every day, the TV meteorologist reports the value of the atmosphere's barometric pressure.

We have included numerous slides on gas pressure in the Beginner's Guide since comprehending what pressure is and how it works is so essential to understanding aerodynamics. It is possible to investigate how static air pressure varies with altitude using an interactive atmosphere simulator. You can see how the pressure changes around a lifting wing using the FoilSim software.

height difference, h, indicates pressure of gas relative to atmospheric pressure.

h= 13mm

barometric pressure =765.2mmHg (atmosphere)

-from the picture, we can see that atmospheric pressure is greater than the gas pressure. so we minus

765.2mm - 13mm= 752.2mmHg

752.2mmHg * (101.3kPa / 760mmHg) = 100.3kPa.

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Complete question:

If the barometer read 765.2 mmHg when the measurement in in the Figure below took place, what is the pressure of the gas in the flask in kilopascals?

A.     7.55 kPa

B. 102.4 kPa

C. 1.007 kPa

D. 752.2 kPa

E. 100.3 kPa

Answer

The freezing point o the solution = 273.714153 K

The boiling point of the solution= 373.1955328 K

Explanation

Given:

Mass of glucose = 5.00 g

Volume of water = 72.8 g

What to find:

The freezing point and the boiling point of the solution.

Step-by-step solution:

Note: (Freezing point of water = 273K, Kf for water =1.87K kg/mol, atomic weight C = 12, H = 1, O = 16).

The freezing point of the solution:

The molecular weight of glucose C6H12O6 = 6(12) + 12(1) + 6(16) = 180 g/mol

The number of moles of glucose = (Mass of glucose/Molecular weight) = 5.00 g/180.0 g/mol = 0.0278 moles

Mass of water = 72.8 g = 0.0728 kg

So molality of glucose = (Moles o glcsose/Volume of solution) = 0.0278 mol/0.0728 kg = 0.3819 mol/kg

The depression in the freezing point, ∆T = Kf x molality = 1.87 K kg/mol x 0.3819 mol/kg = 0.714153 K

Since the freezing point of water = 273 K

Therefore, the freezing point of the solution= 273 K + 0.714153 K = 273.714153K

The boiling point of the solution:

∆T = i x m x Kb

∆T = change in temperature i.e boiling point elevation

i = van't Hoff factor = 1 for glucose since it does not ionize or dissociate. It is a single particle.

m = molality = moles solut/kg solvent = 0.0278 mol/0.0728 kg = 0.3819 mol/kg

Kb = boiling poin constant = 0.512 K kg/mol

∆T = 1 x 0.3819 mol/kg x 0.512 K kg/mol

∆T = 0.1955328 K

Since the freezing point of water = 373 K

Therefore, the boiling point of the solution = 373 K + 0.1955328 K = 373.1955328 K

When 80 ml of 1.0 x 10^(-5) M Ba(OH)2 is added to 20 ml of 1.0 x 10^(-5) M Fe2(SO4)3, no precipitate will form.

To determine if a precipitate will form when 80 ml of 1.0 x 10^(-5) M Ba(OH)2 is added to 20 ml of 1.0 x 10^(-5) M Fe2(SO4)3, we need to consider the solubility of the potential product.

First, let's determine the balanced equation for the reaction between Ba(OH)2 and Fe2(SO4)3:

Ba(OH)2 + Fe2(SO4)3 → BaSO4 + 2Fe(OH)3

According to the balanced equation, the possible precipitate that may form is BaSO4 (barium sulfate).

To determine if it will precipitate, we need to compare the solubility product constant (Ksp) of BaSO4 with the ion product (IP) of the ions present in the solution.

The Ksp expression for BaSO4 is: Ksp = [Ba^2+][SO4^2-]

The IP expression is: IP = [Ba^2+][SO4^2-]

Since both Ba^2+ and SO4^2- come from BaSO4, the concentrations of these ions will be equal. Let's assume x M to represent the concentration of Ba^2+ and SO4^2-.

Therefore, the IP = x * x = x^2

To determine if precipitation occurs, we need to compare the IP with the Ksp.

The Ksp of BaSO4 is approximately 1.1 x 10^(-10) at 25°C.

Since the initial concentrations of Ba^2+ and SO4^2- are both 1.0 x 10^(-5) M in this case, the IP is (1.0 x 10^(-5))^2 = 1.0 x 10^(-10).

Since the IP (1.0 x 10^(-10)) is smaller than the Ksp (1.1 x 10^(-10)), no precipitation will occur. This means that BaSO4 will remain soluble in the solution, and no solid BaSO4 will form.

Therefore, when 80 ml of 1.0 x 10^(-5) M Ba(OH)2 is added to 20 ml of 1.0 x 10^(-5) M Fe2(SO4)3, no precipitate will form.

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A defined medium supports microbial growth and is composed of exact quantities of pure chemicals; generally used for specific experiments when nutrients must be precisely controlled.

A substance that supports microbial growth and is composed of exact quantities of pure chemicals is known as a defined medium.

Defined media are generally used for specific experiments when nutrients must be precisely controlled, as they allow for the accurate measurement and manipulation of the nutrients available to the microorganisms.

This is in contrast to complex media, which are composed of undefined quantities of natural substances and are typically used for the general cultivation of microorganisms.

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Temperature constant, Boyle's law

P₁V₁=P₂V₂

3.15 = P₂.7

P₂=6.43 atm

Answer:

Option C on edge

Explanation:

The conversion of 1 mole of pyruvate to 3 moles of CO₂ via the PDH reaction and Krebs cycle also yields 4 moles of NADH, 1 mole of FADH₂, and 1 mole of GTP.

The Krebs cycle, also known as the citric acid cycle or tricarboxylic acid cycle (TCA cycle), is a series of chemical reactions that occur in the mitochondria of eukaryotic cells and in the cytoplasm of prokaryotic cells. It is a key component of cellular respiration, which is the process by which cells produce energy in the form of ATP (adenosine triphosphate).

The Krebs cycle begins with the acetyl-CoA molecule, which is produced from the breakdown of carbohydrates, fats, and proteins. The acetyl-CoA molecule enters the Krebs cycle and combines with oxaloacetate to form citrate. Through a series of reactions, citrate is converted back into oxaloacetate, producing ATP, NADH, and FADH₂ in the process. These energy-rich molecules are then used in the electron transport chain to produce more ATP.

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Z tons of coke are needed per week to produce 7.61x10^4 tons of pig iron from a modern blast furnace using hematite.

To calculate the amount of coke needed per week to produce 7.61x10^4 tons of pig iron from a modern blast furnace using hematite, we need to consider the stoichiometry of the reaction and the molar mass of the substances involved.

First, let's understand the balanced chemical equation for the production of pig iron from hematite:
Fe2O3 + 3C -> 2Fe + 3CO

From the balanced equation, we can see that 3 moles of carbon (C) react with 1 mole of hematite (Fe2O3) to produce 2 moles of iron (Fe) and 3 moles of carbon monoxide (CO).

Since coke is pure carbon, we can assume that 1 mole of coke is equal to 1 mole of carbon (C). Therefore, the ratio between moles of coke and moles of iron in the reaction is 3:2.

To calculate the amount of coke needed, we can use the following steps:

1. Convert the given amount of pig iron (7.61x10^4 tons) to moles using the molar mass of iron (Fe):
7.61x10^4 tons * (1000 kg/1 ton) * (1 g/1 kg) * (1 mol/55.845 g) = x moles of iron

2. Calculate the moles of coke needed based on the mole ratio:
x moles of iron * (3 moles of coke / 2 moles of iron) = y moles of coke

3. Convert the moles of coke to tons:
y moles of coke * (12.01 g/1 mol) * (1 kg/1000 g) * (1 ton/1000 kg) = z tons of coke


Please note that this calculation assumes 100% yield, meaning that all the reactants are converted into products. In reality, there may be some losses during the process, so the actual amount of coke needed may be slightly higher.

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The answer for the percentage yield is 80%.

What is percentage yield?

Recovery is the ratio of actual recovery to theoretical recovery. It is calculated by dividing the experimental yield by the theoretical yield and multiplying by 100%. If the actual and theoretical yields are equal, the yield rate is 100%. Yields are usually less than 100% because actual yields are often lower than theoretical. This may be due to incomplete or competing reactions and sample loss during recovery. Recoveries can exceed 100%. This means that more samples were recovered from the reaction than expected. This can occur when other reactions that form the product occur. Another potential source of error is overage caused by incomplete removal of moisture or other contaminants from the sample. Yield is always positive.

Therefore, The answer for the percentage yield is 80%.

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The inverse variation equation that relates x and y is y = k/x, where k is a constant of variation. To find the value of k, we can use the given information that y = 3 when x = 13.

Plugging in these values, we get 3 = k/13, which can be solved for k by multiplying both sides by 13, giving us k = 39. Therefore, the explanation of the inverse variation equation is y = 39/x. the inverse variation equation that relates x and y is y = k/x, where k is the constant of variation.

Since y varies inversely with x, we can write the relationship as y = k/x, where k is the constant of variation. To find the value of k, we can use the given information that y = 3 when x = 13 . Step 1: Plug in the given values for y and x into the equation ,3 = k/13 Step 2: Solve for k: k = 3 * 13 ,k = 39 ,Now that we have the value of k, we can write the inverse variation equation relating x and y as ,,y = 39/x .

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Answer:

pharynx

Explanation:

The pharynx, usually called the throat, is part of the respiratory system and digestive system. It carries air, food and fluid down from the nose and mouth. The pharynx is the site of common illnesses, including sore throat and tonsillitis.

Explanation:

Pharynx (Throat) The pharynx, usually called the throat, is part of the respiratory system and digestive system. It carries air, food and fluid down from the nose and mouth. The pharynx is the site of common illnesses, including sore throat and tonsillitis.

2,2,5,5-tetramethyl hex-3-yne cannot be made using acetylide anions because it would require the formation of a highly strained carbon-carbon triple bond.

Acetylide anions are commonly used in organic synthesis to form carbon-carbon triple bonds through reaction with alkenes. However, the formation of a triple bond in 2,2,5,5-tetramethylhex-3-yne would result in a highly strained molecule due to the presence of large groups on adjacent carbon atoms.

The strain in the molecule would make the formation of the triple bond highly energetically unfavorable, preventing the reaction from occurring. In addition, the large groups on the carbon atoms would interfere with the electron-richness of the alkene, making it difficult for the acetylide anion to react and form the desired product. As a result, alternative methods must be used to synthesize 2,2,5,5-tetramethylhex-3-yne.

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Rubbing alcohol can be used as hand sanitizer. Therefore, we need rubbing alcohol when entering the school. Therefore, option (C) is correct.

Alcohol-based hand sanitizer contains at least 60% (v/v) alcohol in water specifically, ethanol or isopropyl alcohol but only if soap and water are not available.

Apply the product to the palm of one hand then rub the hands together. Rub the sanitizer all over the surfaces of hands and fingers and do not go near flame or burning objects during the application of hand sanitizer.

Rubbing alcohol can be isopropyl alcohol or an ethanol-based liquid, with isopropyl alcohol products. Rubbing alcohol has various potential uses for household cleaning and personal care. These uses include eliminating odors and disinfecting surfaces and items within the home.

Rubbing alcohol is used to disinfect hands and surfaces in healthcare settings.

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Answer:

B) Alcohol

Explanation:

It is methanol. Alcohol has group 'OH'

Answer: Im pretty sure its C. or can exactly replicate the procedure

Explanation:

Answer:

it's c I took the test on my but

To assign a priority to an atom that is part of multiple bonds, treat the multiple bonds as an equivalent number of single bonds. Hence, option A is correct.

A carbon atom is a chiral (stereogenic) centre if it is tetrahedral (sp3) and has four different groups (ligands) attached to it.

To assign a priority to an atom that is part of multiple bonds, treat the multiply bonded atom as an equivalent number of singly bonded atoms.

If two isotopes are bonded to the stereogenic centre, the isotope with the higher mass number has a higher priority.

Hence, option A is correct.

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Answer:

The Lewis formalism used for the H2 molecule is H:H or H—H. The former, known as a 'Lewis dot diagram,' indicates a pair of shared electrons between the atomic symbols, while the latter, known as a 'Lewis structure,' uses a dash to indicate the pair of shared electrons that form a covalent bond.

Explanation: